anonymous
  • anonymous
Power Series: I'm having trouble understanding the concept. Here is one problem that I need help with: Find the power series representation for f(x)=(4+x)/(1-x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
Do we write it out like a polynomial?
amistre64
  • amistre64
What is your understanding of a power series?
anonymous
  • anonymous
umm hardly anything at all really. I understand what we are trying to do (at least I think I do) but i dont know how to do it.

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amistre64
  • amistre64
:) from the notes online it looks like it is a "sum" of something... what can you tell me about them.. in general
anonymous
  • anonymous
sums? that its a sequence with all the numbers being added together. so \[\sum_{n=1}^{\infty} 1/n\] would be (1+1/2+1/3....) and so forth.
amistre64
  • amistre64
good..good..... what is your gut telling you about this problem?
anonymous
  • anonymous
take the derivative
amistre64
  • amistre64
I have almost zero concepts about this so your driving.... tell me, how does the derivative help us? We can find the derivative quite easily, but what does that do for us?
anonymous
  • anonymous
umm I don't know. but it makes it into \[f'(x)= 5/((1-x)^2)\]
amistre64
  • amistre64
that is correct for the derivative :) tell me...every time ive seen the big E symbol it was talking about "integration". Would that be any help for us here?
amistre64
  • amistre64
better yet, is there a problem that you already know how to do that you can step me through?
anonymous
  • anonymous
yeah hold on.
anonymous
  • anonymous
\[f(x)=arccot (x)\] so... the derivative of that: \[f'(x)=-1/(1+x^2)\]
anonymous
  • anonymous
meaning the integral of f'(x)=f(x)
amistre64
  • amistre64
usually denoted in the textbooks a F(x) :) but yeah....
anonymous
  • anonymous
and power series f'(x) of that is \[\sum_{n=0}^{\infty} (-1)^nx^(2n) \] (that's x raised to the 2n)
anonymous
  • anonymous
so the integral of the power series=the power series of f(x)
anonymous
  • anonymous
+c
amistre64
  • amistre64
so step one, you found the derivative of arccot(x) right? and worked with it?
anonymous
  • anonymous
yup
amistre64
  • amistre64
then step one here is to find the derivative :) your gut was good....
anonymous
  • anonymous
but I dont know how to deal with the whole being squared instead of just x
amistre64
  • amistre64
5/ (1-x)^2 would it be better in expanded form? x^2 -2x +1 ?
anonymous
  • anonymous
no the standard form for the power series i'm dealing with right now is 1/(1-x) so i have to manipulate it into that form. The five is easy enough to get rid of by simply factoring it out but the (1-x)^2 is harder.
amistre64
  • amistre64
found this, might help if you understand this stuff :) Example 4 Find a power series representation for the following function and determine its interval of convergence. g(x) = 1/ (1-x)^2 Solution....
amistre64
  • amistre64
Solution To do this problem let’s notice that 1/ (1-x)^2 = (d/dx) 1/ (1-x)
anonymous
  • anonymous
crap that means i take the derivative again? -.- thats a double integral.
amistre64
  • amistre64
lol...hold on its using the d/dx form later on.. might be helpful
anonymous
  • anonymous
ok
amistre64
  • amistre64
Then since we’ve got a power series representation for 1/(1-x) all that we’ll need to do is differentiate that power series to get a power series representation for
anonymous
  • anonymous
okay i think i see where this is going
amistre64
  • amistre64
this might be quicker :)
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anonymous
  • anonymous
haha that is :)
amistre64
  • amistre64
it was years before I knew what the "PrtScn" button was for... it captures a screenshot that you can psate into "Paint"
amistre64
  • amistre64
*paste that is
anonymous
  • anonymous
yeah i just figured that out about a year ago myself
amistre64
  • amistre64
any of that jargon help you out with this problem?
anonymous
  • anonymous
i think so I let you know shortly if the answer i get is right.
anonymous
  • anonymous
wrong answer but i think i understand the concept better.
amistre64
  • amistre64
at least one of us does :) good luck
anonymous
  • anonymous
thanks :)

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