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Use MVT to prove that $1-\cos x<x^2$, $x\neq 0$

$$1-\cos x<x^2\implies 1<x^2+\cos x$$ Let $f(x)=x^2+\cos x$,

Notice that $f(-x)=(-x)^2+\cos(-x)=x^2+\cos(x)=f(x) \implies f$ is symmetric about the y-axis. It suffices to show that $x^2+\cos x>1$ for $x>0$ due to this symmetry.

$x>0 \implies (0,x)$

Since $x^2$ and $\cos x$ are differentiable (and thus continuous) $\forall x\in \mathbb R$, MVT yields that in $(0,x)$, $\exists c \in (0,x)$ such that $$\frac{(x^2+\cos x)-(0^2-\cos 0)}{x-0}=2c-\sin c$$ $$\frac{x^2+\cos x-1}{x}=2c-\sin c$$ $-1\le\sin c\le1 \implies -1\le-\sin c\le1 \implies 2c-1\le 2c-\sin c\le 2c+1$ $$\frac{x^2+\cos x-1}{x}\ge 2c-1$$ And now i'm stuck because i can't get $x(2c-1)>0$

Is my proof wrong somewhere?

UPDATE: Thanks for everyone that gave their input. I managed to solve it using $f(x) = 1-\cos x$ directly. However, i'm still curious as to why my initial method fails? is there something that I overlooked?

Martin Sleziak
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Danxe
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  • You know that $f(0)=1$ so all you need to show is that $f'>0$ – rikhavshah Aug 29 '16 at 05:20
  • @David Oh sorry, there was an error in my expression, just fixed it. Thanks! – Danxe Aug 29 '16 at 06:22
  • Why not use the MVT on $1-\cos x$ directly? The derivative is $\sin x$ hence there exists $u$ in $(0,1)$ such that $1-\cos x=x\sin(ux)$ by the MVT. The MVT again shows that $\sin(ux)=\sin(ux)-0=ux\cos(uvx)$ for some $v$ in $(0,1)$. Now, $|\cos(uvx)|\leqslant1$ hence you are done. – Did Aug 29 '16 at 06:29
  • Stronger inequality $1-\frac{x^2}{2}\le \cos(x)\le1-\frac{x^2}{2}+\frac{x^4}{24}$ is shown here: http://math.stackexchange.com/questions/1513020/prove-the-inequalities-1-fracx22-le-cosx-le1-fracx22-fracx4 – Martin Sleziak Aug 29 '16 at 10:14
  • What does the assertion "$x\gt0\implies(0,x)$" mean? – Barry Cipra Aug 30 '16 at 09:41

2 Answers2

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Easier:

Since $\cos'(x) =-\sin(x) $, $\cos(x)-\cos(0) =-\int_0^x \sin(t) dt \ge -\int_0^x t dt =-\frac{x^2}{2} $ so $1-\cos(x) \le \frac{x^2}{2} < x^2 $ for $x \ne 0$.

To show that $\sin(x) \le x$, start with $\sin'(x) = \cos(x) $. then $\sin(x) =\sin(x)-\sin(0) =\int_0^x \cos(t) dt \le\int_0^x dt =x $ since $\cos(t) \le 1$.

marty cohen
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Your bound is not good enough if you want to proceed this way. One fix is to use a better bound than $-1≤\sin c ≤ 1$. Your bound gives you $2c-\sin c>0$ if $c>1$. For $c≤1$, we know that $$\sin c - 0 = c \cos c'$$ for some $c' ∈ (0,c)$ by MVT. Also note that $|\cos c'|\leq 1$. Therefore $-\sin c ≥ -c$, so $2c - \sin c ≥ 2c - c = c > 0$, which proves the result.

In summary, we made the improved piecewise bound $$ 2c- \sin c ≥ \begin{cases} 2c-1 & c>1 \\ c & 0<c<1 \end{cases} ≥ c > 0$$

In fact, we didnt even need that $c≤1$, just that $c>0$.

Calvin Khor
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