Use the Mean Value Theorem to prove that if p > 1 then $\displaystyle (1+x)^p$ $\displaystyle >$ $\displaystyle 1+px$ for

$\displaystyle x$ $\displaystyle \in$ (−1, 0) U (0,$\displaystyle \infty$).

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- Feb 17th 2011, 04:26 AMmaximus101How do you Use the Mean Value Theorem to prove this
Use the Mean Value Theorem to prove that if p > 1 then $\displaystyle (1+x)^p$ $\displaystyle >$ $\displaystyle 1+px$ for

$\displaystyle x$ $\displaystyle \in$ (−1, 0) U (0,$\displaystyle \infty$). - Feb 17th 2011, 06:52 AMSudharaka
Dear maximus101,

Using the Taylor's theorem you could write,

$\displaystyle (1+x)^p=1+px+p(p-1)(1+c)^{p-2}x^2~\text{where}~0<c<x~or~x<c<0$------(A)

$\displaystyle p>1\Rightarrow{p(p-1)>0}$-----(1)

If $\displaystyle 0<c<x~then~1<1+c$-----(2)

If $\displaystyle x<c<0~then~{1+x<1+c<1}$----(3)

$\displaystyle x>-1\Rightarrow{1+x>0}$------(4)

By, (3) and (4); $\displaystyle If~x<c<0\Rightarrow{0<1+c<1}$----(5)

By (2) and (5); for both cases, $\displaystyle 0<c<x~and~x<c<0\Rightarrow{0<1+c}$------(6)

Therefore, by (1) and (6); $\displaystyle p(p-1)(1+c)^{p-2}x^2>0~if~x\neq{0}~and~x>-1$-----(6)

By (A);

$\displaystyle (1+x)^p-1-px=p(p-1)(1+c)^{p-2}x^2>0~if~x\neq{0}~and~x>-1$

$\displaystyle (1+x)^p>1+px~if~x>-1~and~x\neq{0}$ - Feb 19th 2011, 05:24 AMmaximus101
Hey, thank you for this it was very helpful could you explain how I could use the mean value theorem to prove it?

- Feb 19th 2011, 05:50 AMSudharaka
There are several mean value theorems; Rolle's mean value theorem, Cauchy's mean value theorem and the Taylor's mean value theorem. Please refer Mean value theorem - Wikipedia, the free encyclopedia. I have used the Taylor's mean value theorem in the above answer.

- Feb 19th 2011, 06:08 AMmaximus101
- Feb 19th 2011, 06:12 AMSudharaka
- Feb 19th 2011, 06:34 AMmaximus101
ok I worked it out :) thank you