Power series representation

  • Thread starter rent981
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  • #1
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1. Find a power series for F(x)= 3/4x^3-5, where c=1
2. power series = 1/a-r



3. What I did was take a derivative to get a similar function that was easier to solve. I used 1/x^3-1. Then I found a series for that function. Which I got [tex]\sum[/tex](x^3)^n. Then I added back from my orginal problem and got [tex]\sum[/tex](4/3x)^3n does this look at all right?
 

Answers and Replies

  • #2
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Without parentheses it's difficult to tell what your function F is. It should be interpreted as (3/4)*x^3 - 5, but I don't think that's what you had in mind.
 
  • #3
180
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Doesn't seem right to me. How does the 5 affect your result for instance?

From the geometrical series you know that

[tex]\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n[/tex]

Your function

[tex]\frac{3}{4x^3 - 5}[/tex]

is already almost in the same form, all you have to do is manipulate it a little bit.
 
  • #4
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Ok if I rewrite I need to make the first term of the denominator 1. So I divide by -5. So then does the sum from 0 to infinity of (-3/5)(x)^3n look better?
 
  • #5
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Where did the 4 go?
 
  • #6
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oops. Im meant to write [infinity]\sum[/n=0] (-3/5)(4x)^3n.
 
  • #7
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[tex]
\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n (-3/5)(4x)^3n
[/tex]
 
  • #8
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does that look ok?
 
  • #9
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When you divide by five you should do that to the whole denominator. And you have too much x's in there though I suspect that's a typo.
 
  • #10
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Yes that was a typo. So.....[tex]\sum[/tex] (from 0-infinty) (-3/5)(4/5X)^3n would be ok? Sorry this is a new topic for me. Thanks you so much for helping!!!
 
  • #11
180
4
Almost, except that 4/5 is not cubed so it's power is n:

[tex]
\frac{3}{4x^3 - 5} = -\frac{3}{5} \frac{1}{1-\frac{4x^3}{5}} = -\frac{3}{5}\sum_{n=0}^{\infty} (\frac{4x^3}{5})^n
[/tex]
 

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