Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do I prove the inequality $e^{-2x}\leq1-x$ for $0\leq x\leq1/2$?

share|cite|improve this question
(+1) for research effort – The Chaz 2.0 Jan 10 '12 at 21:08
@TheChaz: Out of curiosity: where did you see any research effort? – user6701 Jan 11 '12 at 5:34
@Tim: Nowhere. I downvoted this. – The Chaz 2.0 Jan 11 '12 at 14:33
@TheChaz, Sarcasm does not translate well on the internet. It would have been much more helpful and less confusing had you given your actually reason why you downvoted this question. – user4143 Jan 11 '12 at 23:43

For example, $(1-x)(1+2x)=1+x-2x^2\geq 1$ so $$ e^{2x}\geq 1+2x \geq \frac{1}{1-x} $$

share|cite|improve this answer

Let $f(x)=e^{-2x}-(1-x)$. We have:

  • $f''(x)=4e^{-2x}$, which is always positive. Hence, $f$ is convex.
  • $f(0)=0$
  • $f\left(\dfrac{1}{2}\right)=\dfrac{1}{e}-\dfrac{1}{2}\leq0$.

Since $f$ is convex, its curve is always below $0$ on $\left[0,\dfrac{1}{2}\right]$. This proves that $e^{-2x}\leq 1-x$.

share|cite|improve this answer

A slightly ugly but standard beginning calculus approach is to let $$f(x)=1-x-e^{-2x}.$$ We want to show that $f(x)\ge 0$ in the interval $[0,1/2]$.

A first experimental step might be to use software to graph $y=f(x)$ as $x$ ranges over our interval. If we have a high degree of trust in the graphing software, the picture tells us that $f(x)$ is quite likely to be $\ge 0$ in our interval.

Certainty is better. Use the derivative to study the behaviour of $f(x)$. Note that $f(0)=0$, and $$f'(x)=2e^{-2x} -1.$$ The derivative is positive at $x=0$, and is clearly decreasing. It reaches $0$ at $x=(\ln 2)/2\approx 0.3465$. So $f(x)$ is increasing in the interval $[0,(\ln 2)/2]$, and decreasing from $(\ln 2)/2$ on. By $x=1/2$, $f(x)$ is about $0.13212$, and in particular still positive.

Thus $f(x)$ is $\ge 0$ for $0 \le x\le (\ln 2)/2$, and $f(x) >0$ for $(\ln 2)/2 \le x\le 1/2$. It follows that $f(x)\ge 0$ on the whole interval $[0,1/2]$ (and somewhat beyond $1/2$).

Comment: There are far better ways to prove the inequality. But let's stick to calculusy approaches. To cut down on the negatives, note that equivalently we want to show that $e^{2x}(1-x) \ge 1$ on our interval (we multiplied both sides by the positive number $e^{2x}$). Let $g(x)=e^{2x}(1-x)-1$. Then $g(0)=0$. Also, $g'(x)=(1-2x)e^{2x}$, so $g$ is increasing on the interval $[0,1/2]$, and we are finished.

share|cite|improve this answer
Perhaps it's good to give an example where a software plot is misleading: this function has a value $<0$ at $x=2$, yet the plot seems to be positive for all positive $x$. – Actually, I'm a bit surprised Wolfram|Alpha does this, a good CAS certainly should be able to proove that the plotted function is well-behaved between two evaluation points, and only then simply connect them. – leftaroundabout Jan 10 '12 at 23:22
@leftaroundabout: The SE software seems to have mangled your link. I think you mean something like this instead (with ^ encoded as %5E). – Ilmari Karonen Jan 11 '12 at 12:45
@IlmariKaronen yeah that's what I meant. – leftaroundabout Jan 11 '12 at 15:33

By Taylor expansion, we have $$\ln \frac{1}{1-x} = \sum_{n=1}^{\infty} \frac{x^n}{n},$$ whose convergence radius is $R=1$.

The equation above can also be achieved by integrating both sides of $$\frac{1}{1-x} = \sum_{n=0}^{\infty} {x^n} .$$

For $0\leq x\leq 1/2$ we have the following upper bound $$ \sum_{n=1}^{\infty }\frac{x^{n}}{n}\leq \sum_{n=1}^{\infty }\frac{1}{ n\left( 2^{n}\right) }=\ln 2 \leq 2. $$ Therefore $$ -\ln \left( 1-x\right) =\sum_{n=1}^{\infty }\frac{x^{n}}{n}\leq 2x. $$ The given inequality follows.

share|cite|improve this answer

We may start out in an elementary manner with a very known result that comes from W. Sierpiński inequality or $(e^x)$ Taylor expansion, namely:

$$e^x\geq 1+x $$

We get that: $$e^{-x}\leq\frac1{x+1}; \space e^{-2x}\leq\frac1{2x+1}\leq1-x,\space \space \space 0\leq{x}\leq\frac{1}{2}$$

The proof is complete.

share|cite|improve this answer
In fact your proofs works for $0\le x\le 2$. We have (for $x\ge 0$) the following: $\frac1{1+2x}\le 1-x$ $\Leftrightarrow$ $1\le (1+2x)(1-x)=1+x-2x^2$ $\Leftrightarrow$ $0\le x-x^2=x(x-2)$ $\Leftrightarrow$ $x\in[0,2]$. – Martin Sleziak Jun 19 '12 at 12:58
@Martin Sleziak: thanks for your comment! – I'm an artist Jun 19 '12 at 13:03

For $x > 0$, the Taylor series for $e^{-2x}$ is the alternating series $\sum_n {(-2x)^n \over n!} = 1 - 2x + 2x^2 -...$ If $x < {1 \over 2}$, the terms decrease in magnitude. It's a fact about alternating series that in an alternating series with terms of decreasing magnitude, the overall sum is going to be less than what you get if you add finitely many terms, stopping with a positive term. So we have $$e^{-2x} < 1 - 2x + 2x^2$$ But after a little simple algebra, for $x > 0$ the statement $1 - 2x + 2x^2 < 1 - x$ is equivalent to the statement that $x < {1 \over 2}$. So the inequality follows. (The endpoint case $x = {1 \over 2}$ can be gotten by just plugging in the value).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.