Let \(f(x)\) be a continuous function, whose first and second derivatives are continuous on \([0,2\pi]\) and \(f"(x) \ge 0 \:\: \forall \:x \in [0,2\pi]\). Show that \[\int _0^{2\pi} f(x) \cos x dx \ge 0\]

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Let \(f(x)\) be a continuous function, whose first and second derivatives are continuous on \([0,2\pi]\) and \(f"(x) \ge 0 \:\: \forall \:x \in [0,2\pi]\). Show that \[\int _0^{2\pi} f(x) \cos x dx \ge 0\]

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@myininaya @amistre64 How would you prove it?
@Mr.Math
\[\int_0^{2\pi}f(x)\cos(x)dx=f(x)\sin(x)|_0^{2\pi}-\int_0^{2\pi}f'(x)\sin(x)dx=-\int_0^{2\pi}f'(x)\sin(x)dx\] \[=f'(x)\cos(x)|_0^{2\pi} +\int_0^{2\pi}f''(x)\cos(x)dx=f'(2\pi)-f'(0)+\int_0^{2\pi}f''(x)\cos(x)dx.\] We can see here that \(f'(2\pi)-f'(0)\ge 0\), since \(f'(x)\) is an increasing function. So we're left to show that \(\int_0^{2\pi}f''(x)\cos(x)dx\ge 0\) knowing that \(f''(x)\ge 0\).

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Let \(g(x)=\int f''(x)\cos(x)dx\). The question here is how can we show that \(g(2\pi)\ge g(0)\)? We know that \(g(x)\) is increasing on \((0,\frac{\pi}{2})∪(\frac{3\pi}{2},2\pi)\), and decreasing on \((\frac{\pi}{2}, \frac{3\pi}{2}).\)
f"(x) must be constant, right?
Why?
It doesn't have to be constant.
mr.math do you mean g is positive on those intervals...and g is negative on that interval...
I meant \(g'\) is positive on those interval and negative on that interval.
I know, I just wanted it to be constant I mean not any obvious reason, just like that.
Note that \(g'(x)=f''(x)\cos(x)\), where \(f''(x)\ge 0\). So the sign of \(g'(x)\) (that determines whether g is increasing or decreasing) can be determined by \(\cos(x)\).
I didn't defined it as a constant.
define*
I don't understand why you want \(f''(x)\) to be a constant. If it was then the solution would be obvious since g(2pi)=g(0), and the integral will be 0.
Am I making sense Ishaan?
Of course, you're... it's me, who didn't
This problem is not difficult, but I'm missing something. What am I missing?!
Do you think I should ping Zarkon?
I think I have it. One minute.
You can ping him though, but he should not answer until I finish :P
If I'm not mistaken,\[- \int_0^{2\pi} f'(x) \sin x dx = f'(x) \cos x|_0^{2\pi} - \int _0^{2\pi} f''(x) \cos xdx\]
Correct! I just noticed that :)
So we have to show that \(f'(2\pi)+g(0)\ge g(2\pi)+f'(0).\)
I'm getting tired. @Zarkon should come from his planet now.
Zarkon's offline :(... @satellite73 @eseidl
i am thinking parts
oh i should pay attention. looks like mr.math did parts right?
Yes.
yeah
I think I have it this time. By the squeeze theorem and property of inequality in integrals we have: \[-\int_0^{2\pi}f''(x) dx\le \int_0^{2\pi}f''(x)\cos(x)dx\le \int_0^{2\pi}f''(x)dx.\] Thus \(g(2\pi)-g(0)\le f'(2\pi)-f'(0)\) as required.
that was nice. i was trying to come up with a counter example, and example of a positive function where this integral would be negative. can't seem to do it.
still not sure why i cannot though. why can't it be postive, but larger on the interval \[[\frac{\pi}{2},\frac{3\pi}{2}]\] then on the rest? wouldn't that make the integral negative?
After this proof, I am sure you can't find such an example :P @Ishaan, I will go for few minutes and be back to summarize my solution in one post, so it can be read and understood easily.
i believe you, and understand the first part. but it looks like the second part is hinging on the fact that if \[f''>0\] then \[\int f''(x)\cos(x)dx\geq 0\] and i am not sure exactly why that has to be
It is a given in the problem
Well, what I proved is that a continuous function \(f(x)\) that has a positive second derivative has the property: \[\int_0^{2\pi} f(x)\cos(x)dx\ge 0.\]
ooooooooh ok i see i misinterpreted the last step
@satellite I first said that, but I don't need that statement anymore (I don't know if it's true or not). Read the last two posts.
yes, sorry i see what you wrote, i simply read it wrong.
We can use integration by parts and write \[\int_0^{2\pi}f(x)\cos(x)dx=f(x)\sin(x)|_0^{2\pi}-\int_0^{2\pi}f'(x)\sin(x)dx=-\int_0^{2\pi}f'(x)\sin(x)dx\] \[=f'(x)\cos(x)|_0^{2\pi} +\int_0^{2\pi}f''(x)\cos(x)dx=f'(2\pi)-f'(0)-\int_0^{2\pi}f''(x)\cos(x)dx.\] From here it's sufficient to show that \(f'(2\pi)-f'(0)\ge\int_0^{2\pi}f''(x)\cos(x)dx.\) We know that \(\cos(x)\le 1 \implies f''(x)\cos(x)\le f''(x)\). So \[\int_0^{2\pi}f''(x)\cos(x)dx \le \int_0^{2\pi}f''(x)dx=f'(2\pi)-f'(0). ■\]

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