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It appears to be well-known that $\tanh(x)\le \mathrm{erf}(x)$ on $[0,\infty)$. It's off-handedly mentioned here, for example. Where can I find a formal proof? On the one hand, it's hard to imagine that a "classic" like this wouldn't have been proven already. On the other hand, the Taylor expansions are somewhat involved (tanh involves Bernoulli numbers) and unfortunately, the inequality does not hold termwise in the expansions -- so it's certainly far from obvious.

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Not an answer but related: – Michael Renardy Dec 31 '12 at 15:38
Tastes vary, but this strikes me as a nice example of a fact for which I would much prefer an informal proof to a formal one. It's obvious from the definitions of the two functions that the inequality must hold for large x. Graphing would establish it for small x. – Ben Crowell Dec 31 '12 at 22:02
If proofs by plotting were sufficient, it would have saved us a lot of work here... – Aryeh Kontorovich Jan 1 '13 at 6:58
up vote 17 down vote accepted

Let $f(x)={\mathrm{erf}}(x)-\tanh(x)$. It can be easily seen from Taylor series at $0$ and from asymptotics at $\infty$ that $f(x)>0$ for small $x$ and for large $x$.

Let us prove that $f(x)>0$ by contradiction. Suppose that $f(x)$ is negative for some $x$, then $f'$ must have at least $3$ positive zeros, by Rolle's theorem. This means that the equation $$g(x):=e^{-x^2}(e^{2x}+2+e^{-2x})=2\sqrt{\pi}$$ has at least $3$ positive solutions. But this is not the case because the LHS is monotone. Indeed, differentiating $g$, dividing by $e^{-x^2}$ and replacing $2x$ with $y$ we obtain $$g'(x)=\sinh(y)-y\cosh(y)-y<0,$$ because $\sinh(y) < y \cosh(y)$ as you can see from their Taylor series.

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Beautiful, thank you! – Aryeh Kontorovich Dec 31 '12 at 17:19
Sorry, I can only see that, on your supposition, $f'$ has $2$ (or more) positive zeros. But anyway that's enough for the rest of the argument to work. – John Bentin Dec 31 '12 at 21:20
John, if you have a function that tends to zero from the positive side on both ends of the interval then derivative has ODD number of zeros. Only one if it positive everywhere, and at least 3 if not. – Alexandre Eremenko Dec 31 '12 at 23:36
Yes. But the function $f$ here doesn't tend to zero at both ends of the interval $[0, \infty)$, because $f(0)=\mathrm{erf}\, 0-\mathrm{tanh}\,0=\frac{1}{2}-0=\frac{1}{2}$. – John Bentin Jan 1 '13 at 8:30
Depends how you define erf. If you take the definition given here, which appears to be standard, then erf(0)=0. – Aryeh Kontorovich Jan 1 '13 at 8:38

First, $$\begin{align} 1-\mathrm{erf}(x) &= \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}dt, \cr 1-\tanh(x) &= \int_x^\infty \mathrm{sech}^2 t\;dt . \end{align}$$ Subtract: $$ \mathrm{erf}(x)-\mathrm{tanh}(x) = \int_x^\infty \left(\mathrm{sech}^2 t - \frac{2}{\sqrt{\pi}}e^{-t^2}\right)dt $$ So it suffices to show that this integrand is positive. It is positive for $t>1$ (proof needed), so we establish $\mathrm{erf}(x) > \mathrm{tanh}(x)$ for $x > 1$.

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