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Math Help - Formal power series & Taylor series

  1. #1
    Member Last_Singularity's Avatar
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    Formal power series & Taylor series

    I am honestly stuck on these problems - the subjects are neither covered in our textbook nor did I find anything after about an hour of googling:

    Question 1: Using manipulations with formal power series, find the Taylor series of the following functions at x=0
    (a) \frac{1}{1+x+x^2}
    (b) sin^{-1}(x)
    (c) tanh(x)
    (d) tan(x)

    Question 2: Apply Taylor's formula with Lagrange remainder to estimate the following number with given accuracy (a) cube root of 124 within 0.01 (b) pi within 0.001

    Thanks a lot!
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Last_Singularity View Post
    I am honestly stuck on these problems - the subjects are neither covered in our textbook nor did I find anything after about an hour of googling:

    Question 1: Using manipulations with formal power series, find the Taylor series of the following functions at x=0
    (a) \frac{1}{1+x+x^2}
    Well... 1+x+x^2=1-[-x(x+1)] and you should know the power series for \frac{1}{1-x}

    (d) tan(x)
    Let \tan(x)=\sum_{n \geq 0} a_n x^n
    Differentiate :
    \frac{1}{1+x^2}=\sum_{n \geq 1} na_n x^{n-1}
    Change the indices to get x^n in the summand :
    \frac{1}{1+x^2}=\sum_{n \geq 0} (n+1)a_{n+1} x^n
    And again, note that \frac{1}{1+x^2}=\frac{1}{1-(-x^2)} and identify the coefficients a_n (or a_{n+1}, that doesn't make much difference)



    I really don't know for the other ones... nor am I sure for the methods of these two
    I was just giving it a try
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  3. #3
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    Quote Originally Posted by Last_Singularity View Post
    Question 1: Using manipulations with formal power series, find the Taylor series of the following functions at x=0
    (a) \frac{1}{1+x+x^2}
    (b) sin^{-1}(x)
    (c) tanh(x)
    (d) tan(x)
    For (a), I would write it as \frac{1}{1+x+x^2} = \frac{1-x}{1-x^3} = (1-x)(1-x^3)^{-1}. Now use the binomial series (1-t)^{-1} = 1+t+t^2+t^3+\ldots to get (1-x)(1-x^3)^{-1} = (1-x)(1+x^3+x^6+x^9+\ldots) = 1-x+x^3-x^4+x^6-x^7+\ldots.

    For (b), notice that \tfrac d{dx}(\sin^{-1}x) = (1-x^2)^{-1/2}. Again use a binomial series, this time for (1-t)^{-1/2}, to get the series for (1-x^2)^{-1/2}. Then integrate it to get the series for \sin^{-1}x.

    The series for tan(x) and tanh(x) are both messy. The general term involves Bernoulli numbers (see here). You can get the first few terms by using a method that Moo suggests in the previous comment,namely \tan x = \frac{\sin x}{\cos x} = (x - \tfrac{x^3}{3!} = \ldots)(1 - (\tfrac{x^2}{2!} - \ldots))^{-1} and use the power series for (1-t)^{-1}.

    Quote Originally Posted by Last_Singularity View Post
    Question 2: Apply Taylor's formula with Lagrange remainder to estimate the following number with given accuracy (a) cube root of 124 within 0.01 (b) pi within 0.001
    For the first one, put x=1/125 in the Taylor series for (1-x)^{1/3}.
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