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I am trying to prove that $e^x>1+x^2$ for any $x>0$ for my homework assignment.

However I have run into trouble doing this. I was trying to probe that $\ln {{e}^{x}}>\ln (1+{{x}^{2}})$ is true for $x>0$ and then that would mean that $e^x>1+x^2$ is true because $\ln x$ is a monotone rising function.

However I have come to the following conclusion$$\frac{{{x}^{2}}}{1+{{x}^{2}}}\le \ln (1+{{x}^{2}})\le {{x}^{2}}$$

which means $x\le \frac{{{x}^{2}}}{1+{{x}^{2}}}$ must be true. but it is not.

I am wondering where I made a mistake here - Or perhaps where I made many mistakes?

Maybe there is a much better why to solve this question also?

Thanks a lot :)

Martin Sleziak
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Jason
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  • How do you deduce $x\leq\frac{x^2}{1+x^2}$ from $\frac{x^2}{1+x^2}\leq \ln (1+x^2)\leq x^2$? – Davide Giraudo Jan 25 '12 at 09:59
  • Ah my bad I made a mistake and confused $x>\ln (1+{{x}^{2}})$ with $x<\ln (1+{{x}^{2}})$ - But that still leaves me with $x\le x^2$ no? – Jason Jan 25 '12 at 10:02
  • Think about the derivatives. – Chris Taylor Jan 25 '12 at 10:08
  • What is your definition of $e^x$? – lhf Jan 25 '12 at 10:12
  • @lhf $ln e = 1$ – Jason Jan 25 '12 at 10:17
  • Ok, $\ln e =1 $ defines $e$. How do you define $\ln(x)$ and $a^x$ then? I ask for definitions because if you had defined $e^x$ to be a series then the answer to your question would be immediate. – lhf Jan 25 '12 at 10:18
  • @ChrisTaylor and prove that $e^x$ derivative is always larger? I tried that for a bit but didn't manage very well maybe I missed somthing – Jason Jan 25 '12 at 10:18
  • This is a 'classic' case of using calculus to prove some inequality. What do want to differentiate? – Adam Jan 25 '12 at 10:28
  • Is it possible to use induction on unit intervals (0,1], (1,2],...? If it can be done, is there a good example where it works nicely? – Justahomelessguy Jan 25 '12 at 11:06
  • http://mathoverflow.net/questions/38238/a-principle-of-mathematical-induction-for-partially-ordered-sets-with-infima – Adam Jan 25 '12 at 11:34

6 Answers6

8

Here's how I would proceed. Consider the function $f(x) = e^x - x^2 -1$. Its first derivative is $f'(x) = e^x - 2x$. Let's see for which $x \in [0, +\infty)$ $f'(x) > 0 \ $: differentiate $f$ once more and you obtain $f''(x) = e^x -2 \ > 0 \Leftrightarrow x > \ln(2)$. In other words $x = \ln(2) \ $ is a minimum for $f' \ $, which means that $f'(x) \ge f'(\ln(2)) = 2 - 2\ln(2) > 0 \ $ because $\ln(2) < 1 \ $. Thus $f$ is a strictly monotone increasing function, which yields $f(x) > f(0) = 0 \ $ for all $x \in (0, +\infty) $.

Did
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Emilio Ferrucci
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5

One of many approaches (given you have tagged this as calculus):

For $0 \lt x \le 1$ compare $1+x^2$ with $1+x+x^2/2! + x^3/3! + \cdots$, noting $x \ge x^2$ in this interval

For $1 \lt x$ note $1+1^2 \lt e^1$ and compare $\frac{d}{dx} (1+x^2) = 2x$ with $\frac{d}{dx} e^x = e^x$, and if necessary note $2\times 1 \lt e^1$ and compare $\frac{d^2}{dx^2} (1+x^2) = 2$ with $\frac{d^2}{dx^2} e^x = e^x$.

Henry
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5

Using the series definition of $e^x$, we only have to prove that $1 + x + \dfrac{x^2}2 + \dfrac{x^3}6 \ge 1+x^2$ for all $x\ge0$. This is equivalent to $x^2-3x+6 \ge 0$, which is true since the discriminant is negative.

lhf
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2

By the AM-GM inequality $$ \frac{2t}{1+t^2} < 1 $$ for any $t\in\mathbb{R}^+\setminus\{1\}$, hence by integrating both sides over $[0,x]$ we have $$ \log(1+x^2) < x $$ as wanted.

Jack D'Aurizio
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Define the function $f(x)=e^x-x^2-1$
$f(0)=0$ Prove that the derivative is always positive then the function increases and if at zero is zero then after is positive. (to prove the derivative is positive you can derive it and see that it has a minimum positive.

Martin Sleziak
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alpha.Debi
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You need the elementary inequality for $x>0$

$$\frac{x}{x+1} \leq \log (1+x) \leq x $$

This yields

$$\frac{x^2}{x^2+1} \leq \log (1+x^2) \leq x^2 $$

which is what you want.

Pedro
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