# How to find the Taylor series for the function cosh(z)*cos(z) about the point 0 ?

Apr 12, 2017

cosh(z)cos(z)=sum_(k=0)^oo((cos(k pi/2)cos(k(pi/4))(sqrt(2))^k)/(k!))z^k

#### Explanation:

We know that

$\cosh \left(z\right) = \frac{1}{2} \left({e}^{z} + {e}^{-} z\right)$ and
$\cos \left(z\right) = \frac{1}{2 i} \left({e}^{i z} + {e}^{- i z}\right)$

so

$\cosh \left(z\right) \cos \left(z\right) = \frac{1}{4 i} \left({e}^{\left(1 + i\right) z} + {e}^{\left(1 - i\right) z}\right) + \frac{1}{4 i} \left({e}^{\left(- 1 + i\right) z} + {e}^{\left(- 1 - i\right) z}\right)$

but

$1 + i = \sqrt{2} {e}^{i \phi}$ with $\phi = \arctan 1 = \frac{\pi}{4}$

and

$- 1 + i = \sqrt{2} {e}^{i \left(\pi - \phi\right)}$

so

1/(4i)(e^((1+i)z)+e^((1-i)z))=1/(4i)sum_(k=0)^oo(sqrt 2 z)^k/(k!)(e^(i kphi)+e^(-i kphi))=
=1/2sum_(k=0)^oo(sqrt 2 z)^k/(k!)cos(k pi/4)

analogously we have

1/(4i)(e^((-1+i)z)+e^((-1-i)z))=1/2sum_(k=0)^oo(sqrt 2 z)^k/(k!)cos(k(3pi/4))

and finally

cosh(z)cos(z)=1/2sum_(k=0)^oo(sqrt 2 z)^k/(k!)(cos(k pi/4)+cos(k((3pi)/4)))

NOTE:

Using the identity

$\cos \left(a + b\right) + \cos \left(a - b\right) = 2 \cos \left(a\right) \cos \left(b\right)$

with $a = k \left(\frac{\pi}{2}\right)$ and $b = - k \frac{\pi}{4}$ we have also

cosh(z)cos(z)=sum_(k=0)^oo(sqrt 2 z)^k/(k!)cos(k pi/2)cos(k(pi/4))