Recall the Maclaurin series formula:

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

$\displaystyle =f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...$

Despite being a 5th degree polynomial recall that the Maclaurin series for any polynomial is just the polynomial itself, so this function's Taylor series is identical to itself with two non-zero terms.

The only function that has four or fewer terms is $\displaystyle f(x)= \frac{x^{5}}{3}+2$ as its Maclaurin series is$\displaystyle 2+\frac{1}{3}x^5$.

Using the formula of a binomial series centered at 0:

 $\displaystyle (1+x)^k =\sum_{n=0}^{\infty } \binom{k}{n}x^n$,

where we replace $\displaystyle x$ with $\displaystyle \frac{x}{2}$ and $\displaystyle k=\frac{1}{3}$, we get:

 $\displaystyle \sum_{n=0}^{2}\binom{\frac{1}{3}}{n}\left(\frac{x}{2}\right)^n$ for the first 3 terms.

Then, we find the terms where,

 $\displaystyle T_2(x)=\binom{\frac{1}{3}}{0}(\frac{x}{2})^0 + \binom{\frac{1}{3}}{1}(\frac{x}{2})^1+\binom{\frac{1}{3}}{2}(\frac{x}{2})^2=1+\frac{x}{6}- \frac{x^2}{36}$

By the ratio test, the series $\displaystyle \sum_{n=0}^{\infty }\frac{5^n}{n!}$ converges absolutely: 

$\displaystyle \lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{5^{n+1}}{(n+1)!} \cdot \frac{n!}{5^n} \right |=\lim_{n\rightarrow \infty }\left | \frac{5}{n+1} \right |=0< 1$

Using the root test

$\displaystyle \lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{(n+1)!(x-5)^{n+1}}{n!(x-5)^n} \right |=\left |x-5 \right |\lim_{n\rightarrow \infty }\left | n+1 \right |$ 

and

$\displaystyle \lim_{n\rightarrow \infty }\left | n+1 \right |=\infty$. T

herefore, the series only converges when it is equal to zero.

This occurs when x=5.

To get $\displaystyle f(x)$, we need to find the antiderivative of $\displaystyle f'(x)$ by integrating the third degree polynomial term by term.

We only want up to a third degree polynomial, so we can disregard the fourth order term:

$\displaystyle f(x) = c + (x-2) + (x-2)^2 + \frac{4}{9}(x-2)^3$

Since $\displaystyle f(2) = 7$, substitute $\displaystyle c=7$ for the final $\displaystyle f(x)$.

$\displaystyle f(x)=7 + (x-2) + (x-2)^2 + \frac{4}{9}(x-2)^3$

Recall the important McLaurin series for $\displaystyle e^x$:

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

The series in this question is the same, only replacing x with $\displaystyle \frac{1+t}{2}$ .

So if we have $\displaystyle f(t) = e^{\frac{1+t}{2}}$, we need only evaluate at $\displaystyle t=1$, and thus we obtain $\displaystyle e$.

The general form for the Taylor series (of a function f(x)) about x=a is the following:

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

Because we only want the first three terms, we simply plug in a=1, and then n=0, 1, and 2 for the first three terms (starting at n=0).

The hardest part, then, is finding the zeroth, first, and second derivative of the function given:

$\displaystyle f^{0}(1)=f(1)=6$

$\displaystyle f'(1)=2(1)+3=5$

$\displaystyle f''(1)=2$

The derivative was found using the following rules:

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

Then, simply plug in the remaining information and write out the terms:

$\displaystyle \frac{6(x-1)^0}{0!}+\frac{5(x-1)^1}{1!}+\frac{2(x-1)^2}{2!}=6+5(x-1)+(x-1)^2$

The Taylor series about x=a of any function is given by the following:

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

So, we must find the zeroth, first, second, and third derivatives of the function (for n=0, 1, 2, and 3 which makes the first four terms):

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

$\displaystyle f'(x)=2x+2$

$\displaystyle f''(x)=2$

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

The derivatives were found using the following rule:

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

Now, evaluated at x=a=1, and plugging in the correct n where appropriate, we get the following:

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

which when simplified is equal to

$\displaystyle 4+4(x-1)+(x-1)^2+0$.

The general formula for the Taylor series about x=a for a function is

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

First, we must find the zeroth and first derivative of the function.

The zeroth derivative of a function is just the function itself, so we only have to find the first derivative:

$\displaystyle f'(x)=2$

The derivative was found using the following rule:

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

Now, write the first two terms of the sequence (n=0 and n=1):

$\displaystyle \frac{[2(3)+5](x-3)^0}{0!}+\frac{2(x-3)^1}{1!}=11+2(x-3)$

The general formula for the Taylor series of a given function about x=a is

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

We were asked to find the first three terms, which correspond to n=0, 1, and 2. So first, we need to find the zeroth, first, and second derivative of the given function. The zeroth derivative is just the function itself.

$\displaystyle f'(x)=2e^x$

$\displaystyle f''(x)=2e^x$

The derivatives were found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Now use the above formula to write out the first three terms:

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

Simplified, this becomes

$\displaystyle (2e^1+2)+2e^1(x-1)+e^1(x-1)^2$