Mixing
Showing that $cos^p{Theta} le cos{pTheta}$, for $0<Theta<frac{pi}{2}$ and $0<p<1$, by analyzing...
$begingroup$
Given $0 < Theta < frac{pi}{2}$ and $0 < p < 1$, show that $$cos^p{Theta} le cos{pTheta}$$
Can you please check if my proof is correct? Would also love to know if there're other ways of proving this.
Proof. Let $$f(Theta) = frac{cos^p{Theta}}{cos{pTheta}}$$considering $p$ fixed. The derivative is
$$f'(Theta) = frac{p*cos^{p-1}{Theta}*{(-sin{Theta})*cos{pTheta} -cos^p{Theta}*(-sin{pTheta}*p) } } {cos^2{pTheta}} = frac{p*cos^{p-1}{Theta}*sin{(p-1)Theta}}{cos^2{pTheta}}$$
We have $f(0)=1$, and because $0<p<1$, the term $sin(p-1)Theta$ makes the derivative negative: $f'(Theta) < 0$ on $(0,frac{pi}{2})$.
At $0$ we have $f'(0)=0$, so we cannot immediately deduce that $f$ is decreasing on all of $[0,frac{pi}{2})$.
However, considering that $f(0)=1, f(frac{pi}{2})=0$, and both $f$ and $f'$ are continuous on $[0,frac{pi}{2})$, if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$, with the derivative vanishing at that point, which cannot happen. Therefore $f(Theta) le 1$ throughout $[0,frac{pi}{2}]$, Q.E.D.
calculus trigonometry proof-verification
edited Jan 22 at 13:30
Blue
49k870156
asked Jan 22 at 7:25
AnatolyVorobeyAnatolyVorobey
1,666513
$endgroup$
add a comment |
$begingroup$
Given $0 < Theta < frac{pi}{2}$ and $0 < p < 1$, show that $$cos^p{Theta} le cos{pTheta}$$
Can you please check if my proof is correct? Would also love to know if there're other ways of proving this.
Proof. Let $$f(Theta) = frac{cos^p{Theta}}{cos{pTheta}}$$considering $p$ fixed. The derivative is
$$f'(Theta) = frac{p*cos^{p-1}{Theta}*{(-sin{Theta})*cos{pTheta} -cos^p{Theta}*(-sin{pTheta}*p) } } {cos^2{pTheta}} = frac{p*cos^{p-1}{Theta}*sin{(p-1)Theta}}{cos^2{pTheta}}$$
We have $f(0)=1$, and because $0<p<1$, the term $sin(p-1)Theta$ makes the derivative negative: $f'(Theta) < 0$ on $(0,frac{pi}{2})$.
At $0$ we have $f'(0)=0$, so we cannot immediately deduce that $f$ is decreasing on all of $[0,frac{pi}{2})$.
However, considering that $f(0)=1, f(frac{pi}{2})=0$, and both $f$ and $f'$ are continuous on $[0,frac{pi}{2})$, if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$, with the derivative vanishing at that point, which cannot happen. Therefore $f(Theta) le 1$ throughout $[0,frac{pi}{2}]$, Q.E.D.
calculus trigonometry proof-verification
edited Jan 22 at 13:30
Blue
49k870156
asked Jan 22 at 7:25
AnatolyVorobeyAnatolyVorobey
1,666513
$endgroup$
$begingroup$
if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$. Why???
$endgroup$
– Riemann
Jan 22 at 9:39
$begingroup$
As a continuous function on a compact interval $[0,frac{pi}{2}]$, $f$ must assume its maximum value somewhere in the interval (Extreme Value Theorem); if $f(Theta)>1$ somewhere in $(0,frac{pi}{2})$, then that maximum must lie within $(0, frac{pi}{2})$ because the values on the borders are 0 and 1.
$endgroup$
– AnatolyVorobey
Jan 22 at 9:52
1
$begingroup$
Yes, it is right.
$endgroup$
– Riemann
Jan 22 at 9:56
add a comment |
$begingroup$
Given $0 < Theta < frac{pi}{2}$ and $0 < p < 1$, show that $$cos^p{Theta} le cos{pTheta}$$
Can you please check if my proof is correct? Would also love to know if there're other ways of proving this.
Proof. Let $$f(Theta) = frac{cos^p{Theta}}{cos{pTheta}}$$considering $p$ fixed. The derivative is
$$f'(Theta) = frac{p*cos^{p-1}{Theta}*{(-sin{Theta})*cos{pTheta} -cos^p{Theta}*(-sin{pTheta}*p) } } {cos^2{pTheta}} = frac{p*cos^{p-1}{Theta}*sin{(p-1)Theta}}{cos^2{pTheta}}$$
We have $f(0)=1$, and because $0<p<1$, the term $sin(p-1)Theta$ makes the derivative negative: $f'(Theta) < 0$ on $(0,frac{pi}{2})$.
At $0$ we have $f'(0)=0$, so we cannot immediately deduce that $f$ is decreasing on all of $[0,frac{pi}{2})$.
However, considering that $f(0)=1, f(frac{pi}{2})=0$, and both $f$ and $f'$ are continuous on $[0,frac{pi}{2})$, if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$, with the derivative vanishing at that point, which cannot happen. Therefore $f(Theta) le 1$ throughout $[0,frac{pi}{2}]$, Q.E.D.
calculus trigonometry proof-verification
edited Jan 22 at 13:30
Blue
49k870156
asked Jan 22 at 7:25
AnatolyVorobeyAnatolyVorobey
1,666513
$endgroup$
Given $0 < Theta < frac{pi}{2}$ and $0 < p < 1$, show that $$cos^p{Theta} le cos{pTheta}$$
Can you please check if my proof is correct? Would also love to know if there're other ways of proving this.
Proof. Let $$f(Theta) = frac{cos^p{Theta}}{cos{pTheta}}$$considering $p$ fixed. The derivative is
$$f'(Theta) = frac{p*cos^{p-1}{Theta}*{(-sin{Theta})*cos{pTheta} -cos^p{Theta}*(-sin{pTheta}*p) } } {cos^2{pTheta}} = frac{p*cos^{p-1}{Theta}*sin{(p-1)Theta}}{cos^2{pTheta}}$$
We have $f(0)=1$, and because $0<p<1$, the term $sin(p-1)Theta$ makes the derivative negative: $f'(Theta) < 0$ on $(0,frac{pi}{2})$.
At $0$ we have $f'(0)=0$, so we cannot immediately deduce that $f$ is decreasing on all of $[0,frac{pi}{2})$.
However, considering that $f(0)=1, f(frac{pi}{2})=0$, and both $f$ and $f'$ are continuous on $[0,frac{pi}{2})$, if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$, with the derivative vanishing at that point, which cannot happen. Therefore $f(Theta) le 1$ throughout $[0,frac{pi}{2}]$, Q.E.D.
calculus trigonometry proof-verification
calculus trigonometry proof-verification
edited Jan 22 at 13:30
Blue
49k870156
asked Jan 22 at 7:25
AnatolyVorobeyAnatolyVorobey
1,666513
edited Jan 22 at 13:30
Blue
49k870156
asked Jan 22 at 7:25
AnatolyVorobeyAnatolyVorobey
1,666513
edited Jan 22 at 13:30
Blue
49k870156
edited Jan 22 at 13:30
Blue
49k870156
edited Jan 22 at 13:30
Blue
49k870156
49k870156
asked Jan 22 at 7:25
AnatolyVorobeyAnatolyVorobey
1,666513
asked Jan 22 at 7:25
AnatolyVorobeyAnatolyVorobey
1,666513
asked Jan 22 at 7:25
AnatolyVorobeyAnatolyVorobey
1,666513
1,666513
$begingroup$
if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$. Why???
$endgroup$
– Riemann
Jan 22 at 9:39
$begingroup$
As a continuous function on a compact interval $[0,frac{pi}{2}]$, $f$ must assume its maximum value somewhere in the interval (Extreme Value Theorem); if $f(Theta)>1$ somewhere in $(0,frac{pi}{2})$, then that maximum must lie within $(0, frac{pi}{2})$ because the values on the borders are 0 and 1.
$endgroup$
– AnatolyVorobey
Jan 22 at 9:52
1
$begingroup$
Yes, it is right.
$endgroup$
– Riemann
Jan 22 at 9:56
add a comment |
$begingroup$
if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$. Why???
$endgroup$
– Riemann
Jan 22 at 9:39
$begingroup$
As a continuous function on a compact interval $[0,frac{pi}{2}]$, $f$ must assume its maximum value somewhere in the interval (Extreme Value Theorem); if $f(Theta)>1$ somewhere in $(0,frac{pi}{2})$, then that maximum must lie within $(0, frac{pi}{2})$ because the values on the borders are 0 and 1.
$endgroup$
– AnatolyVorobey
Jan 22 at 9:52
1
$begingroup$
Yes, it is right.
$endgroup$
– Riemann
Jan 22 at 9:56
$begingroup$
if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$. Why???
$endgroup$
– Riemann
Jan 22 at 9:39
$begingroup$
if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$. Why???
$endgroup$
– Riemann
Jan 22 at 9:39
$begingroup$
As a continuous function on a compact interval $[0,frac{pi}{2}]$, $f$ must assume its maximum value somewhere in the interval (Extreme Value Theorem); if $f(Theta)>1$ somewhere in $(0,frac{pi}{2})$, then that maximum must lie within $(0, frac{pi}{2})$ because the values on the borders are 0 and 1.
$endgroup$
– AnatolyVorobey
Jan 22 at 9:52
$begingroup$
As a continuous function on a compact interval $[0,frac{pi}{2}]$, $f$ must assume its maximum value somewhere in the interval (Extreme Value Theorem); if $f(Theta)>1$ somewhere in $(0,frac{pi}{2})$, then that maximum must lie within $(0, frac{pi}{2})$ because the values on the borders are 0 and 1.
$endgroup$
– AnatolyVorobey
Jan 22 at 9:52
1
1
$begingroup$
Yes, it is right.
$endgroup$
– Riemann
Jan 22 at 9:56
$begingroup$
Yes, it is right.
$endgroup$
– Riemann
Jan 22 at 9:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Use Mean Value Theorem you can prove the following lemma:
suppose $f(x)$ is continuous on $[a,b]$ and has derivative in $(a,b)$. If $f'(x)<0$ for $xin(a,b)$, then $f(x)$ is strictly decreasing on $[a,b]$.
So you need only $f'(Theta) < 0$ for $Thetain(0,pi/2)$ in your situation!
Also we can deal this one as follows:
For fix $pin(0,1)$, let $$f(x)=cos px-cos^px,xin[0,pi/2],$$
then $f$ is continuous on $[0,pi/2]$ and is differentiable in $(0,pi/2)$.
Its derivative is
$$f'(x)=psin xleft(cos^{p-1}x-frac{sin px}{sin x}right),xin(0,pi/2).$$
Because $pin(0,1)$, $0<px<x<pi/2$, we have
$$0<frac{sin px}{sin x}<1<frac{1}{cos^{1-p}x}=cos^{p-1}x.$$
So $$f'(x)>0, forall xin(0,pi/2).$$
This implies $$f(x)>f(0)=0,forall xin(0,pi/2).$$
answered Jan 22 at 7:56
RiemannRiemann
3,4651322
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Use Mean Value Theorem you can prove the following lemma:
suppose $f(x)$ is continuous on $[a,b]$ and has derivative in $(a,b)$. If $f'(x)<0$ for $xin(a,b)$, then $f(x)$ is strictly decreasing on $[a,b]$.
So you need only $f'(Theta) < 0$ for $Thetain(0,pi/2)$ in your situation!
Also we can deal this one as follows:
For fix $pin(0,1)$, let $$f(x)=cos px-cos^px,xin[0,pi/2],$$
then $f$ is continuous on $[0,pi/2]$ and is differentiable in $(0,pi/2)$.
Its derivative is
$$f'(x)=psin xleft(cos^{p-1}x-frac{sin px}{sin x}right),xin(0,pi/2).$$
Because $pin(0,1)$, $0<px<x<pi/2$, we have
$$0<frac{sin px}{sin x}<1<frac{1}{cos^{1-p}x}=cos^{p-1}x.$$
So $$f'(x)>0, forall xin(0,pi/2).$$
This implies $$f(x)>f(0)=0,forall xin(0,pi/2).$$
answered Jan 22 at 7:56
RiemannRiemann
3,4651322
$endgroup$
add a comment |
$begingroup$
Use Mean Value Theorem you can prove the following lemma:
suppose $f(x)$ is continuous on $[a,b]$ and has derivative in $(a,b)$. If $f'(x)<0$ for $xin(a,b)$, then $f(x)$ is strictly decreasing on $[a,b]$.
So you need only $f'(Theta) < 0$ for $Thetain(0,pi/2)$ in your situation!
Also we can deal this one as follows:
For fix $pin(0,1)$, let $$f(x)=cos px-cos^px,xin[0,pi/2],$$
then $f$ is continuous on $[0,pi/2]$ and is differentiable in $(0,pi/2)$.
Its derivative is
$$f'(x)=psin xleft(cos^{p-1}x-frac{sin px}{sin x}right),xin(0,pi/2).$$
Because $pin(0,1)$, $0<px<x<pi/2$, we have
$$0<frac{sin px}{sin x}<1<frac{1}{cos^{1-p}x}=cos^{p-1}x.$$
So $$f'(x)>0, forall xin(0,pi/2).$$
This implies $$f(x)>f(0)=0,forall xin(0,pi/2).$$
answered Jan 22 at 7:56
RiemannRiemann
3,4651322
$endgroup$
add a comment |
$begingroup$
Use Mean Value Theorem you can prove the following lemma:
suppose $f(x)$ is continuous on $[a,b]$ and has derivative in $(a,b)$. If $f'(x)<0$ for $xin(a,b)$, then $f(x)$ is strictly decreasing on $[a,b]$.
So you need only $f'(Theta) < 0$ for $Thetain(0,pi/2)$ in your situation!
Also we can deal this one as follows:
For fix $pin(0,1)$, let $$f(x)=cos px-cos^px,xin[0,pi/2],$$
then $f$ is continuous on $[0,pi/2]$ and is differentiable in $(0,pi/2)$.
Its derivative is
$$f'(x)=psin xleft(cos^{p-1}x-frac{sin px}{sin x}right),xin(0,pi/2).$$
Because $pin(0,1)$, $0<px<x<pi/2$, we have
$$0<frac{sin px}{sin x}<1<frac{1}{cos^{1-p}x}=cos^{p-1}x.$$
So $$f'(x)>0, forall xin(0,pi/2).$$
This implies $$f(x)>f(0)=0,forall xin(0,pi/2).$$
answered Jan 22 at 7:56
RiemannRiemann
3,4651322
$endgroup$
Use Mean Value Theorem you can prove the following lemma:
suppose $f(x)$ is continuous on $[a,b]$ and has derivative in $(a,b)$. If $f'(x)<0$ for $xin(a,b)$, then $f(x)$ is strictly decreasing on $[a,b]$.
So you need only $f'(Theta) < 0$ for $Thetain(0,pi/2)$ in your situation!
Also we can deal this one as follows:
For fix $pin(0,1)$, let $$f(x)=cos px-cos^px,xin[0,pi/2],$$
then $f$ is continuous on $[0,pi/2]$ and is differentiable in $(0,pi/2)$.
Its derivative is
$$f'(x)=psin xleft(cos^{p-1}x-frac{sin px}{sin x}right),xin(0,pi/2).$$
Because $pin(0,1)$, $0<px<x<pi/2$, we have
$$0<frac{sin px}{sin x}<1<frac{1}{cos^{1-p}x}=cos^{p-1}x.$$
So $$f'(x)>0, forall xin(0,pi/2).$$
This implies $$f(x)>f(0)=0,forall xin(0,pi/2).$$
answered Jan 22 at 7:56
RiemannRiemann
3,4651322
edited Jan 22 at 11:54
answered Jan 22 at 7:56
RiemannRiemann
3,4651322
answered Jan 22 at 7:56
RiemannRiemann
3,4651322
answered Jan 22 at 7:56
RiemannRiemann
3,4651322
3,4651322
add a comment |
add a comment |
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$begingroup$
if it were the case that $f$ rises above $1$ at some $Theta in (0,frac{pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,frac{pi}{2})$. Why???
$endgroup$
– Riemann
Jan 22 at 9:39
$begingroup$
As a continuous function on a compact interval $[0,frac{pi}{2}]$, $f$ must assume its maximum value somewhere in the interval (Extreme Value Theorem); if $f(Theta)>1$ somewhere in $(0,frac{pi}{2})$, then that maximum must lie within $(0, frac{pi}{2})$ because the values on the borders are 0 and 1.
$endgroup$
– AnatolyVorobey
Jan 22 at 9:52
1
$begingroup$
Yes, it is right.
$endgroup$
– Riemann
Jan 22 at 9:56