how do you show that summation from x=0 to x= infinity for
a^x / x! = exp(a)?
thanks!
Hi alexandrabel90,
We start by assuming we can write $\displaystyle e^a$ as a series of powers of a
$\displaystyle e^a=ba^0+ca+da^2+fa^3+ga^4+ha^5+........$
Since $\displaystyle \frac{d}{da}e^a=e^a$
$\displaystyle f(0)=1,\ f'(0)=1,\ f''(0)=1,\ f'''(0)=1,.....$
we successively differentiate to find all the coefficients, while setting a=0
$\displaystyle 1=b, as\ a^0=1$ and all higher powers are zero
$\displaystyle f'(a)=e^a=c+2da+3fa^2+4ga^3+5ha^4$
$\displaystyle f'(0)=1=c$
$\displaystyle f''(a)=e^a=2d+3(2)fa+4(3)ga^2+5(4)ha^3$
$\displaystyle f''(0)=1=2d\ \Rightarrow\ d=\frac{1}{2!}$
Continuing on like this, all the co-efficients are found