Mixing
elementary but ugly equation, maybe with mathematica
$begingroup$
Fix parameters $1<q<infty$ and $dinmathbb{N}cap[2,infty)$. Let us define a function $f$ by the rule
$$
f(t)=t(1-(d-1)t^q)^{1-1/q}+t^{q-1}(1-(d-1)t^q)^{1/q}-t^q(d-2)-d^{-1}.
$$
I need to prove that there exist a number $0<t<(d-1)^{-1/q}$ such that $f(t)=0$. Ideally, I would like to actually find an exact value for $t$, but this is not absolutely necessary.
This is a fairly elementary problem, but I am nevertheless having difficulty with it.
Technology would be fine. For example, I have Wolfram Mathematica, but unfortunately I don't really know how to use it. Could anyone maybe show me how to do it that way?
If that won't work, an existence proof would be good enough I guess. But I am not seeing any good ideas there, either.
IVT approach fails.
The first and most obvious idea is to use the IVT. However, note that
$$f(0)=-d^{-1}<0$$
and
$$f((d-1)^{-1/q})=-frac{d-2}{d-1}-d^{-1}<0$$
so that the IVT approach will not work on the interval $[0,(d-1)^{-1/q}]$.
The derivative.
To "tighten" the interval so as to make the IVT approach work, we could try maximizing $f$. However, the derivative is given by
$$
f'(t)=(1-(d-1)qt^q)(1-(d-1)t^q)^{-1/q}
+(d-1)(t^{q-2}-qt^{2(q-1)})(1-(d-1)t^q)^{1/q-1}
-(d-2)qt^{q-1}
$$
and finding the roots of this equation is just as difficult as the roots of $f$ itself.
Some special cases.
When $d=2$ and $q=3/2$, solutions are given by
$$t=2^{-2/3}(5^{pm1}(9+4sqrt{5}))^{pm1/3}.$$
When $d=2$ and $q=3$, solutions are given by
$$t=left(frac{1}{10}left(1pm2sqrt{5}right)right)^{1/3}.$$
It would appear that regardless of choice of q, there are solutions when $d=2$ (although I haven't proved it rigorously).
When $d=3$, a solution (not the only one) is given by $t=3^{-1/q}$ (regardless of choice of q).
Thanks guys!
real-analysis analysis
asked Jan 31 at 21:08
Ben WBen W
2,734918
$endgroup$
add a comment |
$begingroup$
Fix parameters $1<q<infty$ and $dinmathbb{N}cap[2,infty)$. Let us define a function $f$ by the rule
$$
f(t)=t(1-(d-1)t^q)^{1-1/q}+t^{q-1}(1-(d-1)t^q)^{1/q}-t^q(d-2)-d^{-1}.
$$
I need to prove that there exist a number $0<t<(d-1)^{-1/q}$ such that $f(t)=0$. Ideally, I would like to actually find an exact value for $t$, but this is not absolutely necessary.
This is a fairly elementary problem, but I am nevertheless having difficulty with it.
Technology would be fine. For example, I have Wolfram Mathematica, but unfortunately I don't really know how to use it. Could anyone maybe show me how to do it that way?
If that won't work, an existence proof would be good enough I guess. But I am not seeing any good ideas there, either.
IVT approach fails.
The first and most obvious idea is to use the IVT. However, note that
$$f(0)=-d^{-1}<0$$
and
$$f((d-1)^{-1/q})=-frac{d-2}{d-1}-d^{-1}<0$$
so that the IVT approach will not work on the interval $[0,(d-1)^{-1/q}]$.
The derivative.
To "tighten" the interval so as to make the IVT approach work, we could try maximizing $f$. However, the derivative is given by
$$
f'(t)=(1-(d-1)qt^q)(1-(d-1)t^q)^{-1/q}
+(d-1)(t^{q-2}-qt^{2(q-1)})(1-(d-1)t^q)^{1/q-1}
-(d-2)qt^{q-1}
$$
and finding the roots of this equation is just as difficult as the roots of $f$ itself.
Some special cases.
When $d=2$ and $q=3/2$, solutions are given by
$$t=2^{-2/3}(5^{pm1}(9+4sqrt{5}))^{pm1/3}.$$
When $d=2$ and $q=3$, solutions are given by
$$t=left(frac{1}{10}left(1pm2sqrt{5}right)right)^{1/3}.$$
It would appear that regardless of choice of q, there are solutions when $d=2$ (although I haven't proved it rigorously).
When $d=3$, a solution (not the only one) is given by $t=3^{-1/q}$ (regardless of choice of q).
Thanks guys!
real-analysis analysis
asked Jan 31 at 21:08
Ben WBen W
2,734918
$endgroup$
add a comment |
$begingroup$
Fix parameters $1<q<infty$ and $dinmathbb{N}cap[2,infty)$. Let us define a function $f$ by the rule
$$
f(t)=t(1-(d-1)t^q)^{1-1/q}+t^{q-1}(1-(d-1)t^q)^{1/q}-t^q(d-2)-d^{-1}.
$$
I need to prove that there exist a number $0<t<(d-1)^{-1/q}$ such that $f(t)=0$. Ideally, I would like to actually find an exact value for $t$, but this is not absolutely necessary.
This is a fairly elementary problem, but I am nevertheless having difficulty with it.
Technology would be fine. For example, I have Wolfram Mathematica, but unfortunately I don't really know how to use it. Could anyone maybe show me how to do it that way?
If that won't work, an existence proof would be good enough I guess. But I am not seeing any good ideas there, either.
IVT approach fails.
The first and most obvious idea is to use the IVT. However, note that
$$f(0)=-d^{-1}<0$$
and
$$f((d-1)^{-1/q})=-frac{d-2}{d-1}-d^{-1}<0$$
so that the IVT approach will not work on the interval $[0,(d-1)^{-1/q}]$.
The derivative.
To "tighten" the interval so as to make the IVT approach work, we could try maximizing $f$. However, the derivative is given by
$$
f'(t)=(1-(d-1)qt^q)(1-(d-1)t^q)^{-1/q}
+(d-1)(t^{q-2}-qt^{2(q-1)})(1-(d-1)t^q)^{1/q-1}
-(d-2)qt^{q-1}
$$
and finding the roots of this equation is just as difficult as the roots of $f$ itself.
Some special cases.
When $d=2$ and $q=3/2$, solutions are given by
$$t=2^{-2/3}(5^{pm1}(9+4sqrt{5}))^{pm1/3}.$$
When $d=2$ and $q=3$, solutions are given by
$$t=left(frac{1}{10}left(1pm2sqrt{5}right)right)^{1/3}.$$
It would appear that regardless of choice of q, there are solutions when $d=2$ (although I haven't proved it rigorously).
When $d=3$, a solution (not the only one) is given by $t=3^{-1/q}$ (regardless of choice of q).
Thanks guys!
real-analysis analysis
asked Jan 31 at 21:08
Ben WBen W
2,734918
$endgroup$
Fix parameters $1<q<infty$ and $dinmathbb{N}cap[2,infty)$. Let us define a function $f$ by the rule
$$
f(t)=t(1-(d-1)t^q)^{1-1/q}+t^{q-1}(1-(d-1)t^q)^{1/q}-t^q(d-2)-d^{-1}.
$$
I need to prove that there exist a number $0<t<(d-1)^{-1/q}$ such that $f(t)=0$. Ideally, I would like to actually find an exact value for $t$, but this is not absolutely necessary.
This is a fairly elementary problem, but I am nevertheless having difficulty with it.
Technology would be fine. For example, I have Wolfram Mathematica, but unfortunately I don't really know how to use it. Could anyone maybe show me how to do it that way?
If that won't work, an existence proof would be good enough I guess. But I am not seeing any good ideas there, either.
IVT approach fails.
The first and most obvious idea is to use the IVT. However, note that
$$f(0)=-d^{-1}<0$$
and
$$f((d-1)^{-1/q})=-frac{d-2}{d-1}-d^{-1}<0$$
so that the IVT approach will not work on the interval $[0,(d-1)^{-1/q}]$.
The derivative.
To "tighten" the interval so as to make the IVT approach work, we could try maximizing $f$. However, the derivative is given by
$$
f'(t)=(1-(d-1)qt^q)(1-(d-1)t^q)^{-1/q}
+(d-1)(t^{q-2}-qt^{2(q-1)})(1-(d-1)t^q)^{1/q-1}
-(d-2)qt^{q-1}
$$
and finding the roots of this equation is just as difficult as the roots of $f$ itself.
Some special cases.
When $d=2$ and $q=3/2$, solutions are given by
$$t=2^{-2/3}(5^{pm1}(9+4sqrt{5}))^{pm1/3}.$$
When $d=2$ and $q=3$, solutions are given by
$$t=left(frac{1}{10}left(1pm2sqrt{5}right)right)^{1/3}.$$
It would appear that regardless of choice of q, there are solutions when $d=2$ (although I haven't proved it rigorously).
When $d=3$, a solution (not the only one) is given by $t=3^{-1/q}$ (regardless of choice of q).
Thanks guys!
real-analysis analysis
real-analysis analysis
asked Jan 31 at 21:08
Ben WBen W
2,734918
asked Jan 31 at 21:08
Ben WBen W
2,734918
edited Feb 2 at 17:00
Ben W
asked Jan 31 at 21:08
Ben WBen W
2,734918
asked Jan 31 at 21:08
Ben WBen W
2,734918
asked Jan 31 at 21:08
Ben WBen W
2,734918
2,734918
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A partial answer
Basically, this is just me carrying out some algebra and numerical experiments to see whether anything obvious falls out. I hope it's of some help.
By replacing $d-1$ with $u$, you get
$$
t(1-ut^q)^{1-1/q}+t^{q-1}(1-ut^q)^{1/q}=t^q(u-1)+(u+1)^{-1}
$$
which has the charm of being symmetric in $u$, and having $u$ be in the set ${1, 2, ldots}$. You're seeking a number
$$
0 < t < u^{-1/q}
$$
making this equation true. Let's put all the $t$ values on one side, OK? Then we have
$$
t(1-ut^q)^{1-1/q}+t^{q-1}(1-ut^q)^{1/q} - t^q(u-1) =(u+1)^{-1}
$$
Let's call the left hand side $f(t)$.
When $t = 0$, the left hand side is...zero. The right hand side is positive.
When $t = u^{-1/q}$, we have
begin{align}
f(t) = f(u^{-1/q})
&= u^{-1/q}(1-(u^{-1/q})^q)^{1-1/q}+(u^{-1/q})^{q-1}(1-u(u^{-1/q})^q)^{1/q} - (u^{-1/q})^q(u-1)\
&= u^{-1/q}(1-u^{-1})^{1-1/q}+(u^{-1/q})^{q}(u^{-1/q})^{-1}(1-u(u^{-1}))^{1/q} - (u^{-1})(u-1)\
&= u^{-1/q}(1-u^{-1})^{1-1/q}+u^{-1}(u^{-1/q})^{-1}(1-1)^{1/q} - (1-u^{-1})\
&= u^{-1/q}(1-u^{-1})^{1-1/q}- (1-u^{-1})\
&= u^{-1/q}(1-u^{-1})(1-u^{-1})^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})u^{-1/q}(1-u^{-1})^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})(u(1-u^{-1}))^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})(u-1)^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})left[(u-1)^{-1/q}- 1 right]\
end{align}
This now looks...awkward. I'd like to claim that at $t = u^{-1/q}$, the left and side is obviously larger than the RHS, and then use the intermediate value theorem. But the case $u = 1$ is worrying me. For in that case, the LHS is $0$, while the RHS is $frac12$. That either means that
i. I messed up the algebra somewhere, or
ii. the IVT approach isn't going to work, at least not for $u = 1$, or
iii. perhaps the claim is false, with $u = 1$ being a good place to look for counterexamples.
In fact, I'd look really closely at the case $u = 1$ (or $d = 2$) for a counterargument. At $t = 0$, your equation reduces to
$$
0(1-(d-1)t^q)^{1-1/q}+0^{q-1}(1-(d-1)t^q)^{1/q}=0^q(d-2)+d^{-1}
$$
so the left hand side of the original equation is $0$ and the right hand side is $frac12$, i.e., we end up with
$$
0 = frac12.
$$
At $t = (d-1)^{-1/q} = 1$, it looks like
$$
1(1-1cdot 1^q)^{1-1/q}+1^{q-1}(1-1 cdot 1^q)^{1/q}=1^q(1)+d^{-1}
$$
which is
$$
0=1 frac12
$$
So the left hand side is some expression that starts at $0$ and ends at $0$, and the right-hand side ranges from $frac12$ to $1.5$. That seems like there's plenty of opportunity for the two sides never to match. I think that if I were willing to invest any more time in this, I'd look closely at the case $d = 2, q = 2$ and draw a couple of graphs to see whether the claim really works out in that case. I have doubts that it does.
Post-comment addition
Here's a brief matlab program that computes the difference $LHS-RHS$ for $t$ in the allowed range, using $q = d = 2$, the case I suspected might be wrong.
function etest()
% t(1-(d-1)t^q)^{1-1/q}+t^{q-1}(1-(d-1)t^q)^{1/q}=t^q(d-2)-d^{-1}
d = 2;
q = 2;
t = linspace(0.00001, (d-1)^(-1/q), 100);
e = t.*(1-(d-1).* t.^q).^(1-1/q)+t.^(q-1).*(1-(d-1).*t.^q).^(1/q) - t.^q * (d-2)+d^(-1);
plot(t, e);
figure(gcf);
The resulting plot looks like this:
It's clear why my intermediate value approach failed here -- the graph is
a "bump" with both ends low, but a high spot in the middle.
That leads to the natural suggestion that you consider finding the point $t_0$ that maximizes $f(t_0)$, and then use IVT on the interval $[0, t_0]$. I leave that to you, however.
answered Feb 1 at 14:01
John HughesJohn Hughes
65.3k24293
$endgroup$
$begingroup$
Thanks, I'll check it out. However, it may be worth noting that if $d=3$ then $c=3^{-1/q}$ works in that special case. There is an uglier solution for $d=2$. So the claim cannot be false for all $d$.
$endgroup$
– Ben W
Feb 1 at 14:04
$begingroup$
See post-comment addition. It sure looks as if you either mis-typed it, or the claim is false.
$endgroup$
– John Hughes
Feb 1 at 14:12
$begingroup$
Clicking on that led me to a query Wolfram can't understand.
$endgroup$
– John Hughes
Feb 1 at 14:33
$begingroup$
Okay, thank you again for the ideas. I have two general comments. (1) You are correct that the IVT approach will not work. In fact, I have determined that it will never work unless we can tighten the interval $(0,(d-1)^{-1/q})$. (2) Something is wrong with your argument that the claim is false. As noted in previous comments, there are solutions $t=sqrt{2pmsqrt{3}}/2$ in the case $d=q=2$. Your plot can't be correct as the domain is only $[-1,1]$. Thank you for the ideas, though.
$endgroup$
– Ben W
Feb 2 at 8:40
$begingroup$
You're correct -- I had a transcription sign-error: when I moved the last term to the left-hand side, I lost a minus sign. See revised version, suggesting an approach that might pay off.
$endgroup$
– John Hughes
Feb 2 at 14:09
add a comment |
$begingroup$
Okay, I figured it out. The IVT works via with $$f((d^q+d-1)^{-1/q})=frac{d+1}{d^q+d-1}.$$
Thanks guys!
answered Feb 2 at 20:43
Ben WBen W
2,734918
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
A partial answer
Basically, this is just me carrying out some algebra and numerical experiments to see whether anything obvious falls out. I hope it's of some help.
By replacing $d-1$ with $u$, you get
$$
t(1-ut^q)^{1-1/q}+t^{q-1}(1-ut^q)^{1/q}=t^q(u-1)+(u+1)^{-1}
$$
which has the charm of being symmetric in $u$, and having $u$ be in the set ${1, 2, ldots}$. You're seeking a number
$$
0 < t < u^{-1/q}
$$
making this equation true. Let's put all the $t$ values on one side, OK? Then we have
$$
t(1-ut^q)^{1-1/q}+t^{q-1}(1-ut^q)^{1/q} - t^q(u-1) =(u+1)^{-1}
$$
Let's call the left hand side $f(t)$.
When $t = 0$, the left hand side is...zero. The right hand side is positive.
When $t = u^{-1/q}$, we have
begin{align}
f(t) = f(u^{-1/q})
&= u^{-1/q}(1-(u^{-1/q})^q)^{1-1/q}+(u^{-1/q})^{q-1}(1-u(u^{-1/q})^q)^{1/q} - (u^{-1/q})^q(u-1)\
&= u^{-1/q}(1-u^{-1})^{1-1/q}+(u^{-1/q})^{q}(u^{-1/q})^{-1}(1-u(u^{-1}))^{1/q} - (u^{-1})(u-1)\
&= u^{-1/q}(1-u^{-1})^{1-1/q}+u^{-1}(u^{-1/q})^{-1}(1-1)^{1/q} - (1-u^{-1})\
&= u^{-1/q}(1-u^{-1})^{1-1/q}- (1-u^{-1})\
&= u^{-1/q}(1-u^{-1})(1-u^{-1})^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})u^{-1/q}(1-u^{-1})^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})(u(1-u^{-1}))^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})(u-1)^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})left[(u-1)^{-1/q}- 1 right]\
end{align}
This now looks...awkward. I'd like to claim that at $t = u^{-1/q}$, the left and side is obviously larger than the RHS, and then use the intermediate value theorem. But the case $u = 1$ is worrying me. For in that case, the LHS is $0$, while the RHS is $frac12$. That either means that
i. I messed up the algebra somewhere, or
ii. the IVT approach isn't going to work, at least not for $u = 1$, or
iii. perhaps the claim is false, with $u = 1$ being a good place to look for counterexamples.
In fact, I'd look really closely at the case $u = 1$ (or $d = 2$) for a counterargument. At $t = 0$, your equation reduces to
$$
0(1-(d-1)t^q)^{1-1/q}+0^{q-1}(1-(d-1)t^q)^{1/q}=0^q(d-2)+d^{-1}
$$
so the left hand side of the original equation is $0$ and the right hand side is $frac12$, i.e., we end up with
$$
0 = frac12.
$$
At $t = (d-1)^{-1/q} = 1$, it looks like
$$
1(1-1cdot 1^q)^{1-1/q}+1^{q-1}(1-1 cdot 1^q)^{1/q}=1^q(1)+d^{-1}
$$
which is
$$
0=1 frac12
$$
So the left hand side is some expression that starts at $0$ and ends at $0$, and the right-hand side ranges from $frac12$ to $1.5$. That seems like there's plenty of opportunity for the two sides never to match. I think that if I were willing to invest any more time in this, I'd look closely at the case $d = 2, q = 2$ and draw a couple of graphs to see whether the claim really works out in that case. I have doubts that it does.
Post-comment addition
Here's a brief matlab program that computes the difference $LHS-RHS$ for $t$ in the allowed range, using $q = d = 2$, the case I suspected might be wrong.
function etest()
% t(1-(d-1)t^q)^{1-1/q}+t^{q-1}(1-(d-1)t^q)^{1/q}=t^q(d-2)-d^{-1}
d = 2;
q = 2;
t = linspace(0.00001, (d-1)^(-1/q), 100);
e = t.*(1-(d-1).* t.^q).^(1-1/q)+t.^(q-1).*(1-(d-1).*t.^q).^(1/q) - t.^q * (d-2)+d^(-1);
plot(t, e);
figure(gcf);
The resulting plot looks like this:
It's clear why my intermediate value approach failed here -- the graph is
a "bump" with both ends low, but a high spot in the middle.
That leads to the natural suggestion that you consider finding the point $t_0$ that maximizes $f(t_0)$, and then use IVT on the interval $[0, t_0]$. I leave that to you, however.
answered Feb 1 at 14:01
John HughesJohn Hughes
65.3k24293
$endgroup$
$begingroup$
Thanks, I'll check it out. However, it may be worth noting that if $d=3$ then $c=3^{-1/q}$ works in that special case. There is an uglier solution for $d=2$. So the claim cannot be false for all $d$.
$endgroup$
– Ben W
Feb 1 at 14:04
$begingroup$
See post-comment addition. It sure looks as if you either mis-typed it, or the claim is false.
$endgroup$
– John Hughes
Feb 1 at 14:12
$begingroup$
Clicking on that led me to a query Wolfram can't understand.
$endgroup$
– John Hughes
Feb 1 at 14:33
$begingroup$
Okay, thank you again for the ideas. I have two general comments. (1) You are correct that the IVT approach will not work. In fact, I have determined that it will never work unless we can tighten the interval $(0,(d-1)^{-1/q})$. (2) Something is wrong with your argument that the claim is false. As noted in previous comments, there are solutions $t=sqrt{2pmsqrt{3}}/2$ in the case $d=q=2$. Your plot can't be correct as the domain is only $[-1,1]$. Thank you for the ideas, though.
$endgroup$
– Ben W
Feb 2 at 8:40
$begingroup$
You're correct -- I had a transcription sign-error: when I moved the last term to the left-hand side, I lost a minus sign. See revised version, suggesting an approach that might pay off.
$endgroup$
– John Hughes
Feb 2 at 14:09
add a comment |
$begingroup$
A partial answer
Basically, this is just me carrying out some algebra and numerical experiments to see whether anything obvious falls out. I hope it's of some help.
By replacing $d-1$ with $u$, you get
$$
t(1-ut^q)^{1-1/q}+t^{q-1}(1-ut^q)^{1/q}=t^q(u-1)+(u+1)^{-1}
$$
which has the charm of being symmetric in $u$, and having $u$ be in the set ${1, 2, ldots}$. You're seeking a number
$$
0 < t < u^{-1/q}
$$
making this equation true. Let's put all the $t$ values on one side, OK? Then we have
$$
t(1-ut^q)^{1-1/q}+t^{q-1}(1-ut^q)^{1/q} - t^q(u-1) =(u+1)^{-1}
$$
Let's call the left hand side $f(t)$.
When $t = 0$, the left hand side is...zero. The right hand side is positive.
When $t = u^{-1/q}$, we have
begin{align}
f(t) = f(u^{-1/q})
&= u^{-1/q}(1-(u^{-1/q})^q)^{1-1/q}+(u^{-1/q})^{q-1}(1-u(u^{-1/q})^q)^{1/q} - (u^{-1/q})^q(u-1)\
&= u^{-1/q}(1-u^{-1})^{1-1/q}+(u^{-1/q})^{q}(u^{-1/q})^{-1}(1-u(u^{-1}))^{1/q} - (u^{-1})(u-1)\
&= u^{-1/q}(1-u^{-1})^{1-1/q}+u^{-1}(u^{-1/q})^{-1}(1-1)^{1/q} - (1-u^{-1})\
&= u^{-1/q}(1-u^{-1})^{1-1/q}- (1-u^{-1})\
&= u^{-1/q}(1-u^{-1})(1-u^{-1})^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})u^{-1/q}(1-u^{-1})^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})(u(1-u^{-1}))^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})(u-1)^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})left[(u-1)^{-1/q}- 1 right]\
end{align}
This now looks...awkward. I'd like to claim that at $t = u^{-1/q}$, the left and side is obviously larger than the RHS, and then use the intermediate value theorem. But the case $u = 1$ is worrying me. For in that case, the LHS is $0$, while the RHS is $frac12$. That either means that
i. I messed up the algebra somewhere, or
ii. the IVT approach isn't going to work, at least not for $u = 1$, or
iii. perhaps the claim is false, with $u = 1$ being a good place to look for counterexamples.
In fact, I'd look really closely at the case $u = 1$ (or $d = 2$) for a counterargument. At $t = 0$, your equation reduces to
$$
0(1-(d-1)t^q)^{1-1/q}+0^{q-1}(1-(d-1)t^q)^{1/q}=0^q(d-2)+d^{-1}
$$
so the left hand side of the original equation is $0$ and the right hand side is $frac12$, i.e., we end up with
$$
0 = frac12.
$$
At $t = (d-1)^{-1/q} = 1$, it looks like
$$
1(1-1cdot 1^q)^{1-1/q}+1^{q-1}(1-1 cdot 1^q)^{1/q}=1^q(1)+d^{-1}
$$
which is
$$
0=1 frac12
$$
So the left hand side is some expression that starts at $0$ and ends at $0$, and the right-hand side ranges from $frac12$ to $1.5$. That seems like there's plenty of opportunity for the two sides never to match. I think that if I were willing to invest any more time in this, I'd look closely at the case $d = 2, q = 2$ and draw a couple of graphs to see whether the claim really works out in that case. I have doubts that it does.
Post-comment addition
Here's a brief matlab program that computes the difference $LHS-RHS$ for $t$ in the allowed range, using $q = d = 2$, the case I suspected might be wrong.
function etest()
% t(1-(d-1)t^q)^{1-1/q}+t^{q-1}(1-(d-1)t^q)^{1/q}=t^q(d-2)-d^{-1}
d = 2;
q = 2;
t = linspace(0.00001, (d-1)^(-1/q), 100);
e = t.*(1-(d-1).* t.^q).^(1-1/q)+t.^(q-1).*(1-(d-1).*t.^q).^(1/q) - t.^q * (d-2)+d^(-1);
plot(t, e);
figure(gcf);
The resulting plot looks like this:
It's clear why my intermediate value approach failed here -- the graph is
a "bump" with both ends low, but a high spot in the middle.
That leads to the natural suggestion that you consider finding the point $t_0$ that maximizes $f(t_0)$, and then use IVT on the interval $[0, t_0]$. I leave that to you, however.
answered Feb 1 at 14:01
John HughesJohn Hughes
65.3k24293
$endgroup$
$begingroup$
Thanks, I'll check it out. However, it may be worth noting that if $d=3$ then $c=3^{-1/q}$ works in that special case. There is an uglier solution for $d=2$. So the claim cannot be false for all $d$.
$endgroup$
– Ben W
Feb 1 at 14:04
$begingroup$
See post-comment addition. It sure looks as if you either mis-typed it, or the claim is false.
$endgroup$
– John Hughes
Feb 1 at 14:12
$begingroup$
Clicking on that led me to a query Wolfram can't understand.
$endgroup$
– John Hughes
Feb 1 at 14:33
$begingroup$
Okay, thank you again for the ideas. I have two general comments. (1) You are correct that the IVT approach will not work. In fact, I have determined that it will never work unless we can tighten the interval $(0,(d-1)^{-1/q})$. (2) Something is wrong with your argument that the claim is false. As noted in previous comments, there are solutions $t=sqrt{2pmsqrt{3}}/2$ in the case $d=q=2$. Your plot can't be correct as the domain is only $[-1,1]$. Thank you for the ideas, though.
$endgroup$
– Ben W
Feb 2 at 8:40
$begingroup$
You're correct -- I had a transcription sign-error: when I moved the last term to the left-hand side, I lost a minus sign. See revised version, suggesting an approach that might pay off.
$endgroup$
– John Hughes
Feb 2 at 14:09
add a comment |
$begingroup$
A partial answer
Basically, this is just me carrying out some algebra and numerical experiments to see whether anything obvious falls out. I hope it's of some help.
By replacing $d-1$ with $u$, you get
$$
t(1-ut^q)^{1-1/q}+t^{q-1}(1-ut^q)^{1/q}=t^q(u-1)+(u+1)^{-1}
$$
which has the charm of being symmetric in $u$, and having $u$ be in the set ${1, 2, ldots}$. You're seeking a number
$$
0 < t < u^{-1/q}
$$
making this equation true. Let's put all the $t$ values on one side, OK? Then we have
$$
t(1-ut^q)^{1-1/q}+t^{q-1}(1-ut^q)^{1/q} - t^q(u-1) =(u+1)^{-1}
$$
Let's call the left hand side $f(t)$.
When $t = 0$, the left hand side is...zero. The right hand side is positive.
When $t = u^{-1/q}$, we have
begin{align}
f(t) = f(u^{-1/q})
&= u^{-1/q}(1-(u^{-1/q})^q)^{1-1/q}+(u^{-1/q})^{q-1}(1-u(u^{-1/q})^q)^{1/q} - (u^{-1/q})^q(u-1)\
&= u^{-1/q}(1-u^{-1})^{1-1/q}+(u^{-1/q})^{q}(u^{-1/q})^{-1}(1-u(u^{-1}))^{1/q} - (u^{-1})(u-1)\
&= u^{-1/q}(1-u^{-1})^{1-1/q}+u^{-1}(u^{-1/q})^{-1}(1-1)^{1/q} - (1-u^{-1})\
&= u^{-1/q}(1-u^{-1})^{1-1/q}- (1-u^{-1})\
&= u^{-1/q}(1-u^{-1})(1-u^{-1})^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})u^{-1/q}(1-u^{-1})^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})(u(1-u^{-1}))^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})(u-1)^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})left[(u-1)^{-1/q}- 1 right]\
end{align}
This now looks...awkward. I'd like to claim that at $t = u^{-1/q}$, the left and side is obviously larger than the RHS, and then use the intermediate value theorem. But the case $u = 1$ is worrying me. For in that case, the LHS is $0$, while the RHS is $frac12$. That either means that
i. I messed up the algebra somewhere, or
ii. the IVT approach isn't going to work, at least not for $u = 1$, or
iii. perhaps the claim is false, with $u = 1$ being a good place to look for counterexamples.
In fact, I'd look really closely at the case $u = 1$ (or $d = 2$) for a counterargument. At $t = 0$, your equation reduces to
$$
0(1-(d-1)t^q)^{1-1/q}+0^{q-1}(1-(d-1)t^q)^{1/q}=0^q(d-2)+d^{-1}
$$
so the left hand side of the original equation is $0$ and the right hand side is $frac12$, i.e., we end up with
$$
0 = frac12.
$$
At $t = (d-1)^{-1/q} = 1$, it looks like
$$
1(1-1cdot 1^q)^{1-1/q}+1^{q-1}(1-1 cdot 1^q)^{1/q}=1^q(1)+d^{-1}
$$
which is
$$
0=1 frac12
$$
So the left hand side is some expression that starts at $0$ and ends at $0$, and the right-hand side ranges from $frac12$ to $1.5$. That seems like there's plenty of opportunity for the two sides never to match. I think that if I were willing to invest any more time in this, I'd look closely at the case $d = 2, q = 2$ and draw a couple of graphs to see whether the claim really works out in that case. I have doubts that it does.
Post-comment addition
Here's a brief matlab program that computes the difference $LHS-RHS$ for $t$ in the allowed range, using $q = d = 2$, the case I suspected might be wrong.
function etest()
% t(1-(d-1)t^q)^{1-1/q}+t^{q-1}(1-(d-1)t^q)^{1/q}=t^q(d-2)-d^{-1}
d = 2;
q = 2;
t = linspace(0.00001, (d-1)^(-1/q), 100);
e = t.*(1-(d-1).* t.^q).^(1-1/q)+t.^(q-1).*(1-(d-1).*t.^q).^(1/q) - t.^q * (d-2)+d^(-1);
plot(t, e);
figure(gcf);
The resulting plot looks like this:
It's clear why my intermediate value approach failed here -- the graph is
a "bump" with both ends low, but a high spot in the middle.
That leads to the natural suggestion that you consider finding the point $t_0$ that maximizes $f(t_0)$, and then use IVT on the interval $[0, t_0]$. I leave that to you, however.
answered Feb 1 at 14:01
John HughesJohn Hughes
65.3k24293
$endgroup$
A partial answer
Basically, this is just me carrying out some algebra and numerical experiments to see whether anything obvious falls out. I hope it's of some help.
By replacing $d-1$ with $u$, you get
$$
t(1-ut^q)^{1-1/q}+t^{q-1}(1-ut^q)^{1/q}=t^q(u-1)+(u+1)^{-1}
$$
which has the charm of being symmetric in $u$, and having $u$ be in the set ${1, 2, ldots}$. You're seeking a number
$$
0 < t < u^{-1/q}
$$
making this equation true. Let's put all the $t$ values on one side, OK? Then we have
$$
t(1-ut^q)^{1-1/q}+t^{q-1}(1-ut^q)^{1/q} - t^q(u-1) =(u+1)^{-1}
$$
Let's call the left hand side $f(t)$.
When $t = 0$, the left hand side is...zero. The right hand side is positive.
When $t = u^{-1/q}$, we have
begin{align}
f(t) = f(u^{-1/q})
&= u^{-1/q}(1-(u^{-1/q})^q)^{1-1/q}+(u^{-1/q})^{q-1}(1-u(u^{-1/q})^q)^{1/q} - (u^{-1/q})^q(u-1)\
&= u^{-1/q}(1-u^{-1})^{1-1/q}+(u^{-1/q})^{q}(u^{-1/q})^{-1}(1-u(u^{-1}))^{1/q} - (u^{-1})(u-1)\
&= u^{-1/q}(1-u^{-1})^{1-1/q}+u^{-1}(u^{-1/q})^{-1}(1-1)^{1/q} - (1-u^{-1})\
&= u^{-1/q}(1-u^{-1})^{1-1/q}- (1-u^{-1})\
&= u^{-1/q}(1-u^{-1})(1-u^{-1})^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})u^{-1/q}(1-u^{-1})^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})(u(1-u^{-1}))^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})(u-1)^{-1/q}- (1-u^{-1})\
&= (1-u^{-1})left[(u-1)^{-1/q}- 1 right]\
end{align}
This now looks...awkward. I'd like to claim that at $t = u^{-1/q}$, the left and side is obviously larger than the RHS, and then use the intermediate value theorem. But the case $u = 1$ is worrying me. For in that case, the LHS is $0$, while the RHS is $frac12$. That either means that
i. I messed up the algebra somewhere, or
ii. the IVT approach isn't going to work, at least not for $u = 1$, or
iii. perhaps the claim is false, with $u = 1$ being a good place to look for counterexamples.
In fact, I'd look really closely at the case $u = 1$ (or $d = 2$) for a counterargument. At $t = 0$, your equation reduces to
$$
0(1-(d-1)t^q)^{1-1/q}+0^{q-1}(1-(d-1)t^q)^{1/q}=0^q(d-2)+d^{-1}
$$
so the left hand side of the original equation is $0$ and the right hand side is $frac12$, i.e., we end up with
$$
0 = frac12.
$$
At $t = (d-1)^{-1/q} = 1$, it looks like
$$
1(1-1cdot 1^q)^{1-1/q}+1^{q-1}(1-1 cdot 1^q)^{1/q}=1^q(1)+d^{-1}
$$
which is
$$
0=1 frac12
$$
So the left hand side is some expression that starts at $0$ and ends at $0$, and the right-hand side ranges from $frac12$ to $1.5$. That seems like there's plenty of opportunity for the two sides never to match. I think that if I were willing to invest any more time in this, I'd look closely at the case $d = 2, q = 2$ and draw a couple of graphs to see whether the claim really works out in that case. I have doubts that it does.
Post-comment addition
Here's a brief matlab program that computes the difference $LHS-RHS$ for $t$ in the allowed range, using $q = d = 2$, the case I suspected might be wrong.
function etest()
% t(1-(d-1)t^q)^{1-1/q}+t^{q-1}(1-(d-1)t^q)^{1/q}=t^q(d-2)-d^{-1}
d = 2;
q = 2;
t = linspace(0.00001, (d-1)^(-1/q), 100);
e = t.*(1-(d-1).* t.^q).^(1-1/q)+t.^(q-1).*(1-(d-1).*t.^q).^(1/q) - t.^q * (d-2)+d^(-1);
plot(t, e);
figure(gcf);
The resulting plot looks like this:
It's clear why my intermediate value approach failed here -- the graph is
a "bump" with both ends low, but a high spot in the middle.
That leads to the natural suggestion that you consider finding the point $t_0$ that maximizes $f(t_0)$, and then use IVT on the interval $[0, t_0]$. I leave that to you, however.
answered Feb 1 at 14:01
John HughesJohn Hughes
65.3k24293
edited Feb 2 at 14:22
answered Feb 1 at 14:01
John HughesJohn Hughes
65.3k24293
answered Feb 1 at 14:01
John HughesJohn Hughes
65.3k24293
answered Feb 1 at 14:01
John HughesJohn Hughes
65.3k24293
65.3k24293
$begingroup$
Thanks, I'll check it out. However, it may be worth noting that if $d=3$ then $c=3^{-1/q}$ works in that special case. There is an uglier solution for $d=2$. So the claim cannot be false for all $d$.
$endgroup$
– Ben W
Feb 1 at 14:04
$begingroup$
See post-comment addition. It sure looks as if you either mis-typed it, or the claim is false.
$endgroup$
– John Hughes
Feb 1 at 14:12
$begingroup$
Clicking on that led me to a query Wolfram can't understand.
$endgroup$
– John Hughes
Feb 1 at 14:33
$begingroup$
Okay, thank you again for the ideas. I have two general comments. (1) You are correct that the IVT approach will not work. In fact, I have determined that it will never work unless we can tighten the interval $(0,(d-1)^{-1/q})$. (2) Something is wrong with your argument that the claim is false. As noted in previous comments, there are solutions $t=sqrt{2pmsqrt{3}}/2$ in the case $d=q=2$. Your plot can't be correct as the domain is only $[-1,1]$. Thank you for the ideas, though.
$endgroup$
– Ben W
Feb 2 at 8:40
$begingroup$
You're correct -- I had a transcription sign-error: when I moved the last term to the left-hand side, I lost a minus sign. See revised version, suggesting an approach that might pay off.
$endgroup$
– John Hughes
Feb 2 at 14:09
add a comment |
$begingroup$
Thanks, I'll check it out. However, it may be worth noting that if $d=3$ then $c=3^{-1/q}$ works in that special case. There is an uglier solution for $d=2$. So the claim cannot be false for all $d$.
$endgroup$
– Ben W
Feb 1 at 14:04
$begingroup$
See post-comment addition. It sure looks as if you either mis-typed it, or the claim is false.
$endgroup$
– John Hughes
Feb 1 at 14:12
$begingroup$
Clicking on that led me to a query Wolfram can't understand.
$endgroup$
– John Hughes
Feb 1 at 14:33
$begingroup$
Okay, thank you again for the ideas. I have two general comments. (1) You are correct that the IVT approach will not work. In fact, I have determined that it will never work unless we can tighten the interval $(0,(d-1)^{-1/q})$. (2) Something is wrong with your argument that the claim is false. As noted in previous comments, there are solutions $t=sqrt{2pmsqrt{3}}/2$ in the case $d=q=2$. Your plot can't be correct as the domain is only $[-1,1]$. Thank you for the ideas, though.
$endgroup$
– Ben W
Feb 2 at 8:40
$begingroup$
You're correct -- I had a transcription sign-error: when I moved the last term to the left-hand side, I lost a minus sign. See revised version, suggesting an approach that might pay off.
$endgroup$
– John Hughes
Feb 2 at 14:09
$begingroup$
Thanks, I'll check it out. However, it may be worth noting that if $d=3$ then $c=3^{-1/q}$ works in that special case. There is an uglier solution for $d=2$. So the claim cannot be false for all $d$.
$endgroup$
– Ben W
Feb 1 at 14:04
$begingroup$
Thanks, I'll check it out. However, it may be worth noting that if $d=3$ then $c=3^{-1/q}$ works in that special case. There is an uglier solution for $d=2$. So the claim cannot be false for all $d$.
$endgroup$
– Ben W
Feb 1 at 14:04
$begingroup$
See post-comment addition. It sure looks as if you either mis-typed it, or the claim is false.
$endgroup$
– John Hughes
Feb 1 at 14:12
$begingroup$
See post-comment addition. It sure looks as if you either mis-typed it, or the claim is false.
$endgroup$
– John Hughes
Feb 1 at 14:12
$begingroup$
Clicking on that led me to a query Wolfram can't understand.
$endgroup$
– John Hughes
Feb 1 at 14:33
$begingroup$
Clicking on that led me to a query Wolfram can't understand.
$endgroup$
– John Hughes
Feb 1 at 14:33
$begingroup$
Okay, thank you again for the ideas. I have two general comments. (1) You are correct that the IVT approach will not work. In fact, I have determined that it will never work unless we can tighten the interval $(0,(d-1)^{-1/q})$. (2) Something is wrong with your argument that the claim is false. As noted in previous comments, there are solutions $t=sqrt{2pmsqrt{3}}/2$ in the case $d=q=2$. Your plot can't be correct as the domain is only $[-1,1]$. Thank you for the ideas, though.
$endgroup$
– Ben W
Feb 2 at 8:40
$begingroup$
Okay, thank you again for the ideas. I have two general comments. (1) You are correct that the IVT approach will not work. In fact, I have determined that it will never work unless we can tighten the interval $(0,(d-1)^{-1/q})$. (2) Something is wrong with your argument that the claim is false. As noted in previous comments, there are solutions $t=sqrt{2pmsqrt{3}}/2$ in the case $d=q=2$. Your plot can't be correct as the domain is only $[-1,1]$. Thank you for the ideas, though.
$endgroup$
– Ben W
Feb 2 at 8:40
$begingroup$
You're correct -- I had a transcription sign-error: when I moved the last term to the left-hand side, I lost a minus sign. See revised version, suggesting an approach that might pay off.
$endgroup$
– John Hughes
Feb 2 at 14:09
$begingroup$
You're correct -- I had a transcription sign-error: when I moved the last term to the left-hand side, I lost a minus sign. See revised version, suggesting an approach that might pay off.
$endgroup$
– John Hughes
Feb 2 at 14:09
add a comment |
$begingroup$
Okay, I figured it out. The IVT works via with $$f((d^q+d-1)^{-1/q})=frac{d+1}{d^q+d-1}.$$
Thanks guys!
answered Feb 2 at 20:43
Ben WBen W
2,734918
$endgroup$
add a comment |
$begingroup$
Okay, I figured it out. The IVT works via with $$f((d^q+d-1)^{-1/q})=frac{d+1}{d^q+d-1}.$$
Thanks guys!
answered Feb 2 at 20:43
Ben WBen W
2,734918
$endgroup$
add a comment |
$begingroup$
Okay, I figured it out. The IVT works via with $$f((d^q+d-1)^{-1/q})=frac{d+1}{d^q+d-1}.$$
Thanks guys!
answered Feb 2 at 20:43
Ben WBen W
2,734918
$endgroup$
Okay, I figured it out. The IVT works via with $$f((d^q+d-1)^{-1/q})=frac{d+1}{d^q+d-1}.$$
Thanks guys!
answered Feb 2 at 20:43
Ben WBen W
2,734918
answered Feb 2 at 20:43
Ben WBen W
2,734918
answered Feb 2 at 20:43
Ben WBen W
2,734918
answered Feb 2 at 20:43
Ben WBen W
2,734918
2,734918
add a comment |
add a comment |
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