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Thread: Produce a taylor series please....

  1. #1
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    Produce a taylor series please....

    for fx=(1/x) centered at c=1
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  2. #2
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    Quote Originally Posted by tiga killa View Post
    for fx=(1/x) centered at c=1

    $\displaystyle f(x)=\frac{1}{x}=\frac{1}{1- [-(x-1)]}$

    This is a geometric series with $\displaystyle r=-(x-1)$

    so

    $\displaystyle f(x)=\sum_{n=0}^{\infty}(-1)^n(x-1)^n$
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  3. #3
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    Quote Originally Posted by tiga killa View Post
    for fx=(1/x) centered at c=1
    Assume that $\displaystyle f(x)$ can be expressed as a polynomial.

    Since it is centred at $\displaystyle c = 1$, it will be of the form

    $\displaystyle x^{-1} = a + b(x - 1) + c(x - 1)^2 + d(x - 1)^3 + e(x - 1)^4 + \dots$.

    At $\displaystyle x = 1$ we find $\displaystyle a = 1$.


    Take the derivative of both sides

    $\displaystyle -x^{-2} = b + 2c(x - 1) + 3d(x - 1)^2 + 4e(x - 1)^3 + 5f(x - 1)^4 + \dots$

    At $\displaystyle x = 1$ we find $\displaystyle b = -1$.


    Take the derivative of both sides

    $\displaystyle 2x^{-3} = 2c + 6d(x - 1) + 12e(x - 1)^2 + 20f(x - 1)^3 + \dots$

    At $\displaystyle x = 1$ we find $\displaystyle c = 1$.


    Take the derivative of both sides

    $\displaystyle -6x^{-4} = 6d + 24e(x - 1) + 60f(x - 1)^2 + \dots$

    At $\displaystyle x = 1$ we find $\displaystyle d = -1$.


    Take the derivative of both sides

    $\displaystyle 24x^{-5} = 24e + 120f(x - 1) + \dots$

    At $\displaystyle x = 1$ we find $\displaystyle e = 1$.


    Take the derivative of both sides

    $\displaystyle -120x^{-6} = 120f + \dots$.

    At $\displaystyle x = 1$ we find $\displaystyle f = -1$.



    Looking at our polynomial we have

    $\displaystyle \frac{1}{x} = 1 - x + x^2 - x^3 + x^4 - x^5 + \dots$

    $\displaystyle \frac{1}{x} = \sum_{n = 0}^{\infty}{(-1)^n(x-1)^n}$.
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    Quote Originally Posted by TheEmptySet View Post
    $\displaystyle f(x)=\frac{1}{x}=\frac{1}{1- [-(x-1)]}$

    This is a geometric series with $\displaystyle r=-(x-1)$

    so

    $\displaystyle f(x)=\sum_{n=0}^{\infty}(-1)^n(x-1)^n$
    Of course, we're assuming that

    $\displaystyle |r| < 1$.

    So $\displaystyle |-(x - 1)| < 1$

    $\displaystyle |x - 1| < 1$

    $\displaystyle -1 < x - 1 < 1$

    $\displaystyle 0 < x < 2$.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    Of course, we're assuming that

    $\displaystyle |r| < 1$.

    So $\displaystyle |-(x - 1)| < 1$

    $\displaystyle |x - 1| < 1$

    $\displaystyle -1 < x - 1 < 1$

    $\displaystyle 0 < x < 2$.
    Of course it has the same radius of convergence as the series you came up with above.
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