Find power series expansion

  • Thread starter andrey21
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  • #1
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Given that the sum of the geometric series is:

1+x+x^(2)+x^(3)+x^(4)....=1/1-x for -1<x<1

Find power series for

1/1+x

Not to sure where to start, any help would be great
 

Answers and Replies

  • #2
tiny-tim
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hi andrey21! :smile:

(try using the X2 icon just above the Reply box :wink:)
1+x+x^(2)+x^(3)+x^(4)....=1/1-x for -1<x<1

come on … how would you convert 1/1-x to 1/1+x ? :wink:
 
  • #3
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:rofl: I realised after posting the question how easy it actally is. I do have a slightly harder one I am having trouble with:

Find the power series of:

3/2-x

Any help would be great
 
  • #4
tiny-tim
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:rofl: I realised after posting the question how easy it actally is.

he he :biggrin:
3/2-x

3/2 1/(1- x/2) :wink:
 
  • #5
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:rofl: I realised after posting the question how easy it actally is. I do have a slightly harder one I am having trouble with:

Find the power series of:

3/2-x
This one is very easy. The power series of 3/2 - x is... 3/2 - x.

On the offchance that you really meant 3/(2 - x), see tiny-tim's post.
 
  • #6
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Ok I know this thread hasn't been replied to in a while but here is my answer: I wanted the power expansion for the following:

3/(2-x) = 3/2 + 3/2x + 9/4 x^2 + 27/8 x^3 + 81/16 x^4 + .....

Is this correct??
 
  • #7
ojs
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No, as you see from the expression that tiny-tim gave you then x can go from -2 to 2, if you try to put x = 1 the left hand side gives you 3, but just 3/2 + 3/2x gives 3 for x = 1, and then your left with all the rest which will then give more then 3 and therefore not a correct expansion.

What is wrong in your series is that when you create it you have to leave the 3/2 outside the 1/(1-x/2), expand the 1/(1-x/2) using the power series for 1/(1-x) and put x/2 in all the places where there is x in the power series, then multiply 3/2 into the series.
 
  • #8
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Ok here is my revised solution:

Expansion of 1/(1-x/2) = 1+ x/2 + x2/4 + x3/8 +...

Now multiplying by 3/2 gives:

3/2 + 3x/4 + 3x2/8 + 3x3/16 + ....
 
  • #9
ojs
79
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That should be correct.
 
  • #10
466
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Thanks ojs
 

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