Evaluating non-elementary integrals

 Objective:

Use Taylor series to evaluate non-elementary integrals

Example

Solution

Practice

Solving differential equations using power series

 Objective: Solve differential equations using power series.

Let's consider the differential equation y'(x) = y. Recall that this is a first order separable equation and its solution is y = Ce^x. For most differential equations there are not any analytical tools to solve them. Power series are an extremely useful tool to solve them. The technique we use is to look for a solution of the form 

Then we determine the coefficients in order to solve the differential equation.

Example

Solution

Using the uniqueness of power series representations, we know that these series can only be equal if their coefficients are equal. Therefore:

Practice

 

  

   

                                                                                                                                                        

Differentiating a power series to find a new power series

 Objective: Differentiate a power series to find a new power series

We showed in an earlier post that it's  possible to differentiate a power series term by term to create a new power series. In the following example we are going to differentiate the binomial series term by term to find a new binomial  series.

Example

Solution

There are 2 ways to solve this problem. First, we can differentiate term by term the first expression to find the binomial series of the second expression. Another easier way is to notice that the second expression is almost the derivative of the first one.

Differentiating the first expression, we have:

We deduct that:

Let's calculate d/dx⎷1+x:

The binomial series for ⎷1+x: is given by:

Practice

Common functions expressed as Taylor series

 Objective: Recognize the Taylor series expansions of common functions

Earlier we showed how to combine some power series to create new power series.

In the following table we summarize the results for the Mclaurin series of the exponential, logarithmic and trigonometric functions.

Example

Deriving Mclaurin series from known series

Find the Mclaurin series of each of the following functions by using one of the series in the table above

Solution

a. The Mclaurin series for cosx is given by:

Let's substitute x by ⎷x:

b

Binomial series

 Objective: write the terms of the binomial series

In previous post I showed how to find the Taylor series for some common function by calculating the coefficients of the Taylor polynomials. In posts relating to binomial series I will show how to use the Taylor series to derive the Taylor series of other functions. Later I will show how to use the Taylor series to solve integrals and differential equations.

 

Example

Solution

The function and the Mclaurin polynomial are graphed below:

The third Mclaurin polynomial provides a good approximation for f(x) = ⎷1 + x for x near 0.

Practice

Convergence of a Taylor series

 If a Taylor series for a function f converges on some interval, how can we determine if that series converges to the function f?Let's remind that a series converges to a particular value if the sequence of its partial sums converges to that value. In order words, the limit of the general term of the partial sum is equal to that value. For a Taylor series of a function f at a, the nth partial sum is given by pn. To determine if the Taylor series converges to f, we need to demonstrate whether

Since the remainder Rn(x) = f(x)-pn(x), the Taylor series converges to f if and only if

Convergence of a Taylor series

To prove that a Taylor series for a function f at a converges to f, we need to prove Rn(x)→0. To do this we use the bound:

Example

Solution

a. The Mclaurin series for e^x is given by 

Let's use the ratio test to determine its convergence:

Let's take the limit:

Therefore the series converges absolutely for all x, and thus, the interval of convergence is (-∞, +∞). 

Let's use the formula for the remainder of a Taylor series:

Since a =0 (Mclaurin series), we have:

❘Rₙ❘≤(M/(n + 1)!❘)x❘ⁿ⁺¹

The exponential function is an increasing function. For any I = [-d.d], the maximum value for M is e^d or M = e^d. 

Substituting M, we have:

Rₙ❘≤ e^d/(n + 1)!❘)x❘ⁿ⁺¹

Taking the limit of both sides we have:

b. Let's prove that the Mclaurin series for sinx converges to sinx. 

The Mclaurin series for sinx is given by:

Let's prove that this series is convergent and determine its convergence.

Let's take the limit of both sides, we have:

 

Let's use the formula for the  Taylor series with remainder:

Since a = 0, we have:  ❘Rₙ❘≤(M/(n + 1)!❘)x❘ⁿ⁺¹. Let's find the maximum value for M. It represents the maximum value of the absolute value des n+1 derivatives of sinx. We have:

For any x or in any interval I = [-d, d], the maximum value for M is 1. Therefore we have:

Practice      

                                                                                                                                            

 

Representing functions with Taylor and Mclaurin series

 In this post, I show how to find a Taylor series for a function and how to find its convergence.

Example

Solution

Instead of applying the general formula of Taylor series, let's say that the Taylor series is of the form 

We already know a =1. All we have to do is to determine fⁿ(a). In order to do that we determine first f'(1), f"(1). f'''(1) then f⁴(n):

The order number derivative is equal to the factorial of the number with the derivative changing sign alternatively. We can generalize by saying fⁿ(1) = (-1)ⁿ n!.

The Taylor series of f at x = 1 then is by substituting fⁿ(1):

 with n as exponent of x-1

To determine the interval convergence of the series, let's determine its convergence. Let's use the ratio test by starting to determine the ratio:

The limit of this ratio when n approaches infinity is equal to the limit of ❘x-1❘. This limit is equal to ❘x-1❘

The series is convergent if ❘x-1❘< 0 i.e to 0 <x<2.  Let's check the convergence of the series at the     endpoints i.e at x = 0 and x = 2.                                                                                                                    

 At x= 0 we have                                                                                                                                                                                        

                                                                                                                             

 Let's use the divergence test by calculating the limit of 1 when n approaches infinity. This limit is equal to 1. The series diverges since the limit is different of 0.                                                                          

At x = 2, we have:                                                                                                                              

The limit of (-1)ⁿ is probably unknown, therefore different of 0.                                         

The series         diverges when x = 2.  The interval of convergence is (0,2).

               Practice