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Taylor series

  1. Aug 27, 2009 #1
    Is it correct that a taylor series does not exist for f(x)=tanh(x)/x and f(x)=ln(1+x)/x. I differentiated to f'''(x) and fn(0) and all equal zero.
     
  2. jcsd
  3. Aug 27, 2009 #2
    Keep in mind that as long as a function is infinitely differentiable at a point x = a, there exists a Taylor series for the function about that point. Whether the series converges to the function away from that point is a different matter. Since [itex]\lim_{x \rightarrow 0} \frac{\tanh (x)}{x} = 1[/itex], I strongly suspect there is a Maclaurin series for tanh(x)/x. Similarly for ln(1+x)/x.

    Using binomial series for 1/(1 + x), integration and division by x, I get

    [tex]\frac{\ln (1+x)}{x}=\sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n+1}[/tex]

    with interval of convergence (-1, 1].

    tanh(x) has no nice series and is not the integral or derivative of a series that is nice, so that one is sticky.

    However, by brute force

    [tex]\frac{\tanh (x)}{x}= 1 - \frac{x^2}{3} + \frac{2x^4}{15} - \frac{17x^6}{315} + O(x^8)[/tex]

    --Elucidus
     
    Last edited: Aug 27, 2009
  4. Aug 28, 2009 #3
    THANK YOU VERY MUCH Elucidus!!!!:cool:
     
  5. Aug 28, 2009 #4
    For hyperbolic tangent here is what math world gives:

    Inline26.gif

    http://mathworld.wolfram.com/HyperbolicTangent.html

    I'm not sure what the [tex]B_{n+1}[/tex] means. I would try and solve it by taking the product of two terms. On for the numerator and and one for the inverse of the denominator. For the inverse of the denominator I'd plug in each of the taylors series terms for an exponential function into the taylor series for 1/x and use the multinomial theorem.
     
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