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Thread: Power series for tanh?

  1. #1
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    Power series for tanh?

    Hi. How do I find the power series for tanh(z) (in the complex plane, if that matters, which it doesn't).

    I know the power series for sinh(z) and cosh(z) (same as sin(t), cos(t), but without the sign changes):
    $\displaystyle \sinh(z)=\sum_{j=0}^\infty \frac{z^{2j+1}}{2j+1!}$
    $\displaystyle \cosh(z)=\sum_{j=0}^\infty \frac{z^{2j}}{2j!}$

    Now what? You can't divide series (or, at a minimum, I don't know how to divide series). So what's next?

    Thanks,
    -J
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by MSUMathStdnt View Post
    Now what? You can't divide series (or, at a minimum, I don't know how to divide series). So what's next?
    There is a similar algorithm to the eucldian division. Look here:

    Division by increasing power order | Mathematical Garden

    Regards.

    Fernando Revilla
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  3. #3
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    You can use the same derivation as any other Taylor series:

    $\displaystyle \displaystyle f(x) = \sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$.


    You have to evaluate $\displaystyle \displaystyle \tanh{x}$ as well as the first, second, third, etc. derivatives at the point $\displaystyle x = a = 0$.

    You should find that when you substitute the values into your Taylor series formula that you can simplify using Bessel numbers.
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  4. #4
    MHF Contributor chisigma's Avatar
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    An 'a little unconventional' approach: the function $\displaystyle y(x)= \tanh x$ is the solution of the following ODE...

    $\displaystyle \displaystyle y^{'} = 1 - y^{2}, y(0)=1$ (1)

    If You suppose that $\displaystyle \tanh x$ is analytic in $\displaystyle x=0$ [and that is true...] is...

    $\displaystyle \displaystyle \tanh x = \sum_{n=0}^{\infty} a_{n}\ x^{n} = \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n}$ (2)

    All right!... now from (1) and (2) You can derive the $\displaystyle a_{n}$ as follows...

    $\displaystyle \displaystyle y(0)=0 \implies a_{0}=0$

    $\displaystyle \displaystyle y^{'}= 1-y^{2} \implies y^{'}(0)= 1 \implies a_{1}=1$

    $\displaystyle \displaystyle y^{''} = -2 y y^{'} \implies y^{''} (0)=0 \implies a_{2}=0$

    $\displaystyle \displaystyle y^{(3)} = -2 y y^{''} -2 y^{'\ 2} \implies y^{(3)} (0)= -2 \implies a_{3}= -\frac{1}{3}$

    $\displaystyle \displaystyle y^{(4)} = -2 y y^{(3)} - 6 y^{'} y^{''} \implies y^{(4)}=0 \implies a_{4}=0$

    $\displaystyle \displaystyle y^{(5)} = -8 y^{'} y^{(3)} -6 y^{''\ 2} -2 y y^{(4)} \implies y^{(5)}(0)= 16 \implies a_{5} = \frac{2}{15}$

    ... and so one...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  5. #5
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    You can find the power series for $\displaystyle \tanh z$ here. The formula for the coefficient of $\displaystyle z^{2j+1}$ involves Bernoulli numbers.
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