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,

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

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 given function is

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

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

$\displaystyle f'(x)=20x+1$

$\displaystyle f''(x)=20$

The derivatives were found using the following rule:
$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Now, follow the above formula to write out the first three terms:

$\displaystyle \frac{11(x-1)^0}{0!}+\frac{21(x-1)^1}{1!}+\frac{20(x-1)^2}{2!}$

which simplified becomes

$\displaystyle 11+21(x-1)+10(x-1)^2$

The Taylor series is defined as

$\displaystyle f(x)=\sum_{k=0}^{\infty }\frac{f^{k}(x_{0})}{k!}(x-x_{0})^{k}$ where the expression within the summation is the nth term of the Taylor polynomial.

To find the Taylor series expansion of $\displaystyle \frac{1}{1-x}$ at $\displaystyle x=0$, we first need to find an expression for the nth term of the Taylor polynomial.

The nth term of the Taylor polynomial is defined as

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

For, $\displaystyle f(x)=\frac{1}{1-x}$ and $\displaystyle x_{0}=0$.

We need to find terms in the taylor polynomial until we can determine the pattern for the nth term.

$\displaystyle f(0)=\frac{1}{1-0}=1$

$\displaystyle f'(x)=\frac{d}{dx}\left [ \frac{1}{1-x} \right ]=\frac{d}{dx}\left [ \left ( 1-x\right )^{-1} ]=-1( 1-x\right )^{-2}(-1)=( 1-x)^{-2}$

$\displaystyle f'(0)=-1( 1-(0) )^{-2}=1=1!$

$\displaystyle f''(x)=\frac{d}{dx}[( 1-x)^{-2}]=-2( 1-x)^{-3}(-1)=2( 1-x)^{-3}$

$\displaystyle f''(0)=2( 1-0)^{-3}=2=2!$

$\displaystyle f'''(x)=\frac{d}{dx}[2( 1-x)^{-3}]=-6( 1-x)^{-4}(-1)=6( 1-x)^{-4}$

$\displaystyle f'''(0)=6( 1-0)^{-4}=6=3!$

Substituting these values into the taylor polynomial we get

$\displaystyle p_{n}x=1+1(x)+\frac{2!}{2!}(x)^{2}+\frac{3!}{3!}(x)^{3}+....+\frac{f^{n}(0)}{n!}(x)^{n}$

$\displaystyle p_{n}x=1+(x)+(x)^{2}+(x)^{3}+....+\frac{f^{n}(0)}{n!}(x)^{n}$

The Taylor polynomial simplifies to 

$\displaystyle p_{n}x=1+(x)+(x)^{2}+(x)^{3}+....+(x)^{n}$

Now that we know the nth term of the Taylor polynomial, we can find the Taylor series.

The Taylor series is defined as

$\displaystyle f(x)=\sum_{k=0}^{\infty }\frac{f^{k}(x_{0})}{k!}(x-x_{0})^{k}$ where the expression within the summation is the nth term of the Taylor polynomial.

For this problem, the taylor series is

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

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

 Use the above to calculate the taylor series expansion of $\displaystyle \frac{1}{1-x^{2}}$ at $\displaystyle x=0$,

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

The Maclaurin series for sine is $\displaystyle \sum_{n=0}^{\infty } \frac{(-1)^n}{(2n+1)!}x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5! } - ...$

Plugging in $\displaystyle \frac{\pi }{ 8 }$ for x gives: 

$\displaystyle .3927 - .0101 + .0001 = .3827$

The Maclaurin series for $\displaystyle e^x$ is $\displaystyle \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2 }{2! } + \frac{x^3}{3!} + ...$

Plugging in 0.78 for x gives: 

$\displaystyle 1 + .78 + .3042 + .079092 = 2.163292$

The Maclaurin series for $\displaystyle e^x$ is $\displaystyle \sum_{n=0}^{\infty } \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

Plugging in x = 0.38 gives: 

$\displaystyle 1 + 0.38 + .0722 + .09145 \overline{3} = 1.461345$

$\displaystyle \begin{align*}&\text{The Maclaurin series is a special case of the Taylor series.}\\&\text{Like the Taylor series, it is used to approximate an unknown}\\&\text{value of a function at a given point, if values of the function}\\&\text{and its derivatives are known at another point. In the case}\\&\text{of the Maclaurin series, this latter point has the value of}\\&\text{zero (note that for many functions, such as the trigonometric}\\&\text{functions, f(0) is readily known):}\\&f(x)\approx\sum_{i=0}^{n}\frac{f^{(i)}(0)}{i!}x^i\approx f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+...+\frac{f^{(n)}(0)}{n!}x^n\\&\text{For the function }-cos(x)\text{ at }x=1.2\\&\text{This expansion becomes:}\\&f(1.2)\approx \frac{-cos((0))}{1}(1.2)^{0}+\frac{sin((0))}{1}(1.2)^{1}+\frac{cos((0))}{2}(1.2)^{2}\\&f(1.2)\approx-0.28\end{align*}$