Proving that $ 1-u = e^{-u - \,u^2/2 - \,u^3/3 -...}$

Question

How can one see that for $-1 < u < 1$ we have the following equality $$ 1-u = e^{-u - \,u^2/2 - \,u^3/3 -...} \,\,\,\,?$$

It's probably easy to prove, however I've tried a couple of things so far (e.g. somehow using the series expansion of exp) but have failed.

The exponent for $e$ is the series expansion for $\log(1-u)$, then note the functions are inverses. — Adam Hughes, Jul 12 '14 at 08:29

score 6 · Accepted Answer · edited Jul 12 '14 at 09:01

6

this holds because for $-1\lt u \lt 1$, $-u-u^2/2-...=\ln(1-u)$ thus

$$e^{-u-u^2/2-...}=e^{\ln(1-u)}=1-u$$

edited Jul 12 '14 at 09:01

answered Jul 12 '14 at 08:30

Ellya

11,783

Great thanks, I wasn't aware that the exponent was the series expansion of log(1-u). – rehband Jul 12 '14 at 08:34
no worries, happy to help :) – Ellya Jul 12 '14 at 08:35

score 4 · Answer 2 · answered Jul 12 '14 at 08:28

4

For $|x|<1,$ using infinite Geometric Series,

$$(1-x)^{-1}=1+x+x^2+\cdots$$

Integrate either sides to find $$-\ln(1-x)=\sum_{r=1}^\infty\frac{x^r}r$$

Now $$\ln y=z\implies y=e^z$$

If $z$ is finite real, real $y>0$

answered Jul 12 '14 at 08:28

lab bhattacharjee

274,582

score 1 · Answer 3 · 2014-07-12T09:15:57.390

1

Compute the Taylor expansion of the RHS: $$\begin{align} f(u)&=e^{-u-u^2/2-u^3/3-...},\\ f'(u)&=(-1-u-u^2-...)\ f(u)=\frac{f(u)}{u-1}\text{ (geometric series)},\\ f''(u)&=\frac{f'(u)(u-1)-f(u)}{(u-1)^2}=0\text{ (!)} \end{align}$$ Hence $$f(0)=1, f'(0)=-1, f''(0)=f'''(0)=...=0\implies f(u)=1-u.$$

edited Jul 12 '14 at 09:15

answered Jul 12 '14 at 08:59

Does not require knowledge of the development of $\ln$. – Jul 12 '14 at 09:13

Proving that $ 1-u = e^{-u - \,u^2/2 - \,u^3/3 -...}$

3 Answers3