3
$\begingroup$

I am trying to show:

$e^x > 1+x+\frac{x^2}{2}$ for $x>0$ using Mean Value Theorem (MVT).

My method is as follow: consider the function $f(x)=e^x - (1+x)$. We know $f'(x) = e^x -1 > 0 $ for $x>0$. Thus, by MVT, exists a $c \in ]0,x[$ such that, $$f'(c)=\frac{f(x)-f(0)}{x-0}=\frac{e^x-(1+x)}{x}$$ Since we know $f'(c) > 0$, the statement is equivalent to, $$ \frac{e^x -(1+x)}{x} > 0 \Rightarrow e^x - 1> x \Leftrightarrow f'(x)>x $$ for $x>0$. Now we can update the original inequality of $f'(c)$, giving $$f'(c)=\frac{e^x-(1+x)}{x} > x \Leftrightarrow e^x > 1+x+x^2 $$ for $x>0$. However, this is not the same as the intended inequality, and is certainly not true, letting $x=1$ yields $e > 3$. What happened here? May someone explain?

Thank you so much!

$\endgroup$
  • $\begingroup$ You can only get $f'(x)>x$, but not $f'(c)>x,\forall c\in]0,x[$. $\endgroup$ – Eclipse Sun Apr 8 '15 at 9:17
4
$\begingroup$

You did wrong in your last step. You proved $f'(x) > x$, but write in the next line $ f'(c) > x$. You have (as $x$ was arbitrary), $f'(c) > c$, giving $$ e^x > 1 + x + cx $$ but this does not help.

$\endgroup$
3
$\begingroup$

Different approach: consider $h(x) = e^x -1 -x -\frac{x^2}{2}$. Clearly $h(0)=0$. Now show that $h'>0 \forall x>0$ thus proving the inequality.

$\endgroup$
1
$\begingroup$

It is universally known that $e^x> 1+x$ for $x> 0$. Thus $$ e^x=\left(e^{x/n}\right)^n> \left(1+\frac{x}{n}\right)^n> 1+n·\frac{x}{n}+\frac{n(n-1)}2·\frac{x^2}{n^2}=1+x+\left(1-\frac1n\right)·\frac{x^2}2. $$ Since this is true for all $n>2$, the requested inequality follows.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.