Determine the Taylor expansion around h = 0 for $\cosh(i\frac{\pi}{2}+ h)$

With regards to the question above:

I calculate that the expansion is as follows: $i(x + \frac{x^3}{3!} + \frac{x^5}{5!}\ldots)$

But I am not sure that can be correct as am confused as to why the expansion is completely imaginary.

Surely we can write $\cosh(i\frac{\pi}{2})$ as the real number $\cos(\frac{\pi}{2})$ which has no imaginary part so i am confused why the expansion is imaginary...

  • $\begingroup$ Your expansion is not correct. Check it over. $\endgroup$ – Lubin May 4 '17 at 18:32
  • $\begingroup$ Use expansion of $e^x$ in $\cosh x=\frac12(e^{x}+e^{-x})$. $\endgroup$ – Nosrati May 4 '17 at 18:38
  • $\begingroup$ But if you differentiate the function and plug into it zero we notice a pattern of 0 i 0 i 0 i 0 i and then you plug them into the general expression of the macluarin expansion you get what i get. Where did i go wrong? $\endgroup$ – David Abraham May 4 '17 at 18:40
  • $\begingroup$ differentiate $\cosh(i\frac{\pi}{2})$.? It's a number. $\endgroup$ – Nosrati May 4 '17 at 18:45
  • $\begingroup$ no cosh(iπ/2 + h) $\endgroup$ – David Abraham May 4 '17 at 18:49

Your answer is totally correct ...

$$\begin{eqnarray*} \cosh(i\frac\pi 2+x)&=&\cosh(i\frac\pi 2)\cosh(x)+\sinh(i\frac\pi 2)\sinh(x) \\&=& \cos(\frac\pi 2)\cosh(x)+i\sin(\frac\pi 2)sinh(x) \\&=&i\sinh(x) \end{eqnarray*}$$ So it is no surprise that your answer is purely imaginary ( because $ \cos(\frac\pi 2)=0$)


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