Mixing
Possible to prove that a particular trigonometric expression is always positive?
$begingroup$
This is a continuation of the an earlier post where the geometric motivation was presented. Here I'd like to ask: is it possible to prove $Delta > 0$ always?
$$begin{align}
Delta &equiv sin(t) sinleft(r+ (2 pi -2 r - t)frac{epsilon}4 right) sinleft( frac{2 - epsilon}2 (pi-r-t)right) \
&quad {} - frac{2 - epsilon}2 sin(r) sinleft(t-frac{epsilon ; t}{4}right) sin(r+t)
end{align}$$
where $$0<r<frac{pi}{4} qquad 0<t<frac{pi}{4} qquad 0<epsilon <1$$
Some relevant posts include this one that renders the final form of $Delta$, which hasn't gotten satisfactory answers.
calculus algebra-precalculus trigonometry optimization
asked Jan 7 at 20:09
RoryHectorRoryHector
7914
$endgroup$
add a comment |
$begingroup$
This is a continuation of the an earlier post where the geometric motivation was presented. Here I'd like to ask: is it possible to prove $Delta > 0$ always?
$$begin{align}
Delta &equiv sin(t) sinleft(r+ (2 pi -2 r - t)frac{epsilon}4 right) sinleft( frac{2 - epsilon}2 (pi-r-t)right) \
&quad {} - frac{2 - epsilon}2 sin(r) sinleft(t-frac{epsilon ; t}{4}right) sin(r+t)
end{align}$$
where $$0<r<frac{pi}{4} qquad 0<t<frac{pi}{4} qquad 0<epsilon <1$$
Some relevant posts include this one that renders the final form of $Delta$, which hasn't gotten satisfactory answers.
calculus algebra-precalculus trigonometry optimization
asked Jan 7 at 20:09
RoryHectorRoryHector
7914
$endgroup$
$begingroup$
What have you tried? where did you get stuck?
$endgroup$
– Neal
Jan 7 at 20:14
2
$begingroup$
@RoryHector: To repeat a comment I made on one of your other questions: It might help if you could provide the context from which this expression arises. There may be a clear geometric reason for it to be positive.
$endgroup$
– Blue
Jan 7 at 20:40
1
$begingroup$
This intrigues me so I did some numerical exploration. It looks like the infimum is $0$ up to floating point error.
$endgroup$
– Neal
Jan 7 at 21:36
2
$begingroup$
@RoryHector: There was no indication that the question here had any bearing on the previous question. (So, you might want to link to the previous one to add context.) Also, you should explain the "complicated" nature of the situation *in the question itself", so that it doesn't get lost in comments that people may not read; there's also the advantage that you can add images to help explain what's going on.
$endgroup$
– Blue
Jan 7 at 23:37
$begingroup$
Of course the traditional way to show that something is always positive is to write it as a sum of squares plus a positive constant.
$endgroup$
– marty cohen
Jan 20 at 16:50
add a comment |
$begingroup$
This is a continuation of the an earlier post where the geometric motivation was presented. Here I'd like to ask: is it possible to prove $Delta > 0$ always?
$$begin{align}
Delta &equiv sin(t) sinleft(r+ (2 pi -2 r - t)frac{epsilon}4 right) sinleft( frac{2 - epsilon}2 (pi-r-t)right) \
&quad {} - frac{2 - epsilon}2 sin(r) sinleft(t-frac{epsilon ; t}{4}right) sin(r+t)
end{align}$$
where $$0<r<frac{pi}{4} qquad 0<t<frac{pi}{4} qquad 0<epsilon <1$$
Some relevant posts include this one that renders the final form of $Delta$, which hasn't gotten satisfactory answers.
calculus algebra-precalculus trigonometry optimization
asked Jan 7 at 20:09
RoryHectorRoryHector
7914
$endgroup$
This is a continuation of the an earlier post where the geometric motivation was presented. Here I'd like to ask: is it possible to prove $Delta > 0$ always?
$$begin{align}
Delta &equiv sin(t) sinleft(r+ (2 pi -2 r - t)frac{epsilon}4 right) sinleft( frac{2 - epsilon}2 (pi-r-t)right) \
&quad {} - frac{2 - epsilon}2 sin(r) sinleft(t-frac{epsilon ; t}{4}right) sin(r+t)
end{align}$$
where $$0<r<frac{pi}{4} qquad 0<t<frac{pi}{4} qquad 0<epsilon <1$$
Some relevant posts include this one that renders the final form of $Delta$, which hasn't gotten satisfactory answers.
calculus algebra-precalculus trigonometry optimization
calculus algebra-precalculus trigonometry optimization
asked Jan 7 at 20:09
RoryHectorRoryHector
7914
asked Jan 7 at 20:09
RoryHectorRoryHector
7914
edited 2 days ago
RoryHector
asked Jan 7 at 20:09
RoryHectorRoryHector
7914
asked Jan 7 at 20:09
RoryHectorRoryHector
7914
asked Jan 7 at 20:09
RoryHectorRoryHector
7914
7914
$begingroup$
What have you tried? where did you get stuck?
$endgroup$
– Neal
Jan 7 at 20:14
2
$begingroup$
@RoryHector: To repeat a comment I made on one of your other questions: It might help if you could provide the context from which this expression arises. There may be a clear geometric reason for it to be positive.
$endgroup$
– Blue
Jan 7 at 20:40
1
$begingroup$
This intrigues me so I did some numerical exploration. It looks like the infimum is $0$ up to floating point error.
$endgroup$
– Neal
Jan 7 at 21:36
2
$begingroup$
@RoryHector: There was no indication that the question here had any bearing on the previous question. (So, you might want to link to the previous one to add context.) Also, you should explain the "complicated" nature of the situation *in the question itself", so that it doesn't get lost in comments that people may not read; there's also the advantage that you can add images to help explain what's going on.
$endgroup$
– Blue
Jan 7 at 23:37
$begingroup$
Of course the traditional way to show that something is always positive is to write it as a sum of squares plus a positive constant.
$endgroup$
– marty cohen
Jan 20 at 16:50
add a comment |
$begingroup$
What have you tried? where did you get stuck?
$endgroup$
– Neal
Jan 7 at 20:14
2
$begingroup$
@RoryHector: To repeat a comment I made on one of your other questions: It might help if you could provide the context from which this expression arises. There may be a clear geometric reason for it to be positive.
$endgroup$
– Blue
Jan 7 at 20:40
1
$begingroup$
This intrigues me so I did some numerical exploration. It looks like the infimum is $0$ up to floating point error.
$endgroup$
– Neal
Jan 7 at 21:36
2
$begingroup$
@RoryHector: There was no indication that the question here had any bearing on the previous question. (So, you might want to link to the previous one to add context.) Also, you should explain the "complicated" nature of the situation *in the question itself", so that it doesn't get lost in comments that people may not read; there's also the advantage that you can add images to help explain what's going on.
$endgroup$
– Blue
Jan 7 at 23:37
$begingroup$
Of course the traditional way to show that something is always positive is to write it as a sum of squares plus a positive constant.
$endgroup$
– marty cohen
Jan 20 at 16:50
$begingroup$
What have you tried? where did you get stuck?
$endgroup$
– Neal
Jan 7 at 20:14
$begingroup$
What have you tried? where did you get stuck?
$endgroup$
– Neal
Jan 7 at 20:14
2
2
$begingroup$
@RoryHector: To repeat a comment I made on one of your other questions: It might help if you could provide the context from which this expression arises. There may be a clear geometric reason for it to be positive.
$endgroup$
– Blue
Jan 7 at 20:40
$begingroup$
@RoryHector: To repeat a comment I made on one of your other questions: It might help if you could provide the context from which this expression arises. There may be a clear geometric reason for it to be positive.
$endgroup$
– Blue
Jan 7 at 20:40
1
1
$begingroup$
This intrigues me so I did some numerical exploration. It looks like the infimum is $0$ up to floating point error.
$endgroup$
– Neal
Jan 7 at 21:36
$begingroup$
This intrigues me so I did some numerical exploration. It looks like the infimum is $0$ up to floating point error.
$endgroup$
– Neal
Jan 7 at 21:36
2
2
$begingroup$
@RoryHector: There was no indication that the question here had any bearing on the previous question. (So, you might want to link to the previous one to add context.) Also, you should explain the "complicated" nature of the situation *in the question itself", so that it doesn't get lost in comments that people may not read; there's also the advantage that you can add images to help explain what's going on.
$endgroup$
– Blue
Jan 7 at 23:37
$begingroup$
@RoryHector: There was no indication that the question here had any bearing on the previous question. (So, you might want to link to the previous one to add context.) Also, you should explain the "complicated" nature of the situation *in the question itself", so that it doesn't get lost in comments that people may not read; there's also the advantage that you can add images to help explain what's going on.
$endgroup$
– Blue
Jan 7 at 23:37
$begingroup$
Of course the traditional way to show that something is always positive is to write it as a sum of squares plus a positive constant.
$endgroup$
– marty cohen
Jan 20 at 16:50
$begingroup$
Of course the traditional way to show that something is always positive is to write it as a sum of squares plus a positive constant.
$endgroup$
– marty cohen
Jan 20 at 16:50
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Given
$$0<r<frac{pi}{4} qquad 0<t<frac{pi}{4} qquad 0<varepsilon <1tag1$$
Easily to see that
$$t<dfracpi2,quad dfrac{varepsilon t}4 <dfracpi{16}.$$
At the same time, sine increases in $left(0,dfracpi2right).$
Therefore,
$$begin{align}
&sin t >sinleft(t-dfrac{varepsilon t}4right),\[4pt]
&Delta > sin tleft(sinleft(r+(2 pi -2 r - t)frac{varepsilon}4 right) sinleft(frac{2 - varepsilon}2 (pi-r-t)right) - frac{2 - varepsilon}2sin rsin(r+t)right).
end{align}$$
On the other hand,
$$sinleft(frac{2-varepsilon}2(pi-r-t)right)
= sinleft((pi-r-t)-(2pi-2r)fracvarepsilon4right)\
= sinleft(r+t+(2pi-2r)fracvarepsilon4right)
= sinleft(r+t+dfrac{varepsilon t}4+(2pi-2r-t)fracvarepsilon4right).$$
So it is sufficiently to prove inequality $delta(varepsilon) >0,$ where
$$delta(varepsilon) = sin(r+varepsilonvarphi)sinleft(r+t+dfrac{varepsilon t}4+varepsilonvarphiright) - frac{2 - varepsilon}2 sin(r)sin(r+t),tag2$$
$$varphi = dfrac{2pi-2r-t}4 inleft(dfrac{5pi}{16},dfracpi2right),quad dfrac{varepsilon t}4 <dfracpi{16},tag3$$
under the conditions $(1).$
Really,
$$delta(varepsilon) = sin(r+varepsilonvarphi)sinleft(r+t+dfrac{varepsilon t}4+varepsilonvarphiright) - frac{2 - varepsilon}2 sin(r)sin(r+t)\
= frac12left(cosleft(t+dfrac{varepsilon t}4right) - cosleft(2r+t+dfrac{varepsilon t}4+2varepsilonvarphiright)- cos(t) + cos(2r+t) +varepsilonsin(r)sin(r+t)right)\
= frac12left(cosleft(t+dfrac{varepsilon t}4right) - cos(t) - cosleft(2r+t+dfrac{varepsilon t}4+2varepsilonvarphiright) + cos(2r+t) +varepsilonsin(r)sin(r+t)right)\
= -sindfrac{varepsilon t}8 sinleft(t+dfrac{varepsilon t}8right) + sinleft(varepsilonvarphi+dfrac{varepsilon t}8right) sinleft(2r+t+varepsilonvarphi+dfrac{varepsilon t}8right) + dfracvarepsilon2sin(r)sin(r+t).$$
Taking in account that
$$varepsilon < 1 < 2-dfrac{r+2t}{pi-r},tag4$$
one can get
$$t+dfrac{varepsilon t}8 < t+dfrac{varepsilon t}8+2r+varepsilonvarphi = pi-t-dfrac{varepsilon t}8-pi+2t+dfrac{varepsilon t}4+2r+varepsilondfrac{2pi-2r-t}4\
= pi-t-dfrac{varepsilon t}8-pi+2t+2r + varepsilondfrac{pi-r}2
< pi-t-dfrac{varepsilon t}8-pi+2t+2r + left(2-dfrac{r+2t}{pi-r}right)dfrac{pi-r}2\
= pi-t-dfrac{varepsilon t}8-pi+2t+2r + pi-r-r-2t
= pi-t-dfrac{varepsilon t}8,$$
$$sinleft(t+dfrac{varepsilon t}8+2r+varepsilonvarphiright) > sinleft(t+dfrac{varepsilon t}8right),$$
$$delta(varepsilon) = sinleft(dfrac{varepsilon t}8+varepsilonvarphiright) sinleft(t+dfrac{varepsilon t}8+2r+varepsilonvarphiright) - sindfrac{varepsilon t}8 sinleft(t+dfrac{varepsilon t}8right) + dfracvarepsilon2sin(r)sin(r+t) > 0.$$
$mathbf{Proved.}$
answered Jan 24 at 2:52
Yuri NegometyanovYuri Negometyanov
11.2k1728
$endgroup$
$begingroup$
Early on you say it is easy to see that $frac{2-epsilon}{2} (pi - r- t) < frac{pi}{2}$. Can you explain how you can say that? I think its clear that $frac{2-epsilon}{2} (pi - r- t) < pi$ but that's it
$endgroup$
– RoryHector
2 days ago
$begingroup$
@RoryHector Thanks! Fixed.
$endgroup$
– Yuri Negometyanov
yesterday
$begingroup$
@RoryHector What about updated varant?
$endgroup$
– Yuri Negometyanov
48 mins ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Given
$$0<r<frac{pi}{4} qquad 0<t<frac{pi}{4} qquad 0<varepsilon <1tag1$$
Easily to see that
$$t<dfracpi2,quad dfrac{varepsilon t}4 <dfracpi{16}.$$
At the same time, sine increases in $left(0,dfracpi2right).$
Therefore,
$$begin{align}
&sin t >sinleft(t-dfrac{varepsilon t}4right),\[4pt]
&Delta > sin tleft(sinleft(r+(2 pi -2 r - t)frac{varepsilon}4 right) sinleft(frac{2 - varepsilon}2 (pi-r-t)right) - frac{2 - varepsilon}2sin rsin(r+t)right).
end{align}$$
On the other hand,
$$sinleft(frac{2-varepsilon}2(pi-r-t)right)
= sinleft((pi-r-t)-(2pi-2r)fracvarepsilon4right)\
= sinleft(r+t+(2pi-2r)fracvarepsilon4right)
= sinleft(r+t+dfrac{varepsilon t}4+(2pi-2r-t)fracvarepsilon4right).$$
So it is sufficiently to prove inequality $delta(varepsilon) >0,$ where
$$delta(varepsilon) = sin(r+varepsilonvarphi)sinleft(r+t+dfrac{varepsilon t}4+varepsilonvarphiright) - frac{2 - varepsilon}2 sin(r)sin(r+t),tag2$$
$$varphi = dfrac{2pi-2r-t}4 inleft(dfrac{5pi}{16},dfracpi2right),quad dfrac{varepsilon t}4 <dfracpi{16},tag3$$
under the conditions $(1).$
Really,
$$delta(varepsilon) = sin(r+varepsilonvarphi)sinleft(r+t+dfrac{varepsilon t}4+varepsilonvarphiright) - frac{2 - varepsilon}2 sin(r)sin(r+t)\
= frac12left(cosleft(t+dfrac{varepsilon t}4right) - cosleft(2r+t+dfrac{varepsilon t}4+2varepsilonvarphiright)- cos(t) + cos(2r+t) +varepsilonsin(r)sin(r+t)right)\
= frac12left(cosleft(t+dfrac{varepsilon t}4right) - cos(t) - cosleft(2r+t+dfrac{varepsilon t}4+2varepsilonvarphiright) + cos(2r+t) +varepsilonsin(r)sin(r+t)right)\
= -sindfrac{varepsilon t}8 sinleft(t+dfrac{varepsilon t}8right) + sinleft(varepsilonvarphi+dfrac{varepsilon t}8right) sinleft(2r+t+varepsilonvarphi+dfrac{varepsilon t}8right) + dfracvarepsilon2sin(r)sin(r+t).$$
Taking in account that
$$varepsilon < 1 < 2-dfrac{r+2t}{pi-r},tag4$$
one can get
$$t+dfrac{varepsilon t}8 < t+dfrac{varepsilon t}8+2r+varepsilonvarphi = pi-t-dfrac{varepsilon t}8-pi+2t+dfrac{varepsilon t}4+2r+varepsilondfrac{2pi-2r-t}4\
= pi-t-dfrac{varepsilon t}8-pi+2t+2r + varepsilondfrac{pi-r}2
< pi-t-dfrac{varepsilon t}8-pi+2t+2r + left(2-dfrac{r+2t}{pi-r}right)dfrac{pi-r}2\
= pi-t-dfrac{varepsilon t}8-pi+2t+2r + pi-r-r-2t
= pi-t-dfrac{varepsilon t}8,$$
$$sinleft(t+dfrac{varepsilon t}8+2r+varepsilonvarphiright) > sinleft(t+dfrac{varepsilon t}8right),$$
$$delta(varepsilon) = sinleft(dfrac{varepsilon t}8+varepsilonvarphiright) sinleft(t+dfrac{varepsilon t}8+2r+varepsilonvarphiright) - sindfrac{varepsilon t}8 sinleft(t+dfrac{varepsilon t}8right) + dfracvarepsilon2sin(r)sin(r+t) > 0.$$
$mathbf{Proved.}$
answered Jan 24 at 2:52
Yuri NegometyanovYuri Negometyanov
11.2k1728
$endgroup$
$begingroup$
Early on you say it is easy to see that $frac{2-epsilon}{2} (pi - r- t) < frac{pi}{2}$. Can you explain how you can say that? I think its clear that $frac{2-epsilon}{2} (pi - r- t) < pi$ but that's it
$endgroup$
– RoryHector
2 days ago
$begingroup$
@RoryHector Thanks! Fixed.
$endgroup$
– Yuri Negometyanov
yesterday
$begingroup$
@RoryHector What about updated varant?
$endgroup$
– Yuri Negometyanov
48 mins ago
add a comment |
$begingroup$
Given
$$0<r<frac{pi}{4} qquad 0<t<frac{pi}{4} qquad 0<varepsilon <1tag1$$
Easily to see that
$$t<dfracpi2,quad dfrac{varepsilon t}4 <dfracpi{16}.$$
At the same time, sine increases in $left(0,dfracpi2right).$
Therefore,
$$begin{align}
&sin t >sinleft(t-dfrac{varepsilon t}4right),\[4pt]
&Delta > sin tleft(sinleft(r+(2 pi -2 r - t)frac{varepsilon}4 right) sinleft(frac{2 - varepsilon}2 (pi-r-t)right) - frac{2 - varepsilon}2sin rsin(r+t)right).
end{align}$$
On the other hand,
$$sinleft(frac{2-varepsilon}2(pi-r-t)right)
= sinleft((pi-r-t)-(2pi-2r)fracvarepsilon4right)\
= sinleft(r+t+(2pi-2r)fracvarepsilon4right)
= sinleft(r+t+dfrac{varepsilon t}4+(2pi-2r-t)fracvarepsilon4right).$$
So it is sufficiently to prove inequality $delta(varepsilon) >0,$ where
$$delta(varepsilon) = sin(r+varepsilonvarphi)sinleft(r+t+dfrac{varepsilon t}4+varepsilonvarphiright) - frac{2 - varepsilon}2 sin(r)sin(r+t),tag2$$
$$varphi = dfrac{2pi-2r-t}4 inleft(dfrac{5pi}{16},dfracpi2right),quad dfrac{varepsilon t}4 <dfracpi{16},tag3$$
under the conditions $(1).$
Really,
$$delta(varepsilon) = sin(r+varepsilonvarphi)sinleft(r+t+dfrac{varepsilon t}4+varepsilonvarphiright) - frac{2 - varepsilon}2 sin(r)sin(r+t)\
= frac12left(cosleft(t+dfrac{varepsilon t}4right) - cosleft(2r+t+dfrac{varepsilon t}4+2varepsilonvarphiright)- cos(t) + cos(2r+t) +varepsilonsin(r)sin(r+t)right)\
= frac12left(cosleft(t+dfrac{varepsilon t}4right) - cos(t) - cosleft(2r+t+dfrac{varepsilon t}4+2varepsilonvarphiright) + cos(2r+t) +varepsilonsin(r)sin(r+t)right)\
= -sindfrac{varepsilon t}8 sinleft(t+dfrac{varepsilon t}8right) + sinleft(varepsilonvarphi+dfrac{varepsilon t}8right) sinleft(2r+t+varepsilonvarphi+dfrac{varepsilon t}8right) + dfracvarepsilon2sin(r)sin(r+t).$$
Taking in account that
$$varepsilon < 1 < 2-dfrac{r+2t}{pi-r},tag4$$
one can get
$$t+dfrac{varepsilon t}8 < t+dfrac{varepsilon t}8+2r+varepsilonvarphi = pi-t-dfrac{varepsilon t}8-pi+2t+dfrac{varepsilon t}4+2r+varepsilondfrac{2pi-2r-t}4\
= pi-t-dfrac{varepsilon t}8-pi+2t+2r + varepsilondfrac{pi-r}2
< pi-t-dfrac{varepsilon t}8-pi+2t+2r + left(2-dfrac{r+2t}{pi-r}right)dfrac{pi-r}2\
= pi-t-dfrac{varepsilon t}8-pi+2t+2r + pi-r-r-2t
= pi-t-dfrac{varepsilon t}8,$$
$$sinleft(t+dfrac{varepsilon t}8+2r+varepsilonvarphiright) > sinleft(t+dfrac{varepsilon t}8right),$$
$$delta(varepsilon) = sinleft(dfrac{varepsilon t}8+varepsilonvarphiright) sinleft(t+dfrac{varepsilon t}8+2r+varepsilonvarphiright) - sindfrac{varepsilon t}8 sinleft(t+dfrac{varepsilon t}8right) + dfracvarepsilon2sin(r)sin(r+t) > 0.$$
$mathbf{Proved.}$
answered Jan 24 at 2:52
Yuri NegometyanovYuri Negometyanov
11.2k1728
$endgroup$
$begingroup$
Early on you say it is easy to see that $frac{2-epsilon}{2} (pi - r- t) < frac{pi}{2}$. Can you explain how you can say that? I think its clear that $frac{2-epsilon}{2} (pi - r- t) < pi$ but that's it
$endgroup$
– RoryHector
2 days ago
$begingroup$
@RoryHector Thanks! Fixed.
$endgroup$
– Yuri Negometyanov
yesterday
$begingroup$
@RoryHector What about updated varant?
$endgroup$
– Yuri Negometyanov
48 mins ago
add a comment |
$begingroup$
Given
$$0<r<frac{pi}{4} qquad 0<t<frac{pi}{4} qquad 0<varepsilon <1tag1$$
Easily to see that
$$t<dfracpi2,quad dfrac{varepsilon t}4 <dfracpi{16}.$$
At the same time, sine increases in $left(0,dfracpi2right).$
Therefore,
$$begin{align}
&sin t >sinleft(t-dfrac{varepsilon t}4right),\[4pt]
&Delta > sin tleft(sinleft(r+(2 pi -2 r - t)frac{varepsilon}4 right) sinleft(frac{2 - varepsilon}2 (pi-r-t)right) - frac{2 - varepsilon}2sin rsin(r+t)right).
end{align}$$
On the other hand,
$$sinleft(frac{2-varepsilon}2(pi-r-t)right)
= sinleft((pi-r-t)-(2pi-2r)fracvarepsilon4right)\
= sinleft(r+t+(2pi-2r)fracvarepsilon4right)
= sinleft(r+t+dfrac{varepsilon t}4+(2pi-2r-t)fracvarepsilon4right).$$
So it is sufficiently to prove inequality $delta(varepsilon) >0,$ where
$$delta(varepsilon) = sin(r+varepsilonvarphi)sinleft(r+t+dfrac{varepsilon t}4+varepsilonvarphiright) - frac{2 - varepsilon}2 sin(r)sin(r+t),tag2$$
$$varphi = dfrac{2pi-2r-t}4 inleft(dfrac{5pi}{16},dfracpi2right),quad dfrac{varepsilon t}4 <dfracpi{16},tag3$$
under the conditions $(1).$
Really,
$$delta(varepsilon) = sin(r+varepsilonvarphi)sinleft(r+t+dfrac{varepsilon t}4+varepsilonvarphiright) - frac{2 - varepsilon}2 sin(r)sin(r+t)\
= frac12left(cosleft(t+dfrac{varepsilon t}4right) - cosleft(2r+t+dfrac{varepsilon t}4+2varepsilonvarphiright)- cos(t) + cos(2r+t) +varepsilonsin(r)sin(r+t)right)\
= frac12left(cosleft(t+dfrac{varepsilon t}4right) - cos(t) - cosleft(2r+t+dfrac{varepsilon t}4+2varepsilonvarphiright) + cos(2r+t) +varepsilonsin(r)sin(r+t)right)\
= -sindfrac{varepsilon t}8 sinleft(t+dfrac{varepsilon t}8right) + sinleft(varepsilonvarphi+dfrac{varepsilon t}8right) sinleft(2r+t+varepsilonvarphi+dfrac{varepsilon t}8right) + dfracvarepsilon2sin(r)sin(r+t).$$
Taking in account that
$$varepsilon < 1 < 2-dfrac{r+2t}{pi-r},tag4$$
one can get
$$t+dfrac{varepsilon t}8 < t+dfrac{varepsilon t}8+2r+varepsilonvarphi = pi-t-dfrac{varepsilon t}8-pi+2t+dfrac{varepsilon t}4+2r+varepsilondfrac{2pi-2r-t}4\
= pi-t-dfrac{varepsilon t}8-pi+2t+2r + varepsilondfrac{pi-r}2
< pi-t-dfrac{varepsilon t}8-pi+2t+2r + left(2-dfrac{r+2t}{pi-r}right)dfrac{pi-r}2\
= pi-t-dfrac{varepsilon t}8-pi+2t+2r + pi-r-r-2t
= pi-t-dfrac{varepsilon t}8,$$
$$sinleft(t+dfrac{varepsilon t}8+2r+varepsilonvarphiright) > sinleft(t+dfrac{varepsilon t}8right),$$
$$delta(varepsilon) = sinleft(dfrac{varepsilon t}8+varepsilonvarphiright) sinleft(t+dfrac{varepsilon t}8+2r+varepsilonvarphiright) - sindfrac{varepsilon t}8 sinleft(t+dfrac{varepsilon t}8right) + dfracvarepsilon2sin(r)sin(r+t) > 0.$$
$mathbf{Proved.}$
answered Jan 24 at 2:52
Yuri NegometyanovYuri Negometyanov
11.2k1728
$endgroup$
Given
$$0<r<frac{pi}{4} qquad 0<t<frac{pi}{4} qquad 0<varepsilon <1tag1$$
Easily to see that
$$t<dfracpi2,quad dfrac{varepsilon t}4 <dfracpi{16}.$$
At the same time, sine increases in $left(0,dfracpi2right).$
Therefore,
$$begin{align}
&sin t >sinleft(t-dfrac{varepsilon t}4right),\[4pt]
&Delta > sin tleft(sinleft(r+(2 pi -2 r - t)frac{varepsilon}4 right) sinleft(frac{2 - varepsilon}2 (pi-r-t)right) - frac{2 - varepsilon}2sin rsin(r+t)right).
end{align}$$
On the other hand,
$$sinleft(frac{2-varepsilon}2(pi-r-t)right)
= sinleft((pi-r-t)-(2pi-2r)fracvarepsilon4right)\
= sinleft(r+t+(2pi-2r)fracvarepsilon4right)
= sinleft(r+t+dfrac{varepsilon t}4+(2pi-2r-t)fracvarepsilon4right).$$
So it is sufficiently to prove inequality $delta(varepsilon) >0,$ where
$$delta(varepsilon) = sin(r+varepsilonvarphi)sinleft(r+t+dfrac{varepsilon t}4+varepsilonvarphiright) - frac{2 - varepsilon}2 sin(r)sin(r+t),tag2$$
$$varphi = dfrac{2pi-2r-t}4 inleft(dfrac{5pi}{16},dfracpi2right),quad dfrac{varepsilon t}4 <dfracpi{16},tag3$$
under the conditions $(1).$
Really,
$$delta(varepsilon) = sin(r+varepsilonvarphi)sinleft(r+t+dfrac{varepsilon t}4+varepsilonvarphiright) - frac{2 - varepsilon}2 sin(r)sin(r+t)\
= frac12left(cosleft(t+dfrac{varepsilon t}4right) - cosleft(2r+t+dfrac{varepsilon t}4+2varepsilonvarphiright)- cos(t) + cos(2r+t) +varepsilonsin(r)sin(r+t)right)\
= frac12left(cosleft(t+dfrac{varepsilon t}4right) - cos(t) - cosleft(2r+t+dfrac{varepsilon t}4+2varepsilonvarphiright) + cos(2r+t) +varepsilonsin(r)sin(r+t)right)\
= -sindfrac{varepsilon t}8 sinleft(t+dfrac{varepsilon t}8right) + sinleft(varepsilonvarphi+dfrac{varepsilon t}8right) sinleft(2r+t+varepsilonvarphi+dfrac{varepsilon t}8right) + dfracvarepsilon2sin(r)sin(r+t).$$
Taking in account that
$$varepsilon < 1 < 2-dfrac{r+2t}{pi-r},tag4$$
one can get
$$t+dfrac{varepsilon t}8 < t+dfrac{varepsilon t}8+2r+varepsilonvarphi = pi-t-dfrac{varepsilon t}8-pi+2t+dfrac{varepsilon t}4+2r+varepsilondfrac{2pi-2r-t}4\
= pi-t-dfrac{varepsilon t}8-pi+2t+2r + varepsilondfrac{pi-r}2
< pi-t-dfrac{varepsilon t}8-pi+2t+2r + left(2-dfrac{r+2t}{pi-r}right)dfrac{pi-r}2\
= pi-t-dfrac{varepsilon t}8-pi+2t+2r + pi-r-r-2t
= pi-t-dfrac{varepsilon t}8,$$
$$sinleft(t+dfrac{varepsilon t}8+2r+varepsilonvarphiright) > sinleft(t+dfrac{varepsilon t}8right),$$
$$delta(varepsilon) = sinleft(dfrac{varepsilon t}8+varepsilonvarphiright) sinleft(t+dfrac{varepsilon t}8+2r+varepsilonvarphiright) - sindfrac{varepsilon t}8 sinleft(t+dfrac{varepsilon t}8right) + dfracvarepsilon2sin(r)sin(r+t) > 0.$$
$mathbf{Proved.}$
answered Jan 24 at 2:52
Yuri NegometyanovYuri Negometyanov
11.2k1728
edited 50 mins ago
answered Jan 24 at 2:52
Yuri NegometyanovYuri Negometyanov
11.2k1728
answered Jan 24 at 2:52
Yuri NegometyanovYuri Negometyanov
11.2k1728
answered Jan 24 at 2:52
Yuri NegometyanovYuri Negometyanov
11.2k1728
11.2k1728
$begingroup$
Early on you say it is easy to see that $frac{2-epsilon}{2} (pi - r- t) < frac{pi}{2}$. Can you explain how you can say that? I think its clear that $frac{2-epsilon}{2} (pi - r- t) < pi$ but that's it
$endgroup$
– RoryHector
2 days ago
$begingroup$
@RoryHector Thanks! Fixed.
$endgroup$
– Yuri Negometyanov
yesterday
$begingroup$
@RoryHector What about updated varant?
$endgroup$
– Yuri Negometyanov
48 mins ago
add a comment |
$begingroup$
Early on you say it is easy to see that $frac{2-epsilon}{2} (pi - r- t) < frac{pi}{2}$. Can you explain how you can say that? I think its clear that $frac{2-epsilon}{2} (pi - r- t) < pi$ but that's it
$endgroup$
– RoryHector
2 days ago
$begingroup$
@RoryHector Thanks! Fixed.
$endgroup$
– Yuri Negometyanov
yesterday
$begingroup$
@RoryHector What about updated varant?
$endgroup$
– Yuri Negometyanov
48 mins ago
$begingroup$
Early on you say it is easy to see that $frac{2-epsilon}{2} (pi - r- t) < frac{pi}{2}$. Can you explain how you can say that? I think its clear that $frac{2-epsilon}{2} (pi - r- t) < pi$ but that's it
$endgroup$
– RoryHector
2 days ago
$begingroup$
Early on you say it is easy to see that $frac{2-epsilon}{2} (pi - r- t) < frac{pi}{2}$. Can you explain how you can say that? I think its clear that $frac{2-epsilon}{2} (pi - r- t) < pi$ but that's it
$endgroup$
– RoryHector
2 days ago
$begingroup$
@RoryHector Thanks! Fixed.
$endgroup$
– Yuri Negometyanov
yesterday
$begingroup$
@RoryHector Thanks! Fixed.
$endgroup$
– Yuri Negometyanov
yesterday
$begingroup$
@RoryHector What about updated varant?
$endgroup$
– Yuri Negometyanov
48 mins ago
$begingroup$
@RoryHector What about updated varant?
$endgroup$
– Yuri Negometyanov
48 mins ago
add a comment |
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$begingroup$
What have you tried? where did you get stuck?
$endgroup$
– Neal
Jan 7 at 20:14
2
$begingroup$
@RoryHector: To repeat a comment I made on one of your other questions: It might help if you could provide the context from which this expression arises. There may be a clear geometric reason for it to be positive.
$endgroup$
– Blue
Jan 7 at 20:40
1
$begingroup$
This intrigues me so I did some numerical exploration. It looks like the infimum is $0$ up to floating point error.
$endgroup$
– Neal
Jan 7 at 21:36
2
$begingroup$
@RoryHector: There was no indication that the question here had any bearing on the previous question. (So, you might want to link to the previous one to add context.) Also, you should explain the "complicated" nature of the situation *in the question itself", so that it doesn't get lost in comments that people may not read; there's also the advantage that you can add images to help explain what's going on.
$endgroup$
– Blue
Jan 7 at 23:37
$begingroup$
Of course the traditional way to show that something is always positive is to write it as a sum of squares plus a positive constant.
$endgroup$
– marty cohen
Jan 20 at 16:50