Taylor Series

A Taylor Series is an expansion of a function into an infinite sum of terms , where each term's exponent is larger and larger, like this:

Example: The Taylor Series for e x

e x = 1 + x + x 2 2! + x 3 3! + x 4 4! + x 5 5! + ...

(Note: ! is the Factorial Function .)

Does it really work?

Using a calculator we get e 2 = (2.71828...) 2 = 7.389056...

But let's try more and more terms of our infinite series:

It starts out really badly, but it then gets better and better!

Try using " 2^n/fact(n) " and n=0 to 20 in the Sigma Calculator and see what you get.

Here are some common Taylor Series:

(There are many more.)

Approximations

We can use the first few terms of a Taylor Series to get an approximate value for a function.

Here we show better and better approximations for cos(x) . The red line is cos(x) , the blue is the approximation ( try plotting it yourself ) :

You can also see the Taylor Series in action at Euler's Formula for Complex Numbers

What is this Magic?

How can we turn a function into a series of power terms like this?

Well, it isn't really magic. First we say we want to have this expansion:

f(x) = c 0 + c 1 (x-a) + c 2 (x-a) 2 + c 3 (x-a) 3 + ...

Then we choose a value "a", and work out the values c 0 , c 1 , c 2 , ... etc

And it is done using derivatives (so we must know the derivative of our function)

Quick review: a derivative gives us the slope of a function at any point.

These basic derivative rules can help us:

  • The derivative of a constant is 0
  • The derivative of ax is a (example: the derivative of 2x is 2 )
  • The derivative of x n is nx n-1 (example: the derivative of x 3 is 3x 2 )

We will use the little mark ’ to mean "derivative of".

OK, let's start:

To get c 0 , choose x=a so all the (x-a) terms become zero, leaving us with:

So c 0 = f(a)

To get c 1 , take the derivative of f(x):

f ’ (x) = c 1 + 2c 2 (x-a) + 3c 3 (x-a) 2 + ...

With x=a all the (x-a) terms become zero:

f ’ (a) = c 1

So c 1 = f ’ (a)

To get c 2 , do the derivative again:

f ’ ’ (x) = 2c 2 + 3×2×c 3 (x-a) + ...

f ’ ’ (a) = 2c 2

So c 2 = f ’ ’ (a)/2

In fact, a pattern is emerging. Each term is

  • the next higher derivative ...
  • ... divided by all the exponents so far multiplied together (for which we can use factorial notation , for example 3! = 3×2×1)

And we get:

f(x) = f(a) + f'(a) 1! (x-a) + f''(a) 2! (x-a) 2 + f'''(a) 3! (x-a) 3 + ...

Now we have a way of finding our own Taylor Series:

For each term: take the next derivative, divide by n!, multiply by (x-a) n

Example: Taylor Series for cos(x)

Start with:

The derivative of cos is −sin , and the derivative of sin is cos , so:

  • f(x) = cos(x)
  • f'(x) = −sin(x)
  • f''(x) = −cos(x)
  • f'''(x) = sin(x)

cos(x) = cos(a) − sin(a) 1! (x-a) − cos(a) 2! (x-a) 2 + sin(a) 3! (x-a) 3 + ...

Now put a=0 , which is nice because cos(0)=1 and sin(0)=0 :

cos(x) = 1 − 0 1! (x-0) − 1 2! (x-0) 2 + 0 3! (x-0) 3 + 1 4! (x-0) 4 + ...

cos(x) = 1 − x 2 /2! + x 4 /4! − ...

Try that for sin(x) yourself, it will help you to learn.

Or try it on another function of your choice.

The key thing is to know the derivatives of your function f(x).

Note: A Maclaurin Series is a Taylor Series where a=0 , so all the examples we have been using so far can also be called Maclaurin Series.

  • 6.4 Working with Taylor Series
  • Introduction
  • 1.1 Approximating Areas
  • 1.2 The Definite Integral
  • 1.3 The Fundamental Theorem of Calculus
  • 1.4 Integration Formulas and the Net Change Theorem
  • 1.5 Substitution
  • 1.6 Integrals Involving Exponential and Logarithmic Functions
  • 1.7 Integrals Resulting in Inverse Trigonometric Functions
  • Key Equations
  • Key Concepts
  • Review Exercises
  • 2.1 Areas between Curves
  • 2.2 Determining Volumes by Slicing
  • 2.3 Volumes of Revolution: Cylindrical Shells
  • 2.4 Arc Length of a Curve and Surface Area
  • 2.5 Physical Applications
  • 2.6 Moments and Centers of Mass
  • 2.7 Integrals, Exponential Functions, and Logarithms
  • 2.8 Exponential Growth and Decay
  • 2.9 Calculus of the Hyperbolic Functions
  • 3.1 Integration by Parts
  • 3.2 Trigonometric Integrals
  • 3.3 Trigonometric Substitution
  • 3.4 Partial Fractions
  • 3.5 Other Strategies for Integration
  • 3.6 Numerical Integration
  • 3.7 Improper Integrals
  • 4.1 Basics of Differential Equations
  • 4.2 Direction Fields and Numerical Methods
  • 4.3 Separable Equations
  • 4.4 The Logistic Equation
  • 4.5 First-order Linear Equations
  • 5.1 Sequences
  • 5.2 Infinite Series
  • 5.3 The Divergence and Integral Tests
  • 5.4 Comparison Tests
  • 5.5 Alternating Series
  • 5.6 Ratio and Root Tests
  • 6.1 Power Series and Functions
  • 6.2 Properties of Power Series
  • 6.3 Taylor and Maclaurin Series
  • 7.1 Parametric Equations
  • 7.2 Calculus of Parametric Curves
  • 7.3 Polar Coordinates
  • 7.4 Area and Arc Length in Polar Coordinates
  • 7.5 Conic Sections
  • A | Table of Integrals
  • B | Table of Derivatives
  • C | Review of Pre-Calculus

Learning Objectives

  • 6.4.1 Write the terms of the binomial series.
  • 6.4.2 Recognize the Taylor series expansions of common functions.
  • 6.4.3 Recognize and apply techniques to find the Taylor series for a function.
  • 6.4.4 Use Taylor series to solve differential equations.
  • 6.4.5 Use Taylor series to evaluate nonelementary integrals.

In the preceding section, we defined Taylor series and showed how to find the Taylor series for several common functions by explicitly calculating the coefficients of the Taylor polynomials. In this section we show how to use those Taylor series to derive Taylor series for other functions. We then present two common applications of power series. First, we show how power series can be used to solve differential equations. Second, we show how power series can be used to evaluate integrals when the antiderivative of the integrand cannot be expressed in terms of elementary functions. In one example, we consider ∫ e − x 2 d x , ∫ e − x 2 d x , an integral that arises frequently in probability theory.

The Binomial Series

Our first goal in this section is to determine the Maclaurin series for the function f ( x ) = ( 1 + x ) r f ( x ) = ( 1 + x ) r for all real numbers r . r . The Maclaurin series for this function is known as the binomial series . We begin by considering the simplest case: r r is a nonnegative integer. We recall that, for r = 0 , 1 , 2 , 3 , 4 , f ( x ) = ( 1 + x ) r r = 0 , 1 , 2 , 3 , 4 , f ( x ) = ( 1 + x ) r can be written as

The expressions on the right-hand side are known as binomial expansions and the coefficients are known as binomial coefficients. More generally, for any nonnegative integer r , r , the binomial coefficient of x n x n in the binomial expansion of ( 1 + x ) r ( 1 + x ) r is given by

For example, using this formula for r = 5 , r = 5 , we see that

We now consider the case when the exponent r r is any real number, not necessarily a nonnegative integer. If r r is not a nonnegative integer, then f ( x ) = ( 1 + x ) r f ( x ) = ( 1 + x ) r cannot be written as a finite polynomial. However, we can find a power series for f . f . Specifically, we look for the Maclaurin series for f . f . To do this, we find the derivatives of f f and evaluate them at x = 0 . x = 0 .

We conclude that the coefficients in the binomial series are given by

We note that if r r is a nonnegative integer, then the ( r + 1 ) st ( r + 1 ) st derivative f ( r + 1 ) f ( r + 1 ) is the zero function, and the series terminates. In addition, if r r is a nonnegative integer, then Equation 6.8 for the coefficients agrees with Equation 6.6 for the coefficients, and the formula for the binomial series agrees with Equation 6.7 for the finite binomial expansion. More generally, to denote the binomial coefficients for any real number r , r , we define

With this notation, we can write the binomial series for ( 1 + x ) r ( 1 + x ) r as

We now need to determine the interval of convergence for the binomial series Equation 6.9 . We apply the ratio test. Consequently, we consider

if and only if | x | < 1 , | x | < 1 , we conclude that the interval of convergence for the binomial series is ( −1 , 1 ) . ( −1 , 1 ) . The behavior at the endpoints depends on r . r . It can be shown that for r ≥ 0 r ≥ 0 the series converges at both endpoints; for −1 < r < 0 , −1 < r < 0 , the series converges at x = 1 x = 1 and diverges at x = −1 ; x = −1 ; and for r < −1 , r < −1 , the series diverges at both endpoints. The binomial series does converge to ( 1 + x ) r ( 1 + x ) r in ( −1 , 1 ) ( −1 , 1 ) for all real numbers r , r , but proving this fact by showing that the remainder R n ( x ) → 0 R n ( x ) → 0 is difficult.

For any real number r , r , the Maclaurin series for f ( x ) = ( 1 + x ) r f ( x ) = ( 1 + x ) r is the binomial series. It converges to f f for | x | < 1 , | x | < 1 , and we write

for | x | < 1 . | x | < 1 .

We can use this definition to find the binomial series for f ( x ) = 1 + x f ( x ) = 1 + x and use the series to approximate 1.5 . 1.5 .

Example 6.17

Finding binomial series.

  • Find the binomial series for f ( x ) = 1 + x . f ( x ) = 1 + x .
  • Use the third-order Maclaurin polynomial p 3 ( x ) p 3 ( x ) to estimate 1.5 . 1.5 . Use Taylor’s theorem to bound the error. Use a graphing utility to compare the graphs of f f and p 3 . p 3 .
  • Here r = 1 2 . r = 1 2 . Using the definition for the binomial series, we obtain 1 + x = 1 + 1 2 x + ( 1 / 2 ) ( − 1 / 2 ) 2 ! x 2 + ( 1 / 2 ) ( − 1 / 2 ) ( − 3 / 2 ) 3 ! x 3 + ⋯ = 1 + 1 2 x − 1 2 ! 1 2 2 x 2 + 1 3 ! 1 · 3 2 3 x 3 − ⋯ + ( −1 ) n + 1 n ! 1 · 3 · 5 ⋯ ( 2 n − 3 ) 2 n x n + ⋯ = 1 + ∑ n = 1 ∞ ( −1 ) n + 1 n ! 1 · 3 · 5 ⋯ ( 2 n − 3 ) 2 n x n . 1 + x = 1 + 1 2 x + ( 1 / 2 ) ( − 1 / 2 ) 2 ! x 2 + ( 1 / 2 ) ( − 1 / 2 ) ( − 3 / 2 ) 3 ! x 3 + ⋯ = 1 + 1 2 x − 1 2 ! 1 2 2 x 2 + 1 3 ! 1 · 3 2 3 x 3 − ⋯ + ( −1 ) n + 1 n ! 1 · 3 · 5 ⋯ ( 2 n − 3 ) 2 n x n + ⋯ = 1 + ∑ n = 1 ∞ ( −1 ) n + 1 n ! 1 · 3 · 5 ⋯ ( 2 n − 3 ) 2 n x n .

Checkpoint 6.16

Find the binomial series for f ( x ) = 1 ( 1 + x ) 2 . f ( x ) = 1 ( 1 + x ) 2 .

Common Functions Expressed as Taylor Series

At this point, we have derived Maclaurin series for exponential, trigonometric, and logarithmic functions, as well as functions of the form f ( x ) = ( 1 + x ) r . f ( x ) = ( 1 + x ) r . In Table 6.1 , we summarize the results of these series. We remark that the convergence of the Maclaurin series for f ( x ) = ln ( 1 + x ) f ( x ) = ln ( 1 + x ) at the endpoint x = 1 x = 1 and the Maclaurin series for f ( x ) = tan −1 x f ( x ) = tan −1 x at the endpoints x = 1 x = 1 and x = −1 x = −1 relies on a more advanced theorem than we present here. (Refer to Abel’s theorem for a discussion of this more technical point.)

Earlier in the chapter, we showed how you could combine power series to create new power series. Here we use these properties, combined with the Maclaurin series in Table 6.1 , to create Maclaurin series for other functions.

Example 6.18

Deriving maclaurin series from known series.

Find the Maclaurin series of each of the following functions by using one of the series listed in Table 6.1 .

  • f ( x ) = cos x f ( x ) = cos x
  • f ( x ) = sinh x f ( x ) = sinh x
  • Using the Maclaurin series for cos x cos x we find that the Maclaurin series for cos x cos x is given by ∑ n = 0 ∞ ( −1 ) n ( x ) 2 n ( 2 n ) ! = ∑ n = 0 ∞ ( −1 ) n x n ( 2 n ) ! = 1 − x 2 ! + x 2 4 ! − x 3 6 ! + x 4 8 ! − ⋯ . ∑ n = 0 ∞ ( −1 ) n ( x ) 2 n ( 2 n ) ! = ∑ n = 0 ∞ ( −1 ) n x n ( 2 n ) ! = 1 − x 2 ! + x 2 4 ! − x 3 6 ! + x 4 8 ! − ⋯ . This series converges to cos x cos x for all x x in the domain of cos x ; cos x ; that is, for all x ≥ 0 . x ≥ 0 .
  • To find the Maclaurin series for sinh x , sinh x , we use the fact that sinh x = e x − e − x 2 . sinh x = e x − e − x 2 . Using the Maclaurin series for e x , e x , we see that the n th n th term in the Maclaurin series for sinh x sinh x is given by x n n ! − ( − x ) n n ! . x n n ! − ( − x ) n n ! . For n n even, this term is zero. For n n odd, this term is 2 x n n ! . 2 x n n ! . Therefore, the Maclaurin series for sinh x sinh x has only odd-order terms and is given by ∑ n = 0 ∞ x 2 n + 1 ( 2 n + 1 ) ! = x + x 3 3 ! + x 5 5 ! + ⋯ . ∑ n = 0 ∞ x 2 n + 1 ( 2 n + 1 ) ! = x + x 3 3 ! + x 5 5 ! + ⋯ .

Checkpoint 6.17

Find the Maclaurin series for sin ( x 2 ) . sin ( x 2 ) .

We also showed previously in this chapter how power series can be differentiated term by term to create a new power series. In Example 6.19 , we differentiate the binomial series for 1 + x 1 + x term by term to find the binomial series for 1 1 + x . 1 1 + x . Note that we could construct the binomial series for 1 1 + x 1 1 + x directly from the definition, but differentiating the binomial series for 1 + x 1 + x is an easier calculation.

Example 6.19

Differentiating a series to find a new series.

Use the binomial series for 1 + x 1 + x to find the binomial series for 1 1 + x . 1 1 + x .

The two functions are related by

so the binomial series for 1 1 + x 1 1 + x is given by

Checkpoint 6.18

Find the binomial series for f ( x ) = 1 ( 1 + x ) 3 / 2 f ( x ) = 1 ( 1 + x ) 3 / 2

In this example, we differentiated a known Taylor series to construct a Taylor series for another function. The ability to differentiate power series term by term makes them a powerful tool for solving differential equations. We now show how this is accomplished.

Solving Differential Equations with Power Series

Consider the differential equation

Recall that this is a first-order separable equation and its solution is y = C e x . y = C e x . This equation is easily solved using techniques discussed earlier in the text. For most differential equations, however, we do not yet have analytical tools to solve them. Power series are an extremely useful tool for solving many types of differential equations. In this technique, we look for a solution of the form y = ∑ n = 0 ∞ c n x n y = ∑ n = 0 ∞ c n x n and determine what the coefficients would need to be. In the next example, we consider an initial-value problem involving y ′ = y y ′ = y to illustrate the technique.

Example 6.20

Power series solution of a differential equation.

Use power series to solve the initial-value problem

Suppose that there exists a power series solution

Differentiating this series term by term, we obtain

If y satisfies the differential equation, then

Using Uniqueness of Power Series on the uniqueness of power series representations, we know that these series can only be equal if their coefficients are equal. Therefore,

Using the initial condition y ( 0 ) = 3 y ( 0 ) = 3 combined with the power series representation

we find that c 0 = 3 . c 0 = 3 . We are now ready to solve for the rest of the coefficients. Using the fact that c 0 = 3 , c 0 = 3 , we have

You might recognize

as the Taylor series for e x . e x . Therefore, the solution is y = 3 e x . y = 3 e x .

Checkpoint 6.19

Use power series to solve y ′ = 2 y , y ( 0 ) = 5 . y ′ = 2 y , y ( 0 ) = 5 .

We now consider an example involving a differential equation that we cannot solve using previously discussed methods. This differential equation

is known as Airy’s equation . It has many applications in mathematical physics, such as modeling the diffraction of light. Here we show how to solve it using power series.

Example 6.21

Power series solution of airy’s equation.

Use power series to solve

with the initial conditions y ( 0 ) = a y ( 0 ) = a and y ′ ( 0 ) = b . y ′ ( 0 ) = b .

We look for a solution of the form

Differentiating this function term by term, we obtain

If y satisfies the equation y ″ = x y , y ″ = x y , then

Using Uniqueness of Power Series on the uniqueness of power series representations, we know that coefficients of the same degree must be equal. Therefore,

More generally, for n ≥ 3 , n ≥ 3 , we have n · ( n − 1 ) c n = c n − 3 . n · ( n − 1 ) c n = c n − 3 . In fact, all coefficients can be written in terms of c 0 c 0 and c 1 . c 1 . To see this, first note that c 2 = 0 . c 2 = 0 . Then

For c 5 , c 6 , c 7 , c 5 , c 6 , c 7 , we see that

Therefore, the series solution of the differential equation is given by

The initial condition y ( 0 ) = a y ( 0 ) = a implies c 0 = a . c 0 = a . Differentiating this series term by term and using the fact that y ′ ( 0 ) = b , y ′ ( 0 ) = b , we conclude that c 1 = b . c 1 = b . Therefore, the solution of this initial-value problem is

Checkpoint 6.20

Use power series to solve y ″ + x 2 y = 0 y ″ + x 2 y = 0 with the initial condition y ( 0 ) = a y ( 0 ) = a and y ′ ( 0 ) = b . y ′ ( 0 ) = b .

Evaluating Nonelementary Integrals

Solving differential equations is one common application of power series. We now turn to a second application. We show how power series can be used to evaluate integrals involving functions whose antiderivatives cannot be expressed using elementary functions.

One integral that arises often in applications in probability theory is ∫ e − x 2 d x . ∫ e − x 2 d x . Unfortunately, the antiderivative of the integrand e − x 2 e − x 2 is not an elementary function. By elementary function, we mean a function that can be written using a finite number of algebraic combinations or compositions of exponential, logarithmic, trigonometric, or power functions. We remark that the term “elementary function” is not synonymous with noncomplicated function. For example, the function f ( x ) = x 2 − 3 x + e x 3 − sin ( 5 x + 4 ) f ( x ) = x 2 − 3 x + e x 3 − sin ( 5 x + 4 ) is an elementary function, although not a particularly simple-looking function. Any integral of the form ∫ f ( x ) d x ∫ f ( x ) d x where the antiderivative of f f cannot be written as an elementary function is considered a nonelementary integral .

Nonelementary integrals cannot be evaluated using the basic integration techniques discussed earlier. One way to evaluate such integrals is by expressing the integrand as a power series and integrating term by term. We demonstrate this technique by considering ∫ e − x 2 d x . ∫ e − x 2 d x .

Example 6.22

Using taylor series to evaluate a definite integral.

  • Express ∫ e − x 2 d x ∫ e − x 2 d x as an infinite series.
  • Evaluate ∫ 0 1 e − x 2 d x ∫ 0 1 e − x 2 d x to within an error of 0.01 . 0.01 .
  • The Maclaurin series for e − x 2 e − x 2 is given by e − x 2 = ∑ n = 0 ∞ ( − x 2 ) n n ! = 1 − x 2 + x 4 2 ! − x 6 3 ! + ⋯ + ( −1 ) n x 2 n n ! + ⋯ = ∑ n = 0 ∞ ( −1 ) n x 2 n n ! . e − x 2 = ∑ n = 0 ∞ ( − x 2 ) n n ! = 1 − x 2 + x 4 2 ! − x 6 3 ! + ⋯ + ( −1 ) n x 2 n n ! + ⋯ = ∑ n = 0 ∞ ( −1 ) n x 2 n n ! . Therefore, ∫ e − x 2 d x = ∫ ( 1 − x 2 + x 4 2 ! − x 6 3 ! + ⋯ + ( −1 ) n x 2 n n ! + ⋯ ) d x = C + x − x 3 3 + x 5 5 · 2 ! − x 7 7 · 3 ! + ⋯ + ( −1 ) n x 2 n + 1 ( 2 n + 1 ) n ! + ⋯ . ∫ e − x 2 d x = ∫ ( 1 − x 2 + x 4 2 ! − x 6 3 ! + ⋯ + ( −1 ) n x 2 n n ! + ⋯ ) d x = C + x − x 3 3 + x 5 5 · 2 ! − x 7 7 · 3 ! + ⋯ + ( −1 ) n x 2 n + 1 ( 2 n + 1 ) n ! + ⋯ .
  • Using the result from part a. we have ∫ 0 1 e − x 2 d x = 1 − 1 3 + 1 10 − 1 42 + 1 216 − ⋯ . ∫ 0 1 e − x 2 d x = 1 − 1 3 + 1 10 − 1 42 + 1 216 − ⋯ . The sum of the first four terms is approximately 0.74 . 0.74 . By the alternating series test, this estimate is accurate to within an error of less than 1 216 ≈ 0.0046296 < 0.01 . 1 216 ≈ 0.0046296 < 0.01 .

Checkpoint 6.21

Express ∫ cos x d x ∫ cos x d x as an infinite series. Evaluate ∫ 0 1 cos x d x ∫ 0 1 cos x d x to within an error of 0.01 . 0.01 .

As mentioned above, the integral ∫ e − x 2 d x ∫ e − x 2 d x arises often in probability theory. Specifically, it is used when studying data sets that are normally distributed, meaning the data values lie under a bell-shaped curve. For example, if a set of data values is normally distributed with mean μ μ and standard deviation σ , σ , then the probability that a randomly chosen value lies between x = a x = a and x = b x = b is given by

(See Figure 6.11 .)

To simplify this integral, we typically let z = x − μ σ . z = x − μ σ . This quantity z z is known as the z z score of a data value. With this simplification, integral Equation 6.10 becomes

In Example 6.23 , we show how we can use this integral in calculating probabilities.

Example 6.23

Using maclaurin series to approximate a probability.

Suppose a set of standardized test scores are normally distributed with mean μ = 100 μ = 100 and standard deviation σ = 50 . σ = 50 . Use Equation 6.11 and the first six terms in the Maclaurin series for e − x 2 / 2 e − x 2 / 2 to approximate the probability that a randomly selected test score is between x = 100 x = 100 and x = 200 . x = 200 . Use the alternating series test to determine how accurate your approximation is.

Since μ = 100 , σ = 50 , μ = 100 , σ = 50 , and we are trying to determine the area under the curve from a = 100 a = 100 to b = 200 , b = 200 , integral Equation 6.11 becomes

The Maclaurin series for e − x 2 / 2 e − x 2 / 2 is given by

Using the first five terms, we estimate that the probability is approximately 0.4922 . 0.4922 . By the alternating series test, we see that this estimate is accurate to within

If you are familiar with probability theory, you may know that the probability that a data value is within two standard deviations of the mean is approximately 95 % . 95 % . Here we calculated the probability that a data value is between the mean and two standard deviations above the mean, so the estimate should be around 47.5 % . 47.5 % . The estimate, combined with the bound on the accuracy, falls within this range.

Checkpoint 6.22

Use the first five terms of the Maclaurin series for e − x 2 / 2 e − x 2 / 2 to estimate the probability that a randomly selected test score is between 100 100 and 150 . 150 . Use the alternating series test to determine the accuracy of this estimate.

Another application in which a nonelementary integral arises involves the period of a pendulum. The integral is

An integral of this form is known as an elliptic integral of the first kind. Elliptic integrals originally arose when trying to calculate the arc length of an ellipse. We now show how to use power series to approximate this integral.

Example 6.24

Period of a pendulum.

The period of a pendulum is the time it takes for a pendulum to make one complete back-and-forth swing. For a pendulum with length L L that makes a maximum angle θ max θ max with the vertical, its period T T is given by

where g g is the acceleration due to gravity and k = sin ( θ max 2 ) k = sin ( θ max 2 ) (see Figure 6.12 ). (We note that this formula for the period arises from a non-linearized model of a pendulum. In some cases, for simplification, a linearized model is used and sin θ sin θ is approximated by θ . ) θ . ) Use the binomial series

to estimate the period of this pendulum. Specifically, approximate the period of the pendulum if

  • you use only the first term in the binomial series, and

We use the binomial series, replacing x x with − k 2 sin 2 θ . − k 2 sin 2 θ . Then we can write the period as

  • Using just the first term in the integrand, the first-order estimate is T ≈ 4 L g ∫ 0 π / 2 d θ = 2 π L g . T ≈ 4 L g ∫ 0 π / 2 d θ = 2 π L g . If θ max θ max is small, then k = sin ( θ max 2 ) k = sin ( θ max 2 ) is small. We claim that when k k is small, this is a good estimate. To justify this claim, consider ∫ 0 π / 2 ( 1 + 1 2 k 2 sin 2 θ + 1 · 3 2 ! 2 2 k 4 sin 4 θ + ⋯ ) d θ . ∫ 0 π / 2 ( 1 + 1 2 k 2 sin 2 θ + 1 · 3 2 ! 2 2 k 4 sin 4 θ + ⋯ ) d θ . Since | sin x | ≤ 1 , | sin x | ≤ 1 , this integral is bounded by ∫ 0 π / 2 ( 1 2 k 2 + 1.3 2 ! 2 2 k 4 + ⋯ ) d θ < π 2 ( 1 2 k 2 + 1 · 3 2 ! 2 2 k 4 + ⋯ ) . ∫ 0 π / 2 ( 1 2 k 2 + 1.3 2 ! 2 2 k 4 + ⋯ ) d θ < π 2 ( 1 2 k 2 + 1 · 3 2 ! 2 2 k 4 + ⋯ ) . Furthermore, it can be shown that each coefficient on the right-hand side is less than 1 1 and, therefore, that this expression is bounded by π k 2 2 ( 1 + k 2 + k 4 + ⋯ ) = π k 2 2 · 1 1 − k 2 , π k 2 2 ( 1 + k 2 + k 4 + ⋯ ) = π k 2 2 · 1 1 − k 2 , which is small for k k small.
  • For larger values of θ max , θ max , we can approximate T T by using more terms in the integrand. By using the first two terms in the integral, we arrive at the estimate T ≈ 4 L g ∫ 0 π / 2 ( 1 + 1 2 k 2 sin 2 θ ) d θ = 2 π L g ( 1 + k 2 4 ) . T ≈ 4 L g ∫ 0 π / 2 ( 1 + 1 2 k 2 sin 2 θ ) d θ = 2 π L g ( 1 + k 2 4 ) .

The applications of Taylor series in this section are intended to highlight their importance. In general, Taylor series are useful because they allow us to represent known functions using polynomials, thus providing us a tool for approximating function values and estimating complicated integrals. In addition, they allow us to define new functions as power series, thus providing us with a powerful tool for solving differential equations.

In the following exercises, use appropriate substitutions to write down the Maclaurin series for the given binomial.

( 1 − x ) 1 / 3 ( 1 − x ) 1 / 3

( 1 + x 2 ) −1 / 3 ( 1 + x 2 ) −1 / 3

( 1 − x ) 1.01 ( 1 − x ) 1.01

( 1 − 2 x ) 2 / 3 ( 1 − 2 x ) 2 / 3

In the following exercises, use the substitution ( b + x ) r = ( b + a ) r ( 1 + x − a b + a ) r ( b + x ) r = ( b + a ) r ( 1 + x − a b + a ) r in the binomial expansion to find the Taylor series of each function with the given center.

x + 2 x + 2 at a = 0 a = 0

x 2 + 2 x 2 + 2 at a = 0 a = 0

x + 2 x + 2 at a = 1 a = 1

2 x − x 2 2 x − x 2 at a = 1 a = 1 ( Hint: 2 x − x 2 = 1 − ( x − 1 ) 2 ) 2 x − x 2 = 1 − ( x − 1 ) 2 )

( x − 8 ) 1 / 3 ( x − 8 ) 1 / 3 at a = 9 a = 9

x x at a = 4 a = 4

x 1 / 3 x 1 / 3 at a = 27 a = 27

x x at x = 9 x = 9

In the following exercises, use the binomial theorem to estimate each number, computing enough terms to obtain an estimate accurate to an error of at most 1 / 1000 . 1 / 1000 .

[T] ( 15 ) 1 / 4 ( 15 ) 1 / 4 using ( 16 − x ) 1 / 4 ( 16 − x ) 1 / 4

[T] ( 1001 ) 1 / 3 ( 1001 ) 1 / 3 using ( 1000 + x ) 1 / 3 ( 1000 + x ) 1 / 3

In the following exercises, use the binomial approximation 1 − x ≈ 1 − x 2 − x 2 8 − x 3 16 − 5 x 4 128 − 7 x 5 256 1 − x ≈ 1 − x 2 − x 2 8 − x 3 16 − 5 x 4 128 − 7 x 5 256 for | x | < 1 | x | < 1 to approximate each number. Compare this value to the value given by a scientific calculator.

[T] 1 2 1 2 using x = 1 2 x = 1 2 in ( 1 − x ) 1 / 2 ( 1 − x ) 1 / 2

[T] 5 = 5 × 1 5 5 = 5 × 1 5 using x = 4 5 x = 4 5 in ( 1 − x ) 1 / 2 ( 1 − x ) 1 / 2

[T] 3 = 3 3 3 = 3 3 using x = 2 3 x = 2 3 in ( 1 − x ) 1 / 2 ( 1 − x ) 1 / 2

[T] 6 6 using x = 5 6 x = 5 6 in ( 1 − x ) 1 / 2 ( 1 − x ) 1 / 2

Integrate the binomial approximation of 1 − x 1 − x to find an approximation of ∫ 0 x 1 − t d t . ∫ 0 x 1 − t d t .

[T] Recall that the graph of 1 − x 2 1 − x 2 is an upper semicircle of radius 1 . 1 . Integrate the binomial approximation of 1 − x 2 1 − x 2 up to order 8 8 from x = −1 x = −1 to x = 1 x = 1 to estimate π 2 . π 2 .

In the following exercises, use the expansion ( 1 + x ) 1 / 3 = 1 + 1 3 x − 1 9 x 2 + 5 81 x 3 − 10 243 x 4 + ⋯ ( 1 + x ) 1 / 3 = 1 + 1 3 x − 1 9 x 2 + 5 81 x 3 − 10 243 x 4 + ⋯ to write the first five terms (not necessarily a quartic polynomial) of each expression.

( 1 + 4 x ) 1 / 3 ; a = 0 ( 1 + 4 x ) 1 / 3 ; a = 0

( 1 + 4 x ) 4 / 3 ; a = 0 ( 1 + 4 x ) 4 / 3 ; a = 0

( 3 + 2 x ) 1 / 3 ; a = −1 ( 3 + 2 x ) 1 / 3 ; a = −1

( x 2 + 6 x + 10 ) 1 / 3 ; a = −3 ( x 2 + 6 x + 10 ) 1 / 3 ; a = −3

Use ( 1 + x ) 1 / 3 = 1 + 1 3 x − 1 9 x 2 + 5 81 x 3 − 10 243 x 4 + ⋯ ( 1 + x ) 1 / 3 = 1 + 1 3 x − 1 9 x 2 + 5 81 x 3 − 10 243 x 4 + ⋯ with x = 1 x = 1 to approximate 2 1 / 3 . 2 1 / 3 .

Use the approximation ( 1 − x ) 2 / 3 = 1 − 2 x 3 − x 2 9 − 4 x 3 81 − 7 x 4 243 − 14 x 5 729 + ⋯ ( 1 − x ) 2 / 3 = 1 − 2 x 3 − x 2 9 − 4 x 3 81 − 7 x 4 243 − 14 x 5 729 + ⋯ for | x | < 1 | x | < 1 to approximate 2 1 / 3 = 2.2 −2 / 3 . 2 1 / 3 = 2.2 −2 / 3 .

Find the 25 th 25 th derivative of f ( x ) = ( 1 + x 2 ) 13 f ( x ) = ( 1 + x 2 ) 13 at x = 0 . x = 0 .

Find the 99 99 th derivative at x = 0 x = 0 of f ( x ) = ( 1 + x 4 ) 25 . f ( x ) = ( 1 + x 4 ) 25 .

In the following exercises, find the Maclaurin series of each function.

f ( x ) = x e 2 x f ( x ) = x e 2 x

f ( x ) = 2 x f ( x ) = 2 x

f ( x ) = sin x x f ( x ) = sin x x

f ( x ) = sin ( x ) x , ( x > 0 ) , f ( x ) = sin ( x ) x , ( x > 0 ) ,

f ( x ) = sin ( x 2 ) f ( x ) = sin ( x 2 )

f ( x ) = e x 3 f ( x ) = e x 3

f ( x ) = cos 2 x f ( x ) = cos 2 x using the identity cos 2 x = 1 2 + 1 2 cos ( 2 x ) cos 2 x = 1 2 + 1 2 cos ( 2 x )

f ( x ) = sin 2 x f ( x ) = sin 2 x using the identity sin 2 x = 1 2 − 1 2 cos ( 2 x ) sin 2 x = 1 2 − 1 2 cos ( 2 x )

In the following exercises, find the Maclaurin series of F ( x ) = ∫ 0 x f ( t ) d t F ( x ) = ∫ 0 x f ( t ) d t by integrating the Maclaurin series of f f term by term. If f f is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero.

F ( x ) = ∫ 0 x e − t 2 d t ; f ( t ) = e − t 2 = ∑ n = 0 ∞ ( −1 ) n t 2 n n ! F ( x ) = ∫ 0 x e − t 2 d t ; f ( t ) = e − t 2 = ∑ n = 0 ∞ ( −1 ) n t 2 n n !

F ( x ) = tan −1 x ; f ( t ) = 1 1 + t 2 = ∑ n = 0 ∞ ( −1 ) n t 2 n F ( x ) = tan −1 x ; f ( t ) = 1 1 + t 2 = ∑ n = 0 ∞ ( −1 ) n t 2 n

F ( x ) = tanh −1 x ; f ( t ) = 1 1 − t 2 = ∑ n = 0 ∞ t 2 n F ( x ) = tanh −1 x ; f ( t ) = 1 1 − t 2 = ∑ n = 0 ∞ t 2 n

F ( x ) = sin −1 x ; f ( t ) = 1 1 − t 2 = ∑ k = 0 ∞ ( 1 2 k ) t 2 k k ! F ( x ) = sin −1 x ; f ( t ) = 1 1 − t 2 = ∑ k = 0 ∞ ( 1 2 k ) t 2 k k !

F ( x ) = ∫ 0 x sin t t d t ; f ( t ) = sin t t = ∑ n = 0 ∞ ( −1 ) n t 2 n ( 2 n + 1 ) ! F ( x ) = ∫ 0 x sin t t d t ; f ( t ) = sin t t = ∑ n = 0 ∞ ( −1 ) n t 2 n ( 2 n + 1 ) !

F ( x ) = ∫ 0 x cos ( t ) d t ; f ( t ) = ∑ n = 0 ∞ ( −1 ) n x n ( 2 n ) ! F ( x ) = ∫ 0 x cos ( t ) d t ; f ( t ) = ∑ n = 0 ∞ ( −1 ) n x n ( 2 n ) !

F ( x ) = ∫ 0 x 1 − cos t t 2 d t ; f ( t ) = 1 − cos t t 2 = ∑ n = 0 ∞ ( −1 ) n t 2 n ( 2 n + 2 ) ! F ( x ) = ∫ 0 x 1 − cos t t 2 d t ; f ( t ) = 1 − cos t t 2 = ∑ n = 0 ∞ ( −1 ) n t 2 n ( 2 n + 2 ) !

F ( x ) = ∫ 0 x ln ( 1 + t ) t d t ; f ( t ) = ∑ n = 0 ∞ ( −1 ) n t n n + 1 F ( x ) = ∫ 0 x ln ( 1 + t ) t d t ; f ( t ) = ∑ n = 0 ∞ ( −1 ) n t n n + 1

In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of f . f .

f ( x ) = sin ( x + π 4 ) = sin x cos ( π 4 ) + cos x sin ( π 4 ) f ( x ) = sin ( x + π 4 ) = sin x cos ( π 4 ) + cos x sin ( π 4 )

f ( x ) = tan x f ( x ) = tan x

f ( x ) = ln ( cos x ) f ( x ) = ln ( cos x )

f ( x ) = e x cos x f ( x ) = e x cos x

f ( x ) = e sin x f ( x ) = e sin x

f ( x ) = sec 2 x f ( x ) = sec 2 x

f ( x ) = tanh x f ( x ) = tanh x

f ( x ) = tan x x f ( x ) = tan x x (see expansion for tan x ) tan x )

In the following exercises, find the radius of convergence of the Maclaurin series of each function.

ln ( 1 + x ) ln ( 1 + x )

1 1 + x 2 1 1 + x 2

tan −1 x tan −1 x

ln ( 1 + x 2 ) ln ( 1 + x 2 )

Find the Maclaurin series of sinh x = e x − e − x 2 . sinh x = e x − e − x 2 .

Find the Maclaurin series of cosh x = e x + e − x 2 . cosh x = e x + e − x 2 .

Differentiate term by term the Maclaurin series of sinh x sinh x and compare the result with the Maclaurin series of cosh x . cosh x .

[T] Let S n ( x ) = ∑ k = 0 n ( −1 ) k x 2 k + 1 ( 2 k + 1 ) ! S n ( x ) = ∑ k = 0 n ( −1 ) k x 2 k + 1 ( 2 k + 1 ) ! and C n ( x ) = ∑ n = 0 n ( −1 ) k x 2 k ( 2 k ) ! C n ( x ) = ∑ n = 0 n ( −1 ) k x 2 k ( 2 k ) ! denote the respective Maclaurin polynomials of degree 2 n + 1 2 n + 1 of sin x sin x and degree 2 n 2 n of cos x . cos x . Plot the errors S n ( x ) C n ( x ) − tan x S n ( x ) C n ( x ) − tan x for n = 1 , .. , 5 n = 1 , .. , 5 and compare them to x + x 3 3 + 2 x 5 15 + 17 x 7 315 − tan x x + x 3 3 + 2 x 5 15 + 17 x 7 315 − tan x on ( − π 4 , π 4 ) . ( − π 4 , π 4 ) .

Use the identity 2 sin x cos x = sin ( 2 x ) 2 sin x cos x = sin ( 2 x ) to find the power series expansion of sin 2 x sin 2 x at x = 0 . x = 0 . ( Hint: Integrate the Maclaurin series of sin ( 2 x ) sin ( 2 x ) term by term.)

If y = ∑ n = 0 ∞ a n x n , y = ∑ n = 0 ∞ a n x n , find the power series expansions of x y ′ x y ′ and x 2 y ″ . x 2 y ″ .

[T] Suppose that y = ∑ k = 0 ∞ a k x k y = ∑ k = 0 ∞ a k x k satisfies y ′ = −2 x y y ′ = −2 x y and y ( 0 ) = 0 . y ( 0 ) = 0 . Show that a 2 k + 1 = 0 a 2 k + 1 = 0 for all k k and that a 2 k + 2 = − a 2 k k + 1 . a 2 k + 2 = − a 2 k k + 1 . Plot the partial sum S 20 S 20 of y y on the interval [ −4 , 4 ] . [ −4 , 4 ] .

[T] Suppose that a set of standardized test scores is normally distributed with mean μ = 100 μ = 100 and standard deviation σ = 10 . σ = 10 . Set up an integral that represents the probability that a test score will be between 90 90 and 110 110 and use the integral of the degree 10 10 Maclaurin polynomial of 1 2 π e − x 2 / 2 1 2 π e − x 2 / 2 to estimate this probability.

[T] Suppose that a set of standardized test scores is normally distributed with mean μ = 100 μ = 100 and standard deviation σ = 10 . σ = 10 . Set up an integral that represents the probability that a test score will be between 70 70 and 130 130 and use the integral of the degree 50 50 Maclaurin polynomial of 1 2 π e − x 2 / 2 1 2 π e − x 2 / 2 to estimate this probability.

[T] Suppose that ∑ n = 0 ∞ a n x n ∑ n = 0 ∞ a n x n converges to a function f ( x ) f ( x ) such that f ( 0 ) = 1 , f ′ ( 0 ) = 0 , f ( 0 ) = 1 , f ′ ( 0 ) = 0 , and f ″ ( x ) = − f ( x ) . f ″ ( x ) = − f ( x ) . Find a formula for a n a n and plot the partial sum S N S N for N = 20 N = 20 on [ −5 , 5 ] . [ −5 , 5 ] .

[T] Suppose that ∑ n = 0 ∞ a n x n ∑ n = 0 ∞ a n x n converges to a function f ( x ) f ( x ) such that f ( 0 ) = 0 , f ′ ( 0 ) = 1 , f ( 0 ) = 0 , f ′ ( 0 ) = 1 , and f ″ ( x ) = − f ( x ) . f ″ ( x ) = − f ( x ) . Find a formula for a n a n and plot the partial sum S N S N for N = 10 N = 10 on [ −5 , 5 ] . [ −5 , 5 ] .

Suppose that ∑ n = 0 ∞ a n x n ∑ n = 0 ∞ a n x n converges to a function y y such that y ″ − y ′ + y = 0 y ″ − y ′ + y = 0 where y ( 0 ) = 1 y ( 0 ) = 1 and y ′ ( 0 ) = 0 . y ′ ( 0 ) = 0 . Find a formula that relates a n + 2 , a n + 1 , a n + 2 , a n + 1 , and a n a n and compute a 0 , ... , a 5 . a 0 , ... , a 5 .

Suppose that ∑ n = 0 ∞ a n x n ∑ n = 0 ∞ a n x n converges to a function y y such that y ″ − y ′ + y = 0 y ″ − y ′ + y = 0 where y ( 0 ) = 0 y ( 0 ) = 0 and y ′ ( 0 ) = 1 . y ′ ( 0 ) = 1 . Find a formula that relates a n + 2 , a n + 1 , a n + 2 , a n + 1 , and a n a n and compute a 1 , ... , a 5 . a 1 , ... , a 5 .

The error in approximating the integral ∫ a b f ( t ) d t ∫ a b f ( t ) d t by that of a Taylor approximation ∫ a b P n ( t ) d t ∫ a b P n ( t ) d t is at most ∫ a b R n ( t ) d t . ∫ a b R n ( t ) d t . In the following exercises, the Taylor remainder estimate R n ≤ M ( n + 1 ) ! | x − a | n + 1 R n ≤ M ( n + 1 ) ! | x − a | n + 1 guarantees that the integral of the Taylor polynomial of the given order approximates the integral of f f with an error less than 1 10 . 1 10 .

  • Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less than 1 100 . 1 100 .
  • Compare the accuracy of the polynomial integral estimate with the remainder estimate.

[T] ∫ 0 π sin t t d t ; P s = 1 − x 2 3 ! + x 4 5 ! − x 6 7 ! + x 8 9 ! ∫ 0 π sin t t d t ; P s = 1 − x 2 3 ! + x 4 5 ! − x 6 7 ! + x 8 9 ! (You may assume that the absolute value of the ninth derivative of sin t t sin t t is bounded by 0.1 . ) 0.1 . )

[T] ∫ 0 2 e − x 2 d x ; p 11 = 1 − x 2 + x 4 2 − x 6 3 ! + ⋯ − x 22 11 ! ∫ 0 2 e − x 2 d x ; p 11 = 1 − x 2 + x 4 2 − x 6 3 ! + ⋯ − x 22 11 ! (You may assume that the absolute value of the 23 rd 23 rd derivative of e − x 2 e − x 2 is less than 2 × 10 14 . ) 2 × 10 14 . )

The following exercises deal with Fresnel integrals .

The Fresnel integrals are defined by C ( x ) = ∫ 0 x cos ( t 2 ) d t C ( x ) = ∫ 0 x cos ( t 2 ) d t and S ( x ) = ∫ 0 x sin ( t 2 ) d t . S ( x ) = ∫ 0 x sin ( t 2 ) d t . Compute the power series of C ( x ) C ( x ) and S ( x ) S ( x ) and plot the sums C N ( x ) C N ( x ) and S N ( x ) S N ( x ) of the first N = 50 N = 50 nonzero terms on [ 0 , 2 π ] . [ 0 , 2 π ] .

[T] The Fresnel integrals are used in design applications for roadways and railways and other applications because of the curvature properties of the curve with coordinates ( C ( t ) , S ( t ) ) . ( C ( t ) , S ( t ) ) . Plot the curve ( C 50 , S 50 ) ( C 50 , S 50 ) for 0 ≤ t ≤ 2 π , 0 ≤ t ≤ 2 π , the coordinates of which were computed in the previous exercise.

Estimate ∫ 0 1 / 4 x − x 2 d x ∫ 0 1 / 4 x − x 2 d x by approximating 1 − x 1 − x using the binomial approximation 1 − x 2 − x 2 8 − x 3 16 − 5 x 4 2128 − 7 x 5 256 . 1 − x 2 − x 2 8 − x 3 16 − 5 x 4 2128 − 7 x 5 256 .

[T] Use Newton’s approximation of the binomial 1 − x 2 1 − x 2 to approximate π π as follows. The circle centered at ( 1 2 , 0 ) ( 1 2 , 0 ) with radius 1 2 1 2 has upper semicircle y = x 1 − x . y = x 1 − x . The sector of this circle bounded by the x x -axis between x = 0 x = 0 and x = 1 2 x = 1 2 and by the line joining ( 1 4 , 3 4 ) ( 1 4 , 3 4 ) corresponds to 1 6 1 6 of the circle and has area π 24 . π 24 . This sector is the union of a right triangle with height 3 4 3 4 and base 1 4 1 4 and the region below the graph between x = 0 x = 0 and x = 1 4 . x = 1 4 . To find the area of this region you can write y = x 1 − x = x × ( binomial expansion of 1 − x ) y = x 1 − x = x × ( binomial expansion of 1 − x ) and integrate term by term. Use this approach with the binomial approximation from the previous exercise to estimate π . π .

Use the approximation T ≈ 2 π L g ( 1 + k 2 4 ) T ≈ 2 π L g ( 1 + k 2 4 ) to approximate the period of a pendulum having length 10 10 meters and maximum angle θ max = π 6 θ max = π 6 where k = sin ( θ max 2 ) . k = sin ( θ max 2 ) . Compare this with the small angle estimate T ≈ 2 π L g . T ≈ 2 π L g .

Suppose that a pendulum is to have a period of 2 2 seconds and a maximum angle of θ max = π 6 . θ max = π 6 . Use T ≈ 2 π L g ( 1 + k 2 4 ) T ≈ 2 π L g ( 1 + k 2 4 ) to approximate the desired length of the pendulum. What length is predicted by the small angle estimate T ≈ 2 π L g ? T ≈ 2 π L g ?

Evaluate ∫ 0 π / 2 sin 4 θ d θ ∫ 0 π / 2 sin 4 θ d θ in the approximation T = 4 L g ∫ 0 π / 2 ( 1 + 1 2 k 2 sin 2 θ + 3 8 k 4 sin 4 θ + ⋯ ) d θ T = 4 L g ∫ 0 π / 2 ( 1 + 1 2 k 2 sin 2 θ + 3 8 k 4 sin 4 θ + ⋯ ) d θ to obtain an improved estimate for T . T .

[T] An equivalent formula for the period of a pendulum with amplitude θ max θ max is T ( θ max ) = 2 2 L g ∫ 0 θ max d θ cos θ − cos ( θ max ) T ( θ max ) = 2 2 L g ∫ 0 θ max d θ cos θ − cos ( θ max ) where L L is the pendulum length and g g is the gravitational acceleration constant. When θ max = π 3 θ max = π 3 we get 1 cos t − 1 / 2 ≈ 2 ( 1 + t 2 2 + t 4 3 + 181 t 6 720 ) . 1 cos t − 1 / 2 ≈ 2 ( 1 + t 2 2 + t 4 3 + 181 t 6 720 ) . Integrate this approximation to estimate T ( π 3 ) T ( π 3 ) in terms of L L and g . g . Assuming g = 9.806 g = 9.806 meters per second squared, find an approximate length L L such that T ( π 3 ) = 2 T ( π 3 ) = 2 seconds.

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Access for free at https://openstax.org/books/calculus-volume-2/pages/1-introduction
  • Authors: Gilbert Strang, Edwin “Jed” Herman
  • Publisher/website: OpenStax
  • Book title: Calculus Volume 2
  • Publication date: Mar 30, 2016
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/calculus-volume-2/pages/1-introduction
  • Section URL: https://openstax.org/books/calculus-volume-2/pages/6-4-working-with-taylor-series

© Feb 5, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

  • Math Article

Taylor Series

Taylor series is the polynomial or a function of an infinite sum of terms. Each successive term will have a larger exponent or higher degree than the preceding term.

The above Taylor series expansion is given for a real values function f(x) where f’(a), f’’(a), f’’’(a), etc., denotes the derivative of the function at point a. If the value of point ‘a’ is zero, then the Taylor series is also called the Maclaurin series.

Taylor’s Series Theorem

Assume that if f(x) be a real or composite function, which is a differentiable function of a neighbourhood number that is also real or composite. Then, the Taylor series describes the following power series :

In terms of sigma notation, the Taylor series can be written as

f (n) (a) = n th derivative of f

n! = factorial of n.

We know that the power series can be defined as

When x = 0,

So, differentiate the given function, it becomes,

f’(x) = a 1 + 2a 2 x + 3a 3 x 2 + 4a 4 x 3 +….

Again, when you substitute x = 0, we get

So, differentiate it again, we get

f”(x) = 2a 2 + 6a 3 x +12a 4 x 2 + …

Now, substitute x=0 in second-order differentiation, we get

f”(0) = 2a 2

Therefore, [f”(0)/2!] = a 2

By generalising the equation, we get

f n (0) / n! = a n

Now substitute the values in the power series we get,

Generalise f in more general form, it becomes

f(x) = b + b 1 (x-a) + b 2 ( x-a) 2 + b 3 (x-a) 3 + ….

Now, x = a, we get

b n = f n (a) / n!

Now, substitute b n in a generalised form

Hence, the Taylor series is proved.

Taylor Series of Sin x

f(x) = sin x 

f’(x) = cos x

f’’(x) = -sin x

f’’’(x) = -cos x

f’’’’(x) = sin x

The taylor series for Sin x at x = 0, is given by:

\(\begin{array}{l}\begin{aligned} \sin x &=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n !} x^{n} \\ &=\frac{1}{1 !} x-\frac{1}{3 !} x^{3}+\frac{1}{5 !} x^{5}-\frac{1}{7 !} x^{7}+\cdots \end{aligned}\end{array} \)

Taylor Series in Several Variables

The Taylor series is also represented in the form of functions of several variables. The general form of the Taylor series in several variables is

Maclaurin Series

If the Taylor Series is centred at 0, then the series is known as the Maclaurin series. It means that,

If a= 0 in the Taylor series, then we get;

This is known as the Maclaurin series.

Example: Maclaurin series of 1/(1-x) is given by:

1+x+x 2 +x 3 +x 4 +…,

Applications of Taylor Series

The uses of the Taylor series are:

  • Taylor series is used to evaluate the value of a whole function in each point if the functional values and derivatives are identified at a single point.
  • The representation of Taylor series reduces many mathematical proofs.
  • The sum of partial series can be used as an approximation of the whole series.
  • Multivariate Taylor series is used in many optimization techniques.
  • This series is used in the power flow analysis of electrical power systems.

Problems and Solutions

Question 1: Determine the Taylor series at x=0 for f(x) = e x

Solution: Given: f(x) = e x

Differentiate the given equation,

f’(x) = e x

f’’(x) =e x

f’’’(x) = e x

At x=0, we get

f’(0) = e 0 =1

f’’(0) = e 0 =1

f’’’(0) = e 0 = 1

When Taylor series at x= 0, then the Maclaurin series is

e x = 1+ x(1) + (x 2 /2!)(1) + (x 3 /3!)(1) + …..

Therefore, e x = 1+ x + (x 2 /2!) + (x 3 /3!)+ …..

Question 2: EValuate the Taylor Series for  f ( x ) = cos ( x )  for  x = 0.

Solution: We need to take the derivatives of the cos x and evaluate them at  x = 0.

f(x) = cos x  ⇒ f(0) = 1

f’(x) = -sin x ⇒ f’(0) = 0

f’’(x) = -cos x ⇒ f’’(0) = -1

f’’’(x) = sin x ⇒ f’’’(0) = 0

f’’’’(x) = cos x ⇒ f’’’’(4) = 1

f (5) (x) = -sin x ⇒ f (5) (0) = 0

f (6) (x) = -cos x ⇒ f (6) (0) = -1

Therefore, according to the Taylor series expansion;

\(\begin{array}{l}\begin{aligned} \cos x &=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n !} x^{n} \\ &=f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\frac{f^{\prime \prime \prime}(0)}{3 !} x^{3}+\frac{f^{(4)}(0)}{4 !} x^{4}+\frac{f^{(5)}(0)}{5 !} x^{5}+\cdots \\ &={1}+{0}-\frac{1}{2 !} x^{2}+{0}+\frac{1}{4 !} x^{4}+{0}-\frac{1}{6 !} x^{6}+\cdots \end{aligned}\end{array} \)

\(\begin{array}{l}\cos x={1}-\frac{1}{2 !} x^{2}+\frac{1}{4 !} x^{4}-\frac{1}{6 !} x^{6}+\cdots\end{array} \)

\(\begin{array}{l}\cos x=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !}\end{array} \)

Question 3: Evaluate the Taylor Series for  f ( x ) = x 3 − 10x 2 + 6  at  x = 3.

Solution: First, we will find the derivatives of the given function.

f(x) =  x 3 − 10x 2 + 6 ⇒ f(3) = -57

f’(x) = 3x 2 − 20x ⇒ f’(3) = 33

f’’(x) = 6x – 20 ⇒ f’’(3) = -2

f’’’(x) = 6 ⇒ f’’’(3) = 6

f’’’’(x) = 0 

Therefore, the required series is:

\(\begin{array}{l}\begin{aligned} x^{3}-10 x^{2}+6 &=\sum_{n=0}^{\infty} \frac{f^{(n)}(3)}{n !}(x-3)^{n} \\ &=f(3)+f^{\prime}(3)(x-3)+\frac{f^{\prime \prime}(3)}{2 !}(x-3)^{2}+\frac{f^{\prime \prime \prime}(3)}{3 !}(x-3)^{3}+0 \\ &=-57-33(x-3)-(x-3)^{2}+(x-3)^{3} \end{aligned}\end{array} \)

Frequently Asked Questions – FAQs

What is a taylor series, what is the use of taylor series, what is a maclaurin series, write the taylor series for tan x, what is taylor series expansion of sec x.

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Taylor Series - Error Bounds

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The Lagrange error bound of a Taylor polynomial gives the worst-case scenario for the difference between the estimated value of the function as provided by the Taylor polynomial and the actual value of the function. This error bound \(\big(R_n(x)\big)\) is the maximum value of the \((n+1)^\text{th}\) term of the Taylor expansion, where \(M\) is an upper bound of the \((n+1)^\text{th}\) derivative for \(a<z<x:\)

\[R_n(x)=\frac{M}{(n+1)!}(x-a)^{n+1}.\]

Calculating Error Bounds

The \(n^\text{th}\) degree Taylor polynomial at \(x=a\) is

\[P_n(x) = f(a) + \frac{f'(a)}{1!}(x-a) +\cdots+ \frac{f^n(a)}{n!}(x-a)^n.\]

Since the Taylor approximation becomes more accurate as more terms are included, the \(P_{n+1}(x)\) polynomial must be more accurate than \(P_n(x):\)

\[\begin{align} P_{n+1}(x) &= f(a) + \frac{f'(a)}{1!}(x-a) +\cdots+ \frac{f^n(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}\\ &= P_n(x) + \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}. \end{align}\]

Since the difference between \(P_n(x)\) and \(P_{n+1}(x)\) is just that last term, the error of \(P_n(x)\) can be no larger than that term. In other words, the error \(R_n\) is

\[R_n(x) = \max\left( \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1} \right).\]

Since \(a\) and \(n\) are constant in this formula, terms depending only on those constants and \(x\) are unaffected by the \(\max\) operator and can be pulled outside:

\[R_n(x) =\frac{\max\big( f^{(n+1)}(a)\big)}{(n+1)!} (x-a)^{n+1}.\]

The largest value obtainable by \(f^{n+1}\) could not possibly exceed the maximum value of that derivative between \(a\) and \(x.\) Call the \(x\) value that provides that maximum value \(z\) and the error becomes

\[R_n(x)=\frac{f^{(n+1)}(z)}{(n+1)!}(x-a)^{n+1}.\]

Let \(M\) be an upper bound on the \((n+1)^\text{th}\) derivative of \(f(x)\) for the interval between \(a\) and \(x\) such that \[\big| f^{(n+1)}(z) \big| \leq M\] for all \(z\in [a, x].\)

The upper bound of the \((n+1)^\text{th}\) derivative on the interval \([a, x]\) will usually occur at \(z=a\) or \(z=x.\) If given a defined interval on which to find the error, test the endpoints of the interval.

What is the upper bound of the third derivative of \(y = \sin(x)\) on the interval \([0, 2\pi]?\) The third derivative of \(y=\sin(x)\) is \(y^{(3)} = -\cos(x),\) which oscillates between -1 and 1. So \(M=1\) and \[\big| f^{(n+1)}(z) \big| \leq 1.\ _\square\]

What is the supremum of the fourth derivative of \(f(x) = e^{ax}\) on the interval \([a, 0]\) for \(-1<a<0?\)

In order to compute the error bound, follow these steps: Step 1: Compute the \((n+1)^\text{th}\) derivative of \(f(x).\) Step 2: Find the upper bound on \(f^{(n+1)}(z)\) for \(z\in [a, x].\) Step 3: Compute \(R_n(x).\)
Find the error bound of the Maclaurin polynomial \(P_3\big(\frac{\pi}{2}\big)\) for \(f(x) = \sin(x).\) The Maclaurin series is just a Taylor series centered at \(a=0.\) Follow the prescribed steps. Step 1: Compute the \((n+1)^\text{th}\) derivative of \(f(x):\) Since \(P_3\) is being investigated, \(n = 3,\) so write down the \(4^\text{th}\) derivative of \(f(x) = \sin(x):\) \[f^{(4)}(x) = \sin(x).\] Step 2: Find the upper bound on \(f^{(n+1)}(z)\) for \(z\in [a, x]:\) The Maclaurin series is centered on \(a = 0\) and \(P_n(x)=P_3\big(\frac{\pi}{2}\big)\) implies \(x = \frac{\pi}{2}:\) \[\begin{align} f^{(4)}(0) &= \sin(0)=0\\ f^{(4)}\left(\frac{\pi}{2}\right) &= \sin\left(\frac{\pi}{2}\right)=1. \end{align}\] So \(M=1.\) Step 3: Compute \(R_n(x):\) \[\begin{align} R_n(x)&=\frac{M}{(n+1)!}(x-a)^{n+1}\\ R_3\left(\frac{\pi}{2}\right)&=\frac{1}{(3+1)!}\left(\frac{\pi}{2}-0\right)^{3+1}\\ &= \frac{\pi^4}{384}.\ _\square \end{align}\]

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Engineering LibreTexts

14.2.7.4: Working with Taylor Series

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Learning Objectives

  • Write the terms of the binomial series.
  • Recognize the Taylor series expansions of common functions.
  • Recognize and apply techniques to find the Taylor series for a function.
  • Use Taylor series to solve differential equations.
  • Use Taylor series to evaluate nonelementary integrals.

In the preceding section, we defined Taylor series and showed how to find the Taylor series for several common functions by explicitly calculating the coefficients of the Taylor polynomials. In this section we show how to use those Taylor series to derive Taylor series for other functions. We then present two common applications of power series. First, we show how power series can be used to solve differential equations. Second, we show how power series can be used to evaluate integrals when the antiderivative of the integrand cannot be expressed in terms of elementary functions. In one example, we consider \( \int e^{−x^2}dx,\) an integral that arises frequently in probability theory.

The Binomial Series

Our first goal in this section is to determine the Maclaurin series for the function \( f(x)=(1+x)^r\) for all real numbers \( r\). The Maclaurin series for this function is known as the binomial series. We begin by considering the simplest case: \( r\) is a nonnegative integer. We recall that, for \( r=0,1,2,3,4,f(x)=(1+x)^r\) can be written as

\[\begin{align} f(x) =(1+x)^0=1 \nonumber \\ f(x) =(1+x)^1=1+x \nonumber\\ f(x) =(1+x)^2=1+2x+x^2 \nonumber \\ f(x) =(1+x)^3=1+3x+3x^2+x^3 \nonumber\\ f(x) =(1+x)^4=1+4x+6x^2+4x^3+x^4\nonumber \end{align}\]

The expressions on the right-hand side are known as binomial expansions and the coefficients are known as binomial coefficients. More generally, for any nonnegative integer \( r\), the binomial coefficient of \( x^n\) in the binomial expansion of \( (1+x)^r\) is given by

\[(^r_n)=\dfrac{r!}{n!(r−n)!}\]

\[f(x)=(1+x)^r= \left(^r_0 \right)1+\left(^r_1\right)x+\left(^r_2\right)x^2+\left(^r_3\right)x^3+⋯+\left(^r_{r−1}\right)x^{r−1}+\left(^r_r\right)x^r=\sum_{n=0}^r\left(^r_n\right)x^n\]

For example, using this formula for \( r=5\), we see that

\[ \begin{align*} f(x) =(1+x)^5 \\ =(^5_0)1+(^5_1)x+(^5_2)x^2+(^5_3)x^3+(^5_4)x^4+(^5_5)x^5 \\ =\dfrac{5!}{0!5!}1+\dfrac{5!}{1!4!}x+\dfrac{5!}{2!3!}x^2+\dfrac{5!}{3!2!}x^3+\dfrac{5!}{4!1!}x^4+\dfrac{5!}{5!0!}x^5 \\ =1+5x+10x^2+10x^3+5x^4+x^5 \end{align*}\]

We now consider the case when the exponent \(r\)

is any real number, not necessarily a nonnegative integer. If \( r\) is not a nonnegative integer, then \( f(x)=(1+x)^r\) cannot be written as a finite polynomial. However, we can find a power series for \( f\). Specifically, we look for the Maclaurin series for \( f\). To do this, we find the derivatives of \(f\) and evaluate them at \( x=0\).

\[ \begin{align*} f(x) =(1+x)^r \\[4pt] f(0) =1 \\[4pt] f′(x) =r(1+x)^{r−1} \\[4pt] f'(0) =r \\[4pt] f''(x) =r(r−1)(1+x)^{r−2}\\[4pt] f''(0) =r(r−1) \\[4pt] f'''(x) =r(r−1)(r−2)(1+x)^{r−3} \\[4pt] f'''(0) =r(r−1)(r−2) \\[4pt] f(n)(x) =r(r−1)(r−2)⋯(r−n+1)(1+x)^{r−n} \\[4pt] f^{(n)}(0) =r(r−1)(r−2)⋯(r−n+1) \end{align*}\]

We conclude that the coefficients in the binomial series are given by

\[\dfrac{f^{(n)}(0)}{n!}=\dfrac{r(r−1)(r−2)⋯(r−n+1)}{n!}\]

We note that if \(r\) is a nonnegative integer, then the \((r+1)st\) derivative \( f^{(r+1)}\) is the zero function, and the series terminates. In addition, if \( r\) is a nonnegative integer, then Equation for the coefficients agrees with Equation for the coefficients, and the formula for the binomial series agrees with Equation for the finite binomial expansion. More generally, to denote the binomial coefficients for any real number \( r\), we define

\[(^r_n)=\dfrac{(r−1)(r−2)⋯(r−n+1)}{n!}\]

With this notation, we can write the binomial series for \( (1+x)^r\) as

\[\sum_{n=0}^∞(^r_n)x^n=1+rx+\dfrac{r(r−1)}{2!}x^2+⋯+\dfrac{r(r−1)⋯(r−n+1)}{n!}x^n+⋯ \label{bin1}\]

We now need to determine the interval of convergence for the binomial series Equation \ref{bin1}. We apply the ratio test. Consequently, we consider

\[\begin{align*} \dfrac{|a_{n+1}|}{|a_n|} =\dfrac{|r(r−1)(r−2)⋯(r−n)|x||^{n+1}}{(n+1)!}⋅\dfrac{n}{|r(r−1)(r−2)⋯(r−n+1)||x|^n} \\[4pt] =\dfrac{|r−n||x|}{|n+1|} \end{align*}\]

\[\lim_{n→∞}\dfrac{|a_{n+1}|}{|a_n|}=|x|<1\]

if and only if \( |x|<1\), we conclude that the interval of convergence for the binomial series is \( (−1,1)\). The behavior at the endpoints depends on \( r\). It can be shown that for \( r≥0\) the series converges at both endpoints; for \( −1<r<0\), the series converges at \( x=1\) and diverges at \( x=−1\); and for \( r<−1\), the series diverges at both endpoints. The binomial series does converge to \( (1+x)^r\) in \( (−1,1)\) for all real numbers \( r\), but proving this fact by showing that the remainder \( R_n(x)→0\) is difficult.

Definition: binomial series

For any real number \( r\), the Maclaurin series for \( f(x)=(1+x)^r\) is the binomial series. It converges to \( f\) for \( |x|<1\), and we write

\[(1+x)^r=\sum_{n=0}^∞(^r_n)x^n=1+rx+\dfrac{r(r−1)}{2!}x^2+⋯+r\dfrac{(r−1)⋯(r−n+1)}{n!}x^n+⋯\]

for \( |x|<1\).

We can use this definition to find the binomial series for \( f(x)=\sqrt{1+x}\) and use the series to approximate \( \sqrt{1.5}\).

Example \(\PageIndex{1}\): Finding Binomial Series

  • Find the binomial series for \( f(x)=\sqrt{1+x}\).
  • Use the third-order Maclaurin polynomial \( p_3(x)\) to estimate \( \sqrt{1.5}\). Use Taylor’s theorem to bound the error. Use a graphing utility to compare the graphs of \( f\) and \( p_3\).

a. Here \( r=\dfrac{1}{2}\). Using the definition for the binomial series, we obtain

\( \sqrt{1+x}=1+\dfrac{1}{2}x+\dfrac{(1/2)(−1/2)}{2!}x^2+\dfrac{(1/2)(−1/2)(−3/2)}{3!}x^3+⋯\)

\( =1+\dfrac{1}{2}x−\dfrac{1}{2!}\dfrac{1}{2^2}x^2+\dfrac{1}{3!}\dfrac{1⋅3}{2^3}x^3−⋯+\dfrac{(−1)^{n+1}}{n!}\dfrac{1⋅3⋅5⋯(2n−3)}{2^n}x^n+⋯\)

\( =1+\sum_{n=1}^∞\dfrac{(−1)^{n+1}}{n!}\dfrac{1⋅3⋅5⋯(2n−3)}{2^n}x^n.\)

b. From the result in part a. the third-order Maclaurin polynomial is

\( p_3(x)=1+\dfrac{1}{2}x−\dfrac{1}{8}x^2+\dfrac{1}{16}x^3\).

\( \sqrt{1.5}=\sqrt{1+0.5}≈1+\dfrac{1}{2}(0.5)−\dfrac{1}{8}(0.5)^2+\dfrac{1}{16}(0.5)^3≈1.2266.\)

From Taylor’s theorem, the error satisfies

\( R_3(0.5)=\dfrac{f^{(4)}(c)}{4!}(0.5)^4\)

for some \( c\) between \( 0\) and \( 0.5\). Since \( f^{(4)}(x)=−\dfrac{15}{2^4(1+x)^{7/2}}\), and the maximum value of \( ∣f^{(4)}(x)∣\) on the interval \( (0,0.5)\) occurs at \( x=0\), we have

\( |R_3(0.5)|≤\dfrac{15}{4!2^4}(0.5)^4≈0.00244.\)

The function and the Maclaurin polynomial \( p_3\) are graphed in Figure.

This graph has two curves. The first one is f(x)= the square root of (1+x) and the second is psub3(x). The curves are very close at y = 1.

Exercise \(\PageIndex{1}\)

Find the binomial series for \( f(x)=\dfrac{1}{(1+x)^2}\).

Use the definition of binomial series for \( r=−2\).

\( \sum_{n=0}^∞(−1)^n(n+1)x^n\)

Common Functions Expressed as Taylor Series

At this point, we have derived Maclaurin series for exponential, trigonometric, and logarithmic functions, as well as functions of the form \( f(x)=(1+x)^r\). In Table, we summarize the results of these series. We remark that the convergence of the Maclaurin series for \( f(x)=ln(1+x)\) at the endpoint \( x=1\) and the Maclaurin series for \( f(x)=tan^{−1}x\) at the endpoints \( x=1\) and \( x=−1\) relies on a more advanced theorem than we present here. (Refer to Abel’s theorem for a discussion of this more technical point.)

Earlier in the chapter, we showed how you could combine power series to create new power series. Here we use these properties, combined with the Maclaurin series in Table, to create Maclaurin series for other functions.

Example \( \PageIndex{2}\): Deriving Maclaurin Series from Known Series

Find the Maclaurin series of each of the following functions by using one of the series listed in Table.

  • \( f(x)=\cos\sqrt{x}\)
  • \( f(x)=\text{sinh} x\)

a. Using the Maclaurin series for \( \cos x\) we find that the Maclaurin series for \( \cos\sqrt{x}\) is given by

\( \sum_{n=0}^∞\dfrac{(−1)^n(\sqrt{x})^{2n}}{(2n)!}=\sum_{n=0}^∞\dfrac{(−1)^nx^n}{(2n)!}=1−\dfrac{x}{2!}+\dfrac{x^2}{4!}−\dfrac{x^3}{6!}+\dfrac{x^4}{8!}−⋯.\)

This series converges to \( cos\sqrt{x}\) for all \( x\) in the domain of \( cos\sqrt{x}\); that is, for all \( x≥0\).

b. To find the Maclaurin series for \( sinhx,\) we use the fact that

\( \text{sinh} x=\dfrac{e^x−e^{−x}}{2}.\)

Using the Maclaurin series for \( e^x\), we see that the n th term in the Maclaurin series for \( sinhx\) is given by

\( \dfrac{x^n}{n!}−\dfrac{(−x)^n}{n!}.\)

For \( n\) even, this term is zero. For \( n\) odd, this term is \( \dfrac{2x^n}{n!}\). Therefore, the Maclaurin series for \( sinhx\) has only odd-order terms and is given by

\( \sum_{n=0}^∞\dfrac{x^{2n+1}}{(2n+1)!}=x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+⋯.\)

Exercise \(\PageIndex{2}\)

Find the Maclaurin series for \( \sin(x^2).\)

Use the Maclaurin series for \( \sin x.\)

\( \sum_{n=0}^∞\dfrac{(−1)^nx^{4n+2}}{(2n+1)!}\)

We also showed previously in this chapter how power series can be differentiated term by term to create a new power series. In Example, we differentiate the binomial series for \( \sqrt{1+x}\) term by term to find the binomial series for \( \dfrac{1}{\sqrt{1+x}}\). Note that we could construct the binomial series for \( \dfrac{1}{\sqrt{1+x}}\) directly from the definition, but differentiating the binomial series for \( \sqrt{1+x}\) is an easier calculation.

Example \( \PageIndex{3}\): Differentiating a Series to Find a New Series

Use the binomial series for \( \sqrt{1+x}\) to find the binomial series for \( \dfrac{1}{\sqrt{1+x}}\).

The two functions are related by

\( \dfrac{d}{dx}\sqrt{1+x}=\dfrac{1}{2\sqrt{1+x}}\),

so the binomial series for \( \dfrac{1}{\sqrt{1+x}}\) is given by

\( \dfrac{1}{\sqrt{1+x}}=2\dfrac{d}{dx}\sqrt{1+x}=1+\sum_{n=1}^∞\dfrac{(−1)^n}{n!}\dfrac{1⋅3⋅5⋯(2n−1)}{2^n}x^n.\)

Exercise \(\PageIndex{3}\)

Find the binomial series for \( f(x)=\dfrac{1}{(1+x)^{3/2}}\)

Differentiate the series for \( \dfrac{1}{\sqrt{1+x}}\)

\( \sum_{n=1}^∞\dfrac{(−1)^n}{n!}\dfrac{1⋅3⋅5⋯(2n−1)}{2^n}x^n\)

In this example, we differentiated a known Taylor series to construct a Taylor series for another function. The ability to differentiate power series term by term makes them a powerful tool for solving differential equations. We now show how this is accomplished.

Solving Differential Equations with Power Series

Consider the differential equation

\[y′(x)=y\]

Recall that this is a first-order separable equation and its solution is \( y=Ce^x\). This equation is easily solved using techniques discussed earlier in the text. For most differential equations, however, we do not yet have analytical tools to solve them. Power series are an extremely useful tool for solving many types of differential equations. In this technique, we look for a solution of the form \( y=\sum_{n=0}^∞c_nx^n\) and determine what the coefficients would need to be. In the next example, we consider an initial-value problem involving \( y′=y\) to illustrate the technique.

Example \( \PageIndex{4}\): Power Series Solution of a Differential Equation

Use power series to solve the initial-value problem

\[ y′=y,y(0)=3\]

Suppose that there exists a power series solution

\( y(x)=\sum_{n=0}^∞c_nx^n=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+⋯\)

Differentiating this series term by term, we obtain

\( y′=c_1+2c_2x+3c_3x^2+4c_4x^3+⋯.\)

If \(y\) satisfies the differential equation, then

\( c_0+c_1x+c_2x^2+c_3x^3+⋯=c_1+2c_2x+3c_3x^2+4c_3x^3+⋯.\)

Using the uniqueness of power series representations, we know that these series can only be equal if their coefficients are equal. Therefore,

\( c_0=c_1,\)

\( c_1=2c_2,\)

\( c_2=3c_3,\)

\( c_3=4c_4,\)

Using the initial condition \( y(0)=3\) combined with the power series representation

\( y(x)=c_0+c_1x+c_2x^2+c_3x^3+⋯\),

we find that \( c_0=3\). We are now ready to solve for the rest of the coefficients. Using the fact that \( c_0=3\), we have

\( c_1=c_0=3=\dfrac{3}{1!}\),

\( c_2=\dfrac{c_1}{2}=\dfrac{3}{2}=\dfrac{3}{2!},\)

\( c_3=\dfrac{c_2}{3}=\dfrac{3}{3⋅2}=\dfrac{3}{3!},\)

\( c_4=\dfrac{c_3}{4}=\dfrac{3}{4⋅3⋅2}=\dfrac{3}{4!}.\)

\( y=3[1+\dfrac{1}{1!}x+\dfrac{1}{2!}x^2+\dfrac{1}{3!}x^3\dfrac{1}{4!}x^4+⋯]=3\sum_{n=0}^∞\dfrac{x^n}{n!}\).

You might recognize

\( \sum_{n=0}^∞\dfrac{x^n}{n!}\)

as the Taylor series for \( e^x\). Therefore, the solution is \( y=3e^x\).

Exercise \(\PageIndex{4}\)

Use power series to solve \( y′=2y,y(0)=5.\)

The equations for the first several coefficients \( c_n\) will satisfy \( c_0=2c_1,c_1=2⋅2c_2,c_2=2⋅3c_3,….\) In general, for all \( n≥0,c_n=2(n+1)C_{n+1}\).

\( y=5e^{2x}\)

We now consider an example involving a differential equation that cannot be solve with "simple" methods. This differential equation

\[y''−xy=0\]

is known as Airy’s equation . It has many applications in mathematical physics, such as modeling the diffraction of light. Here we show how to solve it using power series.

Example \( \PageIndex{5}\): Power Series Solution of Airy’s Equation

Use power series to solve

\[ y''−xy=0\]

with the initial conditions \( y(0)=a\) and \( y'(0)=b.\)

We look for a solution of the form

\( y=\sum_{n=0}^∞c_nx^n=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+⋯\)

Differentiating this function term by term, we obtain

\( y′=c_1+2c_2x+3c_3x^2+4c_4x^3+⋯\)

\( y''=2⋅1c_2+3⋅2c_3x+4⋅3c_4x^2+⋯\)

If \(y\) satisfies the equation \( y''=xy\), then

\( 2⋅1c_2+3⋅2c_3x+4⋅3c_4x^2+⋯=x(c_0+c_1x+c_2x^2+c_3x^3+⋯)\)

Using [link] on the uniqueness of power series representations, we know that coefficients of the same degree must be equal. Therefore,

\( 2⋅1c_2=0\)

\( 3⋅2c_3=c_0\)

\( 4⋅3c_4=c_1\)

\( 5⋅4c_5=c_2\)

More generally, for \( n≥3\), we have \( n⋅(n−1)c_n=c_{n−3}\). In fact, all coefficients can be written in terms of \( c_0\) and \( c_1\). To see this, first note that \( c_2=0\). Then

\( c_3=\dfrac{c_0}{3⋅2}\)

\( c_4=\dfrac{c_1}{4⋅3}\)

For \( c_5,c_6,c_7\), we see that

\( c_5=\dfrac{c_2}{5⋅4}=0\)

\( c_6=\dfrac{c_3}{6⋅5}=\dfrac{c_0}{6⋅5⋅3⋅2}\)

\( c_7=\dfrac{c_4}{7⋅6}=\dfrac{c_1}{7⋅6⋅4⋅3}\)

Therefore, the series solution of the differential equation is given by

\( y=c_0+c_1x+0⋅x^2+\dfrac{c_0}{3⋅2}x^3+\dfrac{c_1}{4⋅3}x^4+0⋅x^5+\dfrac{c_0}{6⋅5⋅3⋅2}x^6+\dfrac{c_1}{7⋅6⋅4⋅3}x^7+⋯\)

The initial condition \( y(0)=a\) implies \( c_0=a\). Differentiating this series term by term and using the fact that \( y′(0)=b\), we conclude that \( c_1=b\).

Therefore, the solution of this initial-value problem is

\( y=a(1+\dfrac{x^3}{3⋅2}+\dfrac{x}{6⋅5⋅3⋅2}+⋯)+b(x+\dfrac{x^4}{4⋅3}+\dfrac{x^7}{7⋅6⋅4⋅3}+⋯)\)

Exercise \(\PageIndex{5}\)

Use power series to solve \( y''+x^2y=0\) with the initial condition \( y(0)=a\) and \( y′(0)=b\)

The coefficients satisfy \( c_0=a,c_1=b,c_2=0,c_3=0,\) and for \( n≥4,n(n−1)c_n=−c_{n−4}\).

\(y=a(1−\dfrac{x^4}{3⋅4}+\dfrac{x^8}{3⋅4⋅7⋅8}−⋯)+b(x−\dfrac{x^5}{4⋅5}+\dfrac{x^9}{4⋅5⋅8⋅9}−⋯)\)

Evaluating Non-elementary Integrals

Solving differential equations is one common application of power series. We now turn to a second application. We show how power series can be used to evaluate integrals involving functions whose antiderivatives cannot be expressed using elementary functions.

One integral that arises often in applications in probability theory is \(\int e^{−x^2}dx\) Unfortunately, the antiderivative of the integrand \(e^{−x^2}\) is not an elementary function. By elementary function, we mean a function that can be written using a finite number of algebraic combinations or compositions of exponential, logarithmic, trigonometric, or power functions. We remark that the term “elementary function” is not synonymous with non-complicated function. For example, the function \(f(x)=\sqrt{x^2−3x}+e^{x^3}−sin(5x+4)\) is an elementary function, although not a particularly simple-looking function. Any integral of the form \( \int f(x)dx\) where the antiderivative of \( f\) cannot be written as an elementary function is considered a non-elementary integral.

Nonelementary integrals cannot be evaluated using the basic integration techniques discussed earlier. One way to evaluate such integrals is by expressing the integrand as a power series and integrating term by term. We demonstrate this technique by considering \( \int e^{−x^2}dx\)

Example \( \PageIndex{6}\): Using Taylor Series to Evaluate a Definite Integral

  • Express \( \int e^{−x^2}dx\) as an infinite series.
  • Evaluate \( \int^1_0 e^{−x^2}dx\) to within an error of \( 0.01\)

a. The Maclaurin series for \( e^{−x^2}\) is given by

\( e^{−x^2}=\sum_{n=0}^∞\dfrac{(−x^2)^n}{n!}=1−x^2+\dfrac{x^4}{2!}−\dfrac{x^6}{3!}+⋯+(−1)^n\dfrac{x^{2n}}{n!}+⋯=\sum_{n=0}^∞(−1)^n\dfrac{x^{2n}}{n!}\)

\( \int e^{−x^2}dx=\int(1−x^2+\dfrac{x^4}{2!}−\dfrac{x^6}{3!}+⋯+(−1)^n\dfrac{x^{2n}}{n!}+⋯)dx=C+x−\dfrac{x^3}{3}+\dfrac{x^5}{5.2!}−\dfrac{x^7}{7.3!}+⋯+(−1)^n\dfrac{x^{2n+1}}{(2n+1)n!}+⋯\)

b. Using the result from part a. we have

\( \int^1_0 e^{−x^2}dx=1−\dfrac{1}{3}+\dfrac{1}{10}−\dfrac{1}{42}+\dfrac{1}{216}−⋯\)

The sum of the first four terms is approximately \( 0.74\). By the alternating series test, this estimate is accurate to within an error of less than \( \dfrac{1}{216}≈0.0046296<0.01\)

Exercise \(\PageIndex{6}\)

Express \( \int cos\sqrt{x}dx\) as an infinite series. Evaluate \( \int^1_0cos\sqrt{x}dx\) to within an error of \( 0.01\)

Use the series found in Example.

\( C+\sum_{n=1}^∞(−1)^{n+1}\dfrac{x^n}{n(2n−2)!}\) The definite integral is approximately \( 0.514\) to within an error of \( 0.01\)

As mentioned above, the integral \( \int e^{−x^2}dx\) arises often in probability theory. Specifically, it is used when studying data sets that are normally distributed, meaning the data values lie under a bell-shaped curve. For example, if a set of data values is normally distributed with mean \( μ\) and standard deviation \( σ\), then the probability that a randomly chosen value lies between \( x=a\) and \( x=b\) is given by

\[\dfrac{1}{σ\sqrt{2π}}\int^b_ae^{−(x−μ)^2/(2σ^2)}dx\]

(See Figure.)

This graph is the normal distribution. It is a bell-shaped curve with the highest point above mu on the x-axis. Also, there is a shaded area under the curve above the x-axis. The shaded area is bounded by alpha on the left and b on the right.

To simplify this integral, we typically let \( z=\dfrac{x−μ}{σ}\). This quantity z is known as the z score of a data value. With this simplification, integral Equation becomes

\[\dfrac{1}{\sqrt{2π}}\int^{(b−μ)/σ}_{(a−μ)/σ}e^{−z^2/2}dz\]

In Example, we show how we can use this integral in calculating probabilities.

Example \( \PageIndex{7}\): Using Maclaurin Series to Approximate a Probability

Suppose a set of standardized test scores are normally distributed with mean \( μ=100\) and standard deviation \( σ=50\). Use Equation and the first six terms in the Maclaurin series for \( e^{−x^2/2}\) to approximate the probability that a randomly selected test score is between \( x=100\) and \( x=200\). Use the alternating series test to determine how accurate your approximation is.

Since \( μ=100,σ=50,\) and we are trying to determine the area under the curve from \( a=100\) to \( b=200\), integral Equation becomes

\( \dfrac{1}{\sqrt{2π}}\int^2_0e^{−z^2/2}dz\)

The Maclaurin series for \( e^{−x^2/2}\) is given by

\( e^{−x^2/2}=\sum_{n=0}^∞\dfrac{(−\dfrac{x^2}{2})^n}{n!}=1−\dfrac{x^2}{2^1⋅1!}+\dfrac{x^4}{2^2⋅2!}−\dfrac{x^6}{2^3⋅3!}+⋯+(−1)^n\dfrac{x^{2n}}{2^n⋅n}!+⋯=\sum_{n=0}^∞(−1)^n\dfrac{x^{2n}}{2^n⋅n!}\).

\( \dfrac{1}{\sqrt{2π}}\int e^{−z^2/2}dz=\dfrac{1}{\sqrt{2π}}\int(1−\dfrac{z^2}{2^1⋅1!}+\dfrac{z^4}{2^2⋅2!}−\dfrac{z^6}{2^3⋅3!}+⋯+(−1)^n\dfrac{z^{2n}}{2^n⋅n!}+⋯)dz\) \( =\dfrac{1}{\sqrt{2π}}(C+z−\dfrac{z^3}{3⋅2^1⋅1!}+\dfrac{z^5}{5⋅2^2⋅2!}−\dfrac{z^7}{7⋅2^3⋅3!}+⋯+(−1)^n\dfrac{z^{2n+1}}{(2n+1)2^n⋅n!}+⋯)\) \( \dfrac{1}{\sqrt{2π}}\int^2_0e^{−z^2/2}dz=\dfrac{1}{\sqrt{2π}}(2−\dfrac{8}{6}+\dfrac{32}{40}−\dfrac{128}{336}+\dfrac{512}{3456}−\dfrac{2^{11}}{11⋅2^5⋅5!}+⋯)\)

Using the first five terms, we estimate that the probability is approximately 0.4922. By the alternating series test, we see that this estimate is accurate to within

\[ \dfrac{1}{\sqrt{2π}}\dfrac{2^{13}}{13⋅2^6⋅6!}≈0.00546\]

If you are familiar with probability theory, you may know that the probability that a data value is within two standard deviations of the mean is approximately \( 95%.\) Here we calculated the probability that a data value is between the mean and two standard deviations above the mean, so the estimate should be around \( 47.5%\). The estimate, combined with the bound on the accuracy, falls within this range.

Exercise \(\PageIndex{7}\)

Use the first five terms of the Maclaurin series for \( e^{−x^2/2}\) to estimate the probability that a randomly selected test score is between \( 100\) and \( 150\). Use the alternating series test to determine the accuracy of this estimate.

Evaluate \( \int^1_0e^{−z^2/2}dz\) using the first five terms of the Maclaurin series for \( e^{−z^2/2}\)

The estimate is approximately \( 0.3414\)

This estimate is accurate to within \( 0.0000094\)

Another application in which a nonelementary integral arises involves the period of a pendulum. The integral is

\[\int^{π/2}_0\dfrac{dθ}{\sqrt{1−k^2sin^2θ}}\]

An integral of this form is known as an elliptic integral of the first kind. Elliptic integrals originally arose when trying to calculate the arc length of an ellipse. We now show how to use power series to approximate this integral.

Example \( \PageIndex{8}\): Period of a Pendulum

The period of a pendulum is the time it takes for a pendulum to make one complete back-and-forth swing. For a pendulum with length \( L\) that makes a maximum angle \( θ_{max}\) with the vertical, its period \( T\) is given by

\( T=4\sqrt{\dfrac{L}{g}}\int^{π/2}_0\dfrac{dθ}{\sqrt{1−k^2sin^2θ}}\)

where \( g\) is the acceleration due to gravity and \( k=sin(\dfrac{θ_{max}}{2})\) (see Figure). (We note that this formula for the period arises from a non-linearized model of a pendulum. In some cases, for simplification, a linearized model is used and \( sinθ\) is approximated by \( θ\).)

This figure is a pendulum. There are three positions of the pendulum shown. When the pendulum is to the far left, it is labeled negative theta max. When the pendulum is in the middle and vertical, it is labeled equilibrium position. When the pendulum is to the far right it is labeled theta max. Also, theta is the angle from equilibrium to the far right position. The length of the pendulum is labeled L.

Use the binomial series

\( \dfrac{1}{\sqrt{1+x}}=1+\sum_{n=1}^∞\dfrac{(−1)^n}{n!}\dfrac{1⋅3⋅5⋯(2n−1)}{2^n}x^n\)

to estimate the period of this pendulum. Specifically, approximate the period of the pendulum if

  • you use only the first term in the binomial series, and
  • you use the first two terms in the binomial series.

We use the binomial series, replacing x with \( −k^2sin^2θ\) Then we can write the period as

\( T=4\sqrt{\dfrac{L}{g}}\int^{π/2}_0(1+\dfrac{1}{2}k^2sin^2θ+\dfrac{1⋅3}{2!2^2}k^4sin^4θ+⋯)dθ\)

a. Using just the first term in the integrand, the first-order estimate is

\( T≈4\sqrt{\dfrac{L}{g}}\int^{π/2}_0dθ=2π\sqrt{\dfrac{L}{g}}\)

If \( θ_{max}\) is small, then \( k=sin(\dfrac{θ_{max}}{2})\) is small. We claim that when \( k\) is small, this is a good estimate. To justify this claim, consider

\( \int^{π/2}_0(1+\dfrac{1}{2}k^2sin^2θ+\dfrac{1⋅3}{2!2^2}k^4sin^4θ+⋯)dθ\)

Since \( |sinx|≤1\), this integral is bounded by

\( \int^{π/2}_0(\dfrac{1}{2}k^2+\dfrac{1.3}{2!2^2}k^4+⋯)dθ<\dfrac{π}{2}(\dfrac{1}{2}k^2+\dfrac{1⋅3}{2!2^2}k^4+⋯)\)

Furthermore, it can be shown that each coefficient on the right-hand side is less than \( 1\) and, therefore, that this expression is bounded by

\( \dfrac{πk^2}{2}(1+k^2+k^4+⋯)=\dfrac{πk^2}{2}⋅\dfrac{1}{1−k^2}\),

which is small for \( k\) small.

b. For larger values of \( θ_{max}\), we can approximate \( T\) by using more terms in the integrand. By using the first two terms in the integral, we arrive at the estimate

\( T≈4\sqrt{\dfrac{L}{g}}\int^{π/2}_0(1+\dfrac{1}{2}k^2sin^2θ)dθ=2π\sqrt{\dfrac{L}{g}}(1+\dfrac{k^2}{4})\)

The applications of Taylor series in this section are intended to highlight their importance. In general, Taylor series are useful because they allow us to represent known functions using polynomials, thus providing us a tool for approximating function values and estimating complicated integrals. In addition, they allow us to define new functions as power series, thus providing us with a powerful tool for solving differential equations.

Key Concepts

  • The binomial series is the Maclaurin series for \( f(x)=(1+x)^r\). It converges for \( |x|<1\).
  • Taylor series for functions can often be derived by algebraic operations with a known Taylor series or by differentiating or integrating a known Taylor series.
  • Power series can be used to solve differential equations.
  • Taylor series can be used to help approximate integrals that cannot be evaluated by other means.

Contributors and Attributions

  • Template:ContribOpenStaxCalc

1 This series is the basis for the small angle approximation used in physics, sometimes without warning you. In this approximation when we have a small angle in our problem we can say that \(\cos(\theta) = 1 - \frac{\theta^2}{2}\) and/or \(\sin(\theta) = \theta\). The students is advised to watch out for this appearing in problem solutions and proofs where the author neglects to inform you about the use of the small angle approximation.

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AP Calculus BC : Taylor Series

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8.4: Taylor Series Examples

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  • Page ID 6519

  • Jeremy Orloff
  • Massachusetts Institute of Technology via MIT OpenCourseWare

The uniqueness of Taylor series along with the fact that they converge on any disk around \(z_0\) where the function is analytic allows us to use lots of computational tricks to find the series and be sure that it converges.

Example \(\PageIndex{1}\)

Use the formula for the coefficients in terms of derivatives to give the Taylor series of \(f(z) = e^z\) around \(z = 0\).

Since \(f'(z) = e^z\), we have \(f^{(n)} (0) = e^0 = 1\). So,

\[e^z = 1 + z + \dfrac{z^2}{2!} + \dfrac{z^3}{3!} + \ ... = \sum_{n = 0}^{\infty} \dfrac{z^n}{n!} \nonumber \]

Example \(\PageIndex{2}\)

Expand \(f(z) = z^8 e^{3z}\) in a Taylor series around \(z = 0\).

Let \(w = 3z\). So,

\[e^{3z} = e^w = \sum_{n = 0}^{\infty} \dfrac{w^n}{n!} = \sum_{k = 0}^{\infty} \dfrac{3^n}{n!} z^n \nonumber \]

\[f(z) = \sum_{n = 0}^{\infty} \dfrac{3^n}{n!} z^{n + 8}. \nonumber \]

Example \(\PageIndex{3}\)

Find the Taylor series of \(\sin (z)\) around \(z = 0\) (Sometimes the Taylor series around 0 is called the Maclaurin series.)

We give two methods for doing this.

\[f^{(n)} (0) = \dfrac{d^n \sin (z)}{dz^n} = \begin{cases} (-1)^m & \text{ for } n = 2m + 1 = \text{ odd}, m = 0, 1, 2,\ ... \\ 0 & \text{ for } n \text{ even} \end{cases} \nonumber \]

Method 2. Using

\[\sin (z) = \dfrac{e^{iz} - e^{-iz}}{2i}, \nonumber \]

\[\begin{align*} \sin (z) &= \dfrac{1}{2i} \left[\sum_{n = 0}^{\infty} \dfrac{(iz)^n}{n!} - \sum_{n = 0}^{\infty} \dfrac{(-iz)^n}{n!}\right] \\[4pt] &= \dfrac{1}{2i} \sum_{n = 0}^{\infty} [(1 - (-1)^n)] \dfrac{i^n z^n}{n!}\end{align*} \]

(We need absolute convergence to add series like this.)

Conclusion:

\[\sin (z) = \sum_{n = 0}^{\infty} (-1)^n \dfrac{z^{2n + 1}}{(2n + 1)!},\nonumber \]

which converges for \(|z| < \infty\).

Example \(\PageIndex{4}\)

Expand the rational function

\[f(z) = \dfrac{1 + 2z^2}{z^3 + z^5}\nonumber \]

around \(z = 0\).

Note that \(f\) has a singularity at 0, so we can’t expect a convergent Taylor series expansion. We’ll aim for the next best thing using the following shortcut.

\[f(z) = \dfrac{1}{z^3} \dfrac{2(1 + z^2) - 1}{1 + z^2} = \dfrac{1}{z^3} [ 2 - \dfrac{1}{1 + z^2}].\nonumber \]

Using the geometric series we have

\[\dfrac{1}{1 + z^2} = \dfrac{1}{1 - (-z^2)} = \sum_{n = 0}^{\infty} (-z^2)^n = 1 - z^2 + z^4 - z^6 +...\nonumber \]

Putting it all together

\[f(z) = \dfrac{1}{z^3} (2 - 1 + z^2 - z^4 + ...) = \left(\dfrac{1}{z^3} + \dfrac{1}{z}\right) - \sum_{n = 0}^{\infty} (-1)^n z^{2n + 1}\nonumber \]

Note: The first terms are called the singular part , i.e. those with negative powers of \(z\). he summation is called the regular or analytic part . Since the geometric series for \(1/(1 + z^2)\) converges for \(|z| < 1\), the entire series is valid in \(0 < |z| < 1\).

Example \(\PageIndex{5}\)

Find the Taylor series for

\[f(z) = \dfrac{e^z}{1 - z}\nonumber \]

around \(z = 0\). Give the radius of convergence.

We start by writing the Taylor series for each of the factors and then multiply them out.

\[\begin{array} {rcl} {f(z)} & = & {\left(1 + z + \dfrac{z^2}{2!} + \dfrac{z^3}{3!} + \ ...\right) (1 + z + z^2 + z^3 + \ ...)} \\ {} & = & {1 + (1 + 1) z + \left(1 + 1 + \dfrac{1}{2!}\right) z^2 + \left(1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!}\right) z^3 + \ ...} \end{array}\nonumber \]

The biggest disk around \(z = 0\) where \(f\) is analytic is \(|z| < 1\). Therefore, by Taylor’s theorem, the radius of convergence is \(R = 1\).

Example \(\PageIndex{6}\)

\[f(z) = \dfrac{1}{1 - z}\nonumber \]

around \(z = 5\). Give the radius of convergence.

We have to manipulate this into standard geometric series form.

\[f(z) = \dfrac{1}{-4(1 + (z - 5)/4)} = -\dfrac{1}{4} \left[1 - \left(\dfrac{z - 5}{4}\right) + \left(\dfrac{z - 5}{4}\right)^2 - \left(\dfrac{z - 5}{4}\right)^3 + \ ...\right]\nonumber \]

Since \(f(z)\) has a singularity at \(z = 1\) the radius of convergence is \(R = 4\). We can also see this by considering the geometric series. The geometric series ratio is \((z - 5)/4\). So the series converges when \(|z - 5|/4 < 1\), i.e. when \(|z - 5| < 4\), i.e. \(R = 4\).

Example \(\PageIndex{7}\)

\[f(z) = \log (1 + z)\nonumber \]

We know that \(f\) is analytic for \(|z| < 1\) and not analytic at \(z = -1\). So, the radius of convergence is \(R = 1\). To find the series representation we take the derivative and use the geometric series.

\[f'(z) = \dfrac{1}{1 + z} = 1 - z + z^2 - z^3 + z^4 - \ ...\nonumber \]

Integrating term by term (allowed by Theorem 8.3.1) we have

\[f(z) = a_0 + z - \dfrac{z^2}{2} + \dfrac{z^3}{3} - \dfrac{z^4}{4} + \ ... = a_0 + \sum_{n = 1}^{\infty} (-1)^{n - 1} \dfrac{z^n}{n}\nonumber \]

Here \(a_0\) is the constant of integration. We find it by evalating at \(z = 0\).

\[f(0) = a_0 = \log (1) = 0.\nonumber \]

Example \(\PageIndex{8}\)

Can the series

\[\sum a_n (z - 2)^n\nonumber \]

converge at \(z = 0\) and diverge at \(z = 3\).

No! We have \(z_0 = 2\). We know the series diverges everywhere outside its radius of convergence. So, if the series converges at \(z = 0\), then the radius of convergence is at least 2. Since \(|3 - z_0| < 2\) we would also have that \(z = 3\) is inside the disk of convergence.

Proof of Taylor’s Theorem

For convenience we restate Taylor’s Theorem \(\PageIndex{1}\).

Theorem \(\PageIndex{1}\): Taylor’s Theorem (Taylor Series)

Suppose \(f(z)\) is an analytic function in a region \(A\). Let \(z_0 \in A\). Then,

\[f(z) = \sum_{n = 0}^{\infty} a_n (z - z_0)^n, \nonumber \]

where the series converges on any disk \(|z - z_0| < r\) contained in \(A\). Furthermore, we have formulas for the coefficients

\[a_n = \dfrac{f^{(n)} (z_0)}{n!} = \dfrac{1}{2\pi i} \int_{\gamma} \dfrac{f(z)}{(z - z_0)^{n + 1}} \ dz \nonumber \]

In order to handle convergence issues we fix \(0 < r_1 < r_2 < r\). We let \(\gamma\) be the circle \(|w - z_0| = r_2\) (traversed counterclockise).

Take \(z\) inside the disk \(|z - z_0| < r_1\). We want to express \(f(z)\) as a power series around \(z_0\). To do this we start with the Cauchy integral formula and then use the geometric series.

As preparation we note that for \(w\) on \(\gamma\) and \(|z - z_0| < r_1\) we have

\[|z - z_0| < r_1 < r_2 = |w - z_0| \nonumber, \nonumber \]

\[\dfrac{|z - z_0|}{|w - z_0|} < 1. \nonumber \]

\[\dfrac{1}{w - z} = \dfrac{1}{w - z_0} \cdot \dfrac{1}{1 - \dfrac{z - z_0}{w - z_0}} = \dfrac{1}{w - z_0} \sum_{n = 0}^{\infty} (\dfrac{z - z_0}{w - z_0})^n = \sum_{n = 0}^{\infty} \dfrac{(z - z_0)^n}{(w - z_0)^{n + 1}} \nonumber \]

Using this and the Cauchy formula gives

\[\begin{array} {rcl} {f(z)} & = & {\dfrac{1}{2\pi i} \int_{\gamma} \dfrac{f(w)}{w - z}\ dw} \\ {} & = & {\dfrac{1}{2\pi i} \int_{\gamma} \sum_{n = 0}^{\infty} \dfrac{f(w)}{(w - z_0)^{n + 1}} (z - z_0)^n\ dw} \\ {} & = & {\sum_{n = 0}^{\infty} (\dfrac{1}{2\pi i} \int_{\gamma} \dfrac{f(w)}{(w - z_0)^{n + 1}} \ dw) (z - z_0)^n} \\ {} & = & {\sum_{n = 0}^{\infty} \dfrac{f^{(n)} (z_0)}{n!} (z - z_0)^n} \end{array} \nonumber \]

The last equality follows from Cauchy’s formula for derivatives. Taken together the last two equalities give Taylor’s formula. QED

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About the lecture:

Over the years, many groups and individuals have put together interesting and thought-provoking research agendas. But few of these have remained as compelling and relevant as  Richard Smalley’s  “Top Ten Problems for Humanity.” Following Smalley’s win of the Nobel Prize for Chemistry for his discovery of buckminsterfullerene (buckyballs), his list of research problems gained widespread attention.

Although Smalley’s agenda was introduced more than twenty years ago, the issues he flagged remain equally daunting today. What is particularly compelling about his list is that it is prioritized, with the solution of each problem in turn leading to the solution of others lower on the list. Smalley’s 10 broad challenges will require efforts that take advantage of the full range of human endeavor, but industrial engineers are uniquely positioned to provide technological and management solutions.

The purpose of this lecture is to motivate creative thought about how industrial engineers can help solve these planetary problems of truly massive scale.

About G. Don Taylor:

G. Don Taylor  is the Executive Vice Provost and the Charles O. Gordon Professor of Industrial and Systems Engineering at Virginia Tech.

Taylor’s research interests focus on the simulation of complex systems and the logistics of material flow and freight transportation. He has served as principal investigator or co-principal investigator on more than 60 externally funded projects. His research has led to the publication of 10 edited books, more than 75 journal articles and book chapters and more than 120 conference papers and technical reports.

Taylor has made more than 200 formal presentations at conferences or seminars and has had research relationships with more than 50 different companies. He is a Fellow and a past-president of the Institute of Industrial and Systems Engineers, or IISE, and is currently the inaugural chairperson of the board of IISE Solutions, Inc., a for-profit subsidiary. He is a recipient of IISE’s prestigious Frank and Lillian Gilbreth Industrial Engineering Award, a Fellow and past member of the board of the World Academy of Productivity Science and is a registered Professional Engineer.

About the series:

The Douglas C. Montgomery Distinguished Lecture Series  is designed to periodically offer talks to raise the overall profile of industrial engineering and provide a forum for top experts to tackle the field’s growing opportunities and challenges.

About Douglas C. Montgomery

Douglas C. Montgomery  is a Regents Professor of industrial engineering and statistics at the School of Computing and Augmented Intelligence at ASU’s Ira A. Fulton Schools of Engineering. His research interests are in industrial statistics. He is an author of 16 books and more than 200 technical papers. He is a recipient of the Shewhart Medal, the Brumbaugh Award, the Hunter Award, the Shewell Award and the Ellis R. Ott Award. Montgomery is also a recipient of the George Box Medal from the European Network For Business and Industrial Statistics, or ENBIS.

Montgomery is a fellow of the American Statistical Association, the American Society for Quality Control, the Royal Statistical Society, the Institute of Industrial Engineers, and an elected member of the International Statistical Institute.

COMMENTS

  1. Calculus II

    For problems 1 & 2 use one of the Taylor Series derived in the notes to determine the Taylor Series for the given function. f (x) = cos(4x) f ( x) = cos ( 4 x) about x = 0 x = 0 Solution f (x) = x6e2x3 f ( x) = x 6 e 2 x 3 about x = 0 x = 0 Solution For problem 3 - 6 find the Taylor Series for each of the following functions.

  2. 10.3E: Exercises for Taylor Polynomials and Taylor Series

    Taylor Polynomials. In exercises 1 - 8, find the Taylor polynomials of degree two approximating the given function centered at the given point. 1) at. 2) at. Answer: 3) at. 4) at. Answer: 5) at.

  3. Taylor Series

    A Taylor Series is an expansion of a function into an infinite sum of terms, where each term's exponent is larger and larger, like this: Example: The Taylor Series for ex ex = 1 + x + x2 2! + x3 3! + x4 4! + x5 5! + ... says that the function: ex is equal to the infinite sum of terms: 1 + x + x2/2! + x3/3! + ... etc

  4. 8.8: Taylor Series

    The Taylor Series of \ (f (x)\), centered at \ (c\) is \ [\sum_ {n=0}^\infty \frac {f\,^ { (n)} (c)} {n!} (x-c)^n.\] Setting \ (c=0\) gives the Maclaurin Series of \ (f (x)\): \ [\sum_ {n=0}^\infty \frac {f\,^ { (n)} (0)} {n!}x^n.\]

  5. Taylor Series Manipulation

    It can be assembled in many creative ways to help us solve problems through the normal operations of function addition, multiplication, and composition. We can also use rules of differentiation and integration to develop new and interesting series. Contents Adding and Multiplying Power Series Composition of Taylor Series

  6. 5.4: Working with Taylor Series

    Figure 5.4.2: If data values are normally distributed with mean μ and standard deviation σ, the probability that a randomly selected data value is between a and b is the area under the curve y = 1 σ√2πe − ( x − μ)2 / ( 2 σ 2) between x = a and x = b. To simplify this integral, we typically let z = x − μ σ.

  7. 6.4 Working with Taylor Series

    6.4.1Write the terms of the binomial series. 6.4.2Recognize the Taylor series expansions of common functions. 6.4.3Recognize and apply techniques to find the Taylor series for a function. 6.4.4Use Taylor series to solve differential equations. 6.4.5Use Taylor series to evaluate nonelementary integrals.

  8. Taylor Series and Maclaurin Series

    This calculus 2 video tutorial explains how to find the Taylor series and the Maclaurin series of a function using a simple formula. It explains how to deri...

  9. Taylor Series (Proof and Examples)

    Maths Math Article Taylor Series Taylor Series Taylor series is the polynomial or a function of an infinite sum of terms. Each successive term will have a larger exponent or higher degree than the preceding term.

  10. Taylor series made easy

    0:00 / 9:06 Taylor series made easy RH 17.2K subscribers Subscribe Subscribed 716 170K views 10 years ago University mathematics The solution to a typical Taylor series exam question...this one...

  11. Taylor Series

    The Lagrange error bound of a Taylor polynomial gives the worst-case scenario for the difference between the estimated value of the function as provided by the Taylor polynomial and the actual value of the function.

  12. Taylor Series Solutions to Initial Value Problems

    This video explains how to determine a Taylor series solution to an initial value problem.https://mathispower4u.com

  13. 5.4: Taylor and Maclaurin Series

    If x = 0, then this series is known as the Maclaurin series for f. Definition 5.4.1: Maclaurin and Taylor series. If f has derivatives of all orders at x = a, then the Taylor series for the function f at a is. ∞ ∑ n = 0f ( n) (a) n! (x − a)n = f(a) + f′ (a)(x − a) + f ″ (a) 2! (x − a)2 + ⋯ + f ( n) (a) n! (x − a)n + ⋯.

  14. 14.2.7.4: Working with Taylor Series

    Figure 14.2.7.4.2: If data values are normally distributed with mean μ and standard deviation σ, the probability that a randomly selected data value is between a and b is the area under the curve y = 1 σ√2πe − ( x − μ)2 / ( 2 σ 2) between x = a and x = b. To simplify this integral, we typically let z = x − μ σ.

  15. Taylor Series

    Track your scores, create tests, and take your learning to the next level! Free practice questions for AP Calculus BC - Taylor Series. Includes full solutions and score reporting.

  16. Taylor Series Calculator

    Given a function f (x) and a point 'a', the n-th order Taylor series of f (x) around 'a' is defined as: T_n (x) = f (a) + f' (a) (x-a) + f'' (a) (x-a)^2 / 2! + ... + f^ (n) (a) (x-a)^n / n! + ... where f^ (n) (a) is the n-th derivative of f (x) evaluated at 'a', and 'n!' is the factorial of n. Show more Related Symbolab blog posts

  17. Taylor Series Solutions to Initial Value Problems

    How to use Taylor Series to solve some differential equations - a calculus course introduction

  18. 8.4: Taylor Series Examples

    Analysis Complex Variables with Applications (Orloff) 8: Taylor and Laurent Series 8.4: Taylor Series Examples Expand/collapse global location 8.4: Taylor Series Examples Page ID

  19. Calculus II

    Here are a set of practice problems for the Series and Sequences chapter of the Calculus II notes. If you'd like a pdf document containing the solutions the download tab above contains links to pdf's containing the solutions for the full book, chapter and section. At this time, I do not offer pdf's for solutions to individual problems.

  20. Differential Equations

    Let's take a look at an example. Example 1 Determine the Taylor series for f (x) = ex f ( x) = e x about x = 0 x = 0 . Of course, it's often easier to find the Taylor series about x = 0 x = 0 but we don't always do that. Example 2 Determine the Taylor series for f (x) = ex f ( x) = e x about x = −4 x = − 4 .

  21. What are the practical applications of the Taylor Series?

    The only way to solve this problem, as far as I know, is using Taylor formula (unless you know how to force computer using more digits, it is possible do that with some programming language, but probably this would be a more complicated and less sure way to solve the problem). Using Taylor formula written before (adding more Taylor terms the ...

  22. Attend the 2024 Douglas C. Montgomery Distinguished Lecture Series with

    2024 Douglas C. Montgomery Distinguished Lecture Series with Dr. G. Don Taylor. Join us for this exciting lecture designed to advance the discussion of important topics in industrial engineering. G. Don Taylor will discuss humanity's most pressing problems and how we can solve them.

  23. Easy way to solve Taylor's series numerical method best example

    724 47K views 1 year ago Numerical Solution of Ordinary Differential Equations (ODE's) In this video explained Easy way to solve Taylor's series numerical method best example. This Taylor's...