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Maclaurin Series

  1. May 1, 2005 #1
    I'm supposed to find the following function as a Maclaurin Series. Please check if I'm correct.

    f(x) = ln(1+x)

    [tex]\sum \frac{\(x^n)((-1)^{n+1})}{n}[/tex]

    and that sum goes from n=1 to [tex]\infty[/tex]

    I also have to find the following functions as power series so please check it for me!

    [tex]f(x) = \frac{1}{1+9x^2} = 9\sum \{(-1)^n}{x^{2n}\\


    \\f(x) = \frac{1+x^2}{1-x^2} = \sum \{(x)^2}+{x^{2n+2}}[/tex]

    those should be two seperate functions but I don't know how to seperate them into two lines. both of these series go from n=0 to [tex]\infty[/tex]
     
    Last edited: May 1, 2005
  2. jcsd
  3. May 1, 2005 #2

    dextercioby

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    You're saying that

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

    which converges for [itex] |x|<1 [/itex].It's okay.

    Daniel.
     
  4. May 1, 2005 #3

    Pyrrhus

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    Yes it's ok.
     
  5. May 1, 2005 #4
    thank you both. Would you mind looking at the two power series questions I just posted...I'll probably be adding more questions throughout the day so if you could help me, it would be greatly appreciated. I'm very new to sequence and series so these questions are probably very "basic" but I do need help, thank you again.
     
  6. May 1, 2005 #5
    Find a Power series representation for the function:

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

    does that look about right to you guys?
     
  7. May 1, 2005 #6

    dextercioby

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    U can write the second as

    [tex]\frac{1+x^{2}}{1-x^{2}}=1+2\frac{1}{\frac{1}{x^{2}}-1} [/tex]

    and it will be easier to write the series.


    Daniel.
     
  8. May 1, 2005 #7
    ah I just realized that I was missing a plus sign.
    [tex]f(x) = \frac{1+x^2}{1-x^2} = \sum \{(x)^2}+{x^{2n+2}}[/tex]

    I'm not too sure of ur method but is my answer correct?
     
  9. May 1, 2005 #8

    Pyrrhus

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    On

    [tex] \frac{x}{4x+1}= x \frac{1}{1 - (-4x)} [/tex]
     
  10. May 1, 2005 #9
    Right that's what I did and I ended up with the series which I displayed.

    [tex]f(x) = x\sum \{(-4x)^n} = 4x\sum {-1^n}{x^n}[/tex]
     
  11. May 1, 2005 #10

    Pyrrhus

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    Should have been

    [tex] \sum_{n=0}^{\infty} (-1)^{n} 2^{2n} x^{n+1} [/tex]
     
  12. May 1, 2005 #11
    Starting with a geometric sum, find the sum of the series:

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

    so with this, I realized that the series is just the derivative of the geometric sum. Since I know that the sum of the geometric series is 1/(1-x), I just took the the derivative of that and got 2/(1-x)^2...does that sound about right? I want to make sure that part is right before I move on. Thank you.
     
  13. May 1, 2005 #12
    oh ok, so I can't just move the four out to the front like I did? I'm just glad that your correction makes sense to me, THANK YOU!
     
  14. May 1, 2005 #13

    Pyrrhus

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    On

    [tex] \frac{1+x^2}{1-x^2} = 1 + \frac{2x^2}{1 -x^2} [/tex]

    That way is easier because

    [tex] 1 + 2x^2 \sum_{n=0}^{\infty} (x^2)^{n} [/tex]
     
  15. May 1, 2005 #14
    I see but the way I did it was to rewrite the function as:

    [tex] (1+x^2) \frac {1}{1-x^2} [/tex]

    and then I worked it from there.
     
  16. May 1, 2005 #15

    Pyrrhus

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    Where you got that two from? in your derivative.

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

    [tex] \frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^2}[/tex]
     
    Last edited: May 1, 2005
  17. May 1, 2005 #16
    oh I made a mistake...sorry...but does what I said look right?
     
  18. May 1, 2005 #17

    Pyrrhus

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    Yes, what you said is true.
     
  19. May 1, 2005 #18
    ok but now here's the problem...it says to use what I found to find the sum of:

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

    [tex] \sum_{n=1}^{\infty} \frac{n}{2^{n}}[/tex]

    I honestly have no idea how to apply what I found to solve those series. If you could lead me in the right direction, it'd be greatly appreciated.
     
  20. May 1, 2005 #19

    Pyrrhus

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    For the first one it's basicly the same except we just use a little trick!, we rewrite it! :wink:

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

    For the second one think of it this way

    [tex] \sum_{n=1}^{\infty} \frac{n}{2^{n}} = \sum_{n=1}^{\infty} n (\frac{1}{2})^{n}}[/tex]
     
  21. May 1, 2005 #20
    ohhh I see. I knew it was going to be some rewriting trick but I wasn't sure what to do. so for the first one I end up with x/(1-x)^2 and the second one, I just substitute x with 1/2 and I end up with 2. Alright, thanks a lot. Expect more questions soon!
     
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