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Math Help - Taylor Series Question

  1. #1
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    Taylor Series Question

    Hi, this question regarding taylor series. i tried solve for 7 hours straight with no ans. differentiate the 1/(x^2 + x +1) up to 4 times without finding any fixed patterns. Experts please help.
    Taylor Series Question-taylor-series.jpg
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  2. #2
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    Re: Taylor Series Question

    Since at x = 0, there is a part where taking out x^n and leaving behind (f(x) / x) ^n, this caused a undefined value since when x = 0, there f(0) / 0 is undefined.
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  3. #3
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    Re: Taylor Series Question

    Quote Originally Posted by hxhx View Post
    Hi, this question regarding taylor series. i tried solve for 7 hours straight with no ans. differentiate the 1/(x^2 + x +1) up to 4 times without finding any fixed patterns. Experts please help.
    Click image for larger version. 

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    The result of this post are not correct for the reason explained in a succesive post...

    Kind regards

    \chi \sigma
    Last edited by chisigma; September 11th 2011 at 09:15 AM.
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  4. #4
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    Re: Taylor Series Question

    this is better
    1-x+x^3-x^4-\frac{x^6}{\left(1+2 (-1)^{2/3}\right) \left((-1)^{2/3}-x\right)}-\frac{x^6}{\left(-1+2 (-1)^{1/3}\right) \left((-1)^{1/3}+x\right)}
    chao
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  5. #5
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    Re: Taylor Series Question

    \displaystyle \begin{align*} \frac{1}{x^2 + x + 1} &= \frac{1}{1 - \left(-x^2 - x\right)}\textrm{ which is the closed form of a geometric series} \\ &= \sum_{n = 0}^{\infty}(-x^2 - x)^n\textrm{ provided }|-x^2 - x| < 1 \end{align*}
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: Taylor Series Question

    Just a little precisation: it was requested the Taylor expansion of f(x)=\frac{1}{1+x+x^{2}} around x=0 [in that case we properly have a McLaurin extension...], so that the result must be in the form...

    \frac{1}{1+x+x^{2}} = \sum_{n=0}^{\infty} a_{n}\ x^{n} (1)

    Kind regards

    \chi \sigma
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    Re: Taylor Series Question

    f(x) = \frac{1}{x^2+x+1} =  \frac{1}{x^2+x+1}\cdot  \frac{1-x}{1-x}
    = \frac{1-x}{1-x^3} = (1-x)(1+x^3 +x^6+x^9+\dots)
    So if
    f(x) = \sum_{n=0}^{\infty} c_n x^n
    then
    c_{18}=1
    c_{19}=-1
    c_{20}=0
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  8. #8
    MHF Contributor chisigma's Avatar
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    Re: Taylor Series Question

    Quote Originally Posted by chisigma View Post
    Here we have the opportunity to use some properties of the Z-Tranform. The function...

    h(z)= \frac{1}{1+z+z^{2}} (1)

    ... can be written as Taylor series h(z)= \sum_{n=0}^{\infty} a_{n}\ z^{n} where the a_{n} satisfy the difference equation...

    a_{n} + a_{n-1} + a_{n-2}=0\ ;\ a_{0}=1\ ,\ a_{1}=0 (2)

    The solutions of the 'characteristic equation' \alpha^{2}+\alpha+1=0 are e^{\pm i\ \frac{\pi}{3}, so that the equation of (2) is...

    a_{n} = c_{1}\ \cos n\ \frac{\pi}{3} + c_{2}\ \sin n\ \frac{\pi}{3} (3)

    Taking into account the 'initial conditions' we arrive to write...

    \frac{1}{1+z+z^{2}} = 1 -z^{2} - z^{3} + z^{5} + z^{6} - z^{8} - z^{9} +... (4)

    Kind regards

    \chi \sigma
    I started with the right way but soon I lost it!... In terms of Zeta-Tranform, difference equation corresponding to the function h(z)=\frac{1}{1+z+z^{2}} is...

    a_{n}+a_{n-1}+a_{n-2}= \delta_{n} (1)

    ... where...

    \delta_{n} =\begin{cases}1 &\text{if }n=0\\ 0 &\text{if } n>0\end{cases} (2)

    ... so that the coefficients of the Taylor series are found symply expanding (1)...

    a_{0}= 1 - a_{-1}-a_{-2}= 1

    a_{1}= 0 - a_{0}-a_{-1}= -1

    a_{2}= 0 - a_{1}-a_{0}= 0

    a_{3}= 0 - a_{2}-a_{1}= 1

    a_{4}= 0 - a_{3}-a_{2}= -1

    ... so that is...

    \frac{1}{1+z+z^{2}}= 1 - z + z^{3} - z^{4} + z^{6}-z^{7}+... (3)

    ... as awkward found...

    Kind regards

    \chi \sigma
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  9. #9
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    Re: Taylor Series Question

    Quote Originally Posted by awkward View Post
    f(x) = \frac{1}{x^2+x+1} =  \frac{1}{x^2+x+1}\cdot  \frac{1-x}{1-x}
    = \frac{1-x}{1-x^3} = (1-x)(1+x^3 +x^6+x^9+\dots)
    So if
    f(x) = \sum_{n=0}^{\infty} c_n x^n
    then
    c_{18}=1
    c_{19}=-1
    c_{20}=0
    I am puzzled about the part on how the C18 C19 C20 became 1, -1, 0 since Cn = (1-x)(x^2)^n, so since x = 0, arent C18 C19 and C20 all zero instead?
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    Re: Taylor Series Question

    Quote Originally Posted by hxhx View Post
    I am puzzled about the part on how the C18 C19 C20 became 1, -1, 0 since Cn = (1-x)(x^2)^n, so since x = 0, arent C18 C19 and C20 all zero instead?
    Wrong.

    What is the expansion of (1-x) (\dots x^{15} + x^{18} + x^{21} \dots) ?
    That is part of the series.
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  11. #11
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    Re: Taylor Series Question

    Quote Originally Posted by awkward View Post
    Wrong.

    What is the expansion of (1-x) (\dots x^{15} + x^{18} + x^{21} \dots) ?
    That is part of the series.
    so the Cn stands for the coefficient of the x^n at nth position of the series? so Cn is not a general form of the coefficent of x^n?
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  12. #12
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    Re: Taylor Series Question

    From your original post:

    f(x) = \sum_{n=0}^{\infty} c_n x^n

    So c_n is the coefficient of x^n in the series for f(x).

    But the problem did not ask for a general form of the coefficient, it simply asked for the value of
    c_{18} + c_{19} + c_{20},
    which can be found without necessarily finding the general coefficient.
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  13. #13
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    Re: Taylor Series Question

    Quote Originally Posted by awkward View Post
    From your original post:

    f(x) = \sum_{n=0}^{\infty} c_n x^n

    So c_n is the coefficient of x^n in the series for f(x).

    But the problem did not ask for a general form of the coefficient, it simply asked for the value of
    c_{18} + c_{19} + c_{20},
    which can be found without necessarily finding the general coefficient.
    So what is the general form of the coefficient in this case?
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  14. #14
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    Re: Taylor Series Question

    Try multiplying out the first few terms of

     (1-x)(1+x^3 +x^6+x^9+\dots)

    and see if there is a pattern.
    Last edited by awkward; September 13th 2011 at 02:22 PM.
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