## Math Genius: Finding the maximum value of \$int_0^1 f^3(x)dx\$

Find the maximum value of $$int_0^1 f^3(x)dx$$ given that $$-1 le f(x) le 1$$ and $$int_0^1 f(x)dx = 0$$

I could not find a way to solve this problem. I tried to use the cauchy-schwarz inequality but could not proceed further

$$int_0^1 f(x) cdot f^2(x) dx le sqrt{left(int_0^1f^4(x)dxright) left( int_0^1 f^2(x) dxright)}$$

Any hints/solutions are appreciated.

Let $$f^+(x)=max{0,f(x)},f^-(x)=max{0,-f(x)},$$ and assume $$A^+={x|f^+(x)>0},A^-={x|f^-(x)>0},$$then $$A^+cap A^- = emptyset$$
$$int_{A^+}f^+(x)=int_{A^-}f^-(x)=a$$for some $$age 0$$ by $$int_{0}^{1}f(x)=0.$$

We have that $$a le m(A^+),$$and,$$ale m(A^-).$$

We now want to find the maximum of $$int_{0}^{1}f^3(x)=int_{A^+}f^+(x)^3-int_{A^-}f^-(x)^3$$

So we just need to find the maximum of $$int_{A^+}f^+(x)^3$$, and the minimum of $$int_{A^-}f^-(x)^3$$.

For the first term, we have $$f^+(x)le 1$$,So $$f^+(x)^3le f^+(x)$$
hence we have $$int_{A^+}f^+(x)^3le int_{A^+}f^+(x) = a.$$
and for the second term we have $$frac{int_{A^-}f^-(x)^3}{m(A^-)}ge left(frac{int_{A^-}f^-(x)}{m(A^-)}right)^3=left(frac{a}{m(A^-)}right)^3$$(You can prove it by Hölder’s inequality)

So we have $$int_{0}^{1}f^3(x)=int_{A^+}f^+(x)^3-int_{A^-}f^-(x)^3le a-frac{a^3}{m(A^-)^2}le a-frac{a^3}{(1-m(A^+))^2}le a-frac{a^3}{(1-a)^2}$$
Since $$2a=int_{A^-}f^-+int_{A^+}f^+le 1$$, so $$ale 1/2.$$ So by a simple computation $$a-frac{a^3}{(1-a)^2}le frac{1}{4}quad ain[0,1/2].$$ When $$a=frac{1}{3}$$, it equals to $$frac{1}{4}.$$

To show $$int_{0}^{1}f(x)^3$$ can attain $$frac{1}{4},$$ consider such $$f(x)$$:$$f(x)=1,0le xle frac{1}{3},f(x)=-frac{1}{2},frac{1}{3}

We solve the problem via approximation by simple functions on uniform partitions of $$[0,1]$$. Consider a partition of $$[0,1]$$ into $$n$$ parts with coefficients $$alpha_i$$. Then the conditions on this simple function correspond to the conditions on the coefficients:
$$sum_{i=1}^nalpha_i=0, quad -1leqalpha_ileq 1, 1leq ileq n.$$
Similarly, the objective function becomes
$$F(alpha) = frac{1}{n}sum_{i=1}^nalpha_i^3$$
Some numerical experiments inform the solution to this problem but I doubt it is that difficult to solve using symmetry and lagrange multipliers or something else. I believe that for a partition of $$n$$ intervals, the optimal coefficients are given by $$alpha_1=1$$ and $$alpha_i = -1/(n-1)$$, or any permutation of the indices. For $$n>2$$, the objective function then gives us
$$F(alpha) = frac{1}{n}left(1-(n-1)frac{1}{(n-1)^3}right) = frac{n-2}{(n-1)^2},$$
which attains a maximum value of $$1/4$$ at $$n=3$$.

Simple functions on uniform partitions are dense in the space of simple functions, and simple functions are dense in $$L^1(0,1)$$. The functional is continuous on the considered domain, so a maximum on a dense subset corresponds to a maximum over the domain, which is a subset of $$L^1(0,1)$$.

Hint: You may use Eulero-Lagrange equations to the Lagrangian $$F(x,z,p)=z^3$$, considering the functional $$mathscr{F}(f)=int_{-1}^1 F(x,f(x),f'(x))dx$$.

## Math Genius: Prove that in all graphs with \$14\$ vertices and \$29\$ edges there are \$4\$ vertices \$A,B,C,D\$ such that they creates the cycle \$AB,BC,CD,DA\$

I tried to make the degrees of vertices small since the bigger the degrees get, the more possible this cycle occurs. Hence, I first drew $$14$$-sided polygon so now we have a graph with $$14$$ vertices and $$14$$ edges. Then I linked every vertices with itself so we have only $$1$$ edge left. However, if I either put this last edge between two adjacent vertices or connect one vertex with itself again, it seems that this cycle does not occur. Does it? Where is the mistake? Is there another way to approach this question or is my reasoning true?

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## Math Genius: \$text{lim inf}_{t downarrow 0}frac{int_{0}^{t}E[X_{s}^{2}]ds}{t^{2}}leq text{lim sup}_{t downarrow 0}frac{int_{0}^{t}E[X_{s}^{2}]ds}{t^{2}}\$

Let $$g(t) = E[X_{t}^{2}]$$, the starting point is the following set of inequalities which hold for all $$t geq 0$$:
begin{align} – int _{0}^{t} 2g(s)ds + t leq g(t) leq int _{0}^{t} 2g(s)ds + t end{align}

Then, we need to prove the following:begin{align} 0 < text{lim inf}_{t downarrow 0}frac{int_{0}^{t}E[X_{s}^{2}]ds}{t^{2}}leq text{lim sup}_{t downarrow 0}frac{int_{0}^{t}E[X_{s}^{2}]ds}{t^{2}} < infty end{align}

An obvious first step is integrating the above inequalities, and Fubini will probably also be useful, since then you can exchange the expectation and integrals. Furtermore, $$X_{t}$$ is also a submartingale, but I am not sure if that is relevant here.

However, even with the above knowledge I still can not picture where the limsup and liminf in the expression come from.

I would really appreciate any help.

EDIT: As per TheBridges’ request, I also give the SDE of $$X_{t}$$ below, though I am not sure if it is relevant towards answering my question.
begin{align} dX_{t} = |X_{t}|dt + dW_{t}, X_{0} = 0 end{align}

To derive the first inequalities at the top, I applied Ito to rewrite $$X_{t}^2$$, and together with the SDE of $$X_{t}$$ you can fairly easily derive these inequalities. But it is still unclear as to how they should be used to derive the second set of inequalities.

## Math Genius: What is the minimal disjunctive normal form of this propositional logic formula?

I have the following formula: $$(neg Aland Bland C)vee (neg Aland Bland neg C)vee (neg Aland neg B)vee (Aland C)vee (Alandneg C)$$

After I did a Karnaugh Map for this formula I found out it is a tautology (in other words – all squares in the map are filled with ones). What is the minimal disjunctive normal form of this formula then?

I’ll show you how to prove the statement is tautology without Karnaugh’s map:

$$(1)$$ distribution:
$$(neg Aland Bland C)lor (neg Aland Bland neg C)equiv(neg Aland B)land(Clorneg C)equiv(neg Aland B)$$
$$(2)$$distribution again:
$$(neg Aland B)lor(neg Alandneg B)equivneg Aland(Blorneg B)equivneg A$$
$$(3)$$distribution once again:
$$(Aland C)lor(Alandneg C)equiv Aland(Clorneg C)equiv A$$

$$underbrace{underbrace{(neg Aland Bland C)lor(neg Aland Bland neg C)}_{neg Aland B}lor(neg Aland neg B)}_{neg A}lorunderbrace{(Aland C)lor(Alandneg C)}_{A}equivneg Alor Aequiv 1$$

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## Math Genius: \$f:(a,+infty)rightarrowmathbb{R} \$ is differentiable function. I need explicit proof of a problem I find obvious

If $$f'(x)>c, forall xin(a,+infty)$$ where $$c>0$$. Prove that $$lim_{xto+infty} f(x) = +infty$$. I would say that this is trivial, how could we prove this explicitly?

Intuition suggests that a function with a positive derivative is strictly increasing. You can prove this using the mean value theorem. Next, you can use the mean value theorem to prove the function is also unbounded from above.

You now have a strictly increasing function with no upper bound, so you know its limit.

This is a direct application of Newton-Leibniz formula
$$f(x) = f(a) + int_a^x f'(t) , dt ge f(a) + c(x-a)$$

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## Math Genius: \$f:(a,+infty)rightarrowmathbb{R} \$ is differentiable function. I need explicit proof of a problem I find obvious

If $$f'(x)>c, forall xin(a,+infty)$$ where $$c>0$$. Prove that $$lim_{xto+infty} f(x) = +infty$$. I would say that this is trivial, how could we prove this explicitly?

Intuition suggests that a function with a positive derivative is strictly increasing. You can prove this using the mean value theorem. Next, you can use the mean value theorem to prove the function is also unbounded from above.

You now have a strictly increasing function with no upper bound, so you know its limit.

This is a direct application of Newton-Leibniz formula
$$f(x) = f(a) + int_a^x f'(t) , dt ge f(a) + c(x-a)$$

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## Math Genius: \$U(mathfrak g)\$ module

Let $$mathfrak g$$ be a Lie algebra and $$U(mathfrak g)$$ is the universal enveloping algebra of $$mathfrak g.$$ Let $$V$$ be a vector space is a unital $$U(mathfrak g)$$ module. I want to show that the map $$Vto V$$, $$vmapsto Xv$$ is linear where $$Xinmathfrak g$$. I can easily show that $$X(v+w)=Xv+Xw$$ which follows from definition. But now how to show that $$X(cv)=cXv$$ for any scalar $$cinmathbb C.$$

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## Math Genius: A total function is representable iff it is weakly representable

The book A Friendly Introduction to Mathematical Logic – 2nd Edition by Christopher C. Leary and Lars Kristiansen gives the following proposition without proof:

Proposition 5.3.6. Suppose that $$f$$ is a total function from $$mathbb{N}^k$$ to $$mathbb{N}$$. Then $$f$$ is representable if and only if $$f$$ is weakly representable.

What/where is the proof of this proposition? After some searching, I’m having trouble finding this exact proposition elsewhere that uses the same definitions of representable and weakly representable defined in the book. I find it odd that the authors chose to omit this proof without any further references.

For ease of reference, here are some definitions from the book:

Definition 5.3.4. Suppose that $$f:mathbb{N}^ktomathbb{N}$$ is a total function. We will say that $$f$$ is a representable function (in $$N$$) if there is an $$mathcal{L}_{NT}$$ formula $$phi(x_1,…,x_{k+1})$$ such that, for all $$a_1,a_2,…a_{k+1}inmathbb{N}$$,

$$text{if }f(a_1,…,a_k)=a_{k+1}text{, then }Nvdashphi(a_1,…,a_{k+1})$$

$$text{if }f(a_1,…,a_k)neq a_{k+1}text{, then }Nvdashlnotphi(a_1,…,a_{k+1})$$.

Definition 5.3.5. Suppose that $$Asubseteqmathbb{N}^k$$ and $$f:Atomathbb{N}$$ is a (possibly) partial function. We will say that $$f$$ is a weakly representable function (in $$N$$) if there is an $$mathcal{L}_{NT}$$ formula $$phi(x_1,…,x_{k+1})$$ such that, for all $$a_1,a_2,…a_{k+1}inmathbb{N}$$,

$$text{if }f(a_1,…,a_k)=a_{k+1}text{, then }Nvdashphi(a_1,…,a_{k+1})$$

$$text{if }f(a_1,…,a_k)neq a_{k+1}text{, then }Nnotvdashphi(a_1,…,a_{k+1}).$$

My secondary question is why are representable functions only defined for total functions (by Definition 5.3.4)? It seems perfectly fine to allow partial functions to be considered representable functions.

This is a cute trick, very similar to Rosser’s improvement to Godel’s incompleteness theorem:

Suppose $$f$$ is weakly representable via $$varphi$$. Consider the following new formula (conflating numerals and numbers for simplicity):

$$psi(x_1,…,x_k,y)equiv$$ “There is some $$z$$ which codes an $$N$$-proof of $$varphi(x_1,…,x_k,y)$$ and there is no $$w which codes a $$N$$-proof of $$varphi(x_1,…,x_k, y’)$$ for any $$y’$$.”

Since $$varphi$$ weakly represents $$f$$ and $$f$$ is total, we get that $$psi$$ represents $$f$$.

• To see that $$N$$ proves $$psi(m_1,…,m_k, f(m_1,…,m_k))$$, let $$u$$ be the smallest code for an $$N$$-proof of $$varphi(m_1,…,m_k, f(m_1,…,m_k))$$ (which exists since $$varphi$$ weakly represents $$f$$). In $$N$$ we can check that $$u$$ is such a proof, and so prove $$psi(m_1,…,m_k, f(m_1,…,m_k))$$.

• More interestingly, suppose $$nnot=f(m_1,…,m_k)$$. Since $$varphi$$ represents $$f$$ and $$f$$ is total, there is some least $$u$$ which codes an $$N$$-proof of $$varphi(m_1,…,m_k, a)$$ for some $$a$$ (namely $$a=f(m_1,…,m_k)$$). $$N$$ can verify this property of $$u$$ (in particular, there are only finitely many possible codes $$), and so prove $$negpsi(m_1,…,m_k, a)$$. Note that $$N$$ does not necessarily prove $$negvarphi(m_1,…,m_k,n)$$ by this analysis, it merely shows that any proof of $$varphi(m_1,…,m_k,n)$$ would have to be longer than the shortest proof of $$varphi(m_1,…,m_k, a)$$.

As to why we focus on total functions, you’re correct, this isn’t strictly necessary. However, the notions are better behaved for total functions. For example, the above argument that weak representability implies representability fails if we look at partial functions. (More computability-theoretically, every total c.e. function is actually a computable function, but this fails for partial functions.)

## Math Genius: Looping Train Tracks

Our son just received some train tracks for his second birthday. We’ve been experimenting with various layouts.

The one below has remained assembled for (almost) half and hour now, and I began to notice that, when going past any point, the train is always heading in the same direction (it seems the train cannot be made to turn around via a loop and return in the opposite direction).

Are there any possible configurations so the train can be made to pass a given point in the A-to-B direction, and later on, the B-to-A direction?

By now, all that remains of the given layout is the photograph.

However, after further reflection on the individual track pieces, there is a rather simple proof that reversal is not possible:

• All straight tracks connect A-to-B
• All Y-switches connect A-to-BB or B-to-AA
• Without any B-to-B or A-to-A connection in either straight tracks or Y-switches, direction cannot be reversed; QED

You can prove it by using a state graph $$G=(X,U)$$ where $$X$$={(0,*,*), (*,0,*), (*,*,0), (1,*,*), (*,1,*), (*,*,1)}. Each state indicates both the position of the train (you have 3: (Upper, bottom left, bottom right)) and its direction (by the value of 0 or 1 for left-to-right and vice versa respectively). For example (1,*,*) means the train is in the upper edge heading right to left. For your photo the state is (*,*,0).

The arcs of $$G$$ indicate if two states are consecutive. For example, from (*,0,*) you can only reach (1,*,*). To be able to pass a given point in a direction, and later on, on the opposite, you must have a path between two vertices with the same position of the two asteriscs and an inverted bit (for example from (*,*,0) to (*,*,1)).
In your situation, this is impossible because all such couple of vertices are in different connected components.

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## Math Genius: How to check if a number is present in the series \$1, 7, 25, 63, 129, 231…\$?

How to check if a number is present in the series $$1, 7, 25, 63, 129, 231…$$?

The nth term of the series is $$(2 n + 1) (2n^2 + 2 n + 3) / 3$$.

if $$N = 7$$, then it is present in the series.

Suppose we seek an $$n$$ solving $$O_n=N$$, with $$n$$th centered octahedral number$$O_n:=frac13(2n+1)(2n^2+2n+3)in(tfrac16(2n)^3,,tfrac16(2n+1)^3).$$Since $$(2n)^3le6Nle(2n+1)^3$$, there’s at most one candidate $$n$$ to check.

This is the sequence A001845 giving the centered octahedral numbers (crystal ball sequence for cubic lattice). For smaller numbers you can simply check the given list there.

Of course, with a formula at hand, you can simply
solve the equation
$$(2n+1)(2n^2+2n+3)/3=k$$
for $$n$$ and see if it has an integral solution. For example, take $$k=20$$. The formula shows that we need $$n<2$$. But for $$n=0,1$$ it is false. So first, estimate $$n$$, and then check for those $$n$$.

$$6N=8n^3+12n^2+16n+6\=(2n+1)^3+5(2n+1)$$
Let $$M =roundleft(sqrt[3]{6N-5sqrt[3]{6N}}right)$$
If $$6N-5M=M^3$$, and $$M$$ is odd, then I think $$N$$ is in the sequence.

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