Taylor Series - AP Calculus BC

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Question

Give the Maclaurin series for the function

$f(x) = \frac{1}{6^{x}}$

up to the third term.

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Answer

Rewrite this function as $f(x) = \frac{1}{6^{x}} =\left ( \frac{1}{6 } \right )^{x} =\left ( e^{\ln\frac{1}{6} } \right \) ^{x}=e ^{-x\ln 6}$ .

The Maclaurin series for $e^{x}$ , taken to the third term, is $e ^{x} = 1 + x + \frac{x^{2}}{2}+...$ .

Substitute $-x \ln 6$ for :

$f(x) =e ^{-x\ln 6}$

$= 1 + (-x\ln 6) + \frac{(-x\ln 6)^{2}}{2}+...$

$= 1 -(\ln 6 ) x +\left [ \frac{( \ln 6)^{2}}{2} \right ]x^{2} +...$

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Question

Give the Maclaurin series for the function

$f(x) = e ^{2x ^{2}}$

up to the third term.

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Answer

The Maclaurin series for $e^{x}$ , taken to the third term, is:

$e ^{x} = 1 + x + \frac{x^{2}}{2}+...$

Substitute $2x^{2}$ for :

$f(x) = e ^{2x ^{2}}$

$e ^{x} = 1 + 2x^{2} + \frac{(2x^{2})^{2}}{2}+...$

$e ^{x} = 1 + 2x^{2} + \frac{4x^{4}}{2}+...$

$e ^{x} = 1 + 2x^{2} + 2x^{4} +...$

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Question

Give the polar form of the equation of a circle with center at and radius .

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Answer

This circle will have equation

$(x-5)^{2} + (y+4) ^{2} = 16^{2}$ .

Rewrite this as follows:

$x ^{2} - 10x + 25 + y ^{2} + 8y + 16 = 256$

$x ^{2} - 10x + 25 + y ^{2} + 8y + 16 - 41 = 256- 41$

$x ^{2}+ y ^{2} - 10x + 8y = 215$

$r ^{2} - 10 r \cos \theta + 8 r \sin \theta = 215$

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Question

Give the Maclaurin series for the function

$f(x) = \cos \frac{4x}{3}$

up to the third term.

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Answer

The Maclaurin series for $\cos x$ is

$\cos x = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} -...$

Substitute $\frac{4x}{3}$ for . The series becomes

$\cos \frac{4x}{3} = 1 - \frac{\left (\frac{4x}{3} \right )^{2}}{2} + \frac{\left (\frac{4x}{3} \right )^{4}}{24} -...$

$= 1 - \frac{\left (\frac{16x^{2}}{9} \right )}{2} + \frac{\left (\frac{256x^{4}}{81} \right )}{24} -...$

$= 1 -\frac{8}{9} x^{2} + \frac{32}{243} x^{4} -...$

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Question

Give the Maclaurin series for the function

$f(x) = \cos \left (2x^{2} \right )$

up to the third term.

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Answer

The Maclaurin series for $\cos x$ is

$\cos x = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} -...$

Substitute $2x^{2}$ for . The series becomes

$\cos \left (2x^{2} \right ) = 1 - \frac{\left (2x^{2} \right )^{2}}{2} + \frac{\left (2x^{2} \right )^{4}}{24} -...$

$= 1 - \frac{ 4x^{4} }{2} + \frac{ 16x^{8} }{24} -...$

$= 1 -2x^{4} + \frac{ 2 }{3}x^{8} -...$

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Question

Give the Maclaurin series for the function

$f(x) = \cos \sqrt {2x}$

up to the third term.

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Answer

The Maclaurin series for $\cos x$ is

$\cos x = 1 - \frac{x^{2}}{2} + \frac{x^{4}}{24} -...$

Substitute $\sqrt{2x}$ for . The series becomes

$\cos \sqrt{2x} = 1 - \frac{\left ( \sqrt{2x} \right )^{2}}{2} + \frac{\left ( \sqrt{2x} \right)^{4}}{24} -...$

$= 1 - \frac{2x}{2} + \frac{4x^{2}}{24} -...$

$= 1 - x + \frac{1}{6}x^{2} -...$

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Question

Give the Maclaurin series for the function

$f(x) = \sqrt{3^x}$

up to the third term.

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Answer

Rewrite this function as

$f(x) = \sqrt{3^x} =\left ( \sqrt{3} \right )^x$

$= \left ( 3 ^{\frac{1}{2}} \right )^x = \left ( e^{\ln 3 ^{\frac{1}{2}}} \right )^x = \left ( e^{\frac{1}{2} \ln 3} \right )^x= \left ( e^{\frac{1}{2} \ln 3 ; \cdot x } \right )$

The Maclaurin series for $e^{x}$ , taken to the third term, is:

$e ^{x} = 1 + x + \frac{x^{2}}{2}+...$

Substitute $\frac{1}{2} \ln 3 ; \cdot x$ for :

$f(x) = \sqrt{3^x} = 1 + \frac{1}{2} \ln 3 ; \cdot x + \frac{(\frac{1}{2} \ln 3 ; \cdot x)^{2}}{2}+...$ $= 1 + \frac{1}{2} \ln 3 ; \cdot x + \frac{\left \(\frac{1}{4} \right\)( \ln 3 )^{2} x^{2}}{2}+...$

$= 1 + \left ( \frac{\ln 3}{2} \right ) x +\left [ \frac{ ( \ln 3 )^{2} }{8} \right ] ; x^{2}+...$

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Question

What is the value of the following infinite series?

$\small \sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n+1}}{3^{2n+1}(2n+1)!}$

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Answer

We can recognize this series as $\small \sin(x)$ since the power series is

$\small \sin x=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+1}}{(2n+1)!}$

with the value $\small \frac{\pi}{3}$ plugged into $\small x$ since

$\small \small \sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n+1}}{3^{2n+1}(2n+1)!}=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\left(\frac{\pi}{3} \right )^{2n+1}$ .

So then we have

$\small \small \sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n+1}}{3^{2n+1}(2n+1)!}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$ .

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Question

Suppose that $\small f(x)=\cos x^3$ . Calculate $\small f^{(48)}(0)$ .

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Answer

Let's find the power series of $\small \cos x^3$ centered at $\small x=0$ to find $\small f^{(48)}(0)$ . We have

$\small \small \small \small \cos x^3= \sum_{n=0}^\infty \frac{(-1)^n(x^3)^{2n}}{(2n)!}=\sum_{n=0}^\infty \frac{(-1)^nx^{6n}}{(2n)!}=1-\frac{x^6}{2!}+\frac{x^{12}}{4!}-...$

This series is much easier to differentiate than the expression $\small \cos x^3$ . We must look at term $\small x^{48}$ , which is the only constant term left after differentiating 48 times. This is the only important term, because when we plug in $\small x=0$ , all of the non-constant terms are zero. So we must have

$\small \small (\cos x^3)^{(48)}|_{x=0}= \left( \frac{x^{48}}{16!}\right )^{(48)}|_{x=0}=\frac{48!}{16!}$

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Question

What is the value of the following infinite series?

$\small \small \small \sum_{n=0}^\infty \frac{1+(-1)^n}{n!}$

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Answer

The infinite series can be computed easily by splitting up the two components of the numerator:

$\small \small \small \small \sum_{n=0}^\infty \frac{1+(-1)^n}{n!}=\sum_{n=0}^\infty \frac{1}{n!}+\sum_{n=0}^\infty \frac{(-1)^n}{n!}$

Now we recall the MacLaurin series for the exponential function $\small e^x$ , which is

$\small e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$

which converges for all $\small x$ . We can see that the two infinite series are $\small e^x$ with $\small x=1,-1$ , respectively. So we have

$\small \small \small \small \small \sum_{n=0}^\infty \frac{1+(-1)^n}{n!}=\sum_{n=0}^\infty \frac{1}{n!}+\sum_{n=0}^\infty \frac{(-1)^n}{n!}=e^1+e^{-1}$

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Question

Find the value of the infinite series.

$\small \sum_{n=0}^{\infty} \frac{\pi^{4n+3}(-1)^n}{(2n+1)}$

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Answer

We can evaluate the series

$\small \sum_{n=0}^{\infty} \frac{\pi^{4n+3}(-1)^n}{(2n+1)}$

by recognizing it as a power series of a known function with a value plugged in for $\small x$ . In particular, it looks similar to $\small \sin x$ :

$\small \sin x =\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}$

After manipulating the series, we get

$\small \small \sum_{n=0}^{\infty} \frac{\pi^{4n+3}(-1)^n}{(2n+1)}=\pi\sum_{n=0}^{\infty} \frac{\pi^{4n+2}(-1)^n}{(2n+1)}=\pi \sum_{n=0}^{\infty} \frac{(\pi^2)^{2n+1}(-1)^n}{(2n+1)}$ .

Now it suffices to evalute $\small \sum_{n=0}^{\infty} \frac{(\pi^2)^{2n+1}(-1)^n}{(2n+1)}$ , which is $\small \sin \pi^2$ .

So the infinite series has value

$\small \small \small \sum_{n=0}^{\infty} \frac{\pi^{4n+3}(-1)^n}{(2n+1)}=\pi \sum_{n=0}^{\infty} \frac{(\pi^2)^{2n+1}(-1)^n}{(2n+1)}=\pi \sin \pi^2$ .

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Question

Find the value of the following infinite series:

$\small \sum_{n=0}^\infty \frac{\pi^{2n+1}(-1)^n}{6^{2n+1}(2n+1)!}$

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Answer

After doing the following manipulation:

$\small \small \sum_{n=0}^\infty \frac{\pi^{2n+1}(-1)^n}{6^{2n+1}(2n+1)!}=\sum_{n=0}^\infty \left(\frac{\pi}{6} \right )^{2n+1}\frac{(-1)^n}{(2n+1)!}$

We can see that this is the power series

$\small \sin x=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$ with $\small \small x=\frac{\pi}{6}$ plugged in.

So we have

$\small \small \small \sum_{n=0}^\infty \frac{\pi^{2n+1}(-1)^n}{6^{2n+1}(2n+1)!}=\sin \frac{\pi}{6}=\frac{1}{2}$

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Question

Find the value of the following series.

$\small \sum_{k=0}^\infty\frac{2^k+(-1)^k}{k!}$

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Answer

We can split up the sum to get

$\small \small \sum_{k=0}^\infty\frac{2^k+(-1)^k}{k!}=\sum_{k=0}^\infty\frac{2^k}{k!}+\sum_{k=0}^\infty\frac{(-1)^k}{k!}$ .

We know that the power series for $\small e^x$ is

$\small \sum_{n=0}^\infty \frac{x^n}{n!}$

and that each sum,

$\small \sum_{k=0}^\infty\frac{2^k}{k!}$

and

$\small \sum_{k=0}^\infty\frac{(-1)^k}{k!}$

are simply $\small e^{x}$ with $\small x=2,-1$ plugged in, respectively.

Thus,

$\small \small \small \sum_{k=0}^\infty\frac{2^k+(-1)^k}{k!}=\sum_{k=0}^\infty\frac{2^k}{k!}+\sum_{k=0}^\infty\frac{(-1)^k}{k!}=e^2+e^{-1}$ .

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Question

Find the value of the infinite series.

$\small \sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}$

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Answer

The series

$\small \sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}$ looks similar to the series for $\small e^x$ , which is $\small e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$

but the series we want to simplify starts at $\small n=1$ , so we can fix this by adding a $\small 1$ and subtracting a $\small 1$ , to leave the value unchanged, i.e.,

$\small \small \sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}=-1+1+\sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}=-1+\sum_{n=0}^\infty \frac{(\ln 5)^n}{n!}$ .

So now we have $\small e^x$ with $\small x=\ln 5$ , which gives us $\small e^{\ln 5}=5$ .

So then we have:

$\small \small \sum_{n=1}^\infty \frac{(\ln 5)^n}{n!}=-1+\sum_{n=0}^\infty \frac{(\ln 5)^n}{n!}=-1+5=4$

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Question

Write out the first two terms of the Maclaurin series of the following function:

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Answer

The Maclaurin series of a function is simply the Taylor series of a function, but about x=0 (so a=0 in the formula):

$\sum_{n=0}^{\infty }\frac{f^{n}(a)(x-a)^{n}}{n!}$

To write out the first two terms (n=0 and n=1), we must find the first derivative of the function (because the zeroth derivative is the function itself):

The derivative was found using the following rule:

$\frac{d}{dx}(x^n)=nx^{n-1}$

Next, use the general form, plugging in n=0 for the first term and n=1 for the second term:

$\frac{(0 + 0+1)(x)^0}{0!}+\frac{(0+0)(x^1)}{1!}=1+0$

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Question

Find the Maclaurin series for the function:

$\frac{1}{e^x}$

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Answer

Write Maclaurin series generated by a function f. The Maclaurin series is centered at for the Taylor series.

$\sum_{b=0}^{\infty}\frac{f^b(a=0)}{b!}x^b=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+...$

Evaluate the function and the derivatives of $f(x)=\frac{1}{e^x}$ at .

$f(x=0)=e^{-x}=e^0=1$

$f'(x=0)= -e^{-x}= -\frac{1}{e^0}= -1$

$f''(x=0)=e^{-x}=e^0=1$

Substitute the values into the power series. The series pattern can be seen as alternating and increasing order.

$1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}... =\sum_{n=0}^{\infty}(-1)^n\frac{x^n}{n!}$

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Question

Find the first three terms of the Maclaurin series for the following function:

$f(x)=\cos(x)+x^2$

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Answer

The Maclaurin series of a function is simply the Taylor series for the function about a=0:

$\sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}$

First, we can find the zeroth, first, and second derivatives of the function (n=0, 1, and 2 are the first three terms).

Plugging these values into the formula we get the following.

$\sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}$

$f^0(0)+f^1(0)\frac{(x-0)^1}{1!}+f^2(0)\frac{(x-0)^2}{2!}$

$1+0+\frac{x^2}{4}$

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Question

Find the Maclaurin Series of the function

up to the fifth degree.

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Answer

The formula for an i-th degree Maclaurin Polynomial is

$f(x)=\sum_{n=0}^{i}\frac{f^{(n)}(0)}{n!}x^n$

For the fifth degree polynomial, we must evaluate the function up to its fifth derivative.

$\begin{matrix} f(0)=sin(0)=0\ f'(0)=cos(0)=1\ f''(0)=-sin(0)=0\ f'''(0)=-cos(0)=-1\ f^{(4)}=sin(0)=0 \ \end{matrix}$

$f^{(5)}(0)=cos(0)=1$

The summation becomes

$\sum_{n=0}^{5}\frac{f^{(n)}(0)}{n!}x^n=\frac{f(0)}{0!}x^0+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\frac{f^{(4)}(0)}{4!}x^4+\frac{f^{(5)}(0)}{5!}x^5$

And substituting for the values of the function and the first five derivative values, we get the Maclaurin Polynomial

$f(x)=sin(x)\approx x-\frac{x^3}{3!}+\frac{x^5}{5!}$

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Question

Write out the first five terms of the Maclaurin series for the following function:

$f(x)=\cos(x)$

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Answer

The Maclaurin series for any function is simply the Taylor series for the function about a=0:

$\sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}$

First, we must find the zeroth through fourth derivative of the function. The zeroth derivative is simply the function itself.

$f'(x)=-\sin(x)$

$f''(x)=-\cos(x)$

$f'''(x)=\sin(x)$

$f^4(x)=\cos(x)$

The following rules were used for the derivatives:

$\frac{d}{dx}\cos(x)=-\sin(x)$ , $\frac{d}{dx}\sin(x)=-\cos(x)$

Next, we simply evaluate all of the derivatives at and then write out all of the terms:

$\frac{1(x)^0}{0!}-\frac{0(x)^1}{1!}-\frac{1(x)^2}{2!}+\frac{0(x)^3}{3!}+\frac{1(x)^4}{4!}$

which simplified is equal to

$1+0-\frac{x^2}{2}+0+\frac{x^4}{24}$

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Question

Write out the first three terms of the Maclaurin series of the following function:

$f(x)=\sin(x)$

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Answer

The Maclaurin series of a function is simply the Taylor series of a function about a=0:

$\sum_{n=0}^{\infty }f^{(n)}(a)\frac{(x-a)^n}{n!}$

Because we were asked to find the first three terms (n=0 to n=2), we must find the zeroth, first, and second derivatives of the function. The zeroth derivative is just the function itself.

$f'(x)=\cos(x)$

$f''(x)=-\sin(x)$

Now plug in into the formula and write out the first three terms (n=0, 1, 2):

$\frac{0(x)^0}{0!}+\frac{1(x)^1}{1!}+\frac{0(x)^2}{2!}=0+x+0$

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