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Mathematics LibreTexts

5.4: 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 non-elementary 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 ex2dx, 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

f(x)=(1+x)0=1,f(x)=(1+x)1=1+x,f(x)=(1+x)2=1+2x+x2,f(x)=(1+x)3=1+3x+3x2+x3f(x)=(1+x)4=1+4x+6x2+4x3+x4.

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 xn in the binomial expansion of (1+x)r is given by

(rn)=r!n!(rn)!

and

f(x)=(1+x)r=(r0)+(r1)x+(r2)x2+(r3)x3++(rr1)xr1+(rr)xr=rn=0(rn)xn.

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

f(x)=(1+x)5=(50)1+(51)x+(52)x2+(53)x3+(54)x4+(55)x5=5!0!5!1+5!1!4!x+5!2!3!x2+5!3!2!x3+5!4!1!x4+5!5!0!x5=1+5x+10x2+10x3+5x4+x5.

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.

f(x)=(1+x)rf(0)=1f(x)=r(1+x)r1f(0)=rf(x)=r(r1)(1+x)r2f(0)=r(r1)f(x)=r(r1)(r2)(1+x)r3f(0)=r(r1)(r2)f(n)(x)=r(r1)(r2)(rn+1)(1+x)rnf(n)(0)=r(r1)(r2)(rn+1)

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

f(n)(0)n!=r(r1)(r2)(rn+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 5.4.3  for the finite binomial expansion. More generally, to denote the binomial coefficients for any real number r, we define

(rn)=(r1)(r2)(rn+1)n!.

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

n=0(rn)xn=1+rx+r(r1)2!x2++r(r1)(rn+1)n!xn+.

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

|an+1||an|=|r(r1)(r2)(rn)|x||n+1(n+1)!n!|r(r1)(r2)(rn+1)||x|n=|rn||x||n+1|.

Since

limn|an+1||an|=|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 r0 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 Rn(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=n=0(rn)xn=1+rx+r(r1)2!x2++r(r1)(rn+1)n!xn+

for |x|<1.

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

Example 5.4.1: Finding Binomial Series
  1. Find the binomial series for f(x)=1+x.
  2. Use the third-order Maclaurin polynomial p3(x) to estimate 1.5. Use Taylor’s theorem to bound the error. Use a graphing utility to compare the graphs of f and p3.
Solution

a. Here r=12. Using the definition for the binomial series, we obtain

1+x=1+12x+(1/2)(1/2)2!x2+(1/2)(1/2)(3/2)3!x3+=1+12x12!122x2+13!1323x3+(1)n+1n!135(2n3)2nxn+=1+12x+n=2(1)n+1n!135(2n3)2nxn.

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

p3(x)=1+12x18x2+116x3.

Therefore,

1.5=1+0.51+12(0.5)18(0.5)2+116(0.5)31.2266.

From Taylor’s theorem, the error satisfies

R3(0.5)=f(4)(c)4!(0.5)4

for some c between 0 and 0.5. Since f(4)(x)=1524(1+x)7/2, and the maximum value of f(4)(x) on the interval (0,0.5) occurs at x=0, we have

|R3(0.5)|154!24(0.5)40.00244.

The function and the Maclaurin polynomial p3 are graphed in Figure 5.4.1.

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.
Figure 5.4.1: The third-order Maclaurin polynomial p3(x) provides a good approximation for f(x)=1+x for x near zero.
Exercise 5.4.1

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

Hint

Use the definition of binomial series for r=2.

Answer

n=0(1)n(n+1)xn

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 5.4.1, 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)=tan1x 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.)

Table 5.4.1: Maclaurin Series for Common Functions
Function Maclaurin Series Interval of Convergence
f(x)=11x n=0xn 1<x<1
f(x)=ex n=0xnn! <x<
f(x)=sinx n=0(1)nx2n+1(2n+1)! <x<
f(x)=cosx n=0(1)nx2n(2n)! <x<
f(x)=ln(1+x) n=0(1)n+1xnn 1<x1
f(x)=tan1x n=0(1)nx2n+12n+1 1x1
f(x)=(1+x)r n=0(rn)xn 1<x<1
Convergence at the endpoints
depends on the value of r 

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 5.4.1, to create Maclaurin series for other functions.

Example 5.4.2: 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 5.4.1.

  1. f(x)=cosx
  2. f(x)=sinhx
Solution

a. Using the Maclaurin series for cosx we find that the Maclaurin series for cosx is given by

n=0(1)n(x)2n(2n)!=n=0(1)nxn(2n)!=1x2!+x24!x36!+x48!.

This series converges to cosx for all x in the domain of cosx; that is, for all x0.

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

sinhx=exex2.

Using the Maclaurin series for ex, we see that the nth term in the Maclaurin series for sinhx is given by

xnn!(x)nn!.

For n even, this term is zero. For n odd, this term is 2xnn!. Therefore, the Maclaurin series for sinhx has only odd-order terms and is given by

n=0x2n+1(2n+1)!=x+x33!+x55!+.

Exercise 5.4.2

Find the Maclaurin series for sin(x2).

Hint

Use the Maclaurin series for sinx.

Answer

n=0(1)nx4n+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 5.4.3, we differentiate the binomial series for 1+x term by term to find the binomial series for 11+x. Note that we could construct the binomial series for 11+x directly from the definition, but differentiating the binomial series for 1+x is an easier calculation.

Example 5.4.3: Differentiating a Series to Find a New Series

Use the binomial series for 1+x to find the binomial series for 11+x.

Solution

The two functions are related by

ddx1+x=121+x,

so the binomial series for 11+x is given by

11+x=2ddx1+x=1+n=1(1)nn!135(2n1)2nxn.

Exercise 5.4.3

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

Hint

Differentiate the series for 11+x

Answer

n=0(1)nn!135(2n+1)2nxn

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=Cex. 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=0cnxn 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 5.4.4: Power Series Solution of a Differential Equation

Use power series to solve the initial-value problem y=y,y(0)=3.

Solution

Suppose that there exists a power series solution

y(x)=n=0cnxn=c0+c1x+c2x2+c3x3+c4x4+.

Differentiating this series term by term, we obtain

y=c1+2c2x+3c3x2+4c4x3+.

If y satisfies the differential equation, then

c0+c1x+c2x2+c3x3+=c1+2c2x+3c3x2+4c3x3+.

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

c0=c1,

c1=2c2,

c2=3c3,

c3=4c4,

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

y(x)=c0+c1x+c2x2+c3x3+,

we find that c0=3. We are now ready to solve for the rest of the coefficients. Using the fact that c0=3, we have

c1=c0=3=31!,c2=c12=32=32!,c3=c23=332=33!,c4=c34=3432=34!.

Therefore,

y=3[1+11!x+12!x2+13!x314!x4+]=3n=0xnn!.

You might recognize

n=0xnn!

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

Exercise 5.4.4

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

Hint

The equations for the first several coefficients cn will satisfy c0=2c1,c1=22c2,c2=23c3,. In general, for all n0,cn=2(n+1)Cn+1.

Answer

y=5e2x

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

yxy=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 5.4.5: Power Series Solution of Airy’s Equation

Use power series to solve yxy=0 with the initial conditions y(0)=a and y(0)=b.

Solution

We look for a solution of the form

y=n=0cnxn=c0+c1x+c2x2+c3x3+c4x4+

Differentiating this function term by term, we obtain

y=c1+2c2x+3c3x2+4c4x3+,y=21c2+32c3x+43c4x2+.

If y satisfies the equation y=xy, then

21c2+32c3x+43c4x2+=x(c0+c1x+c2x2+c3x3+).

Using the Uniqueness of Power Series Theorem (from an earlier Section) on the uniqueness of power series representations, we know that coefficients of the same degree must be equal. Therefore,

21c2=0,

32c3=c0,

43c4=c1,

54c5=c2,

More generally, for n3, we have n(n1)cn=cn3. In fact, all coefficients can be written in terms of c0 and c1. To see this, first note that c2=0. Then

c3=c032,

c4=c143.

For c5,c6,c7, we see that

c5=c254=0,c6=c365=c06532,c7=c476=c17643.

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

y=c0+c1x+0x2+c032x3+c143x4+0x5+c06532x6+c17643x7+.

The initial condition y(0)=a implies c0=a. Differentiating this series term by term and using the fact that y(0)=b, we conclude that c1=b.

Therefore, the solution of this initial-value problem is

y=a(1+x332+x66532+)+b(x+x443+x77643+).

Exercise 5.4.5

Use power series to solve y+x2y=0 with the initial condition y(0)=a and y(0)=b.

Hint

The coefficients satisfy c0=a,c1=b,c2=0,c3=0, and for n4,n(n1)cn=cn4.

Answer

y=a(1x434+x83478)+b(xx545+x94589)

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 ex2dx. Unfortunately, the antiderivative of the integrand ex2 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)=x23x+ex3sin(5x+4) is an elementary function, although not a particularly simple-looking function. Any integral of the form f(x)dx where the antiderivative of f cannot be written as an elementary function is considered a non-elementary integral.

Non-elementary 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 ex2dx.

Example 5.4.6: Using Taylor Series to Evaluate a Definite Integral
  1. Express ex2dx as an infinite series.
  2. Evaluate 10ex2dx to within an error of 0.01.
Solution

a. The Maclaurin series for ex2 is given by

ex2=n=0(x2)nn!=1x2+x42!x63!++(1)nx2nn!+=n=0(1)nx2nn!.

Therefore,

ex2dx=(1x2+x42!x63!++(1)nx2nn!+)dx=C+xx33+x55.2!x77.3!++(1)nx2n+1(2n+1)n!+.

b. Using the result from part a. we have

10ex2dx=113+110142+1216.

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 12160.0046296<0.01.

Exercise 5.4.6

Express cosxdx as an infinite series. Evaluate 10cosxdx to within an error of 0.01.

Hint

Use the series found in Example 5.4.2.

Answer

C+n=1(1)n+1xnn(2n2)! The definite integral is approximately 0.764 to within an error of 0.01.

As mentioned above, the integral ex2dx 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.\label{probeq}

(See Figure \PageIndex{2}.)

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.
Figure \PageIndex{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=\dfrac{1}{σ\sqrt{2π}}e^{−(x−μ)^2/(2σ^2)} between x=a and x=b.

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 \ref{probeq} becomes

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

In Example \PageIndex{7}, 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 \ref{probeq} 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.

Solution

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

\dfrac{1}{\sqrt{2π}}\int ^2_0e^{−z^2/2}\,dz.\label{probeqEx7} 

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

\begin{align*} e^{−x^2/2}&=\sum_{n=0}^∞\dfrac{\left(−\dfrac{x^2}{2}\right)^n}{n!}\\[5pt] &=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}!+⋯\\[5pt] &=\sum_{n=0}^∞(−1)^n\dfrac{x^{2n}}{2^n⋅n!}.\end{align*}

Therefore,

\begin{align*} \dfrac{1}{\sqrt{2π}}\int e^{−z^2/2}\,dz&=\dfrac{1}{\sqrt{2π}}\int \left(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!}+⋯\right)dz\\[5pt] &=\dfrac{1}{\sqrt{2π}}\left(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!}+⋯\right)\end{align*}

We now use this result to evaluate the definite integral from Equation \ref{probeqEx7}:
\dfrac{1}{\sqrt{2π}}\int ^2_0e^{−z^2/2}\,dz=\dfrac{1}{\sqrt{2π}}\left(2−\dfrac{8}{6}+\dfrac{32}{40}−\dfrac{128}{336}+\dfrac{512}{3456}−\dfrac{2^{11}}{11⋅2^5⋅5!}+⋯\right)\nonumber

Using the first five terms, we estimate that the probability is approximately 0.4729. 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.\nonumber

Analysis

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.

Alternative method. If term-by-term integration is done instead, the result of evaluating the definite integral is

\displaystyle \ \dfrac{1}{\sqrt{2π}} \sum_{n=0}^∞\dfrac{ (-1)^n}{2^n n!} \int ^2_0 z^{2n}\; dz =\displaystyle \ \dfrac{1}{\sqrt{2π}} \sum_{n=0}^∞\dfrac{ (-1)^n}{(2n+1)2^n n!} \left[ 2^{2n+1}-0^{2n+1} \right]  =\displaystyle \ \dfrac{1}{\sqrt{2π}} \sum_{n=0}^∞\dfrac{ (-1)^n 2^{n+1}}{ n!(2n+1)}  

=\dfrac{1}{\sqrt{2π}}\left(\dfrac{2}{1\cdot  0! }−\dfrac{4}{3 \cdot  1! }+\dfrac{8}{5 \cdot  2! }−\dfrac{16}{7  \cdot 3!}+\dfrac{32}{ 9 \cdot  4!}−\dfrac{64}{11 \cdot  5! }+⋯\right)
=\dfrac{1}{\sqrt{2π}}\left(2−\dfrac{4}{3}+\dfrac{4}{5}−\dfrac{8}{21}+\dfrac{4}{27}−\dfrac{8}{165}+⋯\right)

Exercise \PageIndex{7}

Given a set of normally distributed standardized test scores with mean μ=100 and standard deviation σ=50, 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.

Hint

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

Answer

The estimate is approximately 0.3414. This estimate is accurate to within 0.0000094.

Another application in which a non-elementary integral arises involves the period of a pendulum. The integral is

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

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^2\sin^2θ}}\nonumber

where g is the acceleration due to gravity and k=\sin\left(\dfrac{θ_{max}}{2}\right) (see Figure \PageIndex{3}). (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 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.
Figure \PageIndex{3}: This pendulum has length L and makes a maximum angle θ_{max} with the vertical.

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\nonumber

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

  1. you use only the first term in the binomial series, and
  2. you use the first two terms in the binomial series.
Solution

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

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

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

T≈4\sqrt{\dfrac{L}{g}}\int ^{π/2}_0\,dθ=2π\sqrt{\dfrac{L}{g}}.\nonumber

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

\int ^{π/2}_0\left(1+\frac{1}{2}k^2\sin^2θ+\dfrac{1⋅3}{2!2^2}k^4\sin^4θ+⋯\right)\,dθ.\nonumber

Since |\sin x|≤1, this integral is bounded by

\int ^{π/2}_0\left(\dfrac{1}{2}k^2+\dfrac{1.3}{2!2^2}k^4+⋯\right)\,dθ\;<\;\dfrac{π}{2}\left(\dfrac{1}{2}k^2+\dfrac{1⋅3}{2!2^2}k^4+⋯\right).\nonumber

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{\frac{L}{g}}\int ^{π/2}_0\left(1+\dfrac{1}{2}k^2\sin^2θ\right)\,dθ=2π\sqrt{\dfrac{L}{g}}\left(1+\dfrac{k^2}{4}\right).\nonumber

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.

Glossary

binomial series
the Maclaurin series for f(x)=(1+x)^r; it is given by \displaystyle (1+x)^r=\sum_{n=0}^∞\binom{r}{n}x^n=1+rx+\dfrac{r(r−1)}{2!}x^2+⋯+\dfrac{r(r−1)⋯(r−n+1)}{n!}x^n+⋯ for |x|<1
non-elementary integral
an integral for which the antiderivative of the integrand cannot be expressed as an elementary function

This page titled 5.4: Working with Taylor Series is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.

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