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If $0<x<$ln $2$ then $x^2\geq e^x-x-1$

I got this problem while reading a proof. Tried to prove it but failed. $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$. So $e^x\geq 1+x$ for all $x$ but how can I involve $x^2$ here? Can anyone please give me a hint.

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  • $\begingroup$ hint: use the standard (Mean Value Theorem) expression for the remainder in Taylor's Theorem. As $e^x$ is always increasing you know that the largest value it gets to on any interval is at the right hand endpoint. $\endgroup$ – lulu Jul 11 '15 at 13:53
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    $\begingroup$ Note: I don't believe you need the lower bound on x. $\endgroup$ – lulu Jul 11 '15 at 13:53
  • $\begingroup$ The statement $x^2>e^x-x-1$ is correct for $x\in ]-\infty,\ln(2)]$ but the proof is not, because you can deduce that $e^x>1+x$ only if $x$ is positive ( so that $\frac{x^2}{2!}+\frac{x^3}{3!}+...\geq0$ but you can not do it when $x$ is negative because you have an $x^3$ there which is negative. Now how you can continue consider the function $f(x)=e^x-x-1-x^2$ and study it's variations, compute $f'$ and find the variations of $f'$ and then find the sign of $f'$ which gives you the variations of $f$ and finally the sign of $f$ (you can even replace $ln(2)$ by a larger constant). $\endgroup$ – Elaqqad Jul 11 '15 at 13:55
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Consider $f(x)=x^2-e^x+x+1$. Note that $f(0)=0$ and $f'(x)=2x-e^x+1$ also satisfies $f'(0)=0$. Moreover, $f''(x)=2-e^x\geq 0$ for $x\in[0,\log(2)]$. All this implies $f'(x)\geq 0$ for $x\in[0,\log(2)]$. The claim follows.

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  • $\begingroup$ the statement is correct even when $x$ is negative $\endgroup$ – Elaqqad Jul 11 '15 at 13:57
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Just another way, by integration $$x<\ln 2 \implies e^x<2\implies \int_0^xe^t \,dt < \int_0^x 2 \, dt \implies e^x-1 < 2 x \implies \int_0^x(e^t-1)\,dt < \int_0^12t\,dt \implies e^x-x-1< x^2$$

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  • $\begingroup$ Ah, but you've brought the lower limit back into play. If, say, x = -2 it is certainly not true that $e^x - 1 < 2x$ $\endgroup$ – lulu Jul 11 '15 at 14:37
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    $\begingroup$ @lulu You are confusing between the OPs question and your tangential discussion on the questions validity for negatives. The above proof is perfectly valid for the domain in the OP's question. $\endgroup$ – Macavity Jul 11 '15 at 15:03
  • $\begingroup$ well...just reading what you wrote, you say that x < ln2 $\Rightarrow$ $e^x - 1$ < 2x which is false as written. If you add the condition x ≤ 0 to your proof then I agree your argument is correct. $\endgroup$ – lulu Jul 11 '15 at 15:06
  • $\begingroup$ @lulu Seems you are still obsessed with negatives, I assume you meant I should add $x>0$ as a condition to start. I think that it's given in the question $0<x<2$, and as such obviously not needing repetition in a concise answer. $\endgroup$ – Macavity Jul 11 '15 at 15:12
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We have to prove that over $I=(0,\log 2)$ we have $e^x\leq 1+x+x^2$.

Since $e^x$ is a convex function, $$ \forall z\in I,\qquad e^z \leq 1+\frac{z}{\log 2} $$ and by integrating both sides over $[0,x]$,

$$ \forall x\in I, \qquad e^{x} \leq 1+x+\frac{x^2}{2\log 2} $$

follows. The last inequality is stronger than the inequality we had to prove, since $2\log 2>1$, or $e<4$.

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As there seems to be some uncertainty regarding the lower bound, I'll write out the proof via the standard formula for the remainder. Suppose x $\in$ (-$\infty$,ln(2)). Then Taylor's Theorem with Remainder tells us that there exists some real number c $\in$(-$\infty$,x] such that $$e^x \;= 1 + x + \frac{e^c}{2}x^2$$ Clearly c < ln(2) so $e^c$ < 2 hence $e^x$ < 1+x + $x^2$. The claim follows. Note that at no point did any lower bound on x appear.

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We can prove the stronger

$$0\le x\le1 \implies x^2\geq e^x-x-1$$

indeed let consider $f(x)=e^x-1-x-x^2\le0,\, x\in[0,1]$ then

  • $f(0)=0$
  • $g(x)=f'(x)=e^x-1-2x \le 0$

indeed

  • $g(0)=0$
  • $g(1)=-2$
  • $g'(x)=e^x-2=0 \implies x=\ln 2$ which is a minimum

therefore the inequality holds.

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