Find the series for $e^{-x}$ (By differentiation of $e^x$)

Question

Given that $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$

Find the series for $e^{-x}$

Note: derive the series for $e^{-x}$ showing your own working.

Trying to work out some questions from my textbook, I don't even know where to start. Help

"Derive" is different from "differentiation". To "derive" the series for $e^{-x}$ one could substitute $-x$ in for $x$ in the series for $e^x$. — Dave, Apr 16 '18 at 05:35

score 0 · Answer 1 · answered Apr 16 '18 at 05:29

0

Hint: Substitute $-x$ instead of $x$.

answered Apr 16 '18 at 05:29

Botond

11,938

score 0 · Accepted Answer · answered Apr 16 '18 at 05:37

0

One thing that will help you out a lot:

If $f(x)=e^x$, and $f(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}...$

Then $f(-x)=e^{-x}$. How does that affect the power series?

answered Apr 16 '18 at 05:37

Saketh Malyala

13,637

So, does this look right to you?
$f(-x)=e^x$$=1-x-\frac{x^2}{2!}-\frac{x^3}{3!}-\frac{x^4}{4!}...$
– uhhhhhhhhhh_ Apr 16 '18 at 06:44
1

@ElliottHawkins no. $(-x)^2=x^2$, and so on. So the $-$ and $+$ will alternate. – Botond Apr 16 '18 at 06:57
@Botond ohhh, right got it. Thankyou! – uhhhhhhhhhh_ Apr 16 '18 at 07:12

Find the series for $e^{-x}$ (By differentiation of $e^x$)

2 Answers2