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

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

We first must find the zeroth, first, and second derivative of the function (for n=0, 1, and 2). The zeroth derivative is the function itself:

$\displaystyle f'(x)=-10\sin(x)$

$\displaystyle f''(x)=-10\cos(x)$

The derivatives were found using the following rules:

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

Now, we just use the formula, with $\displaystyle x=a=0$, to write out the first three terms of the series (n=0, 1, and 2):

$\displaystyle \frac{10(1)(x)^0}{0!}-\frac{10(0)(x)^1}{1!}-\frac{10(1)(x)^2}{2!}=10+0+5x^2$

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

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

We must find the first two terms of the series, corresponding to n=0 and n=1. We need the zeroth and first derivative of the function, the zeroth derivative being the the function itself:

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

The derivative was found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

Now, use the formula above to write out the first two terms:

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

The Maclaurin series for any 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 must find the zeroth through fourth derivative of the function. The zeroth derivative is simply the function itself.

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

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

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

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

The following rules were used for the derivatives:

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

Next, we simply evaluate all of the derivatives at $\displaystyle x=a=0$ and then write out all of the terms:

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

which simplified is equal to

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

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

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

We need to find the zeroth, first, second, and third derivative of the function (n=0, 1, 2, and 3). The zeroth derivative is simply the function itself.

$\displaystyle f'(x)=4x+1$

$\displaystyle f''(x)=4$

$\displaystyle f'''(x)=0$

The derivatives were found using the following rule:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, use the above formula, with $\displaystyle x=a=0$ to write out the first four terms:

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

which simplified becomes

$\displaystyle 1+x+2x^2+0$

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

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

For the first two terms (n=0, 1) we must find the zeroth and first derivative of the function. The zeroth derivative is just the function itself.

$\displaystyle f'(x)=4x+\sin(x)$

Now, using the above formula, write out the first two terms:

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

The formula for an i-th degree Maclaurin Polynomial is

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

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

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

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

The summation becomes

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

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

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

The Maclaurin series for $\displaystyle e^x$ is given by 

$\displaystyle e^x= \sum_{n=0}^{\infty }\frac{x^n}{n!}= 1+x+\frac{x^2}{2}+\frac{x^3}{6}+...$

The Maclaurin Series for $\displaystyle sin(x)$ is given by

$\displaystyle sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}(x)^{2n+1}=x-\frac{x^3}{6}+\frac{x^5}{120}+...$

Since the question asked

$\displaystyle e^x+sin(x)$

To write this sum, we can take out $\displaystyle x^n$ as a common factor for both series:

$\displaystyle \sum_{n=0}^{\infty}x^n\left(\frac{1}{n!}+\frac{(-1)^n*x^{1+n}}{(1+2n)!}\right)$

You might already recognize this as the power series representation for $\displaystyle e^x$. Since $\displaystyle e^x$ is well defined for all $\displaystyle x \in \mathbb{R}$, the radius of convergence is $\displaystyle \infty$.

If we want to find the radius of covergence using convergence tests, we can use the Ratio Test here. We have-

$\displaystyle \lim_{k\to\infty} \left |\frac{a_{k+1}}{a_k} \right | = \lim_{k\to\infty} \left | \frac{x^{k+1}}{(k+1)!} \frac{k!}{x^k} \right | =\lim_{k\to\infty} \left | \frac{x}{k+1}\right | =$

$\displaystyle \lim_{k\to\infty} \frac{\left | x \right |}{k+1} = \left | x \right | \lim_{k\to\infty} \frac{1}{k+1} = \left | x \right | \cdot 0 = 0$

Since this limit equals $\displaystyle 0$ regardless of the value of $\displaystyle x$, and the Ratio Test indicates absolute convergence of a series when the above limit is less than $\displaystyle 1$, $\displaystyle \sum_{k=0}^{\infty} \frac{x^k}{k!}$ converges for all $\displaystyle x \in \mathbb{R}$. Hence the radius of convergence is $\displaystyle \infty$.

The question does not ask which function is the largest at any given point, it asks which grows fastest. For this question we need to look at all of the terms and determine which function has the dominant term. In this case $\displaystyle x^x$ is the dominate term therefore, it will grow the fastest.

Write the correct definition of cosine as a power series.

$\displaystyle cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$

Replace $\displaystyle x$ with the term $\displaystyle x^2$.

$\displaystyle cos(x^2)=1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}+...$

The correct answer is:

$\displaystyle 1-\frac{x^4}{2!}+\frac{x^8}{4!}-\frac{x^{12}}{6!}+...$