Calculus 2 : Maclaurin Series

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #1 : Maclaurin Series

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

Possible Answers:

\displaystyle \small \frac{48!}{12!}

\displaystyle \small 1

\displaystyle \small \frac{48!}{16!}

\displaystyle 0

Correct answer:

\displaystyle \small \frac{48!}{16!}

Explanation:

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

\displaystyle \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 \displaystyle \small \cos x^3. We must look at term \displaystyle \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 \displaystyle \small x=0, all of the non-constant terms are zero. So we must have

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

Example Question #1 : Maclaurin Series

What is the value of the following infinite series?

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

Possible Answers:

\displaystyle \small \frac{\sqrt{3}}{2}

\displaystyle \small \arctan{\pi/3}

\displaystyle \small \frac{1}{2}

\displaystyle \small \pi/4

Correct answer:

\displaystyle \small \frac{\sqrt{3}}{2}

Explanation:

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

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

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

\displaystyle \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

\displaystyle \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}.

Example Question #1 : Maclaurin Series

What is the value of the following infinite series?

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

Possible Answers:

\displaystyle \small e+e^{1/2}

The infinite series diverges.

\displaystyle \small \frac{1}{2}(e+e^{-1})

\displaystyle \small e^2+e^{-2}

\displaystyle \small e+e^{-1}

Correct answer:

\displaystyle \small e+e^{-1}

Explanation:

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

\displaystyle \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 \displaystyle \small e^x, which is 

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

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

\displaystyle \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}

Example Question #4 : Maclaurin Series

Find the value of the infinite series.

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

Possible Answers:

\displaystyle \small \pi \sin \pi^2

\displaystyle \small \pi \cos \pi^4

The series does not converge.

\displaystyle \small \cos \pi^3

\displaystyle \small -\pi\cos \pi^2

Correct answer:

\displaystyle \small \pi \sin \pi^2

Explanation:

We can evaluate the series

\displaystyle \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 \displaystyle \small x. In particular, it looks similar to \displaystyle \small \sin x:

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

After manipulating the series, we get

\displaystyle \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 \displaystyle \small \sum_{n=0}^{\infty} \frac{(\pi^2)^{2n+1}(-1)^n}{(2n+1)}, which is \displaystyle \small \sin \pi^2.

So the infinite series has value

\displaystyle \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.

Example Question #5 : Maclaurin Series

Find the value of the following infinite series:

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

Possible Answers:

\displaystyle \small -\frac{1}{2}

\displaystyle \small \frac{\sqrt{3}}{2}

\displaystyle \small -\frac{\sqrt{3}}{2}

\displaystyle \small \frac{1}{2}

\displaystyle \small \frac{\sqrt{2}}{2}

Correct answer:

\displaystyle \small \frac{1}{2}

Explanation:

After doing the following manipulation:

\displaystyle \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 

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

So we have

\displaystyle \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}

Example Question #6 : Maclaurin Series

Find the value of the following series.

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

Possible Answers:

\displaystyle \small e^{-2}+e^{-1}

Divergent.

\displaystyle \small e^2+e^{-1}

\displaystyle \small e^{1/2}-e^{-1}

\displaystyle \small e^2-e^{-1}

Correct answer:

\displaystyle \small e^2+e^{-1}

Explanation:

We can split up the sum to get 

\displaystyle \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 \displaystyle \small e^x is 

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

and that each sum, 

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

and

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

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

Thus, 

\displaystyle \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}.

Example Question #7 : Maclaurin Series

Find the value of the infinite series.

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

Possible Answers:

Infinite series does not converge.

\displaystyle \small \sqrt{5}

\displaystyle \small 4

\displaystyle \small 5

\displaystyle \small \ln 5

Correct answer:

\displaystyle \small 4

Explanation:

The series 

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

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

\displaystyle \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 \displaystyle \small e^x with \displaystyle \small x=\ln 5, which gives us \displaystyle \small e^{\ln 5}=5.

So then we have:

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

Example Question #1 : Maclaurin Series

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

\displaystyle f(x)=x^3+2x^2+1

Possible Answers:

\displaystyle 0-1

\displaystyle 1+0

\displaystyle 0+0

\displaystyle 0+1

Correct answer:

\displaystyle 1+0

Explanation:

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

\displaystyle \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):

\displaystyle f'(x)=3x^2+4x

The derivative was found using the following rule:

\displaystyle \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:

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

 

Example Question #3091 : Calculus Ii

Find the Maclaurin series for the function:  

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

Possible Answers:

\displaystyle \sum_{n=0}^{\infty}(-1)^n\frac{x^n}{(2n)!}

\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!}\:ln(n)

\displaystyle \sum_{n=0}^{\infty}(-1)^n\frac{x^n}{n!\:ln(n)}

\displaystyle \sum_{n=0}^{\infty}(-1)^n\frac{x^n}{n!}

\displaystyle \sum_{n=0}^{\infty}(-1)^nx^n

Correct answer:

\displaystyle \sum_{n=0}^{\infty}(-1)^n\frac{x^n}{n!}

Explanation:

Write Maclaurin series generated by a function f.  The Maclaurin series is centered at \displaystyle a=0 for the Taylor series.

\displaystyle \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 \displaystyle f(x)=\frac{1}{e^x} at \displaystyle x=0.

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

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

\displaystyle 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.

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

Example Question #10 : Maclaurin Series

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

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

Possible Answers:

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

\displaystyle 0+0+\frac{1}{2}

\displaystyle 0+0+0

\displaystyle 1+0

Correct answer:

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

Explanation:

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

\displaystyle \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). 

\displaystyle f^0=cos(0)+0^2=1

\displaystyle f^1=-sin(0)+2(0)=0

\displaystyle f^2=-cos(0)+2=1

Plugging these values into the formula we get the following.

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

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

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

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