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Math Help - Please check my work

  1. #1
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    Please check my work

    Hello !!

    Give the Taylor series about 0 for f and determine a range of validity for the series

    <br />
f(x)=\frac{1}{(1+\frac{1}{6}x)^2}<br />

    Using the standard taylor series
    <br />
f(x)=\frac{1}{(1-x)}= 1+ x+x^2+x^3+.... for -1<x<1<br />

    Substituting <br />
x = -\frac{1}{6}x<br />
    And taking 2 in the binomial series gives

    <br />
1+(2)*(-\frac{1}{6}x)+\frac{(2)(3)}{2!}(-\frac{1}{6}x)^2+\frac{(2)(3)(4)}{3!}(-\frac{1}{6}x)^3<br />

    Gives

    <br />
1+-\frac{1}{3}x+\frac{1}{12}x^2-\frac{1}{54}x^3<br />

    as

    <br />
f(x)=\frac{1}{(1-x)}= 1+ x+x^2+x^3+.... for -1<x<1<br />

    Then
    <br />
\frac{1}{(1+\frac{1}{6}x)^2}<br />

    is valid for

    <br />
-1<-\frac{1}{6}x<1<br />

    Thanks Macca
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by macca101
    Hello !!

    Give the Taylor series about 0 for f and determine a range of validity for the series

    <br />
f(x)=\frac{1}{(1+\frac{1}{6}x)^2}<br />

    Using the standard taylor series
    <br />
f(x)=\frac{1}{(1-x)}= 1+ x+x^2+x^3+.... for -1<x<1<br />

    Substituting <br />
x = -\frac{1}{6}x<br />
    And taking 2 in the binomial series gives

    <br />
1+(2)*(-\frac{1}{6}x)+\frac{(2)(3)}{2!}(-\frac{1}{6}x)^2+\frac{(2)(3)(4)}{3!}(-\frac{1}{6}x)^3<br />

    Gives

    <br />
1+-\frac{1}{3}x+\frac{1}{12}x^2-\frac{1}{54}x^3<br />

    as

    <br />
f(x)=\frac{1}{(1-x)}= 1+ x+x^2+x^3+.... for -1<x<1<br />

    Then
    <br />
\frac{1}{(1+\frac{1}{6}x)^2}<br />

    is valid for

    <br />
-1<-\frac{1}{6}x<1<br />

    Thanks Macca
    I'm going to try a different, simpler, approach (tell me if you see anything wrong)

    we have the taylor series: f(b)=\frac{1}{(1-b)}= 1+ b+b^2+b^3+.... for -1<b<1

    and we have the series... f(x)=\frac{1}{(1+\frac{1}{6}x)^2}

    now let's solve for: (1+\frac{1}{6}x)^2

    but let's say \neg\frac{1}{6}x=a

    then rewrite the problem: (1-a)^2=1-2a+a^2=1-(2a-a^2)

    and then let's say 2a-a^2=b

    now remembering the taylor series: f(b)=\frac{1}{(1-b)}= 1+ b+b^2+b^3+.... for -1<b<1
    we see that b must be between -1 and 1, let's figure out what x needs to be between:

    -1<b<1\Longrightarrow<br />
-1<2a-a^2<1\Longrightarrow -1<\neg\frac{1}{3}x- (\neg\frac{1}{6}x)^2<1
    and solve from there...

    ~ Q\!u\!i\!c\!k
    Last edited by Quick; July 22nd 2006 at 06:05 AM.
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  3. #3
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Quick
    -1<b<1\Longrightarrow<br />
-1<2a-a^2<1\Longrightarrow -1< \neg\frac{1}{3}x-(\neg\frac{1}{6}x)^2<1
    and solve from there...

    ~ Q\!u\!i\!c\!k
    You can complete the square, watch: (correct me if I'm wrong)

    -1<-\frac{1}{3}x-(\neg\frac{1}{6}x)^2<1

    -1<-\frac{1}{3}x-\frac{1}{36}x^2<1

    1>\frac{1}{3}x+\frac{1}{36}x^2>-1

    -1<\frac{1}{36}x^2+\frac{1}{3}x<1

    -36<x^2+\frac{36}{3}x<36

    -36<x^2+12x<36

    -36+36<x^2+12x+36<36+36

    0<(x+6)^2<72

    \sqrt0<x+6<\sqrt{72}

    \boxed{-6<x<\sqrt{72}-6}

    I hope that's right.
    Actually, I know I'm right
    Last edited by Quick; July 22nd 2006 at 08:38 AM.
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  4. #4
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    How is that a Taylor series?
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  5. #5
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by ThePerfectHacker
    How is that a Taylor series?
    I hope I didn't do anything wrong, (I haven't heard of Taylor series before so I just went off of what macca said)
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  6. #6
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    Quote Originally Posted by ThePerfectHacker
    How is that a Taylor series?
    The Taylor series was found (hopefully) in the first part of my question.


    <br />
f(x)=\frac{1}{(1+\frac{1}{6}x)^2}=1+-\frac{1}{3}x+\frac{1}{12}x^2-\frac{1}{54}x^3+.....<br />

    What I was and sill am struggling with is finding the interval which the series is valid.
    Quick gets it to

    <br />
\boxed{-6<\sqrt{72}-6}<br />

    I have a math cad file that allows me to plot a Taylor series against the original function see below. This indicates to me the series is valid for approximately


    <br />
\boxed{-2<x<2}<br />



    I hope I am making myself clear here I'm not saying quick is wrong (far be it from me) I just don't understand how to find the interval algebraically
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  7. #7
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    Quote Originally Posted by macca101
    The Taylor series was found (hopefully) in the first part of my question.

    <br />
f(x)=\frac{1}{(1+\frac{1}{6}x)^2}=1+-\frac{1}{3}x+\frac{1}{12}x^2-\frac{1}{54}x^3+.....<br />
    You mean geomteric series.
    ---
    I assume you want an infinite series for a function,
    \frac{1}{(1+1/6x)^2}
    Note the infinite geometric,
    \frac{1}{1-t}=1+t+t^2+t^3+... for |t|<1
    Take derivative of both sides,
    \frac{1}{(1-t)^2}=1+2t+3t^2+4t^3+...
    Substitute t=-1/6x
    Thus,
    \frac{1}{(1+1/6x)^2}=1-1/3x+1/12x^2-...
    Good you got that right.

    You are trying to find for what values this series converges.
    You are going to use the generalized ratio test, with
    a_k=(-1)^k\frac{(k+1)}{6^k}
    Thus,
    \left|\frac{a_{k+1}}{a_k}\right|=\frac{k+2}{6^{k+1  }}\cdot \frac{6^k}{k+1}=\frac{1}{6}\cdot \frac{k+2}{k+1}
    Its limit as, k\to\infty is 1/6,
    Therefore the reciprocal of the limit tells you the radius of convergence which is 6.
    Thus,
    |x|<6
    Same as saying,
    -6<x<6
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  8. #8
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    Thanks For That
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