The Taylor series is defined as

 where the expression within the summation is the nth term of the Taylor polynomial.

To find the Taylor series expansion of  at , 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

In this problem,  and .

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

Substituting these values into the taylor polynomial we get

 All the even derivatives disappear and all the odd derivatives oscillate in sign, starting with a positive term.  Mathematically, we represent odd numbers as  or .  We represent oscillating signs as 

The Taylor polynomial simplifies to 

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

We found the taylor series of  at  is

For  at  

To find the Taylor series expansion of  at , 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

In this problem,  and .

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

Substituting these values into the taylor polynomial we get

The Taylor polynomial simplifies to 

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

The Taylor series is defined as

 where the expression within the summation is the nth term of the Taylor polynomial.

The taylor series is

 The taylor series expansion of  at  is then,

The Taylor series about x=a for a function is given by

So, for the first three terms (n=0, 1, 2), we must find the zeroth, first, and second derivative of the function, where the zeroth derivative is just the function itself:

The derivatives were found using the following rules:

, , 

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

which simplified becomes

The Taylor series about  for a function is given by

For the first three terms  we must find the zeroth, first, and second derivative of the function, where the zeroth derivative is just the function itself:

The derivatives were found using the following rules:

, , 

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

which simplified becomes

The Taylor series about  for a function is given by

For the first two terms  we must find the zeroth and first derivative, where the zeroth derivative is just the function itself:

The derivative was found using the following rules:

, , 

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

which simplified becomes

To answer this question we just have to keep taking derivatives and wait to see when we have three values that aren't 0.

first, we see that . that's no good so we take the derivative.

, so our first non-zero term is .

Taking the derivative again we get 

then, 

 so our Maclaurin series looks like this: 

We take the derivative again and get back around:

 so our next term is . And we can see that this is our third non-zero term.

The Taylor series about x=a for a function is given by

So, for the first three terms (n=0, 1, 2), we must find the zeroth, first, and second derivatives. The zeroth derivative is just the function itself.

The derivatives were found using the following rules:

, 

Now, using the above formula, we can write out the first three terms:

which simplifies to

The formula for the Maclaurin Series of a function is defined as follows

 where  is the n-th derivative when 

For the third degree polynomial we solve find the sum with the upper bound 

First we evaluate 

We must also solve for the first, second, and third derivative of 

When  we find the derivative values of

Using these values we find the third degree Maclaurin Polynomial to be

The Taylor series about x=a for any function is given by

We must take the nth derivative of the given function and determine the trend as n goes to infinity. To determine this, we start at n=0 (the zeroth derivative, or the function itself), and go further:

The derivatives were found using the following rules:

, , 

We must now find a pattern for the derivatives we took. The zeroth and first derivatives do not follow a pattern, but the derivatives after do follow a pattern: the sign alternates, the power of x decreases by 1, and the coefficient is the factorial of  starting at the n=2 derivative. 

The derivative can then be expressed as 

for n greater than or equal to 2.

For the Taylor series itself, we must evaluate the derivative at x=a=7. When we do this, and write the pattern elements for the derivatives past n=1, we get

To obtain the Taylor Series for , we start with the Taylor Series for , and substitute  on both sides with , and simplify.

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