0
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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$)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.