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# 1.4E: Exercises

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## Exercise $$\PageIndex{1}$$

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

1. $$\displaystyle (1−x)^{1/3}$$

2. $$\displaystyle (1+x^2)^{−1/3}$$

$$\displaystyle (1+x^2)^{−1/3}=\sum_{n=0}^∞(^{−\frac{1}{3}}_n)x^{2n}$$

3. $$\displaystyle (1−x)^{1.01}$$

4. $$\displaystyle (1−2x)^{2/3}$$

$$\displaystyle (1−2x)^{2/3}=\sum_{n=0}^∞(−1)^n2^n(^{\frac{2}{3}}_n)x^n$$

## Exercise $$\PageIndex{2}$$

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

1. $$\sqrt{x+2}$$ at $$\displaystyle a=0$$

2. $$\displaystyle \sqrt{x^2+2}$$ at $$\displaystyle a=0$$

$$\displaystyle \sqrt{2+x^2}=\sum_{n=0}^∞2^{(1/2)−n}(^{\frac{1}{2}}_n)x^{2n};(∣x^2∣<2)$$

3. $$\displaystyle \sqrt{x+2}$$ at $$\displaystyle a=1$$

4. $$\displaystyle \sqrt{2x−x^2}$$ at $$\displaystyle a=1$$ (Hint: $$\displaystyle 2x−x^2=1−(x−1)^2$$)

$$\displaystyle \sqrt{2x−x^2}=\sqrt{1−(x−1)^2}$$ so $$\displaystyle \sqrt{2x−x^2}=\sum_{n=0}^∞(−1)^n(^{\frac{1}{2}}_n)(x−1)^{2n}$$

5. $$\displaystyle (x−8)^{1/3}$$ at $$\displaystyle a=9$$

6. $$\displaystyle \sqrt{x}$$ at $$\displaystyle a=4$$

$$\displaystyle \sqrt{x}=2\sqrt{1+\frac{x−4}{4}}$$ so $$\displaystyle \sqrt{x}=\sum_{n=0}^∞2^{1−2n}(^{\frac{1}{2}}_n)(x−4)^n$$

7. $$\displaystyle x^{1/3}$$ at $$\displaystyle a=27$$

8. $$\displaystyle \sqrt{x}$$ at $$\displaystyle x=9$$

$$\displaystyle \sqrt{x}=\sum_{n=0}^∞3^{1−3n}(^{\frac{1}{2}}_n)(x−9)^n$$

## Exercise $$\PageIndex{3}$$

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 $$\displaystyle 1/1000.$$

1. $$\displaystyle (15)^{1/4}$$ using $$\displaystyle (16−x)^{1/4}$$

2. $$\displaystyle (1001)^{1/3}$$ using $$\displaystyle (1000+x)^{1/3}$$

$$\displaystyle 10(1+\frac{x}{1000})^{1/3}=\sum_{n=0}^∞10^{1−3n}(^{\frac{1}{3}}_n)x^n$$. Using, for example, a fourth-degree estimate at $$\displaystyle x=1$$ gives $$\displaystyle (1001)^{1/3}≈10(1+(^{\frac{1}{3}}_1)10^{−3}+(^{\frac{1}{3}}_2)10^{−6}+(^{\frac{1}{3}}_3)10^{−9}+(^{\frac{1}{3}}_4)10^{−12})=10(1+\frac{1}{3.10^3}−\frac{1}{9.10^6}+\frac{5}{81.10^9}−\frac{10}{243.10^{12}})=10.00333222...$$ whereas $$\displaystyle (1001)^{1/3}=10.00332222839093....$$ Two terms would suffice for three-digit accuracy.

## Exercise $$\PageIndex{4}$$

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

1. $$\displaystyle \frac{1}{\sqrt{2}}$$ using $$\displaystyle x=\frac{1}{2}$$ in $$\displaystyle (1−x)^{1/2}$$

2. $$\displaystyle \sqrt{5}=5×\frac{1}{\sqrt{5}}$$ using $$\displaystyle x=\frac{4}{5}$$ in $$\displaystyle (1−x)^{1/2}$$

The approximation is $$\displaystyle 2.3152$$; the CAS value is $$\displaystyle 2.23….$$

3. $$\displaystyle \sqrt{3}=\frac{3}{\sqrt{3}}$$ using $$\displaystyle x=\frac{2}{3}$$ in $$\displaystyle (1−x)^{1/2}$$

4. $$\displaystyle \sqrt{6}$$ using $$\displaystyle x=\frac{5}{6}$$ in $$\displaystyle (1−x)^{1/2}$$

The approximation is $$\displaystyle 2.583…$$; the CAS value is $$\displaystyle 2.449….$$

5. Integrate the binomial approximation of $$\displaystyle \sqrt{1−x}$$ to find an approximation of $$\displaystyle ∫^x_0\sqrt{1−t}dt$$.

6. Recall that the graph of $$\displaystyle \sqrt{1−x^2}$$ is an upper semicircle of radius $$\displaystyle 1$$. Integrate the binomial approximation of $$\displaystyle \sqrt{1−x^2}$$ up to order $$\displaystyle 8$$ from $$\displaystyle x=−1$$ to $$\displaystyle x=1$$ to estimate $$\displaystyle \frac{π}{2}$$.

$$\displaystyle \sqrt{1−x^2}=1−\frac{x^2}{2}−\frac{x^4}{8}−\frac{x^6}{16}−\frac{5x^8}{128}+⋯.$$ Thus $$\displaystyle ∫^1_{−1}\sqrt{1−x^2}dx=x−\frac{x^3}{6}−\frac{x^5}{40}−\frac{x^7}{7⋅16}−\frac{5x^9}{9⋅128}+⋯∣^1_{−1}≈2−\frac{1}{3}−\frac{1}{20}−\frac{1}{56}−\frac{10}{9⋅128}+error=1.590...$$ whereas $$\displaystyle \frac{π}{2}=1.570...$$

## Exercise $$\PageIndex{5}$$

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

1. $$\displaystyle (1+4x)^{1/3};a=0$$

2. $$\displaystyle (1+4x)^{4/3};a=0$$

$$\displaystyle (1+x)^{4/3}=(1+x)(1+\frac{1}{3}x−\frac{1}{9}x^2+\frac{5}{81}x^3−\frac{10}{243}x^4+⋯)=1+\frac{4x}{3}+\frac{2x^2}{9}−\frac{4x^3}{81}+\frac{5x^4}{243}+⋯$$

3. $$\displaystyle (3+2x)^{1/3};a=−1$$

4. $$\displaystyle (x^2+6x+10)^{1/3};a=−3$$

$$\displaystyle (1+(x+3)^2)^{1/3}=1+\frac{1}{3}(x+3)^2−\frac{1}{9}(x+3)^4+\frac{5}{81}(x+3)^6−\frac{10}{243}(x+3)^8+⋯$$

5. Use $$\displaystyle (1+x)^{1/3}=1+\frac{1}{3}x−\frac{1}{9}x^2+\frac{5}{81}x^3−\frac{10}{243}x^4+⋯$$ with $$\displaystyle x=1$$ to approximate $$\displaystyle 2^{1/3}$$.

6. Use the approximation $$\displaystyle (1−x)^{2/3}=1−\frac{2x}{3}−\frac{x^2}{9}−\frac{4x^3}{81}−\frac{7x^4}{243}−\frac{14x^5}{729}+⋯$$ for $$\displaystyle |x|<1$$ to approximate $$\displaystyle 2^{1/3}=2.2^{−2/3}$$.

Twice the approximation is $$\displaystyle 1.260…$$ whereas $$\displaystyle 2^{1/3}=1.2599....$$

7. Find the $$\displaystyle 25th$$ derivative of $$\displaystyle f(x)=(1+x^2)^{13}$$ at $$\displaystyle x=0$$.

8. Find the $$\displaystyle 99$$ th derivative of $$\displaystyle f(x)=(1+x^4)^{25}$$.

$$\displaystyle f^{(99)}(0)=0$$

## Exercise $$\PageIndex{6}$$

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

1. $$\displaystyle f(x)=xe^{2x}$$

2. $$\displaystyle f(x)=2^x$$

$$\displaystyle \sum_{n=0}^∞\frac{(ln(2)x)^n}{n!}$$

3. $$\displaystyle f(x)=\frac{sinx}{x}$$

4. $$\displaystyle f(x)=\frac{sin(\sqrt{x})}{\sqrt{x}},(x>0),$$

For $$\displaystyle x>0,sin(\sqrt{x})=\sum_{n=0}^∞(−1)^n\frac{x^{(2n+1)/2}}{\sqrt{x}(2n+1)!}=\sum_{n=0}^∞(−1)^n\frac{x^n}{(2n+1)!}$$.

5. $$\displaystyle f(x)=sin(x^2)$$

6. $$\displaystyle f(x)=e^{x^3}$$

$$\displaystyle e^{x^3}=\sum_{n=0}^∞\frac{x^{3n}}{n!}$$

7. $$\displaystyle f(x)=cos^2x$$ using the identity $$\displaystyle cos^2x=\frac{1}{2}+\frac{1}{2}cos(2x)$$

8. $$\displaystyle f(x)=sin^2x$$ using the identity $$\displaystyle sin^2x=\frac{1}{2}−\frac{1}{2}cos(2x)$$

$$\displaystyle sin^2x=−\sum_{k=1}^∞\frac{(−1)^k2^{2k−1}x^{2k}}{(2k)!}$$

## Exercise $$\PageIndex{7}$$

In the following exercises, find the Maclaurin series of $$\displaystyle F(x)=∫^x_0f(t)dt$$ by integrating the Maclaurin series of $$\displaystyle f$$ term by term. If $$\displaystyle f$$ is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero.

1. $$\displaystyle F(x)=∫^x_0e^{−t^2}dt;f(t)=e^{−t^2}=\sum_{n=0}^∞(−1)^n\frac{t^{2n}}{n!}$$

2. $$\displaystyle F(x)=tan^{−1}x;f(t)=\frac{1}{1+t^2}=\sum_{n=0}^∞(−1)^nt^{2n}$$

$$\displaystyle tan^{−1}x=\sum_{k=0}^∞\frac{(−1)^kx^{2k+1}}{2k+1}$$

3. $$\displaystyle F(x)=tanh^{−1}x;f(t)=\frac{1}{1−t^2}=\sum_{n=0}^∞t^{2n}$$

4. $$\displaystyle F(x)=sin^{−1}x;f(t)=\frac{1}{\sqrt{1−t^2}}=\sum_{k=0}^∞(^{\frac{1}{2}}_k)\frac{t^{2k}}{k!}$$

$$\displaystyle sin^{−1}x=\sum_{n=0}^∞(^{\frac{1}{2}}_n)\frac{x^{2n+1}}{(2n+1)n!}$$

5. $$\displaystyle F(x)=∫^x_0\frac{sint}{t}dt;f(t)=\frac{sint}{t}=\sum_{n=0}^∞(−1)^n\frac{t^{2n}}{(2n+1)!}$$

6. $$\displaystyle F(x)=∫^x_0cos(\sqrt{t})dt;f(t)=\sum_{n=0}^∞(−1)^n\frac{x^n}{(2n)!}$$

$$\displaystyle F(x)=\sum_{n=0}^∞(−1)^n\frac{x^{n+1}}{(n+1)(2n)!}$$

7. $$\displaystyle F(x)=∫^x_0\frac{1−cost}{t^2}dt;f(t)=\frac{1−cost}{t^2}=\sum_{n=0}^∞(−1)^n\frac{t^{2n}}{(2n+2)!}$$

8. $$\displaystyle F(x)=∫^x_0\frac{ln(1+t)}{t}dt;f(t)=\sum_{n=0}^∞(−1)^n\frac{t^n}{n+1}$$

$$\displaystyle F(x)=\sum_{n=1}^∞(−1)^{n+1}\frac{x^n}{n^2}$$

## Exercise $$\PageIndex{8}$$

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

1. $$\displaystyle f(x)=sin(x+\frac{π}{4})=sinxcos(\frac{π}{4})+cosxsin(\frac{π}{4})$$

2. $$\displaystyle f(x)=tanx$$

$$\displaystyle x+\frac{x^3}{3}+\frac{2x^5}{15}+⋯$$

3. $$\displaystyle f(x)=ln(cosx)$$

4. $$\displaystyle f(x)=e^xcosx$$

$$\displaystyle 1+x−\frac{x^3}{3}−\frac{x^4}{6}+⋯$$

5. $$\displaystyle f(x)=e^{sinx}$$

6. $$\displaystyle f(x)=sec^2x$$

$$\displaystyle 1+x^2+\frac{2x^4}{3}+\frac{17x^6}{45}+⋯$$

7. $$\displaystyle f(x)=tanhx$$

8. $$\displaystyle f(x)=\frac{tan\sqrt{x}}{\sqrt{x}}$$ (see expansion for $$\displaystyle tanx$$)

Using the expansion for $$\displaystyle tanx$$ gives $$\displaystyle 1+\frac{x}{3}+\frac{2x^2}{15}$$.

## Exercise $$\PageIndex{9}$$

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

1. $$\displaystyle ln(1+x)$$

2. $$\displaystyle \frac{1}{1+x^2}$$

$$\displaystyle \frac{1}{1+x^2}=\sum_{n=0}^∞(−1)^nx^{2n}$$ so $$\displaystyle R=1$$ by the ratio test.

3. $$\displaystyle tan^{−1}x$$

4. $$\displaystyle ln(1+x^2)$$

$$\displaystyle ln(1+x^2)=\sum_{n=1}^∞\frac{(−1)^{n−1}}{n}x^{2n}$$ so $$\displaystyle R=1$$ by the ratio test.

5. Find the Maclaurin series of $$\displaystyle sinhx=\frac{e^x−e^{−x}}{2}$$.

6. Find the Maclaurin series of $$\displaystyle coshx=\frac{e^x+e^{−x}}{2}$$.

Add series of $$\displaystyle e^x$$ and $$\displaystyle e^{−x}$$ term by term. Odd terms cancel and $$\displaystyle coshx=\sum_{n=0}^∞\frac{x^{2n}}{(2n)!}$$.

## Exercise $$\PageIndex{10}$$

1. Differentiate term by term the Maclaurin series of $$\displaystyle sinhx$$ and compare the result with the Maclaurin series of $$\displaystyle coshx$$.

2. Let $$\displaystyle S_n(x)=\sum_{k=0}^n(−1)^k\frac{x^{2k+1}}{(2k+1)!}$$ and $$\displaystyle C_n(x)=\sum_{n=0}^n(−1)^k\frac{x^{2k}}{(2k)!}$$ denote the respective Maclaurin polynomials of degree $$\displaystyle 2n+1$$ of $$\displaystyle sinx$$ and degree $$\displaystyle 2n$$ of $$\displaystyle cosx$$. Plot the errors $$\displaystyle \frac{S_n(x)}{C_n(x)}−tanx$$ for $$\displaystyle n=1,..,5$$ and compare them to $$\displaystyle x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}−tanx$$ on $$\displaystyle (−\frac{π}{4},\frac{π}{4})$$.

The ratio $$\displaystyle \frac{S_n(x)}{C_n(x)}$$ approximates $$\displaystyle tanx$$ better than does $$\displaystyle p_7(x)=x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}$$ for $$\displaystyle N≥3$$. The dashed curves are $$\displaystyle \frac{S_n}{C_n}−tan$$ for $$\displaystyle n=1,2$$. The dotted curve corresponds to $$\displaystyle n=3$$, and the dash-dotted curve corresponds to $$\displaystyle n=4$$. The solid curve is $$\displaystyle p_7−tanx$$. 3. Use the identity $$\displaystyle 2sinxcosx=sin(2x)$$ to find the power series expansion of $$\displaystyle sin^2x$$ at $$\displaystyle x=0$$. (Hint: Integrate the Maclaurin series of $$\displaystyle sin(2x)$$ term by term.)

4. If $$\displaystyle y=\sum_{n=0}^∞a_nx^n$$, find the power series expansions of $$\displaystyle xy′$$ and $$\displaystyle x^2y''$$.

By the term-by-term differentiation theorem, $$\displaystyle y′=\sum_{n=1}^∞na_nx^{n−1}$$ so $$\displaystyle y′=\sum_{n=1}^∞na_nx^{n−1}xy′=\sum_{n=1}^∞na_nx^n$$, whereas $$\displaystyle y′=\sum_{n=2}^∞n(n−1)a_nx^{n−2}$$ so $$\displaystyle xy''=\sum_{n=2}^∞n(n−1)a_nx^n$$.

5. Suppose that $$\displaystyle y=\sum_{k=0}^∞a^kx^k$$ satisfies $$\displaystyle y′=−2xy$$ and $$\displaystyle y(0)=0$$. Show that $$\displaystyle a_{2k+1}=0$$ for all $$\displaystyle k$$ and that $$\displaystyle a_{2k+2}=\frac{−a_{2k}}{k+1}$$. Plot the partial sum $$\displaystyle S_{20}$$ of $$\displaystyle y$$ on the interval $$\displaystyle [−4,4]$$.

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

The probability is $$\displaystyle p=\frac{1}{\sqrt{2π}}∫^{(b−μ)/σ}_{(a−μ)/σ}e^{−x^2/2}dx$$ where $$\displaystyle a=90$$ and $$\displaystyle b=100$$, that is, $$\displaystyle p=\frac{1}{\sqrt{2π}}∫^1_{−1}e^{−x^2/2}dx=\frac{1}{\sqrt{2π}}∫^1_{−1}\sum_{n=0}^5(−1)^n\frac{x^{2n}}{2^nn!}dx=\frac{2}{\sqrt{2π}}\sum_{n=0}^5(−1)^n\frac{1}{(2n+1)2^nn!}≈0.6827.$$

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

8. Suppose that $$\displaystyle \sum_{n=0}^∞a_nx^n$$ converges to a function $$\displaystyle f(x)$$ such that $$\displaystyle f(0)=1,f′(0)=0$$, and $$\displaystyle f''(x)=−f(x)$$. Find a formula for $$\displaystyle a_n$$ and plot the partial sum $$\displaystyle S_N$$ for $$\displaystyle N=20$$ on $$\displaystyle [−5,5].$$

As in the previous problem one obtains $$\displaystyle a_n=0$$ if $$\displaystyle n$$ is odd and $$\displaystyle a_n=−(n+2)(n+1)a_{n+2}$$ if $$\displaystyle n$$ is even, so $$\displaystyle a_0=1$$ leads to $$\displaystyle a_{2n}=\frac{(−1)^n}{(2n)!}$$. 9. Suppose that $$\displaystyle \sum_{n=0}^∞a_nx^n$$ converges to a function $$\displaystyle f(x)$$ such that $$\displaystyle f(0)=0,f′(0)=1$$, and $$\displaystyle f''(x)=−f(x)$$. Find a formula for an and plot the partial sum $$\displaystyle S_N$$ for $$\displaystyle N=10$$ on $$\displaystyle [−5,5]$$.

10. Suppose that $$\displaystyle \sum_{n=0}^∞a_nx^n$$ converges to a function $$\displaystyle y$$ such that $$\displaystyle y''−y′+y=0$$ where $$\displaystyle y(0)=1$$ and $$\displaystyle y'(0)=0.$$ Find a formula that relates $$\displaystyle a_{n+2},a_{n+1},$$ and an and compute $$\displaystyle a_0,...,a_5$$.

$$\displaystyle y''=\sum_{n=0}^∞(n+2)(n+1)a_{n+2}x^n$$ and $$\displaystyle y′=\sum_{n=0}^∞(n+1)a_{n+1}x^n$$ so $$\displaystyle y''−y′+y=0$$ implies that $$\displaystyle (n+2)(n+1)a_{n+2}−(n+1)a_{n+1}+a_n=0$$ or $$\displaystyle a_n=\frac{a_{n−1}}{n}−\frac{a_{n−2}}{n(n−1)}$$ for all $$\displaystyle n⋅y(0)=a_0=1$$ and $$\displaystyle y′(0)=a_1=0,$$ so $$\displaystyle a_2=\frac{1}{2},a_3=\frac{1}{6},a_4=0$$, and $$\displaystyle a_5=−\frac{1}{120}$$.

11. Suppose that $$\displaystyle \sum_{n=0}^∞a_nx^n$$ converges to a function $$\displaystyle y$$ such that $$\displaystyle y''−y′+y=0$$ where $$\displaystyle y(0)=0$$ and $$\displaystyle y′(0)=1$$. Find a formula that relates $$\displaystyle a_{n+2},a_{n+1}$$, and an and compute $$\displaystyle a_1,...,a_5$$.

The error in approximating the integral $$\displaystyle ∫^b_af(t)dt$$ by that of a Taylor approximation $$\displaystyle ∫^b_aPn(t)dt$$ is at most $$\displaystyle ∫^b_aR_n(t)dt$$. In the following exercises, the Taylor remainder estimate $$\displaystyle R_n≤\frac{M}{(n+1)!}|x−a|^{n+1}$$ guarantees that the integral of the Taylor polynomial of the given order approximates the integral of $$\displaystyle f$$ with an error less than $$\displaystyle \frac{1}{10}$$.

a. Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less than $$\displaystyle \frac{1}{100}$$.

b. Compare the accuracy of the polynomial integral estimate with the remainder estimate.

12. $$\displaystyle ∫^π_0\frac{sint}{t}dt;P_s=1−\frac{x^2}{3!}+\frac{x^4}{5!}−\frac{x^6}{7!}+\frac{x^8}{9!}$$ (You may assume that the absolute value of the ninth derivative of $$\displaystyle \frac{sint}{t}$$ is bounded by $$\displaystyle 0.1$$.)

a. (Proof)

b. We have $$\displaystyle R_s≤\frac{0.1}{(9)!}π^9≈0.0082<0.01.$$ We have $$\displaystyle ∫^π_0(1−\frac{x^2}{3!}+\frac{x^4}{5!}−\frac{x^6}{7!}+\frac{x^8}{9!})dx=π−\frac{π^3}{3⋅3!}+\frac{π^5}{5⋅5!}−\frac{π^7}{7⋅7!}+\frac{π^9}{9⋅9!}=1.852...,$$ whereas $$\displaystyle ∫^π_0\frac{sint}{t}dt=1.85194...$$, so the actual error is approximately $$\displaystyle 0.00006.$$

13. $$\displaystyle ∫^2_0e^{−x^2}dx;p_{11}=1−x^2+\frac{x^4}{2}−\frac{x^6}{3!}+⋯−\frac{x^{22}}{11!}$$ (You may assume that the absolute value of the $$\displaystyle 23rd$$ derivative of $$\displaystyle e^{−x^2}$$ is less than $$\displaystyle 2×10^{14}$$.

## Exercise $$\PageIndex{11}$$

The following exercises deal with Fresnel integrals.

1. The Fresnel integrals are defined by $$\displaystyle C(x)=∫^x_0cos(t^2)dt$$ and $$\displaystyle S(x)=∫^x_0sin(t^2)dt$$. Compute the power series of $$\displaystyle C(x)$$ and $$\displaystyle S(x)$$ and plot the sums $$\displaystyle C_N(x)$$ and $$\displaystyle S_N(x)$$ of the first $$\displaystyle N=50$$ nonzero terms on $$\displaystyle [0,2π]$$.

Since $$\displaystyle cos(t^2)=\sum_{n=0}^∞(−1)^n\frac{t^{4n}}{(2n)!}$$ and $$\displaystyle sin(t^2)=\sum_{n=0}^∞(−1)^n\frac{t^{4n+2}}{(2n+1)!}$$, one has $$\displaystyle S(x)=_sum_{n=0}^∞(−1)^n\frac{x^{4n+3}}{(4n+3)(2n+1)!}$$ and $$\displaystyle C(x)=\sum_{n=0}^∞(−1)^n\frac{x^{4n+1}}{(4n+1)(2n)!}$$. The sums of the first $$\displaystyle 50$$ nonzero terms are plotted below with $$\displaystyle C_{50}(x)$$ the solid curve and $$\displaystyle S_{50}(x)$$ the dashed curve. 2. 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 $$\displaystyle (C(t),S(t))$$. Plot the curve $$\displaystyle (C_{50},S_{50})$$ for $$\displaystyle 0≤t≤2π$$, the coordinates of which were computed in the previous exercise.

## Exercise $$\PageIndex{12}$$

1. Estimate $$\displaystyle ∫^{1/4}_0\sqrt{x−x^2}dx$$ by approximating $$\displaystyle \sqrt{1−x}$$ using the binomial approximation $$\displaystyle 1−\frac{x}{2}−\frac{x^2}{8}−\frac{x^3}{16}−\frac{5x^4}{2128}−\frac{7x^5}{256}$$.

$$\displaystyle ∫^{1/4}_0\sqrt{x}(1−\frac{x}{2}−\frac{x^2}{8}−\frac{x^3}{16}−\frac{5x^4}{128}−\frac{7x^5}{256})dx =\frac{2}{3}2^{−3}−\frac{1}{2}\frac{2}{5}2^{−5}−\frac{1}{8}\frac{2}{7}2^{−7}−\frac{1}{16}\frac{2}{9}2^{−9}−\frac{5}{128}\frac{2}{11}2^{−11}−\frac{7}{256}\frac{2}{13}2^{−13}=0.0767732...$$ whereas $$\displaystyle ∫^{1/4}_0\sqrt{x−x^2}dx=0.076773.$$

2. Use Newton’s approximation of the binomial $$\displaystyle \sqrt{1−x^2}$$ to approximate $$\displaystyle π$$ as follows. The circle centered at $$\displaystyle (\frac{1}{2},0)$$ with radius $$\displaystyle \frac{1}{2}$$ has upper semicircle $$\displaystyle y=\sqrt{x}\sqrt{1−x}$$. The sector of this circle bounded by the $$\displaystyle x$$-axis between $$\displaystyle x=0$$ and $$\displaystyle x=\frac{1}{2}$$ and by the line joining $$\displaystyle (\frac{1}{4},\frac{\sqrt{3}}{4})$$ corresponds to $$\displaystyle \frac{1}{6}$$ of the circle and has area $$\displaystyle \frac{π}{24}$$. This sector is the union of a right triangle with height $$\displaystyle \frac{\sqrt{3}}{4}$$ and base $$\displaystyle \frac{1}{4}$$ and the region below the graph between $$\displaystyle x=0$$ and $$\displaystyle x=\frac{1}{4}$$. To find the area of this region you can write $$\displaystyle y=\sqrt{x}\sqrt{1−x}=\sqrt{x}×(\text{binomial expansion of} \sqrt{1−x})$$ and integrate term by term. Use this approach with the binomial approximation from the previous exercise to estimate $$\displaystyle π$$.

3. Use the approximation $$\displaystyle T≈2π\sqrt{\frac{L}{g}}(1+\frac{k^2}{4})$$ to approximate the period of a pendulum having length $$\displaystyle 10$$ meters and maximum angle $$\displaystyle θ_{max}=\frac{π}{6}$$ where $$\displaystyle k=sin(\frac{θ_{max}}{2})$$. Compare this with the small angle estimate $$\displaystyle T≈2π\sqrt{\frac{L}{g}}$$.

$$\displaystyle T≈2π\sqrt{\frac{10}{9.8}}(1+\frac{sin^2(θ/12)}{4})≈6.453$$ seconds. The small angle estimate is $$\displaystyle T≈2π\sqrt{\frac{10}{9.8}≈6.347}$$. The relative error is around $$\displaystyle 2$$ percent.
4. Suppose that a pendulum is to have a period of $$\displaystyle 2$$ seconds and a maximum angle of $$\displaystyle θ_{max}=\frac{π}{6}$$. Use $$\displaystyle T≈2π\sqrt{\frac{L}{g}}(1+\frac{k^2}{4})$$ to approximate the desired length of the pendulum. What length is predicted by the small angle estimate $$\displaystyle T≈2π\sqrt{\frac{L}{g}}$$?
5. Evaluate $$\displaystyle ∫^{π/2}_0sin^4θdθ$$ in the approximation $$\displaystyle T=4\sqrt{\frac{L}{g}}∫^{π/2}_0(1+\frac{1}{2}k^2sin^2θ+\frac{3}{8}k^4sin^4θ+⋯)dθ$$ to obtain an improved estimate for $$\displaystyle T$$.
$$\displaystyle ∫^{π/2}_0sin^4θdθ=\frac{3π}{16}.$$ Hence $$\displaystyle T≈2π\sqrt{\frac{L}{g}}(1+\frac{k^2}{4}+\frac{9}{256}k^4).$$
6. An equivalent formula for the period of a pendulum with amplitude $$\displaystyle θ_max$$ is $$\displaystyle T(θ_{max})=2\sqrt{2}\sqrt{\frac{L}{g}}∫^{θ_{max}}_0\frac{dθ}{\sqrt{cosθ}−cos(θ_{max})}$$ where $$\displaystyle L$$ is the pendulum length and $$\displaystyle g$$ is the gravitational acceleration constant. When $$\displaystyle θ_{max}=\frac{π}{3}$$ we get $$\displaystyle \frac{1}{\sqrt{cost−1/2}}≈\sqrt{2}(1+\frac{t^2}{2}+\frac{t^4}{3}+\frac{181t^6}{720})$$. Integrate this approximation to estimate $$\displaystyle T(\frac{π}{3})$$ in terms of $$\displaystyle L$$ and $$\displaystyle g$$. Assuming $$\displaystyle g=9.806$$ meters per second squared, find an approximate length $$\displaystyle L$$ such that $$\displaystyle T(\frac{π}{3})=2$$ seconds.