# 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

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arildno
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No, try again. Do EXACTLY as in the 1/(1-q) example, with q=-x^2

arildno said:
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

arildno
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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?

Mathman23 said:
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.

arildno said:
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.

arildno
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Mathman23 said:
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.

assyrian_77 said:
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

arildno
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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

arildno said:
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?

arildno
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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

arildno said:
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.. arildno
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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

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

arildno said:
Try to figure out the rest by yourself

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arildno
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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:
$$\sum_{j=0}^{\infty}(-1)^{j+n}x^{2n+2j}$$

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

/Fred

arildno said:
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:
$$\sum_{j=0}^{\infty}(-1)^{j+n}x^{2n+2j}$$

arildno
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No, draw out the common factor!
$$\sum_{j=0}^{\infty}(-1)^{n+j}x^{2n+2j}=(-1)^{n}x^{2n}\sum_{j=0}^{\infty}(-1)^{j}x^{2j}$$
What is the j-sum equal to?

arildno said:
No, draw out the common factor!

$$\sum_{j=0}^{\infty}(-1)^{n+j}x^{2n+2j}=(-1)^{n}x^{2n}\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = (-1)^{n} 2^{n}$$
What is the j-sum equal to?
??

$$\sum_{j=0}^{\infty}(-1)^{n+j}x^{2n+2j}=(-1)^{n}x^{2n}\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = (-1)^{n} x^{2n}$$

/Fred

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arildno
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Please write out the first few terms of the series $\sum_{j=0}^{\infty}(-1)^{j}x^{2j}$

Last edited:
arildno said:
Please write out the first few terms of the series $\sum_{j=0}^{\infty}(-1)^{n}x^{2j}$
Hello

$\sum_{j=0}^{\infty}(-1)^{n}x^{2j} = (-1)^{n} + (-1)^{n}x^{2} + (-1)^{n}x^{4} + (-1)^{n}x^{6} + (-1)^{n}x^{8}$

/Fred

Ok, what arildno meant was this: $\sum_{j=0}^{\infty}(-1)^{j}x^{2j}$. The index is j everywhere.

arildno
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Oops; sorry about that mistake in the index.

$\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = (-1) + (-1)x^{2} + (-1)x^{4} + (-1)x^{6} + (-1)x^{8}$

/Fred

arildno