What did I do wrong in my Taylor Series Expansion for y=kcosh(x/k)?

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The discussion centers on the Taylor Series Expansion for the function y=kcosh(x/k). The user initially derived the series but failed to include a critical factor of 1/2 in the definition of the hyperbolic cosine function, cosh(x)=e^x + e^{-x}/2. This omission led to incorrect coefficients in the series expansion. The correct expression for y, when k is much greater than x, simplifies to y= k + (1/2k)(x^2).

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y=kcosh(x/k) = k(e^(x/w) + e^(-x/k))

y ~ k[1 + x/k + (1/2!)(x^2/k^2) + (1/3!)(x^3/k^3)] +...+ k[1 - x/k +(1/2!)(x^2/k^2) - (1/3!)(x^3/k^3) +...]

All odd terms except 1 cancel out.

So we are left with

y = k [2 + (2/2!)(x^2/k^2) + (2/4!)(x^4/k^4) + (2/6!)(x^6/k^6) +...]

I've been staring at this for hours and I'm supposed to show that:

y= k + (1/2k)(x^2) for k>>x


What did I do wrong? My error lies somewhere where the twos are bolded.
 
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you lost a 1/2 in your definition of cosh:

[tex]cosh(x)=\frac{e^x+e^{-x}}{2}[/tex]
 
Well, you obviously are missing a factor of two somewhere... but I suppose you figured that out already. The answer lies here:

[tex]\cosh x = \frac{e^x + e^{-x}}{2}[/tex]
 

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