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Prove that $e^x > 1 + (1 + x)\log(1 + x), x > 0$ using power series expansion.

I am a bit puzzled by this statement because the power series for $\log(1+x)$ only converges iff $|x|<1$. Is the problem sound? Should it be $1>x>0$?

5 Answers5

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hint

Let $$f(x)=e^x-1-(x+1)\ln(1+x)$$

For $ x>0 $, $ f $ is twice differentiable at $ [0,x] $, thus by Taylor-Lagrange formula, there exists $ c\in (0,x) $ such that

$$f(x)=f(0)+xf'(0)+\frac{x^2}{2}f''(c)$$

$$=\frac{x^2}{2}\frac{(1+c)e^c-1}{1+c}$$

hamam_Abdallah
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  • I take it from your answer that expanding the terms on the left and the right up to sufficient degrees is a dubious approach? –  Aug 31 '20 at 22:00
  • @crystal_math If you expand both LHS and RHS terms, you will get $c_1$ and $c_2$ which make thins complcate. – hamam_Abdallah Aug 31 '20 at 22:08
  • But my question is how can that work for $x>1$ since $\log(1+x)$'s power series only converges for $|x|<1$? That's my only question on the post –  Aug 31 '20 at 22:11
  • @crystal_math We don't use pwer series until infinity, we just need second order expansion. – hamam_Abdallah Aug 31 '20 at 22:13
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We can make an estimate for $1+(1+x)\log(1+x)$ that does not rely on the power series for $\log(1+x)$, and instead use the power series for $e^x$, which is valid for $x\ge0$.

Since $\frac1{1+t}\le1$ on $[0,x]$, $$ \begin{align} \log(1+x) &=\int_0^x\frac{\mathrm{d}t}{1+t}\tag1\\ &\le x\tag2 \end{align} $$ Furthermore, $$ \begin{align} (1+x)\log(1+x)-x &=x\cdot\frac1x\int_0^x\log(1+t)\,\mathrm{d}t\tag3\\ &\le x\log\left(1+\frac1x\int_0^xt\,\mathrm{d}t\right)\tag4\\[3pt] &=x\log(1+x/2)\tag5\\[3pt] &\le\frac{x^2}2\tag6 \end{align} $$ Explanation:
$(4)$: since $\log(1+t)$ is concave, apply Jensen's Inequality
$(6)$: apply $(2)$

Therefore, by comparing to the Taylor series for $e^x$, $$ \begin{align} 1+(1+x)\log(1+x) &\le1+x+\frac{x^2}2\tag7\\ &\le e^x\tag8 \end{align} $$ Explanation:
$(7)$: apply $(6)$
$(8)$: $e^x=1+x+\frac{x^2}2+\text{positive terms}$

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robjohn
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There are multiple Taylor expansions of a function, you have chosen the expansion about the origin. Consider that any Taylor expansion of $\log(1+x)$ will have the same radius of convergence of $\frac{1}{1+x}$. Well $$\frac{1}{1 + x} = \frac{1}{1+c -(c-x)} = \Big(\frac{1}{1+c}\Big) \frac{1}{1 -(\frac{c-x}{1+c})} = \frac{1}{1+c}\sum_{n=0}^\infty\Big(\frac{c-x}{1+c}\Big)^n $$ The series above converges as long as $|x-c| < |1+c|$ and $c\not=-1$. In effect this means that you can find an expansion anywhere as long as it is away from $c=-1$. In particular you have that $$ \log(1 + x) = \log(1+c) + \frac{1}{1+c}\sum_{n=0}^\infty \frac{1}{n+1}\Big(\frac{c-x}{1+c}\Big)^{n+1} $$ Since your asks about any $x>0$, you can use this approach by first fixing an $x$ and choosing an expansion to meet your needs.

Andrew Shedlock
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Let $$ f(t)=e^t-1, g(t)=(t+1)\log(t+1). $$ Just simply usig Cauchy's MVT, one has, for $x>0$ $$ \frac{f(x)-f(0)}{g(x)-g(0)}=\frac{f'(c)}{g'(c)} $$ where $c\in(0,x)$. Hence $$ \frac{e^x-1}{(x+1)\log(x+1)}=\frac{e^c}{\log(c+1)+1}. \tag1$$ Again using Cauchy's MVT, one has $$ \frac{e^c-1}{\log(1+c)}=(c_1+1)e^{c_1}>1, c_1\in(0,c)$$ which gives $$ e^c>\log(c+1)+1. $$ So one has $$ e^x>1+(x+1)\log(x+1). $$

xpaul
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For $0<x<1$, does the following proof work?

$e^x>1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}$ and $\log(1+x)<x-\dfrac{x^2}{2}+\dfrac{x^3}{3}$ and so,

$$1+(1+x)\log(1+x)<1+x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{3}$$

We observe that $$1+x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{3}<1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}$$

iff

$x^3>x^4$ which is true for $x\in (0,1)$.