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Urgend Geometric series question

  • Thread starter Mathman23
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  • #1
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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
 

Answers and Replies

  • #2
arildno
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No, try again. Do EXACTLY as in the 1/(1-q) example, with q=-x^2
 
  • #3
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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
 
  • #4
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:
[tex]1+q+q^2+q^3+++[/tex]
What do you get?
 
  • #5
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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.
 
  • #6
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arildno said:
No, you simply haven't done it correctly. Try again.
Insert (-x^2) on the q-places in the following expression:
[tex]1+q+q^2+q^3+++[/tex]
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
 
  • #7
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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.
 
  • #8
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.
 
  • #9
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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
 
  • #10
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?
 
  • #11
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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?
 
  • #12
arildno
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No.
Let's see:
We have:
[tex](-1)^{n}x^{2n}+(-1)^{n+1}x^{2n+2}+++=(-1)^{n}x^{2n}(1-x^{2}+++)[/tex]
Try to deduce how the +++++ will look like..:wink:
 
  • #13
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The n + 1 term ???

/Fred

arildno said:
No.
Let's see:
We have:
[tex](-1)^{n}x^{2n}+(-1)^{n+1}x^{2n+2}+++=(-1)^{n}x^{2n}(1-x^{2}+++)[/tex]
Try to deduce how the +++++ will look like..:wink:
 
  • #14
arildno
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What are you talking about?
Try to figure out the rest by yourself
 
  • #15
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I'v have been looking at mathworld, and then arrieved at the following idear:
Then I need to do it for infinity????

/Fred


[tex]S_n = \sum_{k=1} ^ {\infty} (-1)^k \frac{1}{1-x^k} [/tex]


arildno said:
What are you talking about?
Try to figure out the rest by yourself
 
Last edited:
  • #16
arildno
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Try to tackle it this way:
The infinite series starting with the n'th term is of the form:
[tex]\sum_{i=n}^{\infty}(-1)^{i}x^{2i}[/tex]
Now, introduce the new index j=i-n, that is i=j+n
Then, the series can be written in the form:
[tex]\sum_{j=0}^{\infty}(-1)^{j+n}x^{2n+2j}[/tex]
What can you do about this expression?
 
  • #17
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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:
[tex]\sum_{i=n}^{\infty}(-1)^{i}x^{2i}[/tex]
Now, introduce the new index j=i-n, that is i=j+n
Then, the series can be written in the form:
[tex]\sum_{j=0}^{\infty}(-1)^{j+n}x^{2n+2j}[/tex]
What can you do about this expression?
 
  • #18
arildno
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No, draw out the common factor!
[tex]\sum_{j=0}^{\infty}(-1)^{n+j}x^{2n+2j}=(-1)^{n}x^{2n}\sum_{j=0}^{\infty}(-1)^{j}x^{2j}[/tex]
What is the j-sum equal to?
 
  • #19
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arildno said:
No, draw out the common factor!

[tex]\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} [/tex]
What is the j-sum equal to?
??


[tex]\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} [/tex]

/Fred
 
Last edited:
  • #20
arildno
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Please write out the first few terms of the series [itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j}[/itex]
 
Last edited:
  • #21
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arildno said:
Please write out the first few terms of the series [itex]\sum_{j=0}^{\infty}(-1)^{n}x^{2j}[/itex]
Hello

[itex]\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}[/itex]

/Fred
 
  • #22
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Ok, what arildno meant was this: [itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j}[/itex]. The index is j everywhere.
 
  • #23
arildno
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Oops; sorry about that mistake in the index.
 
  • #24
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[itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = (-1) + (-1)x^{2} + (-1)x^{4} + (-1)x^{6} + (-1)x^{8}[/itex]

/Fred
 
  • #25
arildno
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Correct! That series have you seen before in this thread; what does it sum up to?
 

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