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How do I prove that $$e^x \leq x + e^{x^2}$$ for all $x\in\mathbb R$?

My probability book (Grimmett and Stirzaker) says that it's a simple exercise but I don't see it. For $x\leq 0$, we have $$e^x = \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + x + \sum_{k=1}^\infty \frac{x^{2k+1}}{(2k+1)!} \leq \sum_{k=0}^\infty \frac{x^{2k}}{k!} + x = e^{x^2} + x.$$ How do I show it for $x>0$?

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  • $\begingroup$ Specifically what are you confused about here? Are you confused about how they do it for $x\le0$, or do you understand that and want to know how to proceed for the case where $x > 0$? $\endgroup$ – Decaf-Math Sep 28 '18 at 22:21
  • $\begingroup$ @Decaf-Math I want to know how to proceed for $x>0$ $\endgroup$ – Alain Sep 28 '18 at 22:22
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    $\begingroup$ also easy for $x \geq 1$ $\endgroup$ – Will Jagy Sep 28 '18 at 22:24
  • $\begingroup$ A different approach is to consider the function that is the difference between the left hand side and the right hand side of your inequality. This is equal at x equals 0 and then you can just compute the derivative to complete the proof. $\endgroup$ – ericf Sep 28 '18 at 22:30
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    $\begingroup$ @ericf I thought about that too. Say $f(x) = x + e^{x^2} - e^x$. Then $f'(x) = 1 + 2x e^{x^2} - e^x$. How to show that $f'(x)>0$ for $x>0$? $\endgroup$ – Alain Sep 28 '18 at 22:34
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How about this:

For $x\geq 0$, we have $$ x e^{-x}+e^{x^2-x}\geq x(1-x)+1+x^2-x=1$$ where we used the inequality $e^{u} \geq 1+u$ for all $u$. Multiplying on both sides by $e^x$, we find $$x+e^{x^2}\geq e^x$$ For $x<0$, note that $e^{x^2}\geq 1+x^2$, then $$ x+e^{x^2}\geq1+x+x^2=3/4+(x+1/2)^2>0 $$ Consequently, we also have $$ \left(x+e^{x^2}\right)e^{-x}\geq(x+1+x^2)(1-x)=1-x^3, $$ which implies that $$ x+e^{x^2}\geq(1-x^3)e^x>e^x $$

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  • $\begingroup$ Thanks for pointing that out. I think corrected this problem now. $\endgroup$ – minmax Sep 30 '18 at 21:33
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The function $$f(x):=e^{x^2}-e^x$$ has $f(0)=0$, $f'(0)=-1$ and $$f''(x)=(4x^2+2)e^{x^2}-e^x\ .$$ Obviously $f''(x)>0$ when $x\leq0$ or $x\geq1$. For $0\leq x\leq1$ note that $$4x^2+2=\left(2x-{1\over2}\right)^2+{7\over4}+2x>1+(e-1)x\geq e^x\ ,$$ since $e-1<2$ and $\exp$ is convex. It follows that $f$ is convex, and this implies that the line $y=-x$ supports the graph of $f$ at $(0,0)$, i.e. $$-x\leq e^{x^2}-e^x\qquad(-\infty<x<\infty)\ .$$

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  • $\begingroup$ nice use of convexity! should have thought about it when considering $0<x<1$. $\endgroup$ – Alain Sep 30 '18 at 16:43
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The case $x\ge1$ is trivial. For $0<x<1$, we have $$ \frac{e^x-e^{x^2}}{x-x^2}\le e^x\le\frac1{1-x}. $$ The first inequality is true because the LHS, by mean value theorem, is $e^a$ for some $a\in(x^2,x)$. For the second one, note that $f(x):=(1-x)e^x\le1=f(0^+)$ because $f$ is decreasing ($f'(x)=-xe^x<0$) on $(0,1)$.

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  • $\begingroup$ why is LHS equal to $e^a$? $\endgroup$ – Alain Sep 30 '18 at 16:44
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    $\begingroup$ @Alain Mean value theorem. $\endgroup$ – user1551 Sep 30 '18 at 17:55
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We have that

  • for $x\ge1$

$$x\le x^2 \implies e^x\le e^{x^2}\le x+ e^{x^2}$$

  • for $0<x<1$

$$e^{x^2}+x\ge 1+x+x^2 \stackrel{?}\ge e^x =1+x+\frac12x^2+\frac16x^3+\ldots$$

and

$$1+x+x^2 \stackrel{?}\ge 1+x+\frac12x^2+\frac16x^3+\ldots \iff \frac12x^2 \stackrel{?}\ge \frac16x^3+\frac1{24}x^4+\ldots \iff \frac12 \stackrel{?}\ge \frac16x+\frac1{24}x^2+\ldots$$

then to prove $e^x\le x+ e^{x^2}$ it suffices to show that the latter holds for $x=1$ that is

$$\frac12 \ge \frac16+\frac1{24}+\ldots \ge \frac16x+\frac1{24}x^2+\ldots$$

$$\iff 1 \ge \frac13+\frac1{12}+\frac1{60}\ldots=2\sum_{k=3}^\infty \frac{1}{k!}=2e-5$$

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  • $\begingroup$ The first $\iff$ should be $\impliedby$ $\endgroup$ – ℋolo Sep 29 '18 at 2:07
  • $\begingroup$ @Holo Yes of course, I fix that typo. Thanks $\endgroup$ – user Sep 29 '18 at 3:06

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