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Prove that we can't find effective bounds on the point guaranteed by the Mean Value Theorem.
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I wish to show that we cannot find effective bounds on the point that the Mean Value Theorem proves to exist. To prove this loose statement, I aimed at the slightly more specific claim:
For each real number $M$ and each real number $xi$ that lies strictly between $0$ and $1$, construct a function $f$ such that
$$f(0)=0,; f(1)=M,;ftext{ is continuous on }[0,1],; ftext{ is differentiable on }(0,1),;text{ and }xitext{ is the unique point strictly between 0 and 1 such that};f'(xi)=M,.$$
For the $Mneq 0$ and $xineq 1/e$ case, we can show that
$$g(x)=begin{cases}
0&text{ if }x=0,\
1/e&text{ if }x=1\
1&text{ if }x=infty,\
sqrt[1-x]{x}&text{ otherwise}
end{cases}$$
is strictly increasing and continuous on $[0,infty]$. Thus there is a unique positive $alpha$ such that $g(alpha)=xi$. In turn, we can define $f(x)=Mx^alpha$ which will satisfy the claim. For the $Mneq 0$ and $xi=1/e$ case, take the obvious continuous extension of $f(x)=M(x+xln(x))$.
For $M=0$, we first choose $alphageq 1$ and $betageq 1$ such that $frac{alpha}{alpha+beta}=xi$. We then define $f(x)=x^alpha(1-x)^beta$ which will satisfy the claim.
My question however is this:
Can we construct such an $f$ to be a polynomial?
An existential proof isn't desirable here, as I hope to use this family of polynomials as examples. It'd be useful to prove the uniqueness of $xi$ through calculation (but possibly an appeal to monotonicity and the Intermediate Value Theorem).
calculus real-analysis polynomials
asked Oct 30 at 20:29
Robert Wolfe
5,44522361
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up vote
9
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I wish to show that we cannot find effective bounds on the point that the Mean Value Theorem proves to exist. To prove this loose statement, I aimed at the slightly more specific claim:
For each real number $M$ and each real number $xi$ that lies strictly between $0$ and $1$, construct a function $f$ such that
$$f(0)=0,; f(1)=M,;ftext{ is continuous on }[0,1],; ftext{ is differentiable on }(0,1),;text{ and }xitext{ is the unique point strictly between 0 and 1 such that};f'(xi)=M,.$$
For the $Mneq 0$ and $xineq 1/e$ case, we can show that
$$g(x)=begin{cases}
0&text{ if }x=0,\
1/e&text{ if }x=1\
1&text{ if }x=infty,\
sqrt[1-x]{x}&text{ otherwise}
end{cases}$$
is strictly increasing and continuous on $[0,infty]$. Thus there is a unique positive $alpha$ such that $g(alpha)=xi$. In turn, we can define $f(x)=Mx^alpha$ which will satisfy the claim. For the $Mneq 0$ and $xi=1/e$ case, take the obvious continuous extension of $f(x)=M(x+xln(x))$.
For $M=0$, we first choose $alphageq 1$ and $betageq 1$ such that $frac{alpha}{alpha+beta}=xi$. We then define $f(x)=x^alpha(1-x)^beta$ which will satisfy the claim.
My question however is this:
Can we construct such an $f$ to be a polynomial?
An existential proof isn't desirable here, as I hope to use this family of polynomials as examples. It'd be useful to prove the uniqueness of $xi$ through calculation (but possibly an appeal to monotonicity and the Intermediate Value Theorem).
calculus real-analysis polynomials
asked Oct 30 at 20:29
Robert Wolfe
5,44522361
3
Quick idea: consider $f(x) = Mx + p(x)$ where $p(x)$ is a polynomial constructed to satisfy $p(0) = p(1) = 0$, $(forall 0 leq 1 leq x) p''(x) < 0$, and $p'(xi) = 0$ for some arbitrary (unique by construction) $xi$. This should be possible with quartics: it boils down to just solving an underdetermined system on the coefficients of $p$.
– Connor Harris
Oct 30 at 21:09
@ConnorHarris that seems feasible. The algebra is a little bit of a pain though. A solution would be quite slick though.
– Robert Wolfe
Oct 30 at 23:16
One third of a solution: for $1/3leq xi<1/2$ we may use $Mx+x(x-1)(x-c_1)$ and for $1/2<xileq 2/3$ we may use $Mx-x(x-1)(x+c_2)$ where $$c_1=frac{xi(3xi-2)}{2xi-1};text{ and };c_2=frac{xi(3xi-2)}{1-2xi},.$$ For $xi=1/2$ we can get away with $x(1-x)$. But cubics don't seem to carry us the entire way.
– Robert Wolfe
Nov 2 at 3:32
@RobertWolfe I didn't think about the comments before posting an answer. But now that I am seeing them, the $(xpm c)$ in your "cubic" comment is like the $(x+t)$ in my answer, except that I extended to considering $(x+t)^n$ to get past the central third.
– alex.jordan
Nov 2 at 7:29
add a comment |
up vote
9
down vote
favorite
up vote
9
down vote
favorite
I wish to show that we cannot find effective bounds on the point that the Mean Value Theorem proves to exist. To prove this loose statement, I aimed at the slightly more specific claim:
For each real number $M$ and each real number $xi$ that lies strictly between $0$ and $1$, construct a function $f$ such that
$$f(0)=0,; f(1)=M,;ftext{ is continuous on }[0,1],; ftext{ is differentiable on }(0,1),;text{ and }xitext{ is the unique point strictly between 0 and 1 such that};f'(xi)=M,.$$
For the $Mneq 0$ and $xineq 1/e$ case, we can show that
$$g(x)=begin{cases}
0&text{ if }x=0,\
1/e&text{ if }x=1\
1&text{ if }x=infty,\
sqrt[1-x]{x}&text{ otherwise}
end{cases}$$
is strictly increasing and continuous on $[0,infty]$. Thus there is a unique positive $alpha$ such that $g(alpha)=xi$. In turn, we can define $f(x)=Mx^alpha$ which will satisfy the claim. For the $Mneq 0$ and $xi=1/e$ case, take the obvious continuous extension of $f(x)=M(x+xln(x))$.
For $M=0$, we first choose $alphageq 1$ and $betageq 1$ such that $frac{alpha}{alpha+beta}=xi$. We then define $f(x)=x^alpha(1-x)^beta$ which will satisfy the claim.
My question however is this:
Can we construct such an $f$ to be a polynomial?
An existential proof isn't desirable here, as I hope to use this family of polynomials as examples. It'd be useful to prove the uniqueness of $xi$ through calculation (but possibly an appeal to monotonicity and the Intermediate Value Theorem).
calculus real-analysis polynomials
asked Oct 30 at 20:29
Robert Wolfe
5,44522361
I wish to show that we cannot find effective bounds on the point that the Mean Value Theorem proves to exist. To prove this loose statement, I aimed at the slightly more specific claim:
For each real number $M$ and each real number $xi$ that lies strictly between $0$ and $1$, construct a function $f$ such that
$$f(0)=0,; f(1)=M,;ftext{ is continuous on }[0,1],; ftext{ is differentiable on }(0,1),;text{ and }xitext{ is the unique point strictly between 0 and 1 such that};f'(xi)=M,.$$
For the $Mneq 0$ and $xineq 1/e$ case, we can show that
$$g(x)=begin{cases}
0&text{ if }x=0,\
1/e&text{ if }x=1\
1&text{ if }x=infty,\
sqrt[1-x]{x}&text{ otherwise}
end{cases}$$
is strictly increasing and continuous on $[0,infty]$. Thus there is a unique positive $alpha$ such that $g(alpha)=xi$. In turn, we can define $f(x)=Mx^alpha$ which will satisfy the claim. For the $Mneq 0$ and $xi=1/e$ case, take the obvious continuous extension of $f(x)=M(x+xln(x))$.
For $M=0$, we first choose $alphageq 1$ and $betageq 1$ such that $frac{alpha}{alpha+beta}=xi$. We then define $f(x)=x^alpha(1-x)^beta$ which will satisfy the claim.
My question however is this:
Can we construct such an $f$ to be a polynomial?
An existential proof isn't desirable here, as I hope to use this family of polynomials as examples. It'd be useful to prove the uniqueness of $xi$ through calculation (but possibly an appeal to monotonicity and the Intermediate Value Theorem).
calculus real-analysis polynomials
calculus real-analysis polynomials
asked Oct 30 at 20:29
Robert Wolfe
5,44522361
asked Oct 30 at 20:29
Robert Wolfe
5,44522361
edited Oct 31 at 15:16
asked Oct 30 at 20:29
Robert Wolfe
5,44522361
asked Oct 30 at 20:29
Robert Wolfe
5,44522361
asked Oct 30 at 20:29
Robert Wolfe
5,44522361
5,44522361
3
Quick idea: consider $f(x) = Mx + p(x)$ where $p(x)$ is a polynomial constructed to satisfy $p(0) = p(1) = 0$, $(forall 0 leq 1 leq x) p''(x) < 0$, and $p'(xi) = 0$ for some arbitrary (unique by construction) $xi$. This should be possible with quartics: it boils down to just solving an underdetermined system on the coefficients of $p$.
– Connor Harris
Oct 30 at 21:09
@ConnorHarris that seems feasible. The algebra is a little bit of a pain though. A solution would be quite slick though.
– Robert Wolfe
Oct 30 at 23:16
One third of a solution: for $1/3leq xi<1/2$ we may use $Mx+x(x-1)(x-c_1)$ and for $1/2<xileq 2/3$ we may use $Mx-x(x-1)(x+c_2)$ where $$c_1=frac{xi(3xi-2)}{2xi-1};text{ and };c_2=frac{xi(3xi-2)}{1-2xi},.$$ For $xi=1/2$ we can get away with $x(1-x)$. But cubics don't seem to carry us the entire way.
– Robert Wolfe
Nov 2 at 3:32
@RobertWolfe I didn't think about the comments before posting an answer. But now that I am seeing them, the $(xpm c)$ in your "cubic" comment is like the $(x+t)$ in my answer, except that I extended to considering $(x+t)^n$ to get past the central third.
– alex.jordan
Nov 2 at 7:29
add a comment |
3
Quick idea: consider $f(x) = Mx + p(x)$ where $p(x)$ is a polynomial constructed to satisfy $p(0) = p(1) = 0$, $(forall 0 leq 1 leq x) p''(x) < 0$, and $p'(xi) = 0$ for some arbitrary (unique by construction) $xi$. This should be possible with quartics: it boils down to just solving an underdetermined system on the coefficients of $p$.
– Connor Harris
Oct 30 at 21:09
@ConnorHarris that seems feasible. The algebra is a little bit of a pain though. A solution would be quite slick though.
– Robert Wolfe
Oct 30 at 23:16
One third of a solution: for $1/3leq xi<1/2$ we may use $Mx+x(x-1)(x-c_1)$ and for $1/2<xileq 2/3$ we may use $Mx-x(x-1)(x+c_2)$ where $$c_1=frac{xi(3xi-2)}{2xi-1};text{ and };c_2=frac{xi(3xi-2)}{1-2xi},.$$ For $xi=1/2$ we can get away with $x(1-x)$. But cubics don't seem to carry us the entire way.
– Robert Wolfe
Nov 2 at 3:32
@RobertWolfe I didn't think about the comments before posting an answer. But now that I am seeing them, the $(xpm c)$ in your "cubic" comment is like the $(x+t)$ in my answer, except that I extended to considering $(x+t)^n$ to get past the central third.
– alex.jordan
Nov 2 at 7:29
3
3
Quick idea: consider $f(x) = Mx + p(x)$ where $p(x)$ is a polynomial constructed to satisfy $p(0) = p(1) = 0$, $(forall 0 leq 1 leq x) p''(x) < 0$, and $p'(xi) = 0$ for some arbitrary (unique by construction) $xi$. This should be possible with quartics: it boils down to just solving an underdetermined system on the coefficients of $p$.
– Connor Harris
Oct 30 at 21:09
Quick idea: consider $f(x) = Mx + p(x)$ where $p(x)$ is a polynomial constructed to satisfy $p(0) = p(1) = 0$, $(forall 0 leq 1 leq x) p''(x) < 0$, and $p'(xi) = 0$ for some arbitrary (unique by construction) $xi$. This should be possible with quartics: it boils down to just solving an underdetermined system on the coefficients of $p$.
– Connor Harris
Oct 30 at 21:09
@ConnorHarris that seems feasible. The algebra is a little bit of a pain though. A solution would be quite slick though.
– Robert Wolfe
Oct 30 at 23:16
@ConnorHarris that seems feasible. The algebra is a little bit of a pain though. A solution would be quite slick though.
– Robert Wolfe
Oct 30 at 23:16
One third of a solution: for $1/3leq xi<1/2$ we may use $Mx+x(x-1)(x-c_1)$ and for $1/2<xileq 2/3$ we may use $Mx-x(x-1)(x+c_2)$ where $$c_1=frac{xi(3xi-2)}{2xi-1};text{ and };c_2=frac{xi(3xi-2)}{1-2xi},.$$ For $xi=1/2$ we can get away with $x(1-x)$. But cubics don't seem to carry us the entire way.
– Robert Wolfe
Nov 2 at 3:32
One third of a solution: for $1/3leq xi<1/2$ we may use $Mx+x(x-1)(x-c_1)$ and for $1/2<xileq 2/3$ we may use $Mx-x(x-1)(x+c_2)$ where $$c_1=frac{xi(3xi-2)}{2xi-1};text{ and };c_2=frac{xi(3xi-2)}{1-2xi},.$$ For $xi=1/2$ we can get away with $x(1-x)$. But cubics don't seem to carry us the entire way.
– Robert Wolfe
Nov 2 at 3:32
@RobertWolfe I didn't think about the comments before posting an answer. But now that I am seeing them, the $(xpm c)$ in your "cubic" comment is like the $(x+t)$ in my answer, except that I extended to considering $(x+t)^n$ to get past the central third.
– alex.jordan
Nov 2 at 7:29
@RobertWolfe I didn't think about the comments before posting an answer. But now that I am seeing them, the $(xpm c)$ in your "cubic" comment is like the $(x+t)$ in my answer, except that I extended to considering $(x+t)^n$ to get past the central third.
– alex.jordan
Nov 2 at 7:29
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3 Answers
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First, imagine $f$ is some such polynomial for $xi$. Then let $g(x)=f(x)-Mx$. We have $g(0)=g(1)=0$, and $xi$ is the unique number in $(0,1)$ where $g'(x)=0$. So allow me to replace the problem as written with the Mean Value Theorem to one about Rolle's Theorem.
Below is a proof that if you have $xi>frac{1}{2}$, take some integer $n>frac{1-2xi}{xi-1}$, and then take $t=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}$. Then the polynomial $g(x)=(x+t)^nx(1-x)$ satisfies $g(0)=g(1)=0$, and there is a unique number in $(0,1)$ where $g'(x)=0$, and that number is $xi$.
If $xi<frac{1}{2}$, there is a symmetric construction with $t<-1$. And if $xi=frac{1}{2}$, just take $g(x)=x(1-x)$.
For example, with $xi=frac{e}{pi}$, we can take $n=6$, and $t=frac{8(e/pi)^2-7(e/pi)}{-2(e/pi)+1}approx0.09233ldots$. Then $g(x)=(x+t)^6x(1-x)$ is such that $g'$ has only one zero in $(0,1)$, and it is located at $frac{e}{pi}$. See this demonstrated at WolframAlpha.
Explanation
Assume $xi>frac{1}{2}$. Consider $g(x)=(x+t)^nx(1-x)$ for $ninmathbb{N}$ and $tinmathbb{R}_{gt0}$. Then $$
begin{align}
g'(x)&=n(x+t)^{n-1}x(1-x)+(x+t)^n(1-x)-(x+t)^nx\
&=(x+t)^{n-1}big(nx(1-x)+(x+t)(1-x)-(x+t)xbig)\
&=(x+t)^{n-1}big(x^2(-n-2)+x(n+1-2t)+tbig)\
end{align}
$$
The zeros of $g'$ are $-t$ (which is not in $(0,1)$) and $$frac{-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}=frac{Apm B}{C}$$
Since $n,t>0$, it follows that $|B|>|A|$. It follows that one of these two roots is negative (so not in $(0,1)$) and the other is positive. So if the positive root is equal to $xi$, then $g$ satisfies the Rolle's version of the proposition. We have freedom to choose $ninmathbb{N}$, $tinmathbb{R}_{gt0}$, so maybe we can choose them well. In the following, we attempt to solve for $t$ in terms of $xi$ and $n$.
$$begin{align}
xi&=frac{-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}\
-2(n+2)xi&=-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}\
-2(n+2)xi+n+1-2t&=pmsqrt{(n+1-2t)^2+4(n+2)t}
end{align}$$
Squaring both sides:
$$begin{align}
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]+4t^2&=(n+1-2t)^2+4(n+2)t\
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]+4t^2&=(n+1)^2-4(n+1)t+4t^2+4(n+2)t\
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]&=(n+1)^2+4t\
[-2(n+2)xi+n+1]^2-(n+1)^2&=4t[-2(n+2)xi+n+2]\
4(n+2)^2xi^2-4(n+2)(n+1)xi&=4t(-2(n+2)xi+n+2)\
end{align}$$
$$begin{align}
t&=frac{4(n+2)^2xi^2-4(n+2)(n+1)xi}{4(-2(n+2)xi+n+2)}\
&=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}
end{align}$$
We have assumed $xi>frac{1}{2}$, so the denominator is negative. We need $t$ to be positive, so we need the numerator to be negative. Can we choose $n$ to make that happen?
$$begin{align}
(n+2)xi^2-(n+1)xi&<0\
n(xi^2-xi)&<xi-2xi^2\
n(xi-1)&<1-2xi\
n&>frac{1-2xi}{xi-1}
end{align}$$
So yes. If you have $xi>frac{1}{2}$, take some integer $n>frac{1-2xi}{xi-1}$, and then take $t=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}$. Then the polynomial $g(x)=(x+t)^nx(1-x)$ satisfies $g(0)=g(1)=0$, and there is a unique number in $(0,1)$ where $g'(x)=0$, and that number is $xi$.
And restating from the introduction, if $xi<frac{1}{2}$, there is a symmetric construction where $t<-1$. And if $xi=frac{1}{2}$, just take $g(x)=x(1-x)$.
This approach was motivated by starting with $x(1-x)$, and then multiplying by some power of $(x+t)$ that would "warp" the parabola in between $0$ and $1$ without making it wiggle. Something that would stretch $x(1-x)$ vertically, but moreso on the right side than the left side, to push the extremum farther to the right. (Or the other way when $xi<frac{1}{2}$.)
answered Nov 2 at 7:12
alex.jordan
37.9k559118
I'm glad to see that my partial answer generalized in some way. Thanks! I'm still slightly interested to see if someone can post an answer without some case analysis.
– Robert Wolfe
Nov 2 at 21:02
You're welcome, thanks for a question that I found interesting. By case analysis, do you mean $xi>1/2$ versus $xi<1/2$ versus $xi=1/2$?
– alex.jordan
Nov 2 at 21:16
yeah. a small detail.
– Robert Wolfe
Nov 2 at 21:29
it seems that zhw's $(xmapsto 1-x)$ substitution trick could be applied though. that barely qualifies as a case analysis now.
– Robert Wolfe
Nov 2 at 21:37
This is what I meant too, but I could have been more explicit. Applying that reflection effectively turns $t$ in $(0,infty)$ into $tilde{t}$ in $(-infty,-1)$. (Except the whole function is also scaled by $(-1)^n$.) This is where the "$t<-1$" clause came from in the "case" where $xi<1/2$.
– alex.jordan
Nov 2 at 22:26
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Just a few words on solving the nonpolynomial problem: First take $M=0.$ Then $f(x)=x(1-x)$ is a solution for $xi=1/2.$ For other values of $xiin (0,1),$ we look at $g_p(x)=f(x^p)$ for $p>0.$ We have $g_p'(x)=px^{p-1}f'(x^p)=0$ iff $x=(1/2)^{1/p}.$ Thus if $xi$ is given, we take $p=ln(1/2)/ln xi,$ and $g_p$ solves the problem. Finally, if $Mne 0$ and $xi$ is given, we take the same $p$ and the function $M(x+g_p(x))$ solves the problem.
On to polynomial solutions: WLOG $M=0$, for we can use the same idea as above for $Mne 0.$
Let $f(x)=x(1-x)$ as above. If $p$ is a polynomial with $p(0)=0,$ $p(1)=1$ and $p'>0$ on $(0,1),$ then $fcirc p$ is a polynomial that solves the problem for for the unique value $xi in(0,1)$ such that $p(xi)=1/2.$
For $0le b le 1,$ set $p_b(x)=(1-b)x^2 +bx.$ Then $p_b(0)=0,p_b(1)=1,$ and $p_b'>0$ on $(0,1).$ Consider the equation $p_b(x)-1/2=0.$ If $b=1,$ then $x=1/2$ is a solution. For $bin [0,1)$ we have a quadratic equation whose solution in $[0,1]$ is
$$x= frac{(b^2+2(1-b))^{1/2}-b}{2(1-b)}.$$
Verify the right side, as a function of $b,$ strictly decreases on $[0,1)$ from $1/2^{1/2}$ to $1/2.$ Thus for $xiin [1/2,1/2^{1/2}],$ there is a unique $b_{xi}in [0,1]$ such that $p_{b_{xi}}(xi)=1/2.$ Verify that $b_{xi}$ is given by the formula
$$b_{xi} = frac{1/2-xi^2}{xi(1-xi)}.$$
So we've solved the problem for $xiin[1/2,1/2^{1/2}].$ But we've also solved it for $xiin [1/2^{1/2},1/2^{1/4}].$ Just check that that for such $xi,$ $fcirc p_{b_{xi^2}}(x^2)$ does the job. We can keep moving to the right with such intervals and their solutions. We thus obtain solutions for all $xiin [1/2,1).$
What about $xiin (0,1/2]?$ That's easy, now that we've handled the other side. Just check that if $g$ is a solution for $xiin [1/2,1),$ then $1-g(1-x)$ is a solution for $1-xi.$
answered Nov 2 at 21:01
zhw.
70.3k43075
That's a much sleeker answer to the non-polynomial problem. And your polynomial answer almost feels like cheating. Very nice. And very different from alex.jordan's answer.
– Robert Wolfe
Nov 2 at 21:42
+1 Suppose $xi$ is just below $1/2^{1/32}$ (for example). Then this answer constructs a polynomial of degree $32$. My answer produces a polynomial of degree $47$. So this answer is better by that measure. But then if $xi$ is just above $1/2^{1/32}$, my answer still produces a polynomial of degree $47$, and this construction produces an answer of degree $64$, reversing the comparison. This appears to extend to higher powers of $2$ and the two answers are trading the efficiency lead (as measured by lowest polynomial degree) back and forth as $xito1^{-}$.
– alex.jordan
Nov 2 at 22:58
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Robert's comment about cubics is the best we can do.
Proposition: If $f$ is a polynomial of degree at most $3$ that satisfies $f(0) = f(1) = 0$ and has exactly one value $xi in (0, 1)$ for which $f'(xi) = 0$, then $frac{1}{3} leq xi leq frac{2}{3}$.
Proof: Let $f$ have the following form: $$f(x) = x^3 + kx^2 - (1+k) x.$$ (We'll ignore for now the case $xi = frac{1}{2}$, which requires a quadratic; it's trivial to see that for no other value of $xi$ is a quadratic possible.) This is fully general, as wlog we can scale the coefficients of some possible solution $f(x) = ax^3 + bx^2 + cx$ (which must satisfy $a + b + c = 0$) without breaking any condition on $f$. The solutions to $f'(xi) = 3xi^2 + 2k xi - (1+k) = 0$ are thus $$xi = frac{k pm sqrt{k^2 + 3k + 3}}{3}.$$
This can be solved for $k$ by rearranging and squaring to get $(3 xi - k)^2 = k^2 + 3k + 3$, or $$k = frac{1 - 3 xi^2}{2 xi - 1}$$ but the squaring means that $xi$ could be either the upper or the lower solution for any given $k$. Regardless, we know that the fully general formula for a cubic that satisfies $f(0) = f(1) = 0$ and has a not necessarily unique stationary point at $xi$, up to scaling of the coefficients, is $$f(x) = (2 xi - 1) x^3 + (1 - 3 xi^2) x^2 + (3 xi^2 - 2xi) x.$$
We now just need to see which of these cubics have two stationary points in $(0, 1)$. By the Vieta formulas, the solutions to $f'(x) = 3(2 xi - 1) x^2 + 2(1 - 3 xi^2) x + (3 xi^2 - 2xi) = 0$ add up to $frac{2 (3 xi^2 - 1)}{3 (2 xi - 1)}.$ If $xi$ is one solution, then the other solution (call it $xi'$) is begin{align*} xi' &= frac{2 (3 xi^2 - 1)}{3 (2 xi - 1)} - xi \ &= frac{6 xi^2 - 2}{6 xi - 3} - frac{6 xi^2 - 3 xi}{6 xi - 3} \ &= frac{3 xi - 2}{6 xi -3} \
&= frac{3 xi - frac{3}{2}}{6 xi - 3} - frac{frac{1}{2}}{6 xi - 3} \
&= frac{1}{2} - frac{1}{12xi - 6}.end{align*}
Thus, $xi' notin (0, 1)$ if and only if $|12 xi - 6| leq 2$, i.e., if $frac{1}{3} leqxi leq frac{2}{3}.$
answered Nov 5 at 1:55
Connor Harris
4,149723
1
Well, I'd say that finishes up any loose ends. We have two different aesthetic results and a proof of minimality for quartics. Couldn't have hoped for a better resolution.
– Robert Wolfe
Nov 5 at 3:46
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3 Answers
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3 Answers
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down vote
accepted
First, imagine $f$ is some such polynomial for $xi$. Then let $g(x)=f(x)-Mx$. We have $g(0)=g(1)=0$, and $xi$ is the unique number in $(0,1)$ where $g'(x)=0$. So allow me to replace the problem as written with the Mean Value Theorem to one about Rolle's Theorem.
Below is a proof that if you have $xi>frac{1}{2}$, take some integer $n>frac{1-2xi}{xi-1}$, and then take $t=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}$. Then the polynomial $g(x)=(x+t)^nx(1-x)$ satisfies $g(0)=g(1)=0$, and there is a unique number in $(0,1)$ where $g'(x)=0$, and that number is $xi$.
If $xi<frac{1}{2}$, there is a symmetric construction with $t<-1$. And if $xi=frac{1}{2}$, just take $g(x)=x(1-x)$.
For example, with $xi=frac{e}{pi}$, we can take $n=6$, and $t=frac{8(e/pi)^2-7(e/pi)}{-2(e/pi)+1}approx0.09233ldots$. Then $g(x)=(x+t)^6x(1-x)$ is such that $g'$ has only one zero in $(0,1)$, and it is located at $frac{e}{pi}$. See this demonstrated at WolframAlpha.
Explanation
Assume $xi>frac{1}{2}$. Consider $g(x)=(x+t)^nx(1-x)$ for $ninmathbb{N}$ and $tinmathbb{R}_{gt0}$. Then $$
begin{align}
g'(x)&=n(x+t)^{n-1}x(1-x)+(x+t)^n(1-x)-(x+t)^nx\
&=(x+t)^{n-1}big(nx(1-x)+(x+t)(1-x)-(x+t)xbig)\
&=(x+t)^{n-1}big(x^2(-n-2)+x(n+1-2t)+tbig)\
end{align}
$$
The zeros of $g'$ are $-t$ (which is not in $(0,1)$) and $$frac{-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}=frac{Apm B}{C}$$
Since $n,t>0$, it follows that $|B|>|A|$. It follows that one of these two roots is negative (so not in $(0,1)$) and the other is positive. So if the positive root is equal to $xi$, then $g$ satisfies the Rolle's version of the proposition. We have freedom to choose $ninmathbb{N}$, $tinmathbb{R}_{gt0}$, so maybe we can choose them well. In the following, we attempt to solve for $t$ in terms of $xi$ and $n$.
$$begin{align}
xi&=frac{-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}\
-2(n+2)xi&=-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}\
-2(n+2)xi+n+1-2t&=pmsqrt{(n+1-2t)^2+4(n+2)t}
end{align}$$
Squaring both sides:
$$begin{align}
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]+4t^2&=(n+1-2t)^2+4(n+2)t\
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]+4t^2&=(n+1)^2-4(n+1)t+4t^2+4(n+2)t\
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]&=(n+1)^2+4t\
[-2(n+2)xi+n+1]^2-(n+1)^2&=4t[-2(n+2)xi+n+2]\
4(n+2)^2xi^2-4(n+2)(n+1)xi&=4t(-2(n+2)xi+n+2)\
end{align}$$
$$begin{align}
t&=frac{4(n+2)^2xi^2-4(n+2)(n+1)xi}{4(-2(n+2)xi+n+2)}\
&=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}
end{align}$$
We have assumed $xi>frac{1}{2}$, so the denominator is negative. We need $t$ to be positive, so we need the numerator to be negative. Can we choose $n$ to make that happen?
$$begin{align}
(n+2)xi^2-(n+1)xi&<0\
n(xi^2-xi)&<xi-2xi^2\
n(xi-1)&<1-2xi\
n&>frac{1-2xi}{xi-1}
end{align}$$
So yes. If you have $xi>frac{1}{2}$, take some integer $n>frac{1-2xi}{xi-1}$, and then take $t=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}$. Then the polynomial $g(x)=(x+t)^nx(1-x)$ satisfies $g(0)=g(1)=0$, and there is a unique number in $(0,1)$ where $g'(x)=0$, and that number is $xi$.
And restating from the introduction, if $xi<frac{1}{2}$, there is a symmetric construction where $t<-1$. And if $xi=frac{1}{2}$, just take $g(x)=x(1-x)$.
This approach was motivated by starting with $x(1-x)$, and then multiplying by some power of $(x+t)$ that would "warp" the parabola in between $0$ and $1$ without making it wiggle. Something that would stretch $x(1-x)$ vertically, but moreso on the right side than the left side, to push the extremum farther to the right. (Or the other way when $xi<frac{1}{2}$.)
answered Nov 2 at 7:12
alex.jordan
37.9k559118
I'm glad to see that my partial answer generalized in some way. Thanks! I'm still slightly interested to see if someone can post an answer without some case analysis.
– Robert Wolfe
Nov 2 at 21:02
You're welcome, thanks for a question that I found interesting. By case analysis, do you mean $xi>1/2$ versus $xi<1/2$ versus $xi=1/2$?
– alex.jordan
Nov 2 at 21:16
yeah. a small detail.
– Robert Wolfe
Nov 2 at 21:29
it seems that zhw's $(xmapsto 1-x)$ substitution trick could be applied though. that barely qualifies as a case analysis now.
– Robert Wolfe
Nov 2 at 21:37
This is what I meant too, but I could have been more explicit. Applying that reflection effectively turns $t$ in $(0,infty)$ into $tilde{t}$ in $(-infty,-1)$. (Except the whole function is also scaled by $(-1)^n$.) This is where the "$t<-1$" clause came from in the "case" where $xi<1/2$.
– alex.jordan
Nov 2 at 22:26
add a comment |
up vote
5
down vote
accepted
First, imagine $f$ is some such polynomial for $xi$. Then let $g(x)=f(x)-Mx$. We have $g(0)=g(1)=0$, and $xi$ is the unique number in $(0,1)$ where $g'(x)=0$. So allow me to replace the problem as written with the Mean Value Theorem to one about Rolle's Theorem.
Below is a proof that if you have $xi>frac{1}{2}$, take some integer $n>frac{1-2xi}{xi-1}$, and then take $t=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}$. Then the polynomial $g(x)=(x+t)^nx(1-x)$ satisfies $g(0)=g(1)=0$, and there is a unique number in $(0,1)$ where $g'(x)=0$, and that number is $xi$.
If $xi<frac{1}{2}$, there is a symmetric construction with $t<-1$. And if $xi=frac{1}{2}$, just take $g(x)=x(1-x)$.
For example, with $xi=frac{e}{pi}$, we can take $n=6$, and $t=frac{8(e/pi)^2-7(e/pi)}{-2(e/pi)+1}approx0.09233ldots$. Then $g(x)=(x+t)^6x(1-x)$ is such that $g'$ has only one zero in $(0,1)$, and it is located at $frac{e}{pi}$. See this demonstrated at WolframAlpha.
Explanation
Assume $xi>frac{1}{2}$. Consider $g(x)=(x+t)^nx(1-x)$ for $ninmathbb{N}$ and $tinmathbb{R}_{gt0}$. Then $$
begin{align}
g'(x)&=n(x+t)^{n-1}x(1-x)+(x+t)^n(1-x)-(x+t)^nx\
&=(x+t)^{n-1}big(nx(1-x)+(x+t)(1-x)-(x+t)xbig)\
&=(x+t)^{n-1}big(x^2(-n-2)+x(n+1-2t)+tbig)\
end{align}
$$
The zeros of $g'$ are $-t$ (which is not in $(0,1)$) and $$frac{-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}=frac{Apm B}{C}$$
Since $n,t>0$, it follows that $|B|>|A|$. It follows that one of these two roots is negative (so not in $(0,1)$) and the other is positive. So if the positive root is equal to $xi$, then $g$ satisfies the Rolle's version of the proposition. We have freedom to choose $ninmathbb{N}$, $tinmathbb{R}_{gt0}$, so maybe we can choose them well. In the following, we attempt to solve for $t$ in terms of $xi$ and $n$.
$$begin{align}
xi&=frac{-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}\
-2(n+2)xi&=-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}\
-2(n+2)xi+n+1-2t&=pmsqrt{(n+1-2t)^2+4(n+2)t}
end{align}$$
Squaring both sides:
$$begin{align}
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]+4t^2&=(n+1-2t)^2+4(n+2)t\
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]+4t^2&=(n+1)^2-4(n+1)t+4t^2+4(n+2)t\
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]&=(n+1)^2+4t\
[-2(n+2)xi+n+1]^2-(n+1)^2&=4t[-2(n+2)xi+n+2]\
4(n+2)^2xi^2-4(n+2)(n+1)xi&=4t(-2(n+2)xi+n+2)\
end{align}$$
$$begin{align}
t&=frac{4(n+2)^2xi^2-4(n+2)(n+1)xi}{4(-2(n+2)xi+n+2)}\
&=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}
end{align}$$
We have assumed $xi>frac{1}{2}$, so the denominator is negative. We need $t$ to be positive, so we need the numerator to be negative. Can we choose $n$ to make that happen?
$$begin{align}
(n+2)xi^2-(n+1)xi&<0\
n(xi^2-xi)&<xi-2xi^2\
n(xi-1)&<1-2xi\
n&>frac{1-2xi}{xi-1}
end{align}$$
So yes. If you have $xi>frac{1}{2}$, take some integer $n>frac{1-2xi}{xi-1}$, and then take $t=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}$. Then the polynomial $g(x)=(x+t)^nx(1-x)$ satisfies $g(0)=g(1)=0$, and there is a unique number in $(0,1)$ where $g'(x)=0$, and that number is $xi$.
And restating from the introduction, if $xi<frac{1}{2}$, there is a symmetric construction where $t<-1$. And if $xi=frac{1}{2}$, just take $g(x)=x(1-x)$.
This approach was motivated by starting with $x(1-x)$, and then multiplying by some power of $(x+t)$ that would "warp" the parabola in between $0$ and $1$ without making it wiggle. Something that would stretch $x(1-x)$ vertically, but moreso on the right side than the left side, to push the extremum farther to the right. (Or the other way when $xi<frac{1}{2}$.)
answered Nov 2 at 7:12
alex.jordan
37.9k559118
I'm glad to see that my partial answer generalized in some way. Thanks! I'm still slightly interested to see if someone can post an answer without some case analysis.
– Robert Wolfe
Nov 2 at 21:02
You're welcome, thanks for a question that I found interesting. By case analysis, do you mean $xi>1/2$ versus $xi<1/2$ versus $xi=1/2$?
– alex.jordan
Nov 2 at 21:16
yeah. a small detail.
– Robert Wolfe
Nov 2 at 21:29
it seems that zhw's $(xmapsto 1-x)$ substitution trick could be applied though. that barely qualifies as a case analysis now.
– Robert Wolfe
Nov 2 at 21:37
This is what I meant too, but I could have been more explicit. Applying that reflection effectively turns $t$ in $(0,infty)$ into $tilde{t}$ in $(-infty,-1)$. (Except the whole function is also scaled by $(-1)^n$.) This is where the "$t<-1$" clause came from in the "case" where $xi<1/2$.
– alex.jordan
Nov 2 at 22:26
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
First, imagine $f$ is some such polynomial for $xi$. Then let $g(x)=f(x)-Mx$. We have $g(0)=g(1)=0$, and $xi$ is the unique number in $(0,1)$ where $g'(x)=0$. So allow me to replace the problem as written with the Mean Value Theorem to one about Rolle's Theorem.
Below is a proof that if you have $xi>frac{1}{2}$, take some integer $n>frac{1-2xi}{xi-1}$, and then take $t=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}$. Then the polynomial $g(x)=(x+t)^nx(1-x)$ satisfies $g(0)=g(1)=0$, and there is a unique number in $(0,1)$ where $g'(x)=0$, and that number is $xi$.
If $xi<frac{1}{2}$, there is a symmetric construction with $t<-1$. And if $xi=frac{1}{2}$, just take $g(x)=x(1-x)$.
For example, with $xi=frac{e}{pi}$, we can take $n=6$, and $t=frac{8(e/pi)^2-7(e/pi)}{-2(e/pi)+1}approx0.09233ldots$. Then $g(x)=(x+t)^6x(1-x)$ is such that $g'$ has only one zero in $(0,1)$, and it is located at $frac{e}{pi}$. See this demonstrated at WolframAlpha.
Explanation
Assume $xi>frac{1}{2}$. Consider $g(x)=(x+t)^nx(1-x)$ for $ninmathbb{N}$ and $tinmathbb{R}_{gt0}$. Then $$
begin{align}
g'(x)&=n(x+t)^{n-1}x(1-x)+(x+t)^n(1-x)-(x+t)^nx\
&=(x+t)^{n-1}big(nx(1-x)+(x+t)(1-x)-(x+t)xbig)\
&=(x+t)^{n-1}big(x^2(-n-2)+x(n+1-2t)+tbig)\
end{align}
$$
The zeros of $g'$ are $-t$ (which is not in $(0,1)$) and $$frac{-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}=frac{Apm B}{C}$$
Since $n,t>0$, it follows that $|B|>|A|$. It follows that one of these two roots is negative (so not in $(0,1)$) and the other is positive. So if the positive root is equal to $xi$, then $g$ satisfies the Rolle's version of the proposition. We have freedom to choose $ninmathbb{N}$, $tinmathbb{R}_{gt0}$, so maybe we can choose them well. In the following, we attempt to solve for $t$ in terms of $xi$ and $n$.
$$begin{align}
xi&=frac{-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}\
-2(n+2)xi&=-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}\
-2(n+2)xi+n+1-2t&=pmsqrt{(n+1-2t)^2+4(n+2)t}
end{align}$$
Squaring both sides:
$$begin{align}
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]+4t^2&=(n+1-2t)^2+4(n+2)t\
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]+4t^2&=(n+1)^2-4(n+1)t+4t^2+4(n+2)t\
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]&=(n+1)^2+4t\
[-2(n+2)xi+n+1]^2-(n+1)^2&=4t[-2(n+2)xi+n+2]\
4(n+2)^2xi^2-4(n+2)(n+1)xi&=4t(-2(n+2)xi+n+2)\
end{align}$$
$$begin{align}
t&=frac{4(n+2)^2xi^2-4(n+2)(n+1)xi}{4(-2(n+2)xi+n+2)}\
&=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}
end{align}$$
We have assumed $xi>frac{1}{2}$, so the denominator is negative. We need $t$ to be positive, so we need the numerator to be negative. Can we choose $n$ to make that happen?
$$begin{align}
(n+2)xi^2-(n+1)xi&<0\
n(xi^2-xi)&<xi-2xi^2\
n(xi-1)&<1-2xi\
n&>frac{1-2xi}{xi-1}
end{align}$$
So yes. If you have $xi>frac{1}{2}$, take some integer $n>frac{1-2xi}{xi-1}$, and then take $t=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}$. Then the polynomial $g(x)=(x+t)^nx(1-x)$ satisfies $g(0)=g(1)=0$, and there is a unique number in $(0,1)$ where $g'(x)=0$, and that number is $xi$.
And restating from the introduction, if $xi<frac{1}{2}$, there is a symmetric construction where $t<-1$. And if $xi=frac{1}{2}$, just take $g(x)=x(1-x)$.
This approach was motivated by starting with $x(1-x)$, and then multiplying by some power of $(x+t)$ that would "warp" the parabola in between $0$ and $1$ without making it wiggle. Something that would stretch $x(1-x)$ vertically, but moreso on the right side than the left side, to push the extremum farther to the right. (Or the other way when $xi<frac{1}{2}$.)
answered Nov 2 at 7:12
alex.jordan
37.9k559118
First, imagine $f$ is some such polynomial for $xi$. Then let $g(x)=f(x)-Mx$. We have $g(0)=g(1)=0$, and $xi$ is the unique number in $(0,1)$ where $g'(x)=0$. So allow me to replace the problem as written with the Mean Value Theorem to one about Rolle's Theorem.
Below is a proof that if you have $xi>frac{1}{2}$, take some integer $n>frac{1-2xi}{xi-1}$, and then take $t=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}$. Then the polynomial $g(x)=(x+t)^nx(1-x)$ satisfies $g(0)=g(1)=0$, and there is a unique number in $(0,1)$ where $g'(x)=0$, and that number is $xi$.
If $xi<frac{1}{2}$, there is a symmetric construction with $t<-1$. And if $xi=frac{1}{2}$, just take $g(x)=x(1-x)$.
For example, with $xi=frac{e}{pi}$, we can take $n=6$, and $t=frac{8(e/pi)^2-7(e/pi)}{-2(e/pi)+1}approx0.09233ldots$. Then $g(x)=(x+t)^6x(1-x)$ is such that $g'$ has only one zero in $(0,1)$, and it is located at $frac{e}{pi}$. See this demonstrated at WolframAlpha.
Explanation
Assume $xi>frac{1}{2}$. Consider $g(x)=(x+t)^nx(1-x)$ for $ninmathbb{N}$ and $tinmathbb{R}_{gt0}$. Then $$
begin{align}
g'(x)&=n(x+t)^{n-1}x(1-x)+(x+t)^n(1-x)-(x+t)^nx\
&=(x+t)^{n-1}big(nx(1-x)+(x+t)(1-x)-(x+t)xbig)\
&=(x+t)^{n-1}big(x^2(-n-2)+x(n+1-2t)+tbig)\
end{align}
$$
The zeros of $g'$ are $-t$ (which is not in $(0,1)$) and $$frac{-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}=frac{Apm B}{C}$$
Since $n,t>0$, it follows that $|B|>|A|$. It follows that one of these two roots is negative (so not in $(0,1)$) and the other is positive. So if the positive root is equal to $xi$, then $g$ satisfies the Rolle's version of the proposition. We have freedom to choose $ninmathbb{N}$, $tinmathbb{R}_{gt0}$, so maybe we can choose them well. In the following, we attempt to solve for $t$ in terms of $xi$ and $n$.
$$begin{align}
xi&=frac{-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}\
-2(n+2)xi&=-(n+1-2t)pmsqrt{(n+1-2t)^2+4(n+2)t}\
-2(n+2)xi+n+1-2t&=pmsqrt{(n+1-2t)^2+4(n+2)t}
end{align}$$
Squaring both sides:
$$begin{align}
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]+4t^2&=(n+1-2t)^2+4(n+2)t\
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]+4t^2&=(n+1)^2-4(n+1)t+4t^2+4(n+2)t\
[-2(n+2)xi+n+1]^2-4t[-2(n+2)xi+n+1]&=(n+1)^2+4t\
[-2(n+2)xi+n+1]^2-(n+1)^2&=4t[-2(n+2)xi+n+2]\
4(n+2)^2xi^2-4(n+2)(n+1)xi&=4t(-2(n+2)xi+n+2)\
end{align}$$
$$begin{align}
t&=frac{4(n+2)^2xi^2-4(n+2)(n+1)xi}{4(-2(n+2)xi+n+2)}\
&=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}
end{align}$$
We have assumed $xi>frac{1}{2}$, so the denominator is negative. We need $t$ to be positive, so we need the numerator to be negative. Can we choose $n$ to make that happen?
$$begin{align}
(n+2)xi^2-(n+1)xi&<0\
n(xi^2-xi)&<xi-2xi^2\
n(xi-1)&<1-2xi\
n&>frac{1-2xi}{xi-1}
end{align}$$
So yes. If you have $xi>frac{1}{2}$, take some integer $n>frac{1-2xi}{xi-1}$, and then take $t=frac{(n+2)xi^2-(n+1)xi}{-2xi+1}$. Then the polynomial $g(x)=(x+t)^nx(1-x)$ satisfies $g(0)=g(1)=0$, and there is a unique number in $(0,1)$ where $g'(x)=0$, and that number is $xi$.
And restating from the introduction, if $xi<frac{1}{2}$, there is a symmetric construction where $t<-1$. And if $xi=frac{1}{2}$, just take $g(x)=x(1-x)$.
This approach was motivated by starting with $x(1-x)$, and then multiplying by some power of $(x+t)$ that would "warp" the parabola in between $0$ and $1$ without making it wiggle. Something that would stretch $x(1-x)$ vertically, but moreso on the right side than the left side, to push the extremum farther to the right. (Or the other way when $xi<frac{1}{2}$.)
answered Nov 2 at 7:12
alex.jordan
37.9k559118
edited Nov 3 at 6:07
answered Nov 2 at 7:12
alex.jordan
37.9k559118
answered Nov 2 at 7:12
alex.jordan
37.9k559118
answered Nov 2 at 7:12
alex.jordan
37.9k559118
37.9k559118
I'm glad to see that my partial answer generalized in some way. Thanks! I'm still slightly interested to see if someone can post an answer without some case analysis.
– Robert Wolfe
Nov 2 at 21:02
You're welcome, thanks for a question that I found interesting. By case analysis, do you mean $xi>1/2$ versus $xi<1/2$ versus $xi=1/2$?
– alex.jordan
Nov 2 at 21:16
yeah. a small detail.
– Robert Wolfe
Nov 2 at 21:29
it seems that zhw's $(xmapsto 1-x)$ substitution trick could be applied though. that barely qualifies as a case analysis now.
– Robert Wolfe
Nov 2 at 21:37
This is what I meant too, but I could have been more explicit. Applying that reflection effectively turns $t$ in $(0,infty)$ into $tilde{t}$ in $(-infty,-1)$. (Except the whole function is also scaled by $(-1)^n$.) This is where the "$t<-1$" clause came from in the "case" where $xi<1/2$.
– alex.jordan
Nov 2 at 22:26
add a comment |
I'm glad to see that my partial answer generalized in some way. Thanks! I'm still slightly interested to see if someone can post an answer without some case analysis.
– Robert Wolfe
Nov 2 at 21:02
You're welcome, thanks for a question that I found interesting. By case analysis, do you mean $xi>1/2$ versus $xi<1/2$ versus $xi=1/2$?
– alex.jordan
Nov 2 at 21:16
yeah. a small detail.
– Robert Wolfe
Nov 2 at 21:29
it seems that zhw's $(xmapsto 1-x)$ substitution trick could be applied though. that barely qualifies as a case analysis now.
– Robert Wolfe
Nov 2 at 21:37
This is what I meant too, but I could have been more explicit. Applying that reflection effectively turns $t$ in $(0,infty)$ into $tilde{t}$ in $(-infty,-1)$. (Except the whole function is also scaled by $(-1)^n$.) This is where the "$t<-1$" clause came from in the "case" where $xi<1/2$.
– alex.jordan
Nov 2 at 22:26
I'm glad to see that my partial answer generalized in some way. Thanks! I'm still slightly interested to see if someone can post an answer without some case analysis.
– Robert Wolfe
Nov 2 at 21:02
I'm glad to see that my partial answer generalized in some way. Thanks! I'm still slightly interested to see if someone can post an answer without some case analysis.
– Robert Wolfe
Nov 2 at 21:02
You're welcome, thanks for a question that I found interesting. By case analysis, do you mean $xi>1/2$ versus $xi<1/2$ versus $xi=1/2$?
– alex.jordan
Nov 2 at 21:16
You're welcome, thanks for a question that I found interesting. By case analysis, do you mean $xi>1/2$ versus $xi<1/2$ versus $xi=1/2$?
– alex.jordan
Nov 2 at 21:16
yeah. a small detail.
– Robert Wolfe
Nov 2 at 21:29
yeah. a small detail.
– Robert Wolfe
Nov 2 at 21:29
it seems that zhw's $(xmapsto 1-x)$ substitution trick could be applied though. that barely qualifies as a case analysis now.
– Robert Wolfe
Nov 2 at 21:37
it seems that zhw's $(xmapsto 1-x)$ substitution trick could be applied though. that barely qualifies as a case analysis now.
– Robert Wolfe
Nov 2 at 21:37
This is what I meant too, but I could have been more explicit. Applying that reflection effectively turns $t$ in $(0,infty)$ into $tilde{t}$ in $(-infty,-1)$. (Except the whole function is also scaled by $(-1)^n$.) This is where the "$t<-1$" clause came from in the "case" where $xi<1/2$.
– alex.jordan
Nov 2 at 22:26
This is what I meant too, but I could have been more explicit. Applying that reflection effectively turns $t$ in $(0,infty)$ into $tilde{t}$ in $(-infty,-1)$. (Except the whole function is also scaled by $(-1)^n$.) This is where the "$t<-1$" clause came from in the "case" where $xi<1/2$.
– alex.jordan
Nov 2 at 22:26
add a comment |
up vote
2
down vote
Just a few words on solving the nonpolynomial problem: First take $M=0.$ Then $f(x)=x(1-x)$ is a solution for $xi=1/2.$ For other values of $xiin (0,1),$ we look at $g_p(x)=f(x^p)$ for $p>0.$ We have $g_p'(x)=px^{p-1}f'(x^p)=0$ iff $x=(1/2)^{1/p}.$ Thus if $xi$ is given, we take $p=ln(1/2)/ln xi,$ and $g_p$ solves the problem. Finally, if $Mne 0$ and $xi$ is given, we take the same $p$ and the function $M(x+g_p(x))$ solves the problem.
On to polynomial solutions: WLOG $M=0$, for we can use the same idea as above for $Mne 0.$
Let $f(x)=x(1-x)$ as above. If $p$ is a polynomial with $p(0)=0,$ $p(1)=1$ and $p'>0$ on $(0,1),$ then $fcirc p$ is a polynomial that solves the problem for for the unique value $xi in(0,1)$ such that $p(xi)=1/2.$
For $0le b le 1,$ set $p_b(x)=(1-b)x^2 +bx.$ Then $p_b(0)=0,p_b(1)=1,$ and $p_b'>0$ on $(0,1).$ Consider the equation $p_b(x)-1/2=0.$ If $b=1,$ then $x=1/2$ is a solution. For $bin [0,1)$ we have a quadratic equation whose solution in $[0,1]$ is
$$x= frac{(b^2+2(1-b))^{1/2}-b}{2(1-b)}.$$
Verify the right side, as a function of $b,$ strictly decreases on $[0,1)$ from $1/2^{1/2}$ to $1/2.$ Thus for $xiin [1/2,1/2^{1/2}],$ there is a unique $b_{xi}in [0,1]$ such that $p_{b_{xi}}(xi)=1/2.$ Verify that $b_{xi}$ is given by the formula
$$b_{xi} = frac{1/2-xi^2}{xi(1-xi)}.$$
So we've solved the problem for $xiin[1/2,1/2^{1/2}].$ But we've also solved it for $xiin [1/2^{1/2},1/2^{1/4}].$ Just check that that for such $xi,$ $fcirc p_{b_{xi^2}}(x^2)$ does the job. We can keep moving to the right with such intervals and their solutions. We thus obtain solutions for all $xiin [1/2,1).$
What about $xiin (0,1/2]?$ That's easy, now that we've handled the other side. Just check that if $g$ is a solution for $xiin [1/2,1),$ then $1-g(1-x)$ is a solution for $1-xi.$
answered Nov 2 at 21:01
zhw.
70.3k43075
That's a much sleeker answer to the non-polynomial problem. And your polynomial answer almost feels like cheating. Very nice. And very different from alex.jordan's answer.
– Robert Wolfe
Nov 2 at 21:42
+1 Suppose $xi$ is just below $1/2^{1/32}$ (for example). Then this answer constructs a polynomial of degree $32$. My answer produces a polynomial of degree $47$. So this answer is better by that measure. But then if $xi$ is just above $1/2^{1/32}$, my answer still produces a polynomial of degree $47$, and this construction produces an answer of degree $64$, reversing the comparison. This appears to extend to higher powers of $2$ and the two answers are trading the efficiency lead (as measured by lowest polynomial degree) back and forth as $xito1^{-}$.
– alex.jordan
Nov 2 at 22:58
add a comment |
up vote
2
down vote
Just a few words on solving the nonpolynomial problem: First take $M=0.$ Then $f(x)=x(1-x)$ is a solution for $xi=1/2.$ For other values of $xiin (0,1),$ we look at $g_p(x)=f(x^p)$ for $p>0.$ We have $g_p'(x)=px^{p-1}f'(x^p)=0$ iff $x=(1/2)^{1/p}.$ Thus if $xi$ is given, we take $p=ln(1/2)/ln xi,$ and $g_p$ solves the problem. Finally, if $Mne 0$ and $xi$ is given, we take the same $p$ and the function $M(x+g_p(x))$ solves the problem.
On to polynomial solutions: WLOG $M=0$, for we can use the same idea as above for $Mne 0.$
Let $f(x)=x(1-x)$ as above. If $p$ is a polynomial with $p(0)=0,$ $p(1)=1$ and $p'>0$ on $(0,1),$ then $fcirc p$ is a polynomial that solves the problem for for the unique value $xi in(0,1)$ such that $p(xi)=1/2.$
For $0le b le 1,$ set $p_b(x)=(1-b)x^2 +bx.$ Then $p_b(0)=0,p_b(1)=1,$ and $p_b'>0$ on $(0,1).$ Consider the equation $p_b(x)-1/2=0.$ If $b=1,$ then $x=1/2$ is a solution. For $bin [0,1)$ we have a quadratic equation whose solution in $[0,1]$ is
$$x= frac{(b^2+2(1-b))^{1/2}-b}{2(1-b)}.$$
Verify the right side, as a function of $b,$ strictly decreases on $[0,1)$ from $1/2^{1/2}$ to $1/2.$ Thus for $xiin [1/2,1/2^{1/2}],$ there is a unique $b_{xi}in [0,1]$ such that $p_{b_{xi}}(xi)=1/2.$ Verify that $b_{xi}$ is given by the formula
$$b_{xi} = frac{1/2-xi^2}{xi(1-xi)}.$$
So we've solved the problem for $xiin[1/2,1/2^{1/2}].$ But we've also solved it for $xiin [1/2^{1/2},1/2^{1/4}].$ Just check that that for such $xi,$ $fcirc p_{b_{xi^2}}(x^2)$ does the job. We can keep moving to the right with such intervals and their solutions. We thus obtain solutions for all $xiin [1/2,1).$
What about $xiin (0,1/2]?$ That's easy, now that we've handled the other side. Just check that if $g$ is a solution for $xiin [1/2,1),$ then $1-g(1-x)$ is a solution for $1-xi.$
answered Nov 2 at 21:01
zhw.
70.3k43075
That's a much sleeker answer to the non-polynomial problem. And your polynomial answer almost feels like cheating. Very nice. And very different from alex.jordan's answer.
– Robert Wolfe
Nov 2 at 21:42
+1 Suppose $xi$ is just below $1/2^{1/32}$ (for example). Then this answer constructs a polynomial of degree $32$. My answer produces a polynomial of degree $47$. So this answer is better by that measure. But then if $xi$ is just above $1/2^{1/32}$, my answer still produces a polynomial of degree $47$, and this construction produces an answer of degree $64$, reversing the comparison. This appears to extend to higher powers of $2$ and the two answers are trading the efficiency lead (as measured by lowest polynomial degree) back and forth as $xito1^{-}$.
– alex.jordan
Nov 2 at 22:58
add a comment |
up vote
2
down vote
up vote
2
down vote
Just a few words on solving the nonpolynomial problem: First take $M=0.$ Then $f(x)=x(1-x)$ is a solution for $xi=1/2.$ For other values of $xiin (0,1),$ we look at $g_p(x)=f(x^p)$ for $p>0.$ We have $g_p'(x)=px^{p-1}f'(x^p)=0$ iff $x=(1/2)^{1/p}.$ Thus if $xi$ is given, we take $p=ln(1/2)/ln xi,$ and $g_p$ solves the problem. Finally, if $Mne 0$ and $xi$ is given, we take the same $p$ and the function $M(x+g_p(x))$ solves the problem.
On to polynomial solutions: WLOG $M=0$, for we can use the same idea as above for $Mne 0.$
Let $f(x)=x(1-x)$ as above. If $p$ is a polynomial with $p(0)=0,$ $p(1)=1$ and $p'>0$ on $(0,1),$ then $fcirc p$ is a polynomial that solves the problem for for the unique value $xi in(0,1)$ such that $p(xi)=1/2.$
For $0le b le 1,$ set $p_b(x)=(1-b)x^2 +bx.$ Then $p_b(0)=0,p_b(1)=1,$ and $p_b'>0$ on $(0,1).$ Consider the equation $p_b(x)-1/2=0.$ If $b=1,$ then $x=1/2$ is a solution. For $bin [0,1)$ we have a quadratic equation whose solution in $[0,1]$ is
$$x= frac{(b^2+2(1-b))^{1/2}-b}{2(1-b)}.$$
Verify the right side, as a function of $b,$ strictly decreases on $[0,1)$ from $1/2^{1/2}$ to $1/2.$ Thus for $xiin [1/2,1/2^{1/2}],$ there is a unique $b_{xi}in [0,1]$ such that $p_{b_{xi}}(xi)=1/2.$ Verify that $b_{xi}$ is given by the formula
$$b_{xi} = frac{1/2-xi^2}{xi(1-xi)}.$$
So we've solved the problem for $xiin[1/2,1/2^{1/2}].$ But we've also solved it for $xiin [1/2^{1/2},1/2^{1/4}].$ Just check that that for such $xi,$ $fcirc p_{b_{xi^2}}(x^2)$ does the job. We can keep moving to the right with such intervals and their solutions. We thus obtain solutions for all $xiin [1/2,1).$
What about $xiin (0,1/2]?$ That's easy, now that we've handled the other side. Just check that if $g$ is a solution for $xiin [1/2,1),$ then $1-g(1-x)$ is a solution for $1-xi.$
answered Nov 2 at 21:01
zhw.
70.3k43075
Just a few words on solving the nonpolynomial problem: First take $M=0.$ Then $f(x)=x(1-x)$ is a solution for $xi=1/2.$ For other values of $xiin (0,1),$ we look at $g_p(x)=f(x^p)$ for $p>0.$ We have $g_p'(x)=px^{p-1}f'(x^p)=0$ iff $x=(1/2)^{1/p}.$ Thus if $xi$ is given, we take $p=ln(1/2)/ln xi,$ and $g_p$ solves the problem. Finally, if $Mne 0$ and $xi$ is given, we take the same $p$ and the function $M(x+g_p(x))$ solves the problem.
On to polynomial solutions: WLOG $M=0$, for we can use the same idea as above for $Mne 0.$
Let $f(x)=x(1-x)$ as above. If $p$ is a polynomial with $p(0)=0,$ $p(1)=1$ and $p'>0$ on $(0,1),$ then $fcirc p$ is a polynomial that solves the problem for for the unique value $xi in(0,1)$ such that $p(xi)=1/2.$
For $0le b le 1,$ set $p_b(x)=(1-b)x^2 +bx.$ Then $p_b(0)=0,p_b(1)=1,$ and $p_b'>0$ on $(0,1).$ Consider the equation $p_b(x)-1/2=0.$ If $b=1,$ then $x=1/2$ is a solution. For $bin [0,1)$ we have a quadratic equation whose solution in $[0,1]$ is
$$x= frac{(b^2+2(1-b))^{1/2}-b}{2(1-b)}.$$
Verify the right side, as a function of $b,$ strictly decreases on $[0,1)$ from $1/2^{1/2}$ to $1/2.$ Thus for $xiin [1/2,1/2^{1/2}],$ there is a unique $b_{xi}in [0,1]$ such that $p_{b_{xi}}(xi)=1/2.$ Verify that $b_{xi}$ is given by the formula
$$b_{xi} = frac{1/2-xi^2}{xi(1-xi)}.$$
So we've solved the problem for $xiin[1/2,1/2^{1/2}].$ But we've also solved it for $xiin [1/2^{1/2},1/2^{1/4}].$ Just check that that for such $xi,$ $fcirc p_{b_{xi^2}}(x^2)$ does the job. We can keep moving to the right with such intervals and their solutions. We thus obtain solutions for all $xiin [1/2,1).$
What about $xiin (0,1/2]?$ That's easy, now that we've handled the other side. Just check that if $g$ is a solution for $xiin [1/2,1),$ then $1-g(1-x)$ is a solution for $1-xi.$
answered Nov 2 at 21:01
zhw.
70.3k43075
answered Nov 2 at 21:01
zhw.
70.3k43075
answered Nov 2 at 21:01
zhw.
70.3k43075
answered Nov 2 at 21:01
zhw.
70.3k43075
70.3k43075
That's a much sleeker answer to the non-polynomial problem. And your polynomial answer almost feels like cheating. Very nice. And very different from alex.jordan's answer.
– Robert Wolfe
Nov 2 at 21:42
+1 Suppose $xi$ is just below $1/2^{1/32}$ (for example). Then this answer constructs a polynomial of degree $32$. My answer produces a polynomial of degree $47$. So this answer is better by that measure. But then if $xi$ is just above $1/2^{1/32}$, my answer still produces a polynomial of degree $47$, and this construction produces an answer of degree $64$, reversing the comparison. This appears to extend to higher powers of $2$ and the two answers are trading the efficiency lead (as measured by lowest polynomial degree) back and forth as $xito1^{-}$.
– alex.jordan
Nov 2 at 22:58
add a comment |
That's a much sleeker answer to the non-polynomial problem. And your polynomial answer almost feels like cheating. Very nice. And very different from alex.jordan's answer.
– Robert Wolfe
Nov 2 at 21:42
+1 Suppose $xi$ is just below $1/2^{1/32}$ (for example). Then this answer constructs a polynomial of degree $32$. My answer produces a polynomial of degree $47$. So this answer is better by that measure. But then if $xi$ is just above $1/2^{1/32}$, my answer still produces a polynomial of degree $47$, and this construction produces an answer of degree $64$, reversing the comparison. This appears to extend to higher powers of $2$ and the two answers are trading the efficiency lead (as measured by lowest polynomial degree) back and forth as $xito1^{-}$.
– alex.jordan
Nov 2 at 22:58
That's a much sleeker answer to the non-polynomial problem. And your polynomial answer almost feels like cheating. Very nice. And very different from alex.jordan's answer.
– Robert Wolfe
Nov 2 at 21:42
That's a much sleeker answer to the non-polynomial problem. And your polynomial answer almost feels like cheating. Very nice. And very different from alex.jordan's answer.
– Robert Wolfe
Nov 2 at 21:42
+1 Suppose $xi$ is just below $1/2^{1/32}$ (for example). Then this answer constructs a polynomial of degree $32$. My answer produces a polynomial of degree $47$. So this answer is better by that measure. But then if $xi$ is just above $1/2^{1/32}$, my answer still produces a polynomial of degree $47$, and this construction produces an answer of degree $64$, reversing the comparison. This appears to extend to higher powers of $2$ and the two answers are trading the efficiency lead (as measured by lowest polynomial degree) back and forth as $xito1^{-}$.
– alex.jordan
Nov 2 at 22:58
+1 Suppose $xi$ is just below $1/2^{1/32}$ (for example). Then this answer constructs a polynomial of degree $32$. My answer produces a polynomial of degree $47$. So this answer is better by that measure. But then if $xi$ is just above $1/2^{1/32}$, my answer still produces a polynomial of degree $47$, and this construction produces an answer of degree $64$, reversing the comparison. This appears to extend to higher powers of $2$ and the two answers are trading the efficiency lead (as measured by lowest polynomial degree) back and forth as $xito1^{-}$.
– alex.jordan
Nov 2 at 22:58
add a comment |
up vote
1
down vote
Robert's comment about cubics is the best we can do.
Proposition: If $f$ is a polynomial of degree at most $3$ that satisfies $f(0) = f(1) = 0$ and has exactly one value $xi in (0, 1)$ for which $f'(xi) = 0$, then $frac{1}{3} leq xi leq frac{2}{3}$.
Proof: Let $f$ have the following form: $$f(x) = x^3 + kx^2 - (1+k) x.$$ (We'll ignore for now the case $xi = frac{1}{2}$, which requires a quadratic; it's trivial to see that for no other value of $xi$ is a quadratic possible.) This is fully general, as wlog we can scale the coefficients of some possible solution $f(x) = ax^3 + bx^2 + cx$ (which must satisfy $a + b + c = 0$) without breaking any condition on $f$. The solutions to $f'(xi) = 3xi^2 + 2k xi - (1+k) = 0$ are thus $$xi = frac{k pm sqrt{k^2 + 3k + 3}}{3}.$$
This can be solved for $k$ by rearranging and squaring to get $(3 xi - k)^2 = k^2 + 3k + 3$, or $$k = frac{1 - 3 xi^2}{2 xi - 1}$$ but the squaring means that $xi$ could be either the upper or the lower solution for any given $k$. Regardless, we know that the fully general formula for a cubic that satisfies $f(0) = f(1) = 0$ and has a not necessarily unique stationary point at $xi$, up to scaling of the coefficients, is $$f(x) = (2 xi - 1) x^3 + (1 - 3 xi^2) x^2 + (3 xi^2 - 2xi) x.$$
We now just need to see which of these cubics have two stationary points in $(0, 1)$. By the Vieta formulas, the solutions to $f'(x) = 3(2 xi - 1) x^2 + 2(1 - 3 xi^2) x + (3 xi^2 - 2xi) = 0$ add up to $frac{2 (3 xi^2 - 1)}{3 (2 xi - 1)}.$ If $xi$ is one solution, then the other solution (call it $xi'$) is begin{align*} xi' &= frac{2 (3 xi^2 - 1)}{3 (2 xi - 1)} - xi \ &= frac{6 xi^2 - 2}{6 xi - 3} - frac{6 xi^2 - 3 xi}{6 xi - 3} \ &= frac{3 xi - 2}{6 xi -3} \
&= frac{3 xi - frac{3}{2}}{6 xi - 3} - frac{frac{1}{2}}{6 xi - 3} \
&= frac{1}{2} - frac{1}{12xi - 6}.end{align*}
Thus, $xi' notin (0, 1)$ if and only if $|12 xi - 6| leq 2$, i.e., if $frac{1}{3} leqxi leq frac{2}{3}.$
answered Nov 5 at 1:55
Connor Harris
4,149723
1
Well, I'd say that finishes up any loose ends. We have two different aesthetic results and a proof of minimality for quartics. Couldn't have hoped for a better resolution.
– Robert Wolfe
Nov 5 at 3:46
add a comment |
up vote
1
down vote
Robert's comment about cubics is the best we can do.
Proposition: If $f$ is a polynomial of degree at most $3$ that satisfies $f(0) = f(1) = 0$ and has exactly one value $xi in (0, 1)$ for which $f'(xi) = 0$, then $frac{1}{3} leq xi leq frac{2}{3}$.
Proof: Let $f$ have the following form: $$f(x) = x^3 + kx^2 - (1+k) x.$$ (We'll ignore for now the case $xi = frac{1}{2}$, which requires a quadratic; it's trivial to see that for no other value of $xi$ is a quadratic possible.) This is fully general, as wlog we can scale the coefficients of some possible solution $f(x) = ax^3 + bx^2 + cx$ (which must satisfy $a + b + c = 0$) without breaking any condition on $f$. The solutions to $f'(xi) = 3xi^2 + 2k xi - (1+k) = 0$ are thus $$xi = frac{k pm sqrt{k^2 + 3k + 3}}{3}.$$
This can be solved for $k$ by rearranging and squaring to get $(3 xi - k)^2 = k^2 + 3k + 3$, or $$k = frac{1 - 3 xi^2}{2 xi - 1}$$ but the squaring means that $xi$ could be either the upper or the lower solution for any given $k$. Regardless, we know that the fully general formula for a cubic that satisfies $f(0) = f(1) = 0$ and has a not necessarily unique stationary point at $xi$, up to scaling of the coefficients, is $$f(x) = (2 xi - 1) x^3 + (1 - 3 xi^2) x^2 + (3 xi^2 - 2xi) x.$$
We now just need to see which of these cubics have two stationary points in $(0, 1)$. By the Vieta formulas, the solutions to $f'(x) = 3(2 xi - 1) x^2 + 2(1 - 3 xi^2) x + (3 xi^2 - 2xi) = 0$ add up to $frac{2 (3 xi^2 - 1)}{3 (2 xi - 1)}.$ If $xi$ is one solution, then the other solution (call it $xi'$) is begin{align*} xi' &= frac{2 (3 xi^2 - 1)}{3 (2 xi - 1)} - xi \ &= frac{6 xi^2 - 2}{6 xi - 3} - frac{6 xi^2 - 3 xi}{6 xi - 3} \ &= frac{3 xi - 2}{6 xi -3} \
&= frac{3 xi - frac{3}{2}}{6 xi - 3} - frac{frac{1}{2}}{6 xi - 3} \
&= frac{1}{2} - frac{1}{12xi - 6}.end{align*}
Thus, $xi' notin (0, 1)$ if and only if $|12 xi - 6| leq 2$, i.e., if $frac{1}{3} leqxi leq frac{2}{3}.$
answered Nov 5 at 1:55
Connor Harris
4,149723
1
Well, I'd say that finishes up any loose ends. We have two different aesthetic results and a proof of minimality for quartics. Couldn't have hoped for a better resolution.
– Robert Wolfe
Nov 5 at 3:46
add a comment |
up vote
1
down vote
up vote
1
down vote
Robert's comment about cubics is the best we can do.
Proposition: If $f$ is a polynomial of degree at most $3$ that satisfies $f(0) = f(1) = 0$ and has exactly one value $xi in (0, 1)$ for which $f'(xi) = 0$, then $frac{1}{3} leq xi leq frac{2}{3}$.
Proof: Let $f$ have the following form: $$f(x) = x^3 + kx^2 - (1+k) x.$$ (We'll ignore for now the case $xi = frac{1}{2}$, which requires a quadratic; it's trivial to see that for no other value of $xi$ is a quadratic possible.) This is fully general, as wlog we can scale the coefficients of some possible solution $f(x) = ax^3 + bx^2 + cx$ (which must satisfy $a + b + c = 0$) without breaking any condition on $f$. The solutions to $f'(xi) = 3xi^2 + 2k xi - (1+k) = 0$ are thus $$xi = frac{k pm sqrt{k^2 + 3k + 3}}{3}.$$
This can be solved for $k$ by rearranging and squaring to get $(3 xi - k)^2 = k^2 + 3k + 3$, or $$k = frac{1 - 3 xi^2}{2 xi - 1}$$ but the squaring means that $xi$ could be either the upper or the lower solution for any given $k$. Regardless, we know that the fully general formula for a cubic that satisfies $f(0) = f(1) = 0$ and has a not necessarily unique stationary point at $xi$, up to scaling of the coefficients, is $$f(x) = (2 xi - 1) x^3 + (1 - 3 xi^2) x^2 + (3 xi^2 - 2xi) x.$$
We now just need to see which of these cubics have two stationary points in $(0, 1)$. By the Vieta formulas, the solutions to $f'(x) = 3(2 xi - 1) x^2 + 2(1 - 3 xi^2) x + (3 xi^2 - 2xi) = 0$ add up to $frac{2 (3 xi^2 - 1)}{3 (2 xi - 1)}.$ If $xi$ is one solution, then the other solution (call it $xi'$) is begin{align*} xi' &= frac{2 (3 xi^2 - 1)}{3 (2 xi - 1)} - xi \ &= frac{6 xi^2 - 2}{6 xi - 3} - frac{6 xi^2 - 3 xi}{6 xi - 3} \ &= frac{3 xi - 2}{6 xi -3} \
&= frac{3 xi - frac{3}{2}}{6 xi - 3} - frac{frac{1}{2}}{6 xi - 3} \
&= frac{1}{2} - frac{1}{12xi - 6}.end{align*}
Thus, $xi' notin (0, 1)$ if and only if $|12 xi - 6| leq 2$, i.e., if $frac{1}{3} leqxi leq frac{2}{3}.$
answered Nov 5 at 1:55
Connor Harris
4,149723
Robert's comment about cubics is the best we can do.
Proposition: If $f$ is a polynomial of degree at most $3$ that satisfies $f(0) = f(1) = 0$ and has exactly one value $xi in (0, 1)$ for which $f'(xi) = 0$, then $frac{1}{3} leq xi leq frac{2}{3}$.
Proof: Let $f$ have the following form: $$f(x) = x^3 + kx^2 - (1+k) x.$$ (We'll ignore for now the case $xi = frac{1}{2}$, which requires a quadratic; it's trivial to see that for no other value of $xi$ is a quadratic possible.) This is fully general, as wlog we can scale the coefficients of some possible solution $f(x) = ax^3 + bx^2 + cx$ (which must satisfy $a + b + c = 0$) without breaking any condition on $f$. The solutions to $f'(xi) = 3xi^2 + 2k xi - (1+k) = 0$ are thus $$xi = frac{k pm sqrt{k^2 + 3k + 3}}{3}.$$
This can be solved for $k$ by rearranging and squaring to get $(3 xi - k)^2 = k^2 + 3k + 3$, or $$k = frac{1 - 3 xi^2}{2 xi - 1}$$ but the squaring means that $xi$ could be either the upper or the lower solution for any given $k$. Regardless, we know that the fully general formula for a cubic that satisfies $f(0) = f(1) = 0$ and has a not necessarily unique stationary point at $xi$, up to scaling of the coefficients, is $$f(x) = (2 xi - 1) x^3 + (1 - 3 xi^2) x^2 + (3 xi^2 - 2xi) x.$$
We now just need to see which of these cubics have two stationary points in $(0, 1)$. By the Vieta formulas, the solutions to $f'(x) = 3(2 xi - 1) x^2 + 2(1 - 3 xi^2) x + (3 xi^2 - 2xi) = 0$ add up to $frac{2 (3 xi^2 - 1)}{3 (2 xi - 1)}.$ If $xi$ is one solution, then the other solution (call it $xi'$) is begin{align*} xi' &= frac{2 (3 xi^2 - 1)}{3 (2 xi - 1)} - xi \ &= frac{6 xi^2 - 2}{6 xi - 3} - frac{6 xi^2 - 3 xi}{6 xi - 3} \ &= frac{3 xi - 2}{6 xi -3} \
&= frac{3 xi - frac{3}{2}}{6 xi - 3} - frac{frac{1}{2}}{6 xi - 3} \
&= frac{1}{2} - frac{1}{12xi - 6}.end{align*}
Thus, $xi' notin (0, 1)$ if and only if $|12 xi - 6| leq 2$, i.e., if $frac{1}{3} leqxi leq frac{2}{3}.$
answered Nov 5 at 1:55
Connor Harris
4,149723
edited Nov 5 at 3:48
answered Nov 5 at 1:55
Connor Harris
4,149723
answered Nov 5 at 1:55
Connor Harris
4,149723
answered Nov 5 at 1:55
Connor Harris
4,149723
4,149723
1
Well, I'd say that finishes up any loose ends. We have two different aesthetic results and a proof of minimality for quartics. Couldn't have hoped for a better resolution.
– Robert Wolfe
Nov 5 at 3:46
add a comment |
1
Well, I'd say that finishes up any loose ends. We have two different aesthetic results and a proof of minimality for quartics. Couldn't have hoped for a better resolution.
– Robert Wolfe
Nov 5 at 3:46
1
1
Well, I'd say that finishes up any loose ends. We have two different aesthetic results and a proof of minimality for quartics. Couldn't have hoped for a better resolution.
– Robert Wolfe
Nov 5 at 3:46
Well, I'd say that finishes up any loose ends. We have two different aesthetic results and a proof of minimality for quartics. Couldn't have hoped for a better resolution.
– Robert Wolfe
Nov 5 at 3:46
add a comment |
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3
Quick idea: consider $f(x) = Mx + p(x)$ where $p(x)$ is a polynomial constructed to satisfy $p(0) = p(1) = 0$, $(forall 0 leq 1 leq x) p''(x) < 0$, and $p'(xi) = 0$ for some arbitrary (unique by construction) $xi$. This should be possible with quartics: it boils down to just solving an underdetermined system on the coefficients of $p$.
– Connor Harris
Oct 30 at 21:09
@ConnorHarris that seems feasible. The algebra is a little bit of a pain though. A solution would be quite slick though.
– Robert Wolfe
Oct 30 at 23:16
One third of a solution: for $1/3leq xi<1/2$ we may use $Mx+x(x-1)(x-c_1)$ and for $1/2<xileq 2/3$ we may use $Mx-x(x-1)(x+c_2)$ where $$c_1=frac{xi(3xi-2)}{2xi-1};text{ and };c_2=frac{xi(3xi-2)}{1-2xi},.$$ For $xi=1/2$ we can get away with $x(1-x)$. But cubics don't seem to carry us the entire way.
– Robert Wolfe
Nov 2 at 3:32
@RobertWolfe I didn't think about the comments before posting an answer. But now that I am seeing them, the $(xpm c)$ in your "cubic" comment is like the $(x+t)$ in my answer, except that I extended to considering $(x+t)^n$ to get past the central third.
– alex.jordan
Nov 2 at 7:29