# proof $e^{x}<1+x+x^{2}$ (mean value theorem preferred)

How it can be shown that:

$$e^{x}<1+x+x^{2}$$ For all $$x<0.5$$ I tried to use mean value theorem, but I have some problem, any idea or hint if highly appreciated. Clearly $$e^x$$ is continuous over $$\left[x,x+1\right]$$ and differentiable over $$\left(x,x+1\right)$$, hence there exist $$c∈\left(x,x+1\right)$$, such that: $$\frac{f\left(x+1\right)-f\left(x\right)}{x+1-x}=f^{'}\left(c\right)$$ Hence; $$e^{\left(x+1\right)}-e^{\left(x\right)}=e^c$$ or $$\ln\left(e^{\left(x+1\right)}-e^{\left(x\right)}\right)=c$$ but this is not helpful.

• What are your assumptions about $x$? The strong inequality is wrong when $x=0$. – Christian Blatter Dec 5 '19 at 14:43
• $e^x = \sum_{k=2}^{\infty}\frac{x^k}{k!}+x+1< \sum_{k=2}^{\infty}\frac{x^2}{k!}+x+1 = x^2(e^1 - 2)+x+1)<x^2+x+1$ – fGDu94 Dec 5 '19 at 14:43

This is not a complete answer, but I thought it provides a simple method to find some values of x.

That expression is not true for all $$x<0.5$$ . Try $$0$$, and you obtain $$1>1$$ . For $$x<-0.5$$ the inequality holds, because we can write $$e^x -1 < x(x+1)$$ and if we differentiate we obtain $$e^x$$ and $$2x+1$$. Then, for all $$x<-1/2$$, the derivative of $$e^x -1$$ is positive while the one for $$x(x+1)$$ is negative. Since $$e^{(-1/2)} - 1 < 1+1/2+(1/2)^2$$ then the inequality holds for $$x< - 0.5$$.

Actually, it is true for $$x \in ]-∞;0[ \; \cup \; ]0;1.79328[$$ , but I'm not sure how to prove it analytically, neither of where the $$1.79328$$ comes from (aproximated value). I hope my method is close enough to what you wanted.

• the inequality is equivalent to $e^x-x^2-x-1<0$. So the $1.79328$ is one of the roots of $e^x-x^2-x-1$ – Zacharias Zarowski Dec 5 '19 at 16:42
• Yeah, I knew, I just don't know a algebraic expression for it. But one could express it that way, yes. – RicardoMM Dec 5 '19 at 17:55
• Ok I'm also not sure how to find a closed form :) but in this case its enough since its greater than $0.5$ – Zacharias Zarowski Dec 5 '19 at 18:31

Let us consider the function $$g(x) = e^x - 1 - 2x$$ We have that $$g(0)=0$$. Moreover, since $$g(1)=e-3<0$$ while $$g(2)=e^2-5>0$$, by the theorem of zeros, there is a value $$1 such that $$g(a)=0$$. Since the function $$f(x)=e^x$$ is strictly convex, it must be $$e^x<1+2x$$ or each $$0. Therefore $$\int\limits_0^x {e^t dt} < \int\limits_0^x {\left( {1 + 2t} \right)dt}$$ for each $$0. It means that $$e^x - 1 < x + x^2$$ for such values of $$x$$. This prove the inequality for $$0. Now, let be $$x<0$$. Then, again by convexity, it is $$e^t>1+t$$ for each $$t\in \mathbb R$$, $$t\neq 0$$. Therefore $$\int\limits_x^0 {e^t dt} > \int\limits_x^0 {\left( {1 + t} \right)dt}$$ so that $$1 - e^x > - x - \frac{{x^2 }} {2}$$ thus $$1 + x + \frac{{x^2 }} {2} > e^x$$ and, a fortiori, the inequality follows.

You can also prove that $$e^t <1+2t$$ for each $$0< t< 1/2$$ also by means of mean value theorem. Whith $$0 < t <1/2$$ you have that $$\frac{{e^t - e^0 }} {{t - 0}} = e^c < e^{\frac{1} {2}} < 2$$ thus $$e^t< 1+2t$$ if $$0 < t < 1/2$$.

Two partial integrations give the following version of Taylor's theorem: \eqalign{e^x&=1+\int_0^x 1\cdot e^t\>dt=1+(t-x)e^t\biggr|_{t=0}^{t=x}-\int_0^x(t-x)e^t\>dt\cr &= 1+x-{(t-x)^2\over2}e^t\biggr|_{t=0}^{t=x}+\int_0^x{(t-x)^2\over2}e^t\>dt\cr &=1+x+{x^2\over2}+\int_0^x{(t-x)^2\over2}e^t\>dt\ .\cr} We therefore have to show that $$\int_0^x{(t-x)^2\over2}e^t\>dt\leq{x^2\over2}\qquad(-\infty When $$x\leq0$$ this is obviously true. When $$0 we have $$\int_0^x{(t-x)^2\over2}e^t\>dt\leq{e\over2}\int_0^x\tau^2\>d\tau={e\over2}{x^2\over3}<{x^2\over2}\ .$$