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When $x\geq0$ prove that: $$\ln(1+x)\geq x-\frac{x^2}{2}.$$

My effort: From the Taylor series:

$$\ln(1+x)= x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$

I don't know how to continue from here.

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    $\begingroup$ Hint: use Taylor’s theorem with the remainder $\endgroup$ Sep 5 '17 at 17:03
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    $\begingroup$ If the remainder term is negative on the interval, then the sum of the zeroth through $n^{th}$ order terms yields an upper bound. $\endgroup$
    – Albert
    Sep 5 '17 at 17:06
  • $\begingroup$ you have a mistake in the Taylor development. $\endgroup$
    – user48672
    Sep 5 '17 at 17:09
  • $\begingroup$ @user48672 fixed it. Does it deserve a vote up now? $\endgroup$
    – Harrison
    Sep 5 '17 at 17:11
  • $\begingroup$ @ThomasAndrews for $x\geq 0$ man. Thanks a lot. $\endgroup$
    – Harrison
    Sep 5 '17 at 17:12
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Let $f(x)=\ln(1+x)-x+\frac{x^2}{2}$.

Thus, $$f'(x)=\frac{1}{x+1}-1+x=\frac{x^2}{1+x}\geq0.$$

Thus, $f(x)\geq f(0)=0$ and we are done!

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Observe that $$\ln(1+x) = \int_0^{x} \frac{1}{1+t}dt.$$ But, for $t \ge 0$, $$\frac{1}{1+t} \ge 1-t.$$ Therefore, $$\ln(1+x) \ge \int_{0}^x(1-t)dt=x-\frac{x^2}{2}.$$

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  • $\begingroup$ I was looking for an integral proof, but missed this. :) $\endgroup$ Sep 5 '17 at 17:34
  • $\begingroup$ $\frac{1}{1+t}\ge 1-t$, where $t\in\mathbb R$, is true if and only if $t>-1$. But in this case we only need $t\ge 0$ because the definite integral is over $t\in [0,x]$. $\endgroup$
    – user263326
    Sep 5 '17 at 19:13
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You want to show $$1+x\geq e^{x-x^2/2}$$

or $e^{x^2/2}(1+x)\geq e^{x}.$

The right side is $\sum_{k=0}^{\infty} \frac{x^k}{k!}$ and the left side is:

$$(1+x)\sum_{j=0}^{\infty}\frac{x^{2j}}{2^jj!}=\sum_{k=0}^{\infty}\frac{x^{k}}{2^{\lfloor k/2\rfloor}\lfloor k/2\rfloor!}$$

But $2^{\lfloor k/2\rfloor}\lfloor k/2\rfloor!\leq k!$.

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Hint

$$f(x)=\ln(1+x)-x+\frac{x^2}{2}$$

so,

$$f'(x)=\frac{1}{1+x}+x-1=\frac{x^2}{x+1}>0 \text{ for } x> -1$$

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You want to prove $\ln(1+x)\ge x-\frac{x^2}{2}$ for all $x\ge 0$.

To prove it for $0\le x<1$, we'll use Taylor series. It converges because $-1<x\le 1$. See, e.g., here for more information. Or here.

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$

$\ln(1+x)\ge x-\frac{x^2}{2}$ is equivalent to $\frac{x^3}{3}-\frac{x^4}{4}+\cdots\ge 0$.

We'll prove for all $k\ge 3$, $k\in\mathbb Z$, $\frac{x^k}{k}-\frac{x^{k+1}}{k+1}\ge 0$. We could prove this using derivatives, but I'll use a simpler method here. The inequality is equivalent to

$\frac{x^k(k+1-kx)}{k(k+1)}\ge 0$

and we have

$\frac{k+1}{k}>1> x$, $x^k\ge 0$ because $x\ge 0$.

Using derivatives:

Proof: let $f(x)=(k+1)x^k-kx^{k+1}$.

Then $f'(x)=k(k+1)x^{k-1}(1-x)\ge 0$.

$f(x)\ge f(0)=0$.

Also notice that $\left(\frac{x^3}{3}-\frac{x^4}{4}\right)+\frac{x^5}{5}\ge 0$, where also $\frac{x^5}{5}\ge 0$, $\frac{x^{2t+1}}{2t+1}\ge 0$ for all $t\ge 2$, $t\in\mathbb Z$.

Therefore, if $0\le x<1$, then $\frac{x^3}{3}-\frac{x^4}{4}+\cdots \ge 0$, i.e. $\ln(1+x)\ge x-\frac{x^2}{2}$.

By the way, notice that when finding the sum $\frac{x^3}{3}-\frac{x^4}{4}+\cdots$, you have to find the limit of partial sums so that you have to sum the terms in the exact same order $\frac{x^3}{3}$, $\frac{x^3}{3}-\frac{x^4}{4}$, $\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}$, etc. because see the Riemann series theorem: summing the terms in a different order could result in a different number or diverge if the series is conditionally convergent. But in this case, $\frac{x^3}{3}-\frac{x^4}{4}+\cdots$ is not conditionally convergent when $-1<x<1$, because if $-1<x<1$, then it converges to $\ln(1+x)-x+\frac{x^2}{2}$ and $\frac{x^3}{3}+\frac{x^4}{4}+\cdots$ converges to $-\ln(1-x)-x-\frac{x^2}{2}$.

Here's a proof for $x\ge 1$. If $x>1$, then the Taylor series diverges, so we can't use it. $$x-\frac{x^2}{2}=\frac{2x-x^2}{2}=\frac{1}{2}-\frac{(x-1)^2}{2}\le $$

$$\le \frac{1}{2}<\ln 2=\ln(1+1)\le \ln(1+x)$$

because $1=\ln e<\ln 4=\ln 2^2$, i.e. $1<2\ln 2$, because $e<4$ and $\ln x$ is strictly increasing when $x>0$.

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There is an easy way to prove. Note that $$ (1+t)(1-t)=1-t^2\le 1 $$ and hence $$ \frac{1}{1+t}\ge1-t. $$ Integrating from $0$ to $x$, one has $$ \int_0^x\frac{1}{1+t}dt\ge\int_0^x(1-t)dt $$ which gives $$ \ln(1+x)\ge x-\frac12x^2. $$

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  • $\begingroup$ $\frac{1}{1+t}\ge 1-t$ is true if and only if $t>-1$, but we only need $t\ge 0$. Also, this exact same proof has already been posted as an answer here. $\endgroup$
    – user263326
    Sep 6 '17 at 13:22
  • $\begingroup$ @user263326, I did not realize that someone already posted. $\endgroup$
    – xpaul
    Sep 6 '17 at 13:46

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