Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove or disprove:

For every $n\in\mathbb N$ there exist $x_n\in (10^n,+\infty)$ such that $f(x_n)<10^{-n}$ where $f(x)=(x \sin x - \cos^2 x)^2 x^6 + x^2 \cos^8 x.$

share|improve this question
Thus - if this HW, please tag it as such. You should add what you've tried as well. –  nbubis Mar 22 '13 at 21:08
No, this is much harder than homework, although it maybe doesn´t look hard. –  user67878 Mar 22 '13 at 21:13
You still haven't stated where the problem is from, or what you've tried. –  nbubis Mar 22 '13 at 21:20
What could I try, I do not know of a method that could solve this. Well, I did change the variable x to 1/x and plotted it so that I see how it behaves at infinity by seeing how it behaves at zero with this change of variables. wolframalpha.com/input/… –  user67878 Mar 22 '13 at 21:27
I think that this is so hard that maybe I would offer a bounty of 500 if I had enough reputation. –  user67878 Mar 22 '13 at 21:31

1 Answer 1

up vote 1 down vote accepted

We answer the question somewhat reluctantly, since there is undoubtedly a typo.

Let $n=3$, for no good reason. We show that there is no $x_n \gt 1000$ at which our function is $\lt 1/1000$.

For note from $\cos^2 x+\sin^2 x=1$ that one of $\cos^2 x$ or $\sin^2 x$ is $\ge 1/2$.

If $\cos^2 x\ge 1/2$, then our expression is greater than $(1000)^2(1/2)^4$. This is because the second term is greater than $(1000)^2(1/2)^4$, and the first term is non-negative.

If $\sin^2 x \ge 1/2$, then our expression is $\ge (500\sqrt{2}-1)^2(1000)^6$.

So for all $x\gt 1000$, our function is quite large, much larger than $1/1000$.

share|improve this answer
1)If $cos^2x≥1/2$, then our expression is greater than $(1000)^2(1/2)^4$ I agree with this. 2) If $sin^2x≥1/2$, then our expression is $≥(5002√2−1)^2(1000)^6$ I do not agree with this. I do not agree because if $sin^2x≥1/2$ then $sinx$ can be negative and $cosx$ can be arbitrarily small and because of that the expression $(x \sin x - \cos^2 x)^2 x^6$ could be close to zero and, simultaneously, $x^2 \cos^8 x$ could be close to zero, so the proof is not fine. –  user67878 Mar 22 '13 at 22:02
Or the proof is fine but my line of reasoning is not. –  user67878 Mar 22 '13 at 22:13
But if you are correct then I need an explanation of this, I changed the variable from x to 1/x to see how f(x) behaves at infinity by seeing how f(1/x) behaves at zero, and I plotted and got this: wolframalpha.com/input/… Is the process of this change of variable allowed and if it is why do I get this picture? I am really interested why? –  user67878 Mar 22 '13 at 22:18
Now the proof looks fine to me, but my line of reasoning not. –  user67878 Mar 22 '13 at 22:27
@Thus: if $\sin x$ is negative it is less than $\frac 1{\sqrt 2}$. The factor $x$ makes it dominate over the $\cos x$ term, then it gets squared and is greater than 500. It is still large. –  Ross Millikan Mar 22 '13 at 22:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.