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

1 Answer
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(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))#