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Urgend Geometric series question

  • Thread starter Mathman23
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  • #26
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arildno said:
Correct! That series have you seen before in this thread; what does it sum up to?

Hi it sums up to

[itex](-1)^{n} x^{2n}[/itex]

/Fred
 
  • #27
arildno
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No it does not; that is a TERM in the series.
What does the whole series sum up to?
 
  • #28
arildno
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Mathman23 said:
[itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = (-1) + (-1)x^{2} + (-1)x^{4} + (-1)x^{6} + (-1)x^{8}[/itex]

/Fred
OOPS! This is wrong!
When j is even, [itex](-1)^{j}=1[/itex] wheras if j is odd, we have [itex](-1)^{j}=-1[/itex]

Use this info.
 
  • #29
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Hi the series summes up to [tex]\frac{1}{1-x^2}[/tex]

/Fred

arildno said:
No it does not; that is a TERM in the series.
What does the whole series sum up to?
 
  • #30
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Hi

Then

[itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}[/itex]

Because j is even.

/Fred

arildno said:
OOPS! This is wrong!
When j is even, [itex](-1)^{j}=1[/itex] wheras if j is odd, we have [itex](-1)^{j}=-1[/itex]

Use this info.
 
  • #31
arildno
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No, it sums up to 1/(1+x^{2}).
 
  • #32
arildno
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Mathman23 said:
Hi

Then

[itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}[/itex]

Because j is even.

/Fred
No, j alternates being even and odd.
 
  • #33
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Mathman23 said:
Hi

Then

[itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}[/itex]

Because j is even.

/Fred
No, j is not only even! If that would be the case, your expression would be [itex]x^{4} + x^{8} + x^{12}+...[/itex]. Go back and check again.
 
  • #34
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arildno said:
No, it sums up to 1/(1+x^{2}).

Then

[itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}[/itex] Sums up to

[tex]\frac{1}{(1+x^{2})} [/tex]
 
Last edited:
  • #35
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arildno said:
No, it sums up to 1/(1+x^{2}).

Then

[itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j} = x^{2} + x^{4} + x^{6} + x^{8}[/itex] Sums up to [tex][/tex]
 
Last edited:
  • #36
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First of all, [itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j} \neq x^{2} + x^{4} + x^{6} + x^{8}[/itex], it is [itex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j}=1-x^{2} + x^{4} - x^{6} + x^{8}...[/itex].
 
  • #37
arildno
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No. You MUST learn to read properly and stop writing sloppy and nonsensical maths.

We have:
[tex](-1)^{n}x^{2n}\sum_{j=0}^{\infty}(-1)^{j}x^{2j}=\frac{(-1)^{n}x^{2n}}{1+x^{2}}[/tex]
 
Last edited:
  • #38
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And second; you know also this: [itex]\frac{1}{1-q}=1 + q + q^2 + q^3 +...[/itex].

Can you see the connection now?

I strongly suggest you to over the entire problem again and - as arildno is saying - read it properly.
 
  • #39
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arildno said:
No. You MUST learn to read properly and stop writing sloppy and nonsensical maths.

We have:
[tex](-1)^{n}x^{2n}\sum_{j=0}^{\infty}=\frac{(-1)^{n}x^{2n}}{1+x^{2}}[/tex]

Then
[tex]\sum_{j=0}^{\infty}(-1)^{j}x^{2j}=1-x^{2} + x^{4} - x^{6} + x^{8} +\cdots + (-1)^{n}x^{2n} = \sum_{j=0}^{\infty}\frac{(-1)^{n}x^{2n}}{1+x^{2}}[/tex]
 
  • #40
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arildno said:
We have:
[tex](-1)^{n}x^{2n}\sum_{j=0}^{\infty}=\frac{(-1)^{n}x^{2n}}{1+x^{2}}[/tex]
Sorry arildno, but that doesn't make sense.
 
  • #41
arildno
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assyrian_77 said:
Sorry arildno, but that doesn't make sense.
You're right, it doesn't I'll fix it right away.
 

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