Using mean value theorem to show that $\cos (x)>1-x^2/2$

Question

I have a question, by applying the mean value theorem to $f(x)=\frac{x^2}{2}+\cos (x)$, on the interval $[0,x]$, show that $\cos (x)>1-\frac{x^2}{2}$.

We know that $\frac{\text{df}(x)}{\text{dx}}=x-\sin (x)$, for $x>0$. By the MVT, if $x>0$, then $f(x)-f(0)=(x+0) f'(c)$ for some $c>0$.

This is where I get confused:

so, $f(x)>f(0)=1$, but why? Is it my lack of inequality that is showing, or what am I missing? Is $f'(x)\cdot x=1$ or what is going on?

The sign of $f'(c)$ gives the result (and the approach works for $x\lt0$ as well). — Did, Aug 05 '14 at 09:47
@Did Im not sure if I got it? Why exactly equal to 1? and is it the positive f′(c)* positive x = positive __ — ALEXANDER, Aug 05 '14 at 09:57
@ALEXANDER please look at my answer below? Do you understand now how you can prove the inequality? — Varun Iyer, Aug 05 '14 at 10:16
It seems the question the OP really means to ask is why $f(0)=1$... — Did, Aug 05 '14 at 10:44
@Did No that is not what my question is, or at least thats not what I think my question is. Look here, I do see that f(0)=1, but as I stated where I get confused: f(x)>f(0)=1, now since f(0) is 1, then f(x)>1=1, but its the one on the right hand equation that I do not understand where is coming from? why is it equal to 1? and how come? — ALEXANDER, Aug 05 '14 at 10:50
The assertion that $f(x)\gt f(0)=1$ is the conjunction of $f(x)\gt f(0)$ and of $f(0)=1$. If $f(x)\gt f(0)=1$ is not clear to you, either $f(x)\gt f(0)$ is not clear to you or $f(0)=1$ is not clear to you. Pick up your choice. (But why are we trying to elucidate this if you already accepted an answer?) — Did, Aug 05 '14 at 14:15

score 9 · Accepted Answer · edited Aug 05 '14 at 11:09

$ f(x)=x^2/2+\cos(x)$

Note that $f(0)=0^2/2+1=1$ From your equation: $$f(x)-f(0)=(x)f'(c)=x(c-\sin c)$$

Let $g(x)=x-\sin x$ Again you can show that $g'(x)=1-\cos x$ which is always greater than $0$ due to bounded nature of $\cos x$.As $g(0)=0$ and it is an increasing function $\{g'(x)>0\;\forall x>0$}, thus $g(x)>0 \;\forall x>0$.

So $f(x)-f(0)>0\;\forall x>0$ as $x>0$ and $c-\sin c >0\;\forall c>0${as $0<c<x$}.

So $f(x)>f(0)=1$

Varun Iyer · Answer 2 · 2014-08-05T10:32:16.017

3

You started off well.

Notice that, by MVT:

$$f'(c) = \frac{f(x) - f(0)}{x - 0}$$ S0

$$xf'(c) = f(x) - f(0)$$

Notice that x is positive, and since $$f'(x) = x - sin(x)$$

Also, note that $x > \sin(x)$, so $f'(x) > 0$

Therefore,

We can conclude that

$$f(x) > f(0)$$

And

$$\cos(x) > 1- \frac{x^2}{2}$$

edited Aug 05 '14 at 10:32

answered Aug 05 '14 at 10:15

Varun Iyer

6,074

Why not factor out the constant in the derivative – ALEXANDER Aug 05 '14 at 10:16
1

@ALEXANDER which constant are you referring to, $c$? That's simply a value of x that makes this equality true for the MVT. You cannot take it out – Varun Iyer Aug 05 '14 at 10:17
@ALEXANDER all you show is that $f'(x) > 0$, so you can that $f(x) - f(0) > 0$, so that $f(x) > f(0)$ – Varun Iyer Aug 05 '14 at 10:19
You are getting the derivative of x^2/2 to become 2x, but should it not be just x, when you are factoring out the constant – ALEXANDER Aug 05 '14 at 10:19
That I can see, but I do not get where the number 1 is coming from. – ALEXANDER Aug 05 '14 at 10:22
@ALEXANDER oh sorry I fixed the derivative. The $1$ comes from $f(0)$, its $1$ – Varun Iyer Aug 05 '14 at 10:23
$\frac{d\left(\frac{x^2}{2}+\cos (x)\right)}{dx}$=x-Sin[x], or am I wrong? – ALEXANDER Aug 05 '14 at 10:26
@ALEXANDER yes you are correc – Varun Iyer Aug 05 '14 at 10:30
Okey well then your update did not go through – ALEXANDER Aug 05 '14 at 10:31
@ALEXANDER I fixed it now. Do you understand what I did? That's the important part. – Varun Iyer Aug 05 '14 at 10:34
My main problem is not what you have stated, that I get, what I do not get is the part in my question which is f(x)>f(0)=1, not the part f(x)>than f(0) that I understand, i just do not get that it is set equal to 1 – ALEXANDER Aug 05 '14 at 10:39
Your function is $f(x) = \frac{x^2}{2} + \cos x$. $f(0) = 0 + \cos(0) = 1$ – Varun Iyer Aug 05 '14 at 10:57

Using mean value theorem to show that $\cos (x)>1-x^2/2$

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