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Please expand the following at z=0 with steps, if possible: [e^z + e^-z] / 2

Many thanks.

Guest Jun 19, 2017

Best Answer 

 #1
avatar+26839 
+1

ez = 1 + z + z2/2! + z3 /3! + z4/4! + ...

 

e-z = 1 - z + z2/2 - z3/3! + z4/4! - ...

 

So (ez + e-z)/2 = 1 + z2/2! + z4/4! + ...

 

.

Alan  Jun 19, 2017
 #1
avatar+26839 
+1
Best Answer

ez = 1 + z + z2/2! + z3 /3! + z4/4! + ...

 

e-z = 1 - z + z2/2 - z3/3! + z4/4! - ...

 

So (ez + e-z)/2 = 1 + z2/2! + z4/4! + ...

 

.

Alan  Jun 19, 2017
 #2
avatar
0

Thank you very much, Alan.

Guest Jun 19, 2017

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