How to find the sum of the series #sum_"n=0"^oo(z^(4n)/((4n)!))# when |z| < 1 ?

Question

2224 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-12T08:10:55

#sum_(k=0)^oo z^(4k)/(4k!)= (cosh(z)+cosh(i z) )/2#

We have

#e^z = sum_(k=0)^oo z^k/(k!)# and

#e^-z = sum_(k=0)^oo (-1)^kz^k/(k!)#

so

#e^z+e^-z=2sum_(k=0)^oo z^(2k)/(2k!)#

or

#cosh(z)=sum_(k=0)^oo z^(2k)/(2k!)#

now doing

#cosh(z)+cosh(i z) = sum_(k=0)^oo z^(2k)/(2k!)+sum_(k=0)^oo (-1)^kz^(2k)/(2k!) = 2sum_(k=0)^oo z^(4k)/(4k!)#

Finally

#sum_(k=0)^oo z^(4k)/(4k!)= (cosh(z)+cosh(i z) )/2#

or also

#sum_(k=0)^oo z^(4k)/(4k!)= (cosh(z)+cos(z) )/2#