Convergence of Maclaurin Series and Radius of Convergence

[asi123]

Homework Statement

Hey.
I need to develop this function into Maclaurin series and to find the radius of converges, did I do it right?
And another thing, does the series (-1)^n from 1 to infinity (I know, my latex rocks) converges ? I'm pretty sure it's not.

Homework Equations

The Attempt at a Solution

 

[badphysicist]

It looks for most part correct. However, when you integrated you forgot to include the constant term, but that should be easy enough for you pick off. The convergence looks correct. You can pick off the convergence from function you approximating in this case (when does ln behave badly?). Also, (-1)^n does not converge by the limit test (as n->infinity it should go to zero). However, you don't have to worry because the rest of the series converges. You could also use the limit test to test the total convergence of this and find where it breaks down. Then again, there are many tests for convergence.

 

[dynamicsolo]

Your series expression for f(x) is right, up to a point. Don't forget that when you develop a power series by integration that you pick up, yes, an arbitrary constant. So that needs to be evaluated. Pick a value for x that will make this easy to deal with, like x = 0. The entire infinite polynomial becomes zero and you just have f(0) = C , so C = ln 2 .

The way you wrote your series otherwise is all right, though many sources would deal with it differently. Rather than removing a factor of x, many people would instead bring the (1/2) under the summation sign, making the general term

$$(-1)^n \cdot 2^{-(n+1)} \cdot \frac{1}{n+1} \cdot x^{n+1}$$

Since the index on the infinite sum now starts at 1 , they would next perform an "index shift" by re-assigning (n+1) to n . The index now starts at zero again and the Maclaurin series for our function becomes

$$ln(x+2) = ln 2 + \Sigma^{\infty}_{n = 0} (-1)^{n-1} \cdot \frac{x^n}{n 2^{n} }$$

I use a Ratio Test for the radius of convergence, rather than what you used, but I generally agree with your result (though you should write it simply as R = 2 ; radii are non-negative, so you wouldn't write -2 < R < 2 -- I believe you're thinking of the interval of convergence, for which we would still need to test the endpoints...).

The infinite alternating series

1 + (-1) + 1 + (-1) + ...

does not converge by reason of indeterminancy. Since we have a literally infinite reserve of positive and negative 1's, it is possible to use this alternating series to create any integer you like. Thus

0 = 1 + (-1) + 1 + (-1) + ... [infinite pairs of cancelling terms]

3 = 1 + 1 + 1 + 1 + (-1) + 1 + (-1) + ... [infinite pairs of cancelling terms]

-2 = (-1) + (-1) + 1 + (-1) + 1 + (-1) + ... [infinite pairs of cancelling terms] , etc.

We can pull out any finite number of positive or negative 1's to produce any desired integer and still be guaranteed total cancellation of "the rest" of the infinity of terms. Such things are in the nature of infinite sets...

 

[asi123]

Your series expression for f(x) is right, up to a point. Don't forget that when you develop a power series by integration that you pick up, yes, an arbitrary constant. So that needs to be evaluated. Pick a value for x that will make this easy to deal with, like x = 0. The entire infinite polynomial becomes zero and you just have f(0) = C , so C = ln 2 .

The way you wrote your series otherwise is all right, though many sources would deal with it differently. Rather than removing a factor of x, many people would instead bring the (1/2) under the summation sign, making the general term

$$(-1)^n \cdot 2^{-(n+1)} \cdot \frac{1}{n+1} \cdot x^{n+1}$$

Since the index on the infinite sum now starts at 1 , they would next perform an "index shift" by re-assigning (n+1) to n . The index now starts at zero again and the Maclaurin series for our function becomes

$$ln(x+2) = ln 2 + \Sigma^{\infty}_{n = 0} (-1)^{n-1} \cdot \frac{x^n}{n 2^{n} }$$

I use a Ratio Test for the radius of convergence, rather than what you used, but I generally agree with your result (though you should write it simply as R = 2 ; radii are non-negative, so you wouldn't write -2 < R < 2 -- I believe you're thinking of the interval of convergence, for which we would still need to test the endpoints...).

The infinite alternating series

1 + (-1) + 1 + (-1) + ...

does not converge by reason of indeterminancy. Since we have a literally infinite reserve of positive and negative 1's, it is possible to use this alternating series to create any integer you like. Thus

0 = 1 + (-1) + 1 + (-1) + ... [infinite pairs of cancelling terms]

3 = 1 + 1 + 1 + 1 + (-1) + 1 + (-1) + ... [infinite pairs of cancelling terms]

-2 = (-1) + (-1) + 1 + (-1) + 1 + (-1) + ... [infinite pairs of cancelling terms] , etc.

We can pull out any finite number of positive or negative 1's to produce any desired integer and still be guaranteed total cancellation of "the rest" of the infinity of terms. Such things are in the nature of infinite sets...

10x a lot

 

[dynamicsolo]

10x a lot

When I was in school (ha! as opposed to what I do now?), we wrote that as tan(q)>>1 [read "tan q very much"]...