Question

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

Answer 1

`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(z)=1/2(e^z+e^-z)` and
`cos(z)=1/(2i)(e^(iz)+e^(-iz))`

so

`cosh(z)cos(z)=1/(4i)(e^((1+i)z)+e^((1-i)z))+1/(4i)(e^((-1+i)z)+e^((-1-i)z))`

but

`1+i= sqrt(2)e^(iphi)` with `phi=arctan1=pi/4`

and

`-1+i=sqrt(2)e^(i(pi-phi))`

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(a+b)+cos(a-b)=2cos(a)cos(b)`

with `a = k(pi/2)` and `b=-k 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))`