series expansion of e^u?

Question

i don't understand the series expansion of e^u :

e^u=1+u+u^2 /2!+u^3/3!+u^4/4!+............

please explain me.

Answer 1

to answer to that question, you only need to understand taylor series. It's a general theorem that uses consecutive derivations of a function to get a good approximation of that function.

I believe a demonstration is provided in the 2nd link.

Answer 2

Mathematically 2 Hence e=1+1/1!+1/2!+1/3!+.............. infinity
e^x=(1+1/1!+1/2!+1/3!............
=>e^x= 1+x/1!+x^2/2!+x^3/3!+x^4/4!...... 4 all real values of x.

Answer 3

One way to see this is via Taylor series. If you let f(u)=e^u, then every derivative of f(u) is the same as f(u), so all derivatives at u=0 are 1. This gives your series. It does not show that the series converges to the right thing, though.

A better way to see this is to note that the series always converges (has infinite radius of convergence). If you let
y=1+u+u^2/2!+...,
then it's easy to take the derivative term-by-term and see that
y'=y.
The only solutions of this differential equation are
y=Ce^u.
By letting u=0 on both sides, we see that C=1, so
y=e^u.

Answer 4

Mathematically 2 Hence e=1+1/1!+1/2!+1/3!+...................upto infinity
e^x=(1+1/1!+1/2!+1/3!..........)^x
=>e^x= 1+x/1!+x^2/2!+x^3/3!+x^4/4!................+x^n/n!......... 4 all real values of x.