f"(x)>0, show that int(f(x)cosx dx) >0

Art of Problem Solving

AoPS Online

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

f"(x)>0, show that int(f(x)cosx dx) >0

G H J

Reveal topic tags

analytic geometry

L

G H BBookmark kLocked kLocked NReply

Source: ISI(BS) 2009 #2

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2130 posts

Sayan

#1 h May 5, 2012, 9:23 AM • 2 Y

Y by Adventure10, Mango247

Let $f(x)$ be a continuous function, whose first and second derivatives are continuous on $[0,2\pi]$ and $f''(x) \geq 0$ for all $x$ in $[0,2\pi]$ . Show that
$\int_{0}^{2\pi} f(x)\cos x dx \geq 0$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

36 posts

hadikh

#2 h May 5, 2012, 10:00 AM • 12 Y

Y by pratyush, sreelatha, DeepanshuPrasad, Devesh14, Mathfear, ring_r, RANDOMMATHLOVER, Levi_Ackerman1, Adventure10, Mango247, mqoi_KOLA, and 1 other user

Sayan wrote:

Let $f(x)$ be a continuous function, whose first and second derivatives are continuous on $[0,2\pi]$ and $f''(x) \geq 0$ for all $x$ in $[0,2\pi]$ . Show that
$\int_{0}^{2\pi} f(x)\cos x dx \geq 0$

$\int_{0}^{2\pi} f(x)\cos x dx \\ =\int_{0}^{2\pi} \left ( \frac{d}{dx}(f(x)\sin x)-f'(x)\sin x \right ) dx \\ =f(2\pi)\sin 2\pi - f(0)\sin 0 \ +\int_{0}^{2\pi} -f'(x)\sin x dx \\ =0+\int_{0}^{2\pi} \left ( \frac{d}{dx}(f'(x)\cos x)-f''(x)\cos x \right ) dx \\ =f'(2\pi)\cos 2\pi-f'(0)\cos 0 \ -\int_{0}^{2\pi} f''(x)\cos x dx \\ =f'(2\pi)-f'(0)-\int_{0}^{2\pi} f''(x)\cos x dx \\ =\int_{0}^{2\pi} f''(x) dx -\int_{0}^{2\pi} f''(x)\cos x dx \\ =\int_{0}^{2\pi} f''(x)(1-\cos x) dx \geq0$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

951 posts

#3 h May 6, 2012, 8:43 PM • 1 Y

Y by Adventure10

Let $f(x)$ be a continuous and convex function on $[0,2\pi]$ Show that
$\int_{0}^{2\pi} f(x)\cos x dx \geq 0$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

827 posts

#4 h May 7, 2012, 1:44 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Isn't this equivalent to the original problem?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3489 posts

WWW

#5 h May 8, 2012, 3:15 AM • 2 Y

Y by chipgiboy, Adventure10

No, certainly not; check the definition of convexity. (For example $|x|$ is convex on $\mathbb {R}$ but is not differentiable at $0.$ )

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3489 posts

WWW

#6 h May 8, 2012, 6:37 PM • 2 Y

Y by Adventure10, kiyoras_2001

With $f$ only continuous and convex:

Lemma 1. $f$ convex on $[0,2\pi] \implies f(x+\pi)-f(x)$ is an increasing function on $[0,\pi].$

Proof: If $x<y,$ we want to show $f(y+\pi)-f(x+\pi)-(f(y)-f(x)) \ge 0.$ If we divide by $y-x,$ we are comparing slopes of secant lines on the graph of $f$ . Because $f$ is convex, these slopes increase as we move to the right. This gives the result.

Lemma 2. If $g$ decreases on $[0,\pi],$ then $\int_0^{\pi} g(x)\cos x\,dx \ge 0.$

Proof: The integral equals $\int_0^{\pi /2}[g(\pi /2 -x)\cos (\pi /2 -x) + g(\pi /2 +x)\cos (\pi /2 +x)]\,dx.$ Now $\cos (\pi /2 +x) = -\cos (\pi /2 -x)$ , so the last integral equals $\int_0^{\pi /2}[g(\pi /2 -x)-g(\pi /2 +x)]\cos (\pi /2 -x)\,dx.$ Because $g$ is decreasing, the last integrand is $\ge 0$ , giving the lemma.

So now assume we have $f$ continuous and convex on $[0,2\pi ].$ Then $\int_0^{2\pi} f(x)\cos x \, dx = \int_0^{\pi} [f(x)-f(x+\pi )]\cos x \, dx.$ Lemma 1 implies $f(x)-f(x+\pi )$ is decreasing. Lemma 2 then finishes the proof.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1886 posts

Potla

#7 h May 17, 2012, 1:03 AM • 2 Y

Y by Chandrachur, Adventure10

Sayan wrote:

Let $f(x)$ be a continuous function, whose first and second derivatives are continuous on $[0,2\pi]$ and $f''(x) \geq 0$ for all $x$ in $[0,2\pi]$ . Show that
$\int_{0}^{2\pi} f(x)\cos x dx \geq 0$

For the nongeneral problem:
Since $\cos x=-\cos(\pi-x)=-cos(\pi+x)=\cos(2\pi-x),$ therefore note that
$\int_0^{2\pi}f(x)\cos x \ d\, x=\int_0^{\frac{\pi}{2}}(f(x)-f(\pi-x)-f(\pi+x)+f(2\pi -x))\cos x \ d\, x .$
Now, since $\cos$ is positive in our new interval, it is enough to check that $f(x)-f(\pi-x)-f(\pi+x)+f(2\pi -x)\geq 0$ in the required interval. Now, note that from Lagrange's Mean Value Theorem, we can find an $\epsilon_1\in [x, \pi -x ]$ and $\epsilon_2\in[\pi+x, 2\pi-x]$ such that
$\left\{\begin{aligned}&f(x)-f(\pi-x)=(2x-\pi)f'(\epsilon_1)\\& f(pi+x)-f(2\pi-x)=(2x-\pi)f'(\epsilon_2).\end{aligned}\right.$
Therefore, we get
$f(x)-f(\pi-x)-f(\pi+x)+f(2\pi-x)=(\pi-2x)(f'(\epsilon_2)-f'(\epsilon_1))\geq 0;$
Where the last step follows from $f''(x)\geq 0\implies f'(x)$ is increasing, and $\epsilon_2 \geq \pi+x\geq \pi -x \geq \epsilon_1.\Box$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

253 posts

#8 h Jun 25, 2013, 5:15 PM • 2 Y

Y by Adventure10, Mango247

it will be
$ \left\{\begin{aligned}&f(x)-f(\pi-x)=(2x-\pi)f'(\epsilon_1)\\& f(\pi+x)-f(2\pi-x)=(2x-\pi)f'(\epsilon_2).\end{aligned}\right. $

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

337 posts

#9 h Aug 24, 2020, 3:58 PM

Y by

Potla wrote:

Sayan wrote:

Let $f(x)$ be a continuous function, whose first and second derivatives are continuous on $[0,2\pi]$ and $f''(x) \geq 0$ for all $x$ in $[0,2\pi]$ . Show that
$\int_{0}^{2\pi} f(x)\cos x dx \geq 0$

For the nongeneral problem:
Since $\cos x=-\cos(\pi-x)=-cos(\pi+x)=\cos(2\pi-x),$ therefore note that
$\int_0^{2\pi}f(x)\cos x \ d\, x=\int_0^{\frac{\pi}{2}}(f(x)-f(\pi-x)-f(\pi+x)+f(2\pi -x))\cos x \ d\, x .$ Now, since $\cos$ is positive in our new interval, it is enough to check that $f(x)-f(\pi-x)-f(\pi+x)+f(2\pi -x)\geq 0$ in the required interval. Now, note that from Lagrange's Mean Value Theorem, we can find an $\epsilon_1\in [x, \pi -x ]$ and $\epsilon_2\in[\pi+x, 2\pi-x]$ such that
$f(x)-f(\pi-x)=(2x-\pi)f'(\epsilon_1)\text{ }\&\text{ } f(\pi+x)-f(2\pi-x)=(2x-\pi)f'(\epsilon_2).$ Therefore, we get
$f(x)-f(\pi-x)-f(\pi+x)+f(2\pi-x)=(\pi-2x)(f'(\epsilon_2)-f'(\epsilon_1))\geq 0;$
Where the last step follows from $f''(x)\geq 0\implies f'(x)$ is increasing, and $\epsilon_2 \geq \pi+x\geq \pi -x \geq \epsilon_1.\Box$

FTFY
personal opinion

This post has been edited 1 time. Last edited by stranger_02, Aug 24, 2020, 4:00 PM
Reason: personal opinion

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

75 posts

#10 h Nov 22, 2024, 11:26 AM

Y by

hadikh wrote:

Sayan wrote:

Let $f(x)$ be a continuous function, whose first and second derivatives are continuous on $[0,2\pi]$ and $f''(x) \geq 0$ for all $x$ in $[0,2\pi]$ . Show that
$\int_{0}^{2\pi} f(x)\cos x dx \geq 0$

$\int_{0}^{2\pi} f(x)\cos x dx \\ =\int_{0}^{2\pi} \left ( \frac{d}{dx}(f(x)\sin x)-f'(x)\sin x \right ) dx \\ =f(2\pi)\sin 2\pi - f(0)\sin 0 \ +\int_{0}^{2\pi} -f'(x)\sin x dx \\ =0+\int_{0}^{2\pi} \left ( \frac{d}{dx}(f'(x)\cos x)-f''(x)\cos x \right ) dx \\ =f'(2\pi)\cos 2\pi-f'(0)\cos 0 \ -\int_{0}^{2\pi} f''(x)\cos x dx \\ =f'(2\pi)-f'(0)-\int_{0}^{2\pi} f''(x)\cos x dx \\ =\int_{0}^{2\pi} f''(x) dx -\int_{0}^{2\pi} f''(x)\cos x dx \\ =\int_{0}^{2\pi} f''(x)(1-\cos x) dx \geq0$

wow i was stuck at the second last line for days...

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

13 posts

#11 h Mar 27, 2025, 3:58 PM

Y by

Sh!t i was so dumb thinking the derivative of $cosx$ is $sinx$ all the time while trying this problem :wallbash_red:

:wallbash_red:

:wallbash_red:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

775 posts

#12 h Mar 28, 2025, 4:07 PM

Y by

Let's take a Laplace Transform approach. Let $f(0)=A, f'(0)=B$ (for $A,B \in \mathbb{R}$ ) and let $f''(x)=g(x)$ (i) where $g:[0,2\pi] \rightarrow [0,\infty)$ is a differentiable function whose Laplace Transform $G(s)$ exists. Taking the Laplace Transform of (i) yields:

$s^2F(s) - sf(0)-f'(0) = G(s) \Rightarrow F(s) = \frac{G(s)+As+B}{s^2} = G(s) \cdot \frac{1}{s^2} + \frac{A}{s} +\frac{B}{s^2}$ (ii);

of which the inverse Laplace Transform of (ii) produces:

$f(x) = L^{-1}[F(s)] = \int_{0}^{x}\tau \cdot g(x-\tau) d\tau + (A+Bx)$ (iii).

If $\int_{0}^{2\pi} f(x)\cos(x) dx \ge 0$ , then:

$f(x)\sin(x)|_{0}^{2\pi} - \int_{0}^{2\pi} f'(x)\sin(x) dx = -\int_{0}^{2\pi} [g(0)x+B]\sin(x) dx$ ;

or $[g(0)x+B]\cos(x)|_{0}^{2\pi} - \int_{0}^{2\pi} g(0)\cos(x)dx$ ;

or $2\pi g(0)-g(0)\sin(x)|_{0}^{2\pi} = \textcolor{red}{2\pi g(0) \ge 0}$ .

QED

This post has been edited 2 times. Last edited by Mathzeus1024, Mar 28, 2025, 4:18 PM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a