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Math Help - Taylor series formula

  1. #1
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    Taylor series formula

    I need some help with figuring out the Taylor series formula.

    I have two equations:
    f(x)=sin(x/2) at c=0
    f(x)=ln(2x) at c=2

    I need to find the equation for the Taylor series formula for each of them. So for the first one, I worked out the Taylor polynomial to be x/2-x^3/48.... I know n=1 from that. But how do I write the rest of the formula? Since it's Maclaurin, I know the formula is [f^(n)(0)(x)^n]/n!, but what numbers do I plug in?

    The second is ln(4)+((x-2)/2)-(((x-2)^2)/8)+(((x-2)^3)/24)... So n=0, and the other formula applies: [f^(n)(c)(x-c)^n]/n!, but I have the problem with the numbers again.. I need this to get the interval of convergence via the Ratio test.

    I would appreciate your help
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Pandora View Post
    I need some help with figuring out the Taylor series formula.

    I have two equations:
    f(x)=sin(x/2) at c=0
    f(x)=ln(2x) at c=2

    I need to find the equation for the Taylor series formula for each of them. So for the first one, I worked out the Taylor polynomial to be x/2-x^3/48.... I know n=1 from that. But how do I write the rest of the formula? Since it's Maclaurin, I know the formula is [f^(n)(0)(x)^n]/n!, but what numbers do I plug in?

    The second is ln(4)+((x-2)/2)-(((x-2)^2)/8)+(((x-2)^3)/24)... So n=0, and the other formula applies: [f^(n)(c)(x-c)^n]/n!, but I have the problem with the numbers again.. I need this to get the interval of convergence via the Ratio test.

    I would appreciate your help
    If you know certain Taylor series already, for example \sin(x)=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1} and \ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}x^n, then you can get the Taylor series of other functions very easily, like this:

    \sin\left(\frac{x}{2}\right)=\sum_{n=0}^\infty\fra  c{(-1)^n}{(2n+1)!}\left(\frac{x}{2}\right)^{2n+1}=\sum  _{n=0}^\infty\frac{(-1)^n}{2^{2n+1}(2n+1)!}x^n

    and

    \ln(2x)=\ln\big(2(x-2+2)\big)=\ln\left(4\left(1+\frac{x-2}{2}\right)\right)=\ln(4)+\sum_{n=1}^\infty\frac{  (-1)^{n-1}}{n}\left(\frac{x-2}{2}\right)^n
    =\ln(4)+\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n 2^n}(x-2)^n
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