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Math Help - Maclaurin series

  1. #1
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    Maclaurin series

    Find a Maclaurin series expansion for each function.

    f(x) = cos x

    f(x) = 1/(1-x)

    f(x) = x sin x
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  2. #2
    o_O
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    These are pretty standard but if the intent is to "discover" these on your own, just directly apply the formula for the Maclaurin series (a taylor series centred around a = 0):

    f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^{2} + ... + \frac{f^{(n)}(a)}{n!}(x-a)^{n} + ...

    Taking a = 0: f(x) = f(0) + f'(a)(x) + \frac{f''(a)}{2!}x^{2} + ... + \frac{f^{(n)}(0)}{n!}x^{n} + ...

    For the f(x) = cos x, you would need to find:
    f(0) = \cos (0), f'(0) = -\sin (0), f''(0) = -\cos (0) ..

    Then generalize a pattern and that should be your Maclaurin series (Of course, if you wanted to prove it you can do it with induction to verify your series).

    Also, a google search for maclaurin series would probably give you these results.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Rabbit016 View Post
    Find a Maclaurin series expansion for each function.

    f(x) = cos x

    f(x) = 1/(1-x)

    f(x) = x sin x
    Using o_O's above formula we see that

    cos(x)\cos(0)-\sin(0)x-\frac{\cos(0)x^2}{2!}+\frac{\sin(0)x^3}{3!}+\frac{  \cos(0)x^4}{4!}+...= 1-\frac{x^2}{2!}+\frac{x^4}{4!}-...=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}

    this is applicable \forall{x}\in\mathbb{R}
    Because applying the use of the root test for finding intervals of convergence we get

    \lim_{n\to\infty}\bigg|\frac{x^{2n}}{(2n)!}\bigg|^  {\frac{1}{n}}=0<1

    Therefore we run into an identity verifying that cos(x) is analytic across the reals

    for

    \frac{1}{1-x}

    Either recognize this as the converse of the sum of an infinite geometric seires or notice that the temrs are given by

    \frac{1}{1-x}=1+x+x^2+x^3+...=\sum_{n=0}^{\infty}x^n

    Using the Root test again we get

    \lim_{n\to\infty}|x^n|^{\frac{1}{n}}=\lim_{n\to\in  fty}|x|=|x|<1\Rightarrow{-1<x<1}

    Now seeing that at x=1 and x=-1 we have a divergent series this function is analytic only on |x|<1


    For the third I will leave it up to you to show the termiwise derivation but

    \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}

    \therefore{x\cdot\sin(x)=x\cdot\sum_{n=0}^{\infty}  \frac{(-1)^nx^{2n+1}}{(2n+1)!}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+2}}{(2n+1)!}}

    and applying the ratio or root test again we see this is also analytic across the reals
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