Approximating erf by tanh - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T19:14:24Z http://mathoverflow.net/feeds/question/117735 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/117735/approximating-erf-by-tanh Approximating erf by tanh Aryeh Kontorovich 2012-12-31T15:18:47Z 2012-12-31T17:57:10Z <p>It appears to be well-known that $\tanh(x)\le \mathrm{erf}(x)$ on $[0,\infty)$. It's off-handedly mentioned <a href="http://web.cs.dal.ca/~jborwein/tanh-sinh.pdf" rel="nofollow">here</a>, 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.</p> http://mathoverflow.net/questions/117735/approximating-erf-by-tanh/117743#117743 Answer by Alexandre Eremenko for Approximating erf by tanh Alexandre Eremenko 2012-12-31T17:17:41Z 2012-12-31T17:17:41Z <p>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 $x$.</p> <p>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 $$g'(x)=\sinh(y)-y\cosh(y)-y&lt;0,$$ because $\sinh(y) &lt; y \cosh(y)$ as you can see from their Taylor series.</p> http://mathoverflow.net/questions/117735/approximating-erf-by-tanh/117745#117745 Answer by Gerald Edgar for Approximating erf by tanh Gerald Edgar 2012-12-31T17:37:42Z 2012-12-31T17:57:10Z <p>First, \begin{align} 1-\mathrm{erf}(x) &amp;= \frac{2}{\sqrt{\pi}}\int_x^\infty e^{-t^2}dt, \cr 1-\tanh(x) &amp;= \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 $x > 1$.</p>