# Calculus 2 : Taylor Series

## Example Questions

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### Example Question #1 : Maclaurin Series

For which of the following functions can the Maclaurin series representation be expressed in four or fewer non-zero terms?

Explanation:

Recall the Maclaurin series formula:

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  as its Maclaurin series is.

### Example Question #2 : Maclaurin Series

Let

Find the the first three terms of the Taylor Series for  centered at .

Explanation:

Using the formula of a binomial series centered at 0:

,

where we replace  with  and , we get:

for the first 3 terms.

Then, we find the terms where,

### Example Question #1 : Taylor's Theorem

Determine the convergence of the Taylor Series for  at  where .

Divergent.

Does not exist.

Inconclusive.

Absolutely Convergent.

Conditionally Convergent.

Absolutely Convergent.

Explanation:

By the ratio test, the series  converges absolutely:

### Example Question #2 : Taylor's Theorem

Find the interval of convergence for  of the Taylor Series .

Explanation:

Using the root test

and

. T

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

This occurs when x=5.

### Example Question #1 : Taylor Series

Suppose that the derivative of a function, denoted , can be approximated by the third degree Taylor polynomial, centered at :

If , find the third degree Taylor polynomial for  centered at .

Explanation:

To get , we need to find the antiderivative of  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:

Since , substitute  for the final .

### Example Question #1 : Taylor And Maclaurin Series

Consider the following series:

What is the evaluation of this series at ?

Explanation:

Recall the important McLaurin series for :

The series in this question is the same, only replacing x with  .

So if we have , we need only evaluate at , and thus we obtain .

### Example Question #3 : Maclaurin Series

Write the first three terms of the Taylor series for the following function about :

Explanation:

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

.

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:

The derivative was found using the following rules:

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

### Example Question #4 : Maclaurin Series

Write out the first four terms of the Taylor series about  for the following function:

Explanation:

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

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):

The derivatives were found using the following rule:

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

which when simplified is equal to

.

### Example Question #1 : Taylor's Theorem

Find the first two terms of the Taylor series about  for the following function:

Explanation:

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

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:

The derivative was found using the following rule:

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

### Example Question #5 : Taylor's Theorem

Write out the first three terms of the Taylor series for the following function about :

Explanation:

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

.

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.

The derivatives were found using the following rules:

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

Simplified, this becomes

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