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Prove $2+2cos(2pitheta) leq 4exp(-2|theta |^2)$.
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I'm struggling to prove the following easy looking inequality from page 7 of Maynard's Primes with Restricted Digits .
Let $theta in mathbbR$, and let $|theta|$ denote the distance to the integer nearest to $theta$. I want to prove $$2+2cos(2pitheta) leq 4exp(-2|theta |^2).$$ From the double angle formula we arrive at the equivalent $$|cos(pi theta) | leq e^theta,$$ and since both sides are 1-periodic, we need only check that $$0 leq e^-theta^2 - cos(pi theta) qquad textfor $theta in left[0,frac12right]$,$$ and $$0 leq e^-(1-theta)^2 + cos(pi theta) qquad textfor $theta in left[frac12,1right]$.$$ It seems like these aren't as easy as "take the derivative a few times." Any suggestions?
inequality upper-lower-bounds
asked Jul 19 at 3:07
Dzoooks
740214
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up vote
2
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favorite
I'm struggling to prove the following easy looking inequality from page 7 of Maynard's Primes with Restricted Digits .
Let $theta in mathbbR$, and let $|theta|$ denote the distance to the integer nearest to $theta$. I want to prove $$2+2cos(2pitheta) leq 4exp(-2|theta |^2).$$ From the double angle formula we arrive at the equivalent $$|cos(pi theta) | leq e^theta,$$ and since both sides are 1-periodic, we need only check that $$0 leq e^-theta^2 - cos(pi theta) qquad textfor $theta in left[0,frac12right]$,$$ and $$0 leq e^-(1-theta)^2 + cos(pi theta) qquad textfor $theta in left[frac12,1right]$.$$ It seems like these aren't as easy as "take the derivative a few times." Any suggestions?
inequality upper-lower-bounds
asked Jul 19 at 3:07
Dzoooks
740214
My instinct says try using complex number...
– Karn Watcharasupat
Jul 19 at 3:11
@stevengregory Yupp, the second one is false.
– amsmath
Jul 19 at 3:40
@amsmath Yes, the second one is incorrectly written. It should be "$0leq e^-(1-theta)^2+cos(pitheta)$" which is true
– Spot
Jul 19 at 4:25
@amsmath - No. I used the wrong formula for $|theta|$. So my comment is meaningless. To avoid confusion, I am going to delete my previous comment.
– steven gregory
Jul 19 at 8:30
@Spot Woops! Corrected now.
– Dzoooks
Jul 19 at 16:21
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm struggling to prove the following easy looking inequality from page 7 of Maynard's Primes with Restricted Digits .
Let $theta in mathbbR$, and let $|theta|$ denote the distance to the integer nearest to $theta$. I want to prove $$2+2cos(2pitheta) leq 4exp(-2|theta |^2).$$ From the double angle formula we arrive at the equivalent $$|cos(pi theta) | leq e^theta,$$ and since both sides are 1-periodic, we need only check that $$0 leq e^-theta^2 - cos(pi theta) qquad textfor $theta in left[0,frac12right]$,$$ and $$0 leq e^-(1-theta)^2 + cos(pi theta) qquad textfor $theta in left[frac12,1right]$.$$ It seems like these aren't as easy as "take the derivative a few times." Any suggestions?
inequality upper-lower-bounds
asked Jul 19 at 3:07
Dzoooks
740214
I'm struggling to prove the following easy looking inequality from page 7 of Maynard's Primes with Restricted Digits .
Let $theta in mathbbR$, and let $|theta|$ denote the distance to the integer nearest to $theta$. I want to prove $$2+2cos(2pitheta) leq 4exp(-2|theta |^2).$$ From the double angle formula we arrive at the equivalent $$|cos(pi theta) | leq e^theta,$$ and since both sides are 1-periodic, we need only check that $$0 leq e^-theta^2 - cos(pi theta) qquad textfor $theta in left[0,frac12right]$,$$ and $$0 leq e^-(1-theta)^2 + cos(pi theta) qquad textfor $theta in left[frac12,1right]$.$$ It seems like these aren't as easy as "take the derivative a few times." Any suggestions?
inequality upper-lower-bounds
asked Jul 19 at 3:07
Dzoooks
740214
edited Jul 19 at 16:21
asked Jul 19 at 3:07
Dzoooks
740214
asked Jul 19 at 3:07
Dzoooks
740214
asked Jul 19 at 3:07
Dzoooks
740214
740214
My instinct says try using complex number...
– Karn Watcharasupat
Jul 19 at 3:11
@stevengregory Yupp, the second one is false.
– amsmath
Jul 19 at 3:40
@amsmath Yes, the second one is incorrectly written. It should be "$0leq e^-(1-theta)^2+cos(pitheta)$" which is true
– Spot
Jul 19 at 4:25
@amsmath - No. I used the wrong formula for $|theta|$. So my comment is meaningless. To avoid confusion, I am going to delete my previous comment.
– steven gregory
Jul 19 at 8:30
@Spot Woops! Corrected now.
– Dzoooks
Jul 19 at 16:21
add a comment |Â
My instinct says try using complex number...
– Karn Watcharasupat
Jul 19 at 3:11
@stevengregory Yupp, the second one is false.
– amsmath
Jul 19 at 3:40
@amsmath Yes, the second one is incorrectly written. It should be "$0leq e^-(1-theta)^2+cos(pitheta)$" which is true
– Spot
Jul 19 at 4:25
@amsmath - No. I used the wrong formula for $|theta|$. So my comment is meaningless. To avoid confusion, I am going to delete my previous comment.
– steven gregory
Jul 19 at 8:30
@Spot Woops! Corrected now.
– Dzoooks
Jul 19 at 16:21
My instinct says try using complex number...
– Karn Watcharasupat
Jul 19 at 3:11
My instinct says try using complex number...
– Karn Watcharasupat
Jul 19 at 3:11
@stevengregory Yupp, the second one is false.
– amsmath
Jul 19 at 3:40
@stevengregory Yupp, the second one is false.
– amsmath
Jul 19 at 3:40
@amsmath Yes, the second one is incorrectly written. It should be "$0leq e^-(1-theta)^2+cos(pitheta)$" which is true
– Spot
Jul 19 at 4:25
@amsmath Yes, the second one is incorrectly written. It should be "$0leq e^-(1-theta)^2+cos(pitheta)$" which is true
– Spot
Jul 19 at 4:25
@amsmath - No. I used the wrong formula for $|theta|$. So my comment is meaningless. To avoid confusion, I am going to delete my previous comment.
– steven gregory
Jul 19 at 8:30
@amsmath - No. I used the wrong formula for $|theta|$. So my comment is meaningless. To avoid confusion, I am going to delete my previous comment.
– steven gregory
Jul 19 at 8:30
@Spot Woops! Corrected now.
– Dzoooks
Jul 19 at 16:21
@Spot Woops! Corrected now.
– Dzoooks
Jul 19 at 16:21
add a comment |Â
3 Answers
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up vote
2
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accepted
In fact, because both sides of $|cospitheta|le e^-VertthetaVert^2$ are symmetric around $frac12$ ($f(theta)=f(1-theta)$), we only need to check $0le e^-theta^2-cospitheta$ for $thetain[0,1/2]$. This can be rewritten as
$$e^-theta^2gecospithetatag1$$
The idea of the proof of $(1)$ is to find two polynomials $p(x),q(x)$ with $cospi xle q(x)le p(x)le e^-x^2$ with $xin[0,1/2]$ and then prove $q(x)le p(x)$ over the same range of $x$. These polynomials are truncations of the Maclaurin series of the two sides of $(1)$, to $x^2$ for the LHS and to $x^4$ for the RHS, so we have
$$e^-x^2ge 1-fracx^22ge1-frac(pi x)^22+frac(pi x)^424gecospi x$$
$$-fracx^22ge-frac(pi x)^22+frac(pi x)^424$$
$$-frac12ge-fracpi^22+fracpi^4x^224$$
$$fracpi^2-12cdotfrac24pi^4ge x^2$$
$$xlesqrtfracpi^2-12cdotfrac24pi^4=1.045dots$$
Since this contains $[0,1/2]$, $q(x)le p(x)$ over the range in question and $(1)$ is proved.
answered Jul 19 at 6:04
Parcly Taxel
33.6k136588
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up vote
2
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Let's prove the first inequality, namely
$$
tag1 cos (pi theta) leq e^-theta^2 text for theta in [0,1/2].
$$
First, we get rid of the exponential by observing that for any $xgeq 0$ we have $e^-x geq 1-x$, hence it is enough to show
$$
tag2 cos (pi theta) leq 1-theta^2 text for theta in [0,1/2].
$$
Consider the function $f(theta) = 1-theta^2 - cos(pi theta)$. We will simply analyse it through differentiation.
We have
$$f'(x) = -2theta +pi sin(pi theta),$$
and
$$
f''(theta) = - 2 + pi^2 cos(pi theta).
$$
Set $theta_0 = frac1pi arccos frac2pi^2$ and notice that $f''(theta) > 0$ for all $theta in [0, theta_0)$ and $f''(theta) <0 $ when $theta> theta_0$. This shows that $f$ is convex in $[0,theta_0)$ and is concave in $(theta_0, 1/2]$.
Since $f'(0)= 0$ and $f''>0$ in $[0,theta_0)$ we see that $f'$ must be positive in $(0,theta_0)$ and hence $f$ is increasing in $[0,theta_0)$; in particular
$f(theta) geq 0$ in $[0,theta_0)$ and $f(theta_0) >0$.
Finally, $f$ is concave in $(theta_0, 1/2]$ and is positive at the both endpoints of this interval, hence it cannot become $0$ in $(theta_0, 1/2]$, since concavity forces $f$ to stay above the straight line joining the points of its graph at $theta_0$ and $1/2$.
I haven't tried the second inequality, but should be something similar.
answered Jul 19 at 5:45
Hayk
1,39129
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up vote
2
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To look at Parly Taxels answer from a different direction.
Prove that
$$|cos(pi theta) | leq e^theta
quad textwhere theta in [0, 0.5]tag1.$$
and where $|theta| = left|theta - lfloortheta + 0.5 rfloor right|$ is the distance to the integer nearest to $theta$.
It's easy to check this for $theta = 0$ and for $theta = 0.5$. So we can ignore those numbers from here on. We can now rewrite eqn$(1)$ as
$$cos(pi theta) leq e^-theta^2
quad textwhere theta in (0, 0.5)tag2.$$
Since $ln$ is strictly increasing on $(0, infty)$, eqn$(2.)$ is equivalent to
$$ln(cos(pi theta)) leq -theta^2
quad textwhere theta in (0, 0.5)tag3.$$
Let $f(theta) = -ln(cos(pi theta)) - theta^2$.
Taking the derivative of both sides wrt $theta$, we find
$$f'(theta) = pi tan(pi theta) - 2theta$$
Since $pi theta$ is in the first quadrant, we know that
$tan(pi theta) > pi theta$. It follows that
$pi tan(pi theta) > pi^2 theta > 2 theta$ Hence $f'(theta) > 0$ and $f(theta)$ is strictly increasing on $(0, 0.5)$.
It follows that eqn$(3.)$, eqn$(2.)$, and therefore eqn$(1.)$ is true.
answered Jul 19 at 9:25
steven gregory
16.4k22055
For the readers, see math.stackexchange.com/questions/98998/… for a geometric proof that $tan(x) > x$ for $x$ in first quadrant.
– Dzoooks
Jul 19 at 16:40
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
In fact, because both sides of $|cospitheta|le e^-VertthetaVert^2$ are symmetric around $frac12$ ($f(theta)=f(1-theta)$), we only need to check $0le e^-theta^2-cospitheta$ for $thetain[0,1/2]$. This can be rewritten as
$$e^-theta^2gecospithetatag1$$
The idea of the proof of $(1)$ is to find two polynomials $p(x),q(x)$ with $cospi xle q(x)le p(x)le e^-x^2$ with $xin[0,1/2]$ and then prove $q(x)le p(x)$ over the same range of $x$. These polynomials are truncations of the Maclaurin series of the two sides of $(1)$, to $x^2$ for the LHS and to $x^4$ for the RHS, so we have
$$e^-x^2ge 1-fracx^22ge1-frac(pi x)^22+frac(pi x)^424gecospi x$$
$$-fracx^22ge-frac(pi x)^22+frac(pi x)^424$$
$$-frac12ge-fracpi^22+fracpi^4x^224$$
$$fracpi^2-12cdotfrac24pi^4ge x^2$$
$$xlesqrtfracpi^2-12cdotfrac24pi^4=1.045dots$$
Since this contains $[0,1/2]$, $q(x)le p(x)$ over the range in question and $(1)$ is proved.
answered Jul 19 at 6:04
Parcly Taxel
33.6k136588
add a comment |Â
up vote
2
down vote
accepted
In fact, because both sides of $|cospitheta|le e^-VertthetaVert^2$ are symmetric around $frac12$ ($f(theta)=f(1-theta)$), we only need to check $0le e^-theta^2-cospitheta$ for $thetain[0,1/2]$. This can be rewritten as
$$e^-theta^2gecospithetatag1$$
The idea of the proof of $(1)$ is to find two polynomials $p(x),q(x)$ with $cospi xle q(x)le p(x)le e^-x^2$ with $xin[0,1/2]$ and then prove $q(x)le p(x)$ over the same range of $x$. These polynomials are truncations of the Maclaurin series of the two sides of $(1)$, to $x^2$ for the LHS and to $x^4$ for the RHS, so we have
$$e^-x^2ge 1-fracx^22ge1-frac(pi x)^22+frac(pi x)^424gecospi x$$
$$-fracx^22ge-frac(pi x)^22+frac(pi x)^424$$
$$-frac12ge-fracpi^22+fracpi^4x^224$$
$$fracpi^2-12cdotfrac24pi^4ge x^2$$
$$xlesqrtfracpi^2-12cdotfrac24pi^4=1.045dots$$
Since this contains $[0,1/2]$, $q(x)le p(x)$ over the range in question and $(1)$ is proved.
answered Jul 19 at 6:04
Parcly Taxel
33.6k136588
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
In fact, because both sides of $|cospitheta|le e^-VertthetaVert^2$ are symmetric around $frac12$ ($f(theta)=f(1-theta)$), we only need to check $0le e^-theta^2-cospitheta$ for $thetain[0,1/2]$. This can be rewritten as
$$e^-theta^2gecospithetatag1$$
The idea of the proof of $(1)$ is to find two polynomials $p(x),q(x)$ with $cospi xle q(x)le p(x)le e^-x^2$ with $xin[0,1/2]$ and then prove $q(x)le p(x)$ over the same range of $x$. These polynomials are truncations of the Maclaurin series of the two sides of $(1)$, to $x^2$ for the LHS and to $x^4$ for the RHS, so we have
$$e^-x^2ge 1-fracx^22ge1-frac(pi x)^22+frac(pi x)^424gecospi x$$
$$-fracx^22ge-frac(pi x)^22+frac(pi x)^424$$
$$-frac12ge-fracpi^22+fracpi^4x^224$$
$$fracpi^2-12cdotfrac24pi^4ge x^2$$
$$xlesqrtfracpi^2-12cdotfrac24pi^4=1.045dots$$
Since this contains $[0,1/2]$, $q(x)le p(x)$ over the range in question and $(1)$ is proved.
answered Jul 19 at 6:04
Parcly Taxel
33.6k136588
In fact, because both sides of $|cospitheta|le e^-VertthetaVert^2$ are symmetric around $frac12$ ($f(theta)=f(1-theta)$), we only need to check $0le e^-theta^2-cospitheta$ for $thetain[0,1/2]$. This can be rewritten as
$$e^-theta^2gecospithetatag1$$
The idea of the proof of $(1)$ is to find two polynomials $p(x),q(x)$ with $cospi xle q(x)le p(x)le e^-x^2$ with $xin[0,1/2]$ and then prove $q(x)le p(x)$ over the same range of $x$. These polynomials are truncations of the Maclaurin series of the two sides of $(1)$, to $x^2$ for the LHS and to $x^4$ for the RHS, so we have
$$e^-x^2ge 1-fracx^22ge1-frac(pi x)^22+frac(pi x)^424gecospi x$$
$$-fracx^22ge-frac(pi x)^22+frac(pi x)^424$$
$$-frac12ge-fracpi^22+fracpi^4x^224$$
$$fracpi^2-12cdotfrac24pi^4ge x^2$$
$$xlesqrtfracpi^2-12cdotfrac24pi^4=1.045dots$$
Since this contains $[0,1/2]$, $q(x)le p(x)$ over the range in question and $(1)$ is proved.
answered Jul 19 at 6:04
Parcly Taxel
33.6k136588
edited Jul 19 at 17:43
answered Jul 19 at 6:04
Parcly Taxel
33.6k136588
answered Jul 19 at 6:04
Parcly Taxel
33.6k136588
answered Jul 19 at 6:04
Parcly Taxel
33.6k136588
33.6k136588
add a comment |Â
add a comment |Â
up vote
2
down vote
Let's prove the first inequality, namely
$$
tag1 cos (pi theta) leq e^-theta^2 text for theta in [0,1/2].
$$
First, we get rid of the exponential by observing that for any $xgeq 0$ we have $e^-x geq 1-x$, hence it is enough to show
$$
tag2 cos (pi theta) leq 1-theta^2 text for theta in [0,1/2].
$$
Consider the function $f(theta) = 1-theta^2 - cos(pi theta)$. We will simply analyse it through differentiation.
We have
$$f'(x) = -2theta +pi sin(pi theta),$$
and
$$
f''(theta) = - 2 + pi^2 cos(pi theta).
$$
Set $theta_0 = frac1pi arccos frac2pi^2$ and notice that $f''(theta) > 0$ for all $theta in [0, theta_0)$ and $f''(theta) <0 $ when $theta> theta_0$. This shows that $f$ is convex in $[0,theta_0)$ and is concave in $(theta_0, 1/2]$.
Since $f'(0)= 0$ and $f''>0$ in $[0,theta_0)$ we see that $f'$ must be positive in $(0,theta_0)$ and hence $f$ is increasing in $[0,theta_0)$; in particular
$f(theta) geq 0$ in $[0,theta_0)$ and $f(theta_0) >0$.
Finally, $f$ is concave in $(theta_0, 1/2]$ and is positive at the both endpoints of this interval, hence it cannot become $0$ in $(theta_0, 1/2]$, since concavity forces $f$ to stay above the straight line joining the points of its graph at $theta_0$ and $1/2$.
I haven't tried the second inequality, but should be something similar.
answered Jul 19 at 5:45
Hayk
1,39129
add a comment |Â
up vote
2
down vote
Let's prove the first inequality, namely
$$
tag1 cos (pi theta) leq e^-theta^2 text for theta in [0,1/2].
$$
First, we get rid of the exponential by observing that for any $xgeq 0$ we have $e^-x geq 1-x$, hence it is enough to show
$$
tag2 cos (pi theta) leq 1-theta^2 text for theta in [0,1/2].
$$
Consider the function $f(theta) = 1-theta^2 - cos(pi theta)$. We will simply analyse it through differentiation.
We have
$$f'(x) = -2theta +pi sin(pi theta),$$
and
$$
f''(theta) = - 2 + pi^2 cos(pi theta).
$$
Set $theta_0 = frac1pi arccos frac2pi^2$ and notice that $f''(theta) > 0$ for all $theta in [0, theta_0)$ and $f''(theta) <0 $ when $theta> theta_0$. This shows that $f$ is convex in $[0,theta_0)$ and is concave in $(theta_0, 1/2]$.
Since $f'(0)= 0$ and $f''>0$ in $[0,theta_0)$ we see that $f'$ must be positive in $(0,theta_0)$ and hence $f$ is increasing in $[0,theta_0)$; in particular
$f(theta) geq 0$ in $[0,theta_0)$ and $f(theta_0) >0$.
Finally, $f$ is concave in $(theta_0, 1/2]$ and is positive at the both endpoints of this interval, hence it cannot become $0$ in $(theta_0, 1/2]$, since concavity forces $f$ to stay above the straight line joining the points of its graph at $theta_0$ and $1/2$.
I haven't tried the second inequality, but should be something similar.
answered Jul 19 at 5:45
Hayk
1,39129
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let's prove the first inequality, namely
$$
tag1 cos (pi theta) leq e^-theta^2 text for theta in [0,1/2].
$$
First, we get rid of the exponential by observing that for any $xgeq 0$ we have $e^-x geq 1-x$, hence it is enough to show
$$
tag2 cos (pi theta) leq 1-theta^2 text for theta in [0,1/2].
$$
Consider the function $f(theta) = 1-theta^2 - cos(pi theta)$. We will simply analyse it through differentiation.
We have
$$f'(x) = -2theta +pi sin(pi theta),$$
and
$$
f''(theta) = - 2 + pi^2 cos(pi theta).
$$
Set $theta_0 = frac1pi arccos frac2pi^2$ and notice that $f''(theta) > 0$ for all $theta in [0, theta_0)$ and $f''(theta) <0 $ when $theta> theta_0$. This shows that $f$ is convex in $[0,theta_0)$ and is concave in $(theta_0, 1/2]$.
Since $f'(0)= 0$ and $f''>0$ in $[0,theta_0)$ we see that $f'$ must be positive in $(0,theta_0)$ and hence $f$ is increasing in $[0,theta_0)$; in particular
$f(theta) geq 0$ in $[0,theta_0)$ and $f(theta_0) >0$.
Finally, $f$ is concave in $(theta_0, 1/2]$ and is positive at the both endpoints of this interval, hence it cannot become $0$ in $(theta_0, 1/2]$, since concavity forces $f$ to stay above the straight line joining the points of its graph at $theta_0$ and $1/2$.
I haven't tried the second inequality, but should be something similar.
answered Jul 19 at 5:45
Hayk
1,39129
Let's prove the first inequality, namely
$$
tag1 cos (pi theta) leq e^-theta^2 text for theta in [0,1/2].
$$
First, we get rid of the exponential by observing that for any $xgeq 0$ we have $e^-x geq 1-x$, hence it is enough to show
$$
tag2 cos (pi theta) leq 1-theta^2 text for theta in [0,1/2].
$$
Consider the function $f(theta) = 1-theta^2 - cos(pi theta)$. We will simply analyse it through differentiation.
We have
$$f'(x) = -2theta +pi sin(pi theta),$$
and
$$
f''(theta) = - 2 + pi^2 cos(pi theta).
$$
Set $theta_0 = frac1pi arccos frac2pi^2$ and notice that $f''(theta) > 0$ for all $theta in [0, theta_0)$ and $f''(theta) <0 $ when $theta> theta_0$. This shows that $f$ is convex in $[0,theta_0)$ and is concave in $(theta_0, 1/2]$.
Since $f'(0)= 0$ and $f''>0$ in $[0,theta_0)$ we see that $f'$ must be positive in $(0,theta_0)$ and hence $f$ is increasing in $[0,theta_0)$; in particular
$f(theta) geq 0$ in $[0,theta_0)$ and $f(theta_0) >0$.
Finally, $f$ is concave in $(theta_0, 1/2]$ and is positive at the both endpoints of this interval, hence it cannot become $0$ in $(theta_0, 1/2]$, since concavity forces $f$ to stay above the straight line joining the points of its graph at $theta_0$ and $1/2$.
I haven't tried the second inequality, but should be something similar.
answered Jul 19 at 5:45
Hayk
1,39129
answered Jul 19 at 5:45
Hayk
1,39129
answered Jul 19 at 5:45
Hayk
1,39129
answered Jul 19 at 5:45
Hayk
1,39129
1,39129
add a comment |Â
add a comment |Â
up vote
2
down vote
To look at Parly Taxels answer from a different direction.
Prove that
$$|cos(pi theta) | leq e^theta
quad textwhere theta in [0, 0.5]tag1.$$
and where $|theta| = left|theta - lfloortheta + 0.5 rfloor right|$ is the distance to the integer nearest to $theta$.
It's easy to check this for $theta = 0$ and for $theta = 0.5$. So we can ignore those numbers from here on. We can now rewrite eqn$(1)$ as
$$cos(pi theta) leq e^-theta^2
quad textwhere theta in (0, 0.5)tag2.$$
Since $ln$ is strictly increasing on $(0, infty)$, eqn$(2.)$ is equivalent to
$$ln(cos(pi theta)) leq -theta^2
quad textwhere theta in (0, 0.5)tag3.$$
Let $f(theta) = -ln(cos(pi theta)) - theta^2$.
Taking the derivative of both sides wrt $theta$, we find
$$f'(theta) = pi tan(pi theta) - 2theta$$
Since $pi theta$ is in the first quadrant, we know that
$tan(pi theta) > pi theta$. It follows that
$pi tan(pi theta) > pi^2 theta > 2 theta$ Hence $f'(theta) > 0$ and $f(theta)$ is strictly increasing on $(0, 0.5)$.
It follows that eqn$(3.)$, eqn$(2.)$, and therefore eqn$(1.)$ is true.
answered Jul 19 at 9:25
steven gregory
16.4k22055
For the readers, see math.stackexchange.com/questions/98998/… for a geometric proof that $tan(x) > x$ for $x$ in first quadrant.
– Dzoooks
Jul 19 at 16:40
add a comment |Â
up vote
2
down vote
To look at Parly Taxels answer from a different direction.
Prove that
$$|cos(pi theta) | leq e^theta
quad textwhere theta in [0, 0.5]tag1.$$
and where $|theta| = left|theta - lfloortheta + 0.5 rfloor right|$ is the distance to the integer nearest to $theta$.
It's easy to check this for $theta = 0$ and for $theta = 0.5$. So we can ignore those numbers from here on. We can now rewrite eqn$(1)$ as
$$cos(pi theta) leq e^-theta^2
quad textwhere theta in (0, 0.5)tag2.$$
Since $ln$ is strictly increasing on $(0, infty)$, eqn$(2.)$ is equivalent to
$$ln(cos(pi theta)) leq -theta^2
quad textwhere theta in (0, 0.5)tag3.$$
Let $f(theta) = -ln(cos(pi theta)) - theta^2$.
Taking the derivative of both sides wrt $theta$, we find
$$f'(theta) = pi tan(pi theta) - 2theta$$
Since $pi theta$ is in the first quadrant, we know that
$tan(pi theta) > pi theta$. It follows that
$pi tan(pi theta) > pi^2 theta > 2 theta$ Hence $f'(theta) > 0$ and $f(theta)$ is strictly increasing on $(0, 0.5)$.
It follows that eqn$(3.)$, eqn$(2.)$, and therefore eqn$(1.)$ is true.
answered Jul 19 at 9:25
steven gregory
16.4k22055
For the readers, see math.stackexchange.com/questions/98998/… for a geometric proof that $tan(x) > x$ for $x$ in first quadrant.
– Dzoooks
Jul 19 at 16:40
add a comment |Â
up vote
2
down vote
up vote
2
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To look at Parly Taxels answer from a different direction.
Prove that
$$|cos(pi theta) | leq e^theta
quad textwhere theta in [0, 0.5]tag1.$$
and where $|theta| = left|theta - lfloortheta + 0.5 rfloor right|$ is the distance to the integer nearest to $theta$.
It's easy to check this for $theta = 0$ and for $theta = 0.5$. So we can ignore those numbers from here on. We can now rewrite eqn$(1)$ as
$$cos(pi theta) leq e^-theta^2
quad textwhere theta in (0, 0.5)tag2.$$
Since $ln$ is strictly increasing on $(0, infty)$, eqn$(2.)$ is equivalent to
$$ln(cos(pi theta)) leq -theta^2
quad textwhere theta in (0, 0.5)tag3.$$
Let $f(theta) = -ln(cos(pi theta)) - theta^2$.
Taking the derivative of both sides wrt $theta$, we find
$$f'(theta) = pi tan(pi theta) - 2theta$$
Since $pi theta$ is in the first quadrant, we know that
$tan(pi theta) > pi theta$. It follows that
$pi tan(pi theta) > pi^2 theta > 2 theta$ Hence $f'(theta) > 0$ and $f(theta)$ is strictly increasing on $(0, 0.5)$.
It follows that eqn$(3.)$, eqn$(2.)$, and therefore eqn$(1.)$ is true.
answered Jul 19 at 9:25
steven gregory
16.4k22055
To look at Parly Taxels answer from a different direction.
Prove that
$$|cos(pi theta) | leq e^theta
quad textwhere theta in [0, 0.5]tag1.$$
and where $|theta| = left|theta - lfloortheta + 0.5 rfloor right|$ is the distance to the integer nearest to $theta$.
It's easy to check this for $theta = 0$ and for $theta = 0.5$. So we can ignore those numbers from here on. We can now rewrite eqn$(1)$ as
$$cos(pi theta) leq e^-theta^2
quad textwhere theta in (0, 0.5)tag2.$$
Since $ln$ is strictly increasing on $(0, infty)$, eqn$(2.)$ is equivalent to
$$ln(cos(pi theta)) leq -theta^2
quad textwhere theta in (0, 0.5)tag3.$$
Let $f(theta) = -ln(cos(pi theta)) - theta^2$.
Taking the derivative of both sides wrt $theta$, we find
$$f'(theta) = pi tan(pi theta) - 2theta$$
Since $pi theta$ is in the first quadrant, we know that
$tan(pi theta) > pi theta$. It follows that
$pi tan(pi theta) > pi^2 theta > 2 theta$ Hence $f'(theta) > 0$ and $f(theta)$ is strictly increasing on $(0, 0.5)$.
It follows that eqn$(3.)$, eqn$(2.)$, and therefore eqn$(1.)$ is true.
answered Jul 19 at 9:25
steven gregory
16.4k22055
edited Jul 19 at 20:24
answered Jul 19 at 9:25
steven gregory
16.4k22055
answered Jul 19 at 9:25
steven gregory
16.4k22055
answered Jul 19 at 9:25
steven gregory
16.4k22055
16.4k22055
For the readers, see math.stackexchange.com/questions/98998/… for a geometric proof that $tan(x) > x$ for $x$ in first quadrant.
– Dzoooks
Jul 19 at 16:40
add a comment |Â
For the readers, see math.stackexchange.com/questions/98998/… for a geometric proof that $tan(x) > x$ for $x$ in first quadrant.
– Dzoooks
Jul 19 at 16:40
For the readers, see math.stackexchange.com/questions/98998/… for a geometric proof that $tan(x) > x$ for $x$ in first quadrant.
– Dzoooks
Jul 19 at 16:40
For the readers, see math.stackexchange.com/questions/98998/… for a geometric proof that $tan(x) > x$ for $x$ in first quadrant.
– Dzoooks
Jul 19 at 16:40
add a comment |Â
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My instinct says try using complex number...
– Karn Watcharasupat
Jul 19 at 3:11
@stevengregory Yupp, the second one is false.
– amsmath
Jul 19 at 3:40
@amsmath Yes, the second one is incorrectly written. It should be "$0leq e^-(1-theta)^2+cos(pitheta)$" which is true
– Spot
Jul 19 at 4:25
@amsmath - No. I used the wrong formula for $|theta|$. So my comment is meaningless. To avoid confusion, I am going to delete my previous comment.
– steven gregory
Jul 19 at 8:30
@Spot Woops! Corrected now.
– Dzoooks
Jul 19 at 16:21