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Let $f(x)=x\cos x+\sin x$. Show that for every $M>0$ and every $k\geq 1$, there is $x_0>M$ such that

$f(x)\geq k$, $\forall x\in [x_0,x_0+\frac{1}{k}]$.

If we replace $x$ by $2\pi n$ then

$|f(2\pi n)=|2\pi n|$.

Can this help?

Can I get a help to start it?

Unknown
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  • Your idea seems very relevant! Indeed, consider that $\sin x\ge 0$ for $x\in[2\pi n,2\pi n+1]$ so $f(x)\ge x\cos x$. What else might you be able to say, for that interval. (Also, note that $1/k\le 1$ for all $k\ge 1$.) – yearning4pi Dec 17 '21 at 03:54

2 Answers2

2

Let $x_0=2\pi n$ where $n\in\Bbb Z^+$ is large enough that

$2\pi n(\cos 1)-1\ge k$ and $2\pi n>M.$

If $x\in [x_0,x_0+1/k]$ then $\cos x=\cos (x-2\pi n)\ge \cos 1>0$ so $$f(x)=x\cos x+\sin x\ge 2\pi n\cos x-|\sin x|\ge$$ $$\ge 2\pi n(\cos 1)-|\sin x|\ge$$ $$\ge 2\pi n(\cos 1)-1\ge k.$$

DanielWainfleet
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0

Let $M>0$ (large). Choose $M < x= 2k \pi + l$ , $k \geq 1$ and $l \in (0,\frac{1}{k})$.

We know that

$$\cos{(2k \pi + l)}= \cos{(2k \pi )} \cos{(l)} - \sin{( l)} \sin{(2k \pi)} = \cos(l).$$

and

$$\sin{(2k \pi + l)}= \sin{(2k \pi )} \cos{(l)} - \sin{( l)} \cos{(2k \pi)} = \sin(l).$$

Therefore,

$$f(2 k \pi + l) = (2 k \pi + l) \cos(2 k \pi + l) + \sin(2 k \pi + l)$$

$$f(x)= f(2 k \pi + l) = (2 k \pi + l) \cos(l) + \sin(l) >k\quad x=2k\pi + l \in [x,x+\frac{1}{k}].$$

Mr. Proof
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