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W11 - Examples

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Power series as functions

Geometric series: algebra meets calculus

Consider the geometric series as a power series functions:

Take the derivative of both sides of the function:

This means satisfies the identity:

Now compute the derivative of the series:

On the other hand, compute the square of the series:

So we find that the same relationship holds, namely , for the closed formula and the series formula for this function.

Manipulating geometric series: algebra

Find power series that represent the following functions:

(a) (b) (c) (d)

Solution

(a)

Rewrite in format .

Introduce double negative:

Choose .

Plug into geometric series.

Geometric series in :

Plug in :

Simplify:

Final answer:

(b)

Rewrite in format .

Rewrite:

Choose .

Plug into geometric series.

Geometric series in :

Plug in :

Final answer:

(c)

Rewrite in format .

Rewrite:

Choose . Here .

Plug into geometric series.

Geometric series in :

Plug in :

Obtain:

Multiply by .

Distribute:

Final answer:

(d)

Rewrite in format .

Rewrite:

Choose . Here .

Plug into geometric series.

Geometric series in :

Plug in :

Obtain:

Multiply by .

Distribute:

Final answer:

Manipulating geometric series: calculus

Find a power series that represents .

Solution

Differentiate to obtain similarity to geometric sum formula.

Differentiate :

Find power series of differentiated function.

Power series by modifying with :

Integrate series to find original function.

Integrate both sides:

Use known point to solve for :

Final answer:

Recognizing and manipulating geometric series: Part I

(a) Evaluate . (Hint: consider the series of .)

(b) Find a series approximation for .

Solution

(a) Evaluate . (Hint: consider the series of .)

Find the series representation of following the hint.

Notice that .

We know the series of :

Notice that ; this is the desired function when .

Integrate the series term-by-term:

Solve for using , so and thus . So:

Notice the similar formula.

The series formula looks similar to the formula .

Choose to recreate the desired series.

We obtain equality by setting because .

Final answer is .

(b) Find a series approximation for .

Observe that . Therefore we can use the series

Plug into the series for . Plug in and simplify:

Recognizing and manipulating geometric series: Part II

(a) Find a series representing using differentiation.

(b) Find a series representing .

Solution

(a) Find a series representing .

Notice that .

Obtain the series for .

Let :

Integrate the series for by terms.

Set up the strategy. We know:

and:

Integrate term-by-term:

Conclude that:

Solve for by testing at .

Plugging in, obtain:

so .

Final answer is .

(b) Find a series representing .

Find a series representing the integrand.

Integrand is .

Rewrite integrand in format of geometric series sum:

Write the series:

Integrate the integrand series by terms.

Integrate term-by-term:

This is our final answer.

Taylor and Maclaurin series

Maclaurin series of e to the x

What is the Maclaurin series of ?

Solution Because , we find that for all .

So for all . Therefore for all by the Derivative-Coefficient identity.

Thus:

Maclaurin series of cos x

Find the Maclaurin series representation of .

Solution Use the Derivative-Coefficient Identity to solve for the coefficients:

0
1
2
3
4
5

By studying the generating pattern of the coefficients, we find for the series:

Maclaurin series from other Maclaurin series

(a) Find the Maclaurin series of using the Maclaurin series of .

(b) Find the Maclaurin series of using the Maclaurin series of .

(c) Using (b), find the value of .

Solution

(a)

Remember that

Differentiate

Differentiate term-by-term:

Take negative because :

Final answer is

(b)

Recall the series

Compute the series for .

Set :

Compute the product.

Product of series:

(c)

Derivatives at are calculable from series coefficients.

Suppose we know the series

Then .

It may be easier to compute for a given than to compute the derivative functions and then evaluate them.

Compute .

Write the series such that it reveals the coefficients:

Coefficient with corresponds to the term with , not necessarily the term (e.g. if the first term is as here).

Compute :

Compute .

Use Derivative-Coefficient Identity:

Computing a Taylor series

Find the first five terms of the Taylor series of centered at .

Solution A Taylor series is just a Maclaurin series that isn’t centered at .

The general format looks like this:

The coefficients satisfy . (Notice the .)

We find the coefficients by computing the derivatives and evaluating at :

By dividing by we can write out the first terms of the series:

Applications of Taylor series

Taylor polynomial approximations

Let and let be the Taylor polynomials expanded around .

By considering the alternating series error bound, find the first for which must have error less than .

Solution

Write the Maclaurin series of because we are expanding around .

Alternating sign, odd function:

Notice this series is alternating, so AST error bound formula applies.

AST error bound formula is:

Here the series is and is the error.

Notice that is part of the terms in this formula.

Implement error bound to set up equation for .

Find such that , and therefore by the AST error bound formula:

Plug in .

From the series of we obtain for :

We seek the first time it happens that .

Solve for the first time .

Equations to solve:

Method: list the values:

The first time is below happens when .

Interpret result and state the answer.

When , the term at is less than .

Therefore the sum of prior terms is accurate to an error of less than .

The sum of prior terms equals .

Since because there is no term, the same sum is .

The final answer is .

It would be wrong to infer at the beginning that the answer is , or to solve to get .

Taylor polynomials to approximate a definite integral

Approximate using a Taylor polynomial with an error no greater than .

Solution

Write the series of the integrand.

Plug into the series of :

Compute definite integral by terms.

Antiderivative by terms:

Plug in bounds for definite integral:

Notice AST, apply error formula.

Compute some terms:

So we can guarantee an error less than by summing the first terms through .

Final answer is .