How is the taylor series for #e^z# derived?

1 Answer
Oct 9, 2015

See explanation...

Explanation:

Here's one way of thinking about it.

Let #e(z) = sum_(n=0)^oo z^n/(n!)#

(i.e. the Taylor series for #e^z#, but pretend we don't know that yet).

Then we find:

#e(0) = 1#

#e(1) = sum_(n=0)^oo 1/(n!) = e#

#d/(dz) e(z) = sum_(n=0)^oo n*(z^(n-1))/(n!) =sum_(n=1)^oo (z^(n-1)/((n-1)!))#

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

Compare with #e^0 = 1#, #e^1 = e# and #d/(dz) e^z = e^z#

Essentially this #e(z)# is doing a very good impression of #e^z#

More formally:

Let #f(z) = e^z#

Then using #d/(dz) e^z = e^z# and #e^0 = 1#, the Taylor expansion at #0# is:

#sum_(n=0)^oo (f^((n))(0) z^n/(n!))=sum_(n=0)^oo e^0 z^n/(n!)=sum_(n=0)^oo z^n/(n!)#