- #1
Alem2000
- 117
- 0
I was looking at my math text and I saw that [tex]e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}[/tex] and it sugests that you prove it to yourself. So this is what I don't get if [tex] f(x)=e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}[/tex] then [tex]\frac{d}{dx}f(x)[/tex] should equal that..now I am not qeustioning that its true I am just having a hard time proving it to myself this is what I did
[tex]\frac{d}{dx}(e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!})=\sum_{n=0}^{\infty}\frac{nx^n-1}{n!}[/tex] and after that I just set them equal. Now I will warn you I am horrible at factorials so I didnt simplify and this is what turns out [tex]\sum_{n=0}^{\infty}\frac{nx^n-1}{n!}=\sum_{n=0}^{\infty}\frac{x^n}{n!}[/tex] now can somone tell me what I need to correct or simplify?
[tex]\frac{d}{dx}(e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!})=\sum_{n=0}^{\infty}\frac{nx^n-1}{n!}[/tex] and after that I just set them equal. Now I will warn you I am horrible at factorials so I didnt simplify and this is what turns out [tex]\sum_{n=0}^{\infty}\frac{nx^n-1}{n!}=\sum_{n=0}^{\infty}\frac{x^n}{n!}[/tex] now can somone tell me what I need to correct or simplify?