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

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

Substitute \(\displaystyle -x \ln 6\) for \(\displaystyle x\):

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

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

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

This circle will have equation

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

Rewrite this as follows:

\(\displaystyle x ^{2} - 10x + 25 + y ^{2} + 8y + 16 = 256\)

\(\displaystyle x ^{2} - 10x + 25 + y ^{2} + 8y + 16 - 41 = 256- 41\)

\(\displaystyle x ^{2}+ y ^{2} - 10x + 8y = 215\)

\(\displaystyle r ^{2} - 10 r \cos \theta + 8 r \sin \theta = 215\)

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

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

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

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

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

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

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

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

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

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

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, 

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

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:

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

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:

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