Not able to simplify this summation formula?

  • Thread starter musicgold
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  • #1
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Hi,

Please see the attached pdf file. Equation 1 and equation 2 are equivalent.
Can someone please help me understand how to simplify equation 1 to get to equation 2?

Thanks.
 

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  • #2
Mentallic
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The summation variable is n, so x can be considered a constant hence it can be pulled out the front.

What is

[tex]1+r+r^2+...+r^n[/tex]

equal to?
 
  • #3
271
7
The summation variable is n, so x can be considered a constant hence it can be pulled out the front.

What is

[tex]1+r+r^2+...+r^n[/tex]

equal to?
Please see the new attached file.
This is how far I could go.
 

Attachments

  • #4
HallsofIvy
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The point is that both of your series are geometric series. What you do in your last post is essentially repeating the proof that the sum of the geometric series, [itex]\sum_{n=0}^\infty r^n[/itex] is [itex]\frac{1}{1- r}[/itex] except that you have [itex]r= \frac{1}{\gamma}[/itex].
 
  • #5
Mentallic
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The point is that both of your series are geometric series. What you do in your last post is essentially repeating the proof that the sum of the geometric series, [itex]\sum_{n=0}^\infty r^n[/itex] is [itex]\frac{1}{1- r}[/itex] except that you have [itex]r= \frac{1}{\gamma}[/itex].
There is no mention of infinite sums.

musicgold, so what is

[tex]\sum_{i=1}^{n}\frac{1}{1+y}[/tex]

And hence, let n=10. Also,

[tex]\frac{1-\frac{1}{\gamma^{n+1}}}{1-\frac{1}{\gamma}}[/tex]

Can be simplified further. At least get rid of the fraction within the denominator.
 
  • #6
271
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Please see the attached file.

I think, I am close, but not sure how to get rid of the 'y' in the encircled term.
What am I missing?
 

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  • #7
Mentallic
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There is no mention of infinite sums.

musicgold, so what is

[tex]\sum_{i=1}^{n}\frac{1}{1+y}[/tex]
Sorry, this was supposed to be

[tex]\sum_{i=1}^{n}\frac{1}{(1+y)^i}[/tex]

Please see the attached file.

I think, I am close, but not sure how to get rid of the 'y' in the encircled term.
What am I missing?
We're looking for

[tex]\sum_{n=1}^{10}\frac{1}{(1+y)^n}[/tex]

while you're finding

[tex]\sum_{n=0}^{10}\frac{1}{(1+y)^n}[/tex]

In your second attachment when you found

[tex]s=\sum\frac{1}{\gamma^n}=1+\frac{1}{\gamma}+\frac{1}{\gamma^2}+...[/tex]

You began the sum with n=0 when you should've began with n=1.
 
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  • #8
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Mentallic,

Yes. I was able to solve it with that correction. See attached. Thank you very much.
 

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  • #9
Mentallic
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Good work :smile:
 

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