## Urgend Geometric series question

Hi

I have the following problem:

show that

1/(1+x^2)) = 1-x^2 + x^4 + (-1)^n*(x^2n-2) + (-1)^n * (x^2n)/(1+x^2)

I that know this arctan function can be expanded as a geometric series by using:

1 + q + q^2 + q^3 + .... + = 1/(1-q)

Then by putting q = -x^2. I get:

1/(1-(-x^2) = 1 - x^2 - (-x^2) - (-x^2)^3 + .... +

My question is how do I proceed from this to get the desired result???

Sincerley and Best Regards

Fred
 Recognitions: Gold Member Homework Help Science Advisor No, try again. Do EXACTLY as in the 1/(1-q) example, with q=-x^2

 Quote by arildno No, try again. Do EXACTLY as in the 1/(1-q) example, with q=-x^2
I still get first result, but has it something to do that the series converges? And then I then add a sum to the first result?

/Fred

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## Urgend Geometric series question

No, you simply haven't done it correctly. Try again.
Insert (-x^2) on the q-places in the following expression:
$$1+q+q^2+q^3+++$$
What do you get?

 Quote by Mathman23 Then by putting q = -x^2. I get: 1/(1-(-x^2) = 1 - x^2 - (-x^2) - (-x^2)^3 + .... +
This is wrong. Try again.

 Quote by arildno No, you simply haven't done it correctly. Try again. Insert (-x^2) on the q-places in the following expression: $$1+q+q^2+q^3+++$$ What do you get?
Hello by inserting q = -x^2 into the above I get:

1/(1-x^2) = 1 - x^2 + x^4 - x^6 + .... +

but how do I proceed from there ?

Sincerley

Fred
 Now you got it right and should be able to "see" how the general term looks like. Just see how the sign changes and how the exponent increases.

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 Quote by Mathman23 Hello by inserting q = -x^2 into the above I get: 1/(1-x^2) = 1 - x^2 + x^4 - x^6 + .... + but how do I proceed from there ? Sincerley Fred
Great! One flaw though:
You've been working with 1/(1-(-x^2)), not 1/(1-x^2)!

Now, find the n'th term, and scrutinize the infinite part of the series starting with the n'th term.

 Quote by assyrian_77 Now you got it right and should be able to "see" how the general term looks like. Just see how the sign changes and how the exponent increases.

Hello I can see that the the function changes in the following way

(-1)^n * (x^2)^n, then

1/(1-x^2) = 1/(1-x^2) = 1 - x^2 + x^4 - x^6 + .... + (-1)^n * (x^2)^n

Do I then rewrite the above result to get the result required ???

Sincerely
Fred
 Recognitions: Gold Member Homework Help Science Advisor You have now found the general term of the series expansion of 1/(1-(-x^2)) For all N>=n, draw out the common factor (-1)^(n)x^(2n) What are you left with then?

The n-1 term ???

/Fred

 Quote by arildno You have now found the general term of the series expansion of 1/(1-(-x^2)) For all N>=n, draw out the common factor (-1)^(n)x^(2n) What are you left with then?
 Recognitions: Gold Member Homework Help Science Advisor No. Let's see: We have: $$(-1)^{n}x^{2n}+(-1)^{n+1}x^{2n+2}+++=(-1)^{n}x^{2n}(1-x^{2}+++)$$ Try to deduce how the +++++ will look like..

The n + 1 term ???

/Fred

 Quote by arildno No. Let's see: We have: $$(-1)^{n}x^{2n}+(-1)^{n+1}x^{2n+2}+++=(-1)^{n}x^{2n}(1-x^{2}+++)$$ Try to deduce how the +++++ will look like..
 Recognitions: Gold Member Homework Help Science Advisor What are you talking about? Try to figure out the rest by yourself

I'v have been looking at mathworld, and then arrieved at the following idear:
Then I need to do it for infinity????

/Fred

$$S_n = \sum_{k=1} ^ {\infty} (-1)^k \frac{1}{1-x^k}$$

 Quote by arildno What are you talking about? Try to figure out the rest by yourself
 Recognitions: Gold Member Homework Help Science Advisor Try to tackle it this way: The infinite series starting with the n'th term is of the form: $$\sum_{i=n}^{\infty}(-1)^{i}x^{2i}$$ Now, introduce the new index j=i-n, that is i=j+n Then, the series can be written in the form: $$\sum_{j=0}^{\infty}(-1)^{j+n}x^{2n+2j}$$ What can you do about this expression?

I rewrite it as a product of a two Sums ??

/Fred

 Quote by arildno Try to tackle it this way: The infinite series starting with the n'th term is of the form: $$\sum_{i=n}^{\infty}(-1)^{i}x^{2i}$$ Now, introduce the new index j=i-n, that is i=j+n Then, the series can be written in the form: $$\sum_{j=0}^{\infty}(-1)^{j+n}x^{2n+2j}$$ What can you do about this expression?