Trigonometric inequality

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December 20th 2008, 12:13 PM
Mathstud28

Trigonometric inequality

This one is fun. Find what values of $x$ make $\cos(\sin(x))\geqslant\sin(\cos(x))$ true. Next find all values that make $\sin(\cos(x))\geqslant\cos(\sin(x))$ true.

Note: No calculus.
December 20th 2008, 12:59 PM
NonCommAlg

Quote:

Originally Posted by Mathstud28

This one is fun. Find what values of $x$ make $\cos(\sin(x))\geqslant\sin(\cos(x))$ true. Next find all values that make $\sin(\cos(x))\geqslant\cos(\sin(x))$ true.

Note: No calculus.

it's a pretty problem! it's always true that $\sin(\cos x) < \cos (\sin x).$ first note that $-\sqrt{2} \leq \cos x \pm \sin x \leq \sqrt{2},$ because $(\cos x \pm \sin x)^2 = 1 \pm \sin(2x) \leq 2.$ but $\sqrt{2} < \frac{\pi}{2}.$ thus:

$(1)\ \ \ \frac{-\pi}{2} < \frac{\cos x + \sin x - \frac{\pi}{2}}{2} < 0,$

$(2) \ \ \ 0 < \frac{\cos x - \sin x + \frac{\pi}{2}}{2} < \frac{\pi}{2}.$

now we have: $\sin(\cos x) \ - \ \cos(\sin x)= 2 \sin \left(\frac{\cos x + \sin x - \frac{\pi}{2}}{2} \right) \cos \left(\frac{\cos x - \sin x + \frac{\pi}{2}}{2} \right) < 0,$ by $(1)$ and $(2). \ \ \Box$
December 20th 2008, 01:05 PM
Mathstud28

Quote:

Originally Posted by NonCommAlg

it's a pretty problem! it's always true that $\sin(\cos x) < \cos (\sin x).$ first note that $-\sqrt{2} \leq \cos x \pm \sin x \leq \sqrt{2},$ because $(\cos x \pm \sin x)^2 = 1 \pm \sin(2x) \leq 2.$ but $\sqrt{2} < \frac{\pi}{2}.$ thus:

$(1)\ \ \ \frac{-\pi}{2} < \frac{\cos x + \sin x - \frac{\pi}{2}}{2} < 0,$

$(2) \ \ \ 0 < \frac{\cos x - \sin x + \frac{\pi}{2}}{2} < \frac{\pi}{2}.$

now we have: $\sin(\cos x) \ - \ \cos(\sin x)= 2 \sin \left(\frac{\cos x + \sin x - \frac{\pi}{2}}{2} \right) \cos \left(\frac{\cos x - \sin x + \frac{\pi}{2}}{2} \right) < 0,$ by $(1)$ and $(2). \ \ \Box$

Aha! Similar to my solution...I admit though it was easier for me because I was given that $\sin(\cos(x))<\cos(\sin(x))$ and was just asked to prove it.
December 20th 2008, 01:21 PM
NonCommAlg

try your inequality's cousin: (Wink) prove that: $(\sin x)^{\cos x} < (\cos x)^{\sin x},$ for all $0 \leq x < \frac{\pi}{4}.$

i'd like to see different approaches. of course, using calculus probably would be the most convenient one and you may use it. but are there other ways to do it? (Wondering)
December 20th 2008, 02:23 PM
Mathstud28

Quote:

Originally Posted by NonCommAlg

try your inequality's cousin: (Wink) prove that: $(\sin x)^{\cos x} < (\cos x)^{\sin x},$ for all $0 \leq x < \frac{\pi}{4}.$

i'd like to see different approaches. of course, using calculus probably would be the most convenient one and you may use it. but are there other ways to do it? (Wondering)

Besides the obvious way of doing it with maxs and stuff how about this method. It uses some shady postulates (that may be untrue), but this is for fun so why not just suggest it?

Suppose the inequality presented is true, then so must be the inequality $\cos(x)\ln(\sin(x))<\sin(x)\ln(\cos(x))$ , or alternatively $-\sin(x)\ln(\cos(x))<-\cos(x)\ln(\sin(x))$ . It can be verified that both these functions are monotonically increasing, positive, and continuous on the specified interval.

Now here is where the geometrically logical but probably incorrect "lemma" I am using comes into play. It goes someting to the tune that if $f,g>0~a$ and $f,g$ are monotonic as well as continuous on the interval $[a,b]$ , then $f(x)<g(x)~~~a\leqslant x \leqslant b\quad\Longleftrightarrow\quad \int_a^b f(x)dx<\int_a^b g(x)dx$ . Now suppose that this is true, it can be easily verified that $-\int_0^{\frac{\pi}{4}}\cos(x)\ln(\sin(x))dx=\frac{ \sqrt{2}\ln(2)}{4}+\frac{\sqrt{2}}{2}$ $>-\int_0^{\frac{\pi}{4}}\sin(x)\ln(\cos(x))dx=1-\frac{\sqrt{2}\ln(2)}{4}-\frac{\sqrt{2}}{2}$

Now supposing that my "made up" lemma is correct, this proves the inequality.

I have learned to not always trust my geometric intuition...so Im not too sure about this (Worried)
December 20th 2008, 03:00 PM
NonCommAlg

Quote:

Originally Posted by Mathstud28

Besides the obvious way of doing it with maxs and stuff how about this method. It uses some shady postulates (that may be untrue), but this is for fun so why not just suggest it?

Suppose the inequality presented is true, then so must be the inequality $\cos(x)\ln(\sin(x))<\sin(x)\ln(\cos(x))$ , or alternatively $-\sin(x)\ln(\cos(x))<-\cos(x)\ln(\sin(x))$ . It can be verified that both these functions are monotonically increasing, positive, and continuous on the specified interval.

Now here is where the geometrically logical but probably incorrect "lemma" I am using comes into play. It goes someting to the tune that if $f,g>0~a$ and $f,g$ are monotonic as well as continuous on the interval $[a,b]$ , then $f(x)<g(x)~~~a\leqslant x \leqslant b\quad\Longleftrightarrow\quad \int_a^b f(x)dx<\int_a^b g(x)dx$ . Now suppose that this is true, it can be easily verified that $-\int_0^{\frac{\pi}{4}}\cos(x)\ln(\sin(x))dx=\frac{ \sqrt{2}\ln(2)}{4}+\frac{\sqrt{2}}{2}$ $>-\int_0^{\frac{\pi}{4}}\sin(x)\ln(\cos(x))dx=1-\frac{\sqrt{2}\ln(2)}{4}-\frac{\sqrt{2}}{2}$

Now supposing that my "made up" lemma is correct, this proves the inequality.

I have learned to not always trust my geometric intuition...so Im not too sure about this (Worried)

unfortunately your made up lemma is not always true, e.g. $f(x)=\frac{x+1}{4}, \ g(x)=x.$ then $\int_0^1 f(x) \ dx < \int_0^1 g(x) \ dx,$ but neither $f < g$ nor $f > g$ on the unit interval.

the trouble here comes from this fact that $f$ and $g$ intersect! so you need to assume that $f,g$ do not intersect in the interval. also i don't think we need $f,g$ to be positive.
December 20th 2008, 03:04 PM
Mathstud28

Quote:

Originally Posted by NonCommAlg

unfortunately your made up lemma is not always true, e.g. $f(x)=\frac{x+1}{4}, \ g(x)=x.$ then $\int_0^1 f(x) \ dx < \int_0^1 g(x) \ dx,$ but neither $f < g$ nor $f > g$ on the unit interval.

the trouble here comes from this fact that $f$ and $g$ intersect! so you need to assume that $f,g$ do not intersect in the interval. also i don't think we need $f,g$ to be positive.

Dangit! I forgot to say that. I knew that they cannot intersect...and no we do not need them positive we need $fg>0$ . I have come up with a "proof" of my lemma if anyone wants to see it.

EDIT: Wait, they cannot intersect except possibly at the interval because then $f<g$ after that point.

For example $x<\frac{x+1}{4}~~0<x<\frac{1}{3}$ but it is reversed afterwards...but you applied my lemma to the interval $[0,1]$ , and $x\not<\frac{x+1}{4}~~\forall x \in[0.1]$

EDIT EDIT: We dont even need $fg>0$ we must just have $|f|<|g|$

EDIT EDIT EDIT: I am busy now but I will come back later and write out this lemma in a clear manner...I will then attempt to prove it
December 20th 2008, 03:15 PM
NonCommAlg

Quote:

Originally Posted by Mathstud28

Dangit! I forgot to say that. I knew that they cannot intersect...and no we do not need them positive we need $fg>0$ . I have come up with a "proof" of my lemma if anyone wants to see it.

EDIT: Wait, they cannot intersect except possibly at the interval because then $f<g$ after that point.

For example $x<\frac{x+1}{4}~~0<x<\frac{1}{3}$ but it is reversed afterwards...but you applied my lemma to the interval $[0,1]$ , and $x\not<\frac{x+1}{4}~~\forall x \in[0.1]$

my example was a counter-example to this side of your inequality: $\int_a^b f < \int_a^b g \Longrightarrow f < g.$ in my example $\int_0^1 f < \int_0^1 g$ but $f \not< g$ in [0,1].

you still have a lot of things to do: first you need to show that $\cos(x) \ln(\sin(x)) \neq \sin(x) \ln(\cos(x))$ on the interval and then monotonocity of the functions! (Evilgrin)
December 20th 2008, 03:30 PM
bkarpuz

Quote:

Originally Posted by Mathstud28

This one is fun. Find what values of $x$ make $\cos(\sin(x))\geqslant\sin(\cos(x))$ true. Next find all values that make $\sin(\cos(x))\geqslant\cos(\sin(x))$ true.

Note: No calculus.

Quote:

Originally Posted by NonCommAlg

try your inequality's cousin: prove that: $(\sin x)^{\cos x} < (\cos x)^{\sin x},$ for all $0 \leq x < \frac{\pi}{4}.$

i'd like to see different approaches. of course, using calculus probably would be the most convenient one and you may use it. but are there other ways to do it?

I have the following solution using calculus.
Let $f(t):=\cos(\sin(t))-\sin(\cos(t))$ for $t\in\mathbb{R}$ .
Then, we see that $f$ is $2\pi$ periodic, hence it suffices to prove $f\geq0$ on $[0,2\pi]$ .
On the other hand, we have $f(t)=f(2\pi-t)$ for all $t\in[0,2\pi]$ , which indicates that it suffices to prove $f(t)\geq0$ for all $t\in[0,\pi]$ .
Clearly, $\cos$ is positive (and decreasing) on $[0,\pi/2]$ , $0\leq\sin(t)\leq1<\pi/2$ for all $t\in[\pi/2,\pi]$ , and $\sin$ is negative (and increasing) on $[-\pi/2,0]$ , $-\pi/2<-1\geq\cos(t)\leq0$ .
Therefore, $f>0$ on $[\pi/2,\pi]$ .
To complete the proof we have to prove $f\geq0$ on $[0,\pi/2]$ .

Similar reasoning to the discussion above about increasing and decreasing natures of the functions $\cos$ and $\sin$ together with the fact $\sin(t)\leq t$ for all $t\in[0,\infty)$ , we get $f(t)\geq\cos(t)-\sin(\cos(t))\geq\cos(t)-\cos(t)=0$ for all $t\in[0,\pi/2]$ , and the proof is hence completed. $\rule{0.3cm}{0.3cm}$

$\rule{9.6cm}{0.05cm}$ $\rule{9.7cm}{0.05cm}$
http://img126.imageshack.us/img126/9408/89794076ui5.jpg http://img126.imageshack.us/img126/8669/26882070yl6.jpg http://img126.imageshack.us/img126/2069/50509447vr7.jpg
.................Graph of $f$ .....................................Graph of $f^{\prime}$ .....................................Graph of $f^{\prime\prime}$
$\rule{9.6cm}{0.05cm}$ $\rule{9.7cm}{0.05cm}$
December 20th 2008, 09:17 PM
Mathstud28

So the statement is this: Suppose that $f,g$ posses the following charcteristics on $[a,b]$ : they are positive, they are continuous, they do not intersect, and they are monotonic. Then on $[a,b]$ it is true that $f<g~\Longleftrightarrow~\int_a^b f(x)~dx<\int_a^b g(x)~dx$

First let us prove that $f<g~\implies~\int_a^b f(x)~dx<\int_a^b g(x)~dx$ . Consider any partition $P_n$ of $[a,b]$ consisting of the set of points $\left\{x_i\right\}~1\leqslant i\leqslant n$ . Now as in the usual way define $\Delta x_i=x_i-x_{i-1}$ , $(Mf)_i=\sup_{x_{i-1}<x<x_i}f(x)$ , and $U\left(P_n,f\right)=\sum_{i=1}^{n}(Mf)_i\cdot\Delt a x_i$ . Now it is clear that since $f(x)<g(x)$ that $\sup_{x_{i-1}< x<x_i}f(x)<\sup_{x_{i-1}<x<x_i}g(x)$ and since $\Delta x_i>0$ this implies that $\sup_{x_{i-1}< x<x_i}f(x)\cdot\Delta x_i<\sup_{x_{i-1}<x<x_i}g(x)\cdot \Delta x_i$ . Finally we can conclude that $U\left(P_n,f\right)=\sum_{i=1}^{n}(Mf)_i\cdot\Delt a x_i<\sum_{i=1}^n (Mg)_i\cdot \Delta x_i=U\left(P_n,g\right)$ . And since $f,g$ are continuous, thus Riemann integrable, $\int_a^b f(x)~dx=\inf_{n\in\mathbb{N}}\left\{U\left(P_n,f\r ight)\right\}<\inf_{n\in\mathbb{N}}\left\{U\left(P _n,g\right)\right\}=\int_a^b g(x)~dx$

Now let us prove that $\int_a^b f(x)~dx<\int_a^b g(x)~dx~\implies~ f(x)<g(x)$ . Define $P_n,\Delta x_i, M_i, U\left(P_n,f\right)$ as before.

1. Now it is clear that either $(Mf)_i<(Mg)_i$ or $(Mf)_i>(Mg)_i$ for all $i$ . To see this first define $h(x)=f(x)-g(x)$ it is clear that $h$ is continuous. Then suppose that there were two values $1\leqslant j,k \leqslant n$ such that $(Mf)_j<(Mg)_j$ and $(Mf)_k>(Mg)_k$ then there exists a $\xi\in[x_{j-1},x_j]$ such that $f(\xi)<g(\xi)\implies h(\xi)<0$ and there exists a $\xi_1\in[x_{k-1},x_k]$ such that $f(\xi_1)>g(\xi_1)\implies h(\xi_1)>0$ . Now because $h$ is continuous and $[a,b]$ connected this implies there exists a $x\in[a,b]$ such that $h(x)=0\implies f(x)=g(x)$ which contradicts that the functions do not intersect.

2. So from the fact that $\inf_{n\in\mathbb{N}}\left\{U\left(P_n,f\right)\ri ght\}<\inf_{n\in\mathbb{N}}\left\{U\left(P_n,g\rig ht)\right\}$ we can see that $(Mf)_i<(Mg)_i$

3. So all that is left to do is prove that $(Mf)_i<(Mg)_i\implies f(x)<g(x)~\forall x\in[x_{i-1},x_i]$ . To do this once again define $h(x)=f(x)-g(x)$ . Let $\xi_f$ be the point such that $f(\xi_f)=\sup_{x_{i-1}\leqslant x \leqslant x_i}f(x)$ , and let $\xi_g$ be defined similarly. Now since $[x_{i-1},x_i]$ is compact it follows that $\xi_f,\xi_g\in[x_{i-1},x_i]$ . Now consider when $f,g$ are monotonically increasing, it is clear now that $\xi_f,\xi_g=b$ . So $h(b)=f(\xi)=\sup_{x_{i-1}\leqslant x \leqslant x_i}f(x)-f(\xi)=\sup_{x_{i-1}\leqslant x \leqslant x_i}g(x)<0$ . So now suppose there was a point $y\in[x_{i-1},x_i]$ such that $f(y)>g(y)$ , then at that point $h(y)>0$ and by the connectedness of $[x_{i-1},x_i]$ and the continuity of $h(x)$ there must be a point in $[x_{i-1},x_i]$ such that $h=0\implies f=g$ , but this contradicts the two functions not intersecting. The proof is done similarly for $f,g$ being monotonically decreasing.

4. Now since the interval $[x_{-1},x_i]$ was arbitrary in 3. this completes the proof $\blacksquare$
December 20th 2008, 11:15 PM
NonCommAlg

Quote:

Originally Posted by Mathstud28

So the statement is this: Suppose that $f,g$ posses the following charcteristics on $[a,b]$ : they are positive, they are continuous, they do not intersect, and they are monotonic. Then on $[a,b]$ it is true that $f<g~\Longleftrightarrow~\int_a^b f(x)~dx<\int_a^b g(x)~dx$

First let us prove that $f<g~\implies~\int_a^b f(x)~dx<\int_a^b g(x)~dx$ . Consider any partition $P_n$ of $[a,b]$ consisting of the set of points $\left\{x_i\right\}~1\leqslant i\leqslant n$ . Now as in the usual way define $\Delta x_i=x_i-x_{i-1}$ , $(Mf)_i=\sup_{x_{i-1}<x<x_i}f(x)$ , and $U\left(P_n,f\right)=\sum_{i=1}^{n}(Mf)_i\cdot\Delt a x_i$ . Now it is clear that since $f(x)<g(x)$ that $\sup_{x_{i-1}< x<x_i}f(x)<\sup_{x_{i-1}<x<x_i}g(x)$ and since $\Delta x_i>0$ this implies that $\sup_{x_{i-1}< x<x_i}f(x)\cdot\Delta x_i<\sup_{x_{i-1}<x<x_i}g(x)\cdot \Delta x_i$ . Finally we can conclude that $U\left(P_n,f\right)=\sum_{i=1}^{n}(Mf)_i\cdot\Delt a x_i<\sum_{i=1}^n (Mg)_i\cdot \Delta x_i=U\left(P_n,g\right)$ . And since $f,g$ are continuous, thus Riemann integrable, $\int_a^b f(x)~dx=\inf_{n\in\mathbb{N}}\left\{U\left(P_n,f\r ight)\right\}<\inf_{n\in\mathbb{N}}\left\{U\left(P _n,g\right)\right\}=\int_a^b g(x)~dx$

Now let us prove that $\int_a^b f(x)~dx<\int_a^b g(x)~dx~\implies~ f(x)<g(x)$ . Define $P_n,\Delta x_i, M_i, U\left(P_n,f\right)$ as before.

1. Now it is clear that either $(Mf)_i<(Mg)_i$ or $(Mf)_i>(Mg)_i$ for all $i$ . To see this first define $h(x)=f(x)-g(x)$ it is clear that $h$ is continuous. Then suppose that there were two values $1\leqslant j,k \leqslant n$ such that $(Mf)_j<(Mg)_j$ and $(Mf)_k>(Mg)_k$ then there exists a $\xi\in[x_{j-1},x_j]$ such that $f(\xi)<g(\xi)\implies h(\xi)<0$ and there exists a $\xi_1\in[x_{k-1},x_k]$ such that $f(\xi_1)>g(\xi_1)\implies h(\xi_1)>0$ . Now because $h$ is continuous and $[a,b]$ connected this implies there exists a $x\in[a,b]$ such that $h(x)=0\implies f(x)=g(x)$ which contradicts that the functions do not intersect.

2. So from the fact that $\inf_{n\in\mathbb{N}}\left\{U\left(P_n,f\right)\ri ght\}<\inf_{n\in\mathbb{N}}\left\{U\left(P_n,g\rig ht)\right\}$ we can see that $(Mf)_i<(Mg)_i$

3. So all that is left to do is prove that $(Mf)_i<(Mg)_i\implies f(x)<g(x)~\forall x\in[x_{i-1},x_i]$ . To do this once again define $h(x)=f(x)-g(x)$ . Let $\xi_f$ be the point such that $f(\xi_f)=\sup_{x_{i-1}\leqslant x \leqslant x_i}f(x)$ , and let $\xi_g$ be defined similarly. Now since $[x_{i-1},x_i]$ is compact it follows that $\xi_f,\xi_g\in[x_{i-1},x_i]$ . Now consider when $f,g$ are monotonically increasing, it is clear now that $\xi_f,\xi_g=b$ . So $h(b)=f(\xi)=\sup_{x_{i-1}\leqslant x \leqslant x_i}f(x)-f(\xi)=\sup_{x_{i-1}\leqslant x \leqslant x_i}g(x)<0$ . So now suppose there was a point $y\in[x_{i-1},x_i]$ such that $f(y)>g(y)$ , then at that point $h(y)>0$ and by the connectedness of $[x_{i-1},x_i]$ and the continuity of $h(x)$ there must be a point in $[x_{i-1},x_i]$ such that $h=0\implies f=g$ , but this contradicts the two functions not intersecting. The proof is done similarly for $f,g$ being monotonically decreasing.

4. Now since the interval $[x_{-1},x_i]$ was arbitrary in 3. this completes the proof $\blacksquare$

ok, i didn't read your proof but i'm sure it's a good practice for you since you're studying Rudin! (Nod) first of all, you don't need to assume $f,g$ are positive or monotonic. "continuous" and

"not intersecting" are only conditions we need: let $h=g-f.$ suppose first that $h > 0$ on the interval. it's not hard to see that the integral of a positive continuous function is positive.*

thus $\int_a^b h > 0.$ conversely, suppose $\int_a^b h > 0.$ since $f, g$ do not intersect, we have $h \neq 0$ everywhere on [a,b]. so by the intermediate value theorem, either $h > 0$ or $h < 0,$ everywhere

on [a,b]. but if $h < 0,$ then $-h > 0$ and hence by * we'll have $\int_a^b (-h)> 0,$ and hence $\int_a^b h < 0,$ which is a contradiction. Q.E.D.

* in general, if $h$ is continuous, non-negative and not identically 0 on [a,b], then $\int_a^b h > 0.$ Hint: since $h$ is not identically 0 over [a,b], there exists a subinterval of [a,b] over which: $h > 0.$
December 20th 2008, 11:23 PM
Mathstud28

Quote:

Originally Posted by NonCommAlg

ok, i didn't read your proof but i'm sure it's a good practice for you since you're studying Rudin!

Yeah, I am not trying to be easy, I am trying to be as rigorous as possible...now this may not always be the best way...but it helps me learn all the material since I end up using three fifths of it in one proof.

And I understand your proof, but the reason it is so short is that a lot of the stuff you just stated I proved...now of course for a mathematician such as yourself this is obvious...but I thought for us other folks it would be best to show it.

Thanks for your time NonCommAlg (Nod)