# Sandbox for drafts of long, complex posts

This sandbox is intended for saving drafts of long, complex posts, especially posts whose composition takes a long time. It serves to localize to one thread the front-page "bumps" caused by edits to drafts of such posts, so that they may be easily ignored. Also, it helps to guard against losing longly-composed posts due to system crashes.

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## Proper Use of the Sandbox

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The sandbox has been "wiki locked" to prevent the creation of new answers. There are more than enough existing answers for users to edit over, and this will greatly reduce the frequency at which we request that the answers be disassociated from specific users.

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## locked by user642796Jan 12 '17 at 6:11

This question’s answers are a collaborative effort. If you see something that can be improved, just edit the answer to improve it! No additional answers can be added here.

• I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. – Asaf Karagila Jul 18 '12 at 8:35
• (+1) For thinking outside the (sand)box. – cardinal Jul 18 '12 at 19:40
• At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! – Grace Note Oct 5 '12 at 14:45
• To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. – leo Dec 17 '12 at 18:03
• PSA: Between the creation of this sandbox (in July 2012) and today (December 2015), technology has advanced. Something like StackEdit (or others, it's simply the only one I know) essentially solves all the limitations of this sandbox. You can have multiple concurrent drafts, you don't have to worry about polluting meta's front page, you can leave your draft untouched for days and expect it to still be there, you don't have to explicitly clear up your draft when you're done... Maybe someday we can get rid of this outdated crutch. – Najib Idrissi Dec 2 '15 at 14:07

$$\begin{array}{r} &&&&&&&U&X\\ &&&&-&-&-&-&- \\ L&M&N&)&R&S&T&U&N \\ &&&&R&T&Y&X \\ &&&&-&-&-&- \\ &&&&&T&Y&Y&N \\ &&&&&T&Y&Y&J \\ &&&&&-&-&-&- \\ &&&&&&&&Y \end{array}$$

Out of the ten letters, $$\{L,M,N,R,S,T,U,X,Y\}$$

• L, R, T, U can't be $$0$$ because they are the first digit of a number.
• X, Y can't be zero or because of the column $$\quad \begin{array}{c}U\\X\\-\\Y\end{array}\quad$$
• Similarly J (and Y) can't be $$0$$.
• N can't be $$0$$ because UN mod $$10$$ equals X and X $$\ne 0$$.
• S can't be $$0$$ because of $$\quad \begin{array}{cC}R&S\\ R&T\\ -&-\\ &T \end{array}$$

Thus $$M = 0$$.

Since the domain of our variables is the set of digits base ten, we will use ABC (for example) to mean $$100A+10B+C$$ and we will use $$A \cdot B \cdot C$$ to mean $$A$$ times $$B$$ times $$C$$.

Then, because $$M=0$$,

• $$U \cdot N = YX$$
• $$X \cdot N = YJ$$
• $$U \cdot L = RT$$
• $$X \cdot L = TY$$

As @RossMillikan said, $$T \in \{1,2,3,4\}$$. Note that, if $$Y \ge 5$$, then $$S$$ has to be odd. And if $$Y \le 4$$, then $$S$$ has to ber even. That implies these possibilities.

$$\begin{array}{c} T & S & Y \\ \hline 1 & 3 & 5 \\ 2 & 4 & 1 \\ 2 & 5 & 6 & *\\ 3 & 6 & 1 & *\\ 3 & 7 & 6 \\ 4 & 8 & 2 \\ 4 & 9 & 7 & * \end{array}$$

Because $$M=0$$, then $$XL = 10T+Y$$ needs to be a product of two digits. So we can remove the starred items.

$$\begin{array}{c} T & S & Y \\ \hline 1 & 3 & 5 \\ 2 & 4 & 1 \\ 3 & 7 & 6 \\ 4 & 8 & 2 \end{array}$$

Next we look at $$X \cdot LMN = TYYJ$$

We see that $$X \ge 2$$, $$X \cdot L = TY$$, and $$X \cdot N = YJ$$

When $$TY=15$$, then $$X \in \{3,5\}$$. But $$X \ne 5$$ since $$Y=5$$, so we must have $$X=3$$. But then $$L=5$$. So $$(T,S,Y) \ne (1,3,5)$$.

With that sort of reasoning, I found the only solution is

$$\begin{array}{c} T & S & Y & X & L & M & N & J \\ \hline 4 & 8 & 2 & 7 & 6 & 0 & 4 & 1 \end{array}$$

Hence the solution is

$$\begin{array}{r} &&&&&&&9&7\\ &&&&-&-&-&-&- \\ 6&0&3&)&5&8&4&9&3 \\ &&&&5&4&2&7 \\ &&&&-&-&-&- \\ &&&&&4&2&2&3 \\ &&&&&4&2&2&1 \\ &&&&&-&-&-&- \\ &&&&&&&&2 \end{array}$$

The integral corresponding to this entry is: $$$$I= \int_{0}^{\infty}\frac{x\: d{x}}{(ax^2+2bx+c)^{n}}$$$$ The value of this integral, as has been mentioned in the tables, is- \begin{align} \int_{0}^{\infty}\frac{x\: d{x}}{(ax^2+2bx+c)^{n}} & = \frac{(-1)^n}{(n-1){!}{!}}\frac{\partial^{n-2}}{\partial c^{n-2}}\Big(\frac{1}{2(ac-b^2)}-\frac{b}{2(ac-b^2)^{\frac{3}{2}}} cot^{-1}(\frac{b}{\sqrt{ac-b^2}})\Big)\, [ac>b^2] \\ & = \frac{(-1)^n}{(n-1){!}{!}}\frac{\partial^{n-2}}{\partial c^{n-2}}\Big(\frac{1}{2(ac-b^2)}+\frac{b}{4(b^2-ac)^{\frac{3}{2}}} \ln\Big(\frac{b+\sqrt{b^2-ac}}{b-\sqrt{b^2-ac}}\Big)\Big)\ [b^2>ac>0] \\ & = \frac{a^{n-2}}{2(n-1)(2n-1)b^{2n-2}}\qquad [ac=b^2] \end{align} For the subsequent discussion, please refer to the rules in this paper.

We use rule $$P_2$$, to expand the denominator as a bracket-series: $$$$(ax^2+2bx+c)^{-n}=\sum_{n_1,n_2,n_3=0}^{\infty}\phi_{n_1n_2n_3}a^{n_1}(2b)^{n_2}c^{n_3}x^{2n_1+n_2}\frac{}{\Gamma(n)}$$$$ Substituting the above expansion in the first expression for I, and employing the definition that $$ \mapsto \int_{0}^{\infty}x^{a-1}dx$$ , we have: $$$$\label{eq:14} I=\sum_{n_1,n_2,n_3=0}^{\infty}\phi_{n_1n_2n_3}a^{n_1}(2b)^{n_2}c^{n_3}\frac{<2n_1+n_2+2>}{\Gamma(n)}$$$$ Following is the system of linear equations we are supposed to solve: \begin{align*} 2n_1 + n_2 + 2 &= 0 \\ n + n_1 + n_2 + n_3 &= 0 \end{align*} There are 3 variables and 2 equations, thus, there are $${{no. of sums}\choose {index}}={3\choose 1}=3$$ different solutions, taking one variable to be free at a time.

Following are the solutions obtained:

• Region $$|\frac{ac}{b^2}|<1$$:$$I_1=\frac{c^{2-n} \Gamma (1-n) \, _2\tilde{F}_1\left(1,\frac{3}{2};3-n;\frac{a c}{b^2}\right)}{4 b^2}$$
• Region $$|\frac{b^2}{ac}|<1$$:$$I_2=\frac{\sqrt{a} c^{1-n} \Gamma (n-1) \, _2F_1\left(1,n-1;\frac{1}{2};\frac{b^2}{a c}\right)-\sqrt{\pi } b c^{\frac{1}{2}-n} \Gamma \left(n-\frac{1}{2}\right) \left(1-\frac{b^2}{a c}\right)^{\frac{1}{2}-n}}{2 a^{3/2} \Gamma (n)}$$
• Region $$|\frac{ac}{b^2}|<1$$:$$I_3=\frac{4^{1-n} a^{n-2} b^{2-2 n} \Gamma (2-n) \Gamma (2 n-2) \left(1-\frac{a c}{b^2}\right)^{\frac{1}{2}-n}}{\Gamma (n)}$$

It could be seen that both $$I_1$$ and $$I_3$$ converge in the same region, so we should add them up, as per rule $$E_3$$: $$I_4=-\frac{\Gamma (1-n) \left(2 a^{n-2} b^{4-2 n} \Gamma \left(n-\frac{1}{2}\right) \left(1-\frac{a c}{b^2}\right)^{\frac{1}{2}-n}-\sqrt{\pi } c^{2-n} \, _2\tilde{F}_1\left(1,\frac{3}{2};3-n;\frac{a c}{b^2}\right)\right)}{4 \sqrt{\pi } b^2}$$

The solution for this region, $$|\frac{ac}{b^2}|<1$$, thus obtained, must correspond to $$I$$, for a suitable choice of the parametric-triad (a,b,c). Also, it's known from the book of Gradshteyn and Ryzhik that for $$(0, the integral could be expressed in the following closed form: $$\int_{0}^{\infty}\frac{x\: d{x}}{(ax^2+2bx+c)^{n}} = \frac{(-1)^n}{(n-1){!}{!}}\frac{\partial^{n-2}}{\partial c^{n-2}}\Big(\frac{1}{2(ac-b^2)}+\frac{b}{4(b^2-ac)^{\frac{3}{2}}} \ln\Big(\frac{b+\sqrt{b^2-ac}}{b-\sqrt{b^2-ac}}\Big)\Big)\: [b^2>ac>0]$$

However, a derivation of this closed-form starting with the solution obtained using the rules of the Method of Brackets: $$I_4$$, turns out to be too non-trivial a task to accomplish, so a numerical consistency check is called for. However, this method has its own set of nasty complications.

It is known that the gamma function has poles of order 1 at all non-positive integers. Owing to this, $$I_1$$ (and so, $$I_4$$) fails to provide a suitable expression, $$\forall$$ n $$\in$$ $$\mathbb{N}$$.

Also, the hypergeometric $$_2{F}_1$$ happens to have the following standard expansion: \begin{align} _2{F}_1(a,b;c;z) &= \sum_{k=0}^{\infty}\frac{(a)_k(b)_k}{(c)_kk{!}}z^k \\ &= \sum_{k=0}^{\infty}\frac{\Gamma(a+k)\Gamma(b+k)\Gamma(c)}{\Gamma(a)\Gamma(b)\Gamma(c+k)\Gamma(k+1)}z^n \end{align} which is convergent in $$|z|<1$$, as long as Re(c$$-$$a$$-$$b)$$>$$0.

Here, a$$=$$1, b$$=\frac{3}{2}$$, c$$=$$ 3$$-$$n, and z $$=$$ $$\frac{ac}{b^2}$$. Now, $$\Gamma$$(3$$-$$n) is a simple pole $$\forall$$ n $$\in$$ $$\mathbb{N}\setminus\{1,2\}$$. Thus, a numerical check for $$I_1$$ turns out to be impossible.

For $$|\frac{ac}{b^2}|<1$$, the other contributing term is $$I_3$$. For reasons explained before, this expression suffers from an indeterminate form at n$$=$$0, and happens to have simple poles $$\forall$$ n $$\in$$ $$\mathbb{N}$$. To elucidate this point, note that the factor $$\Gamma$$(2-n) has a simple pole $$\forall$$ n $$\in$$ $$\mathbb{N}\setminus\{1\}$$. However, $$\Gamma$$(2n-2) has a simple pole at n$$=$$1, thus, there's a simple pole for $$I_3$$ $$\forall$$ n $$\in$$ $$\mathbb{N}$$.

A numerical consistency check for relevant expressions, and the integral itself, seems a hopeless endeavor for $$|\frac{ac}{b^2}|<1$$. So, let's proceed to the region $$|\frac{b^2}{ac}|<1$$, and see for ourselves what fate awaits us.

$$I_2$$ is a sum of two terms, the first of which happens to have a simple pole at n$$=$$1, and overall has an indeterminate form at n$$=$$0.

Let's show explicitly that the first term of $$I_2$$ indeed has a simple pole at n$$=$$1. Ignoring the constant factors, we can write: \begin{align} \frac{\Gamma(n-1)\; _2{F}_1(1,n-1;\frac{1}{2};\frac{b^2}{ac})}{\Gamma(n)} &= \sum_{k=0}^{\infty}\frac{\Gamma(k+1)\Gamma(k+n-1)\Gamma(\frac{1}{2})}{\Gamma(1)\Gamma(n-1)\Gamma(k+\frac{1}{2})\Gamma(k+1)}\Big(\frac{b^2}{ac}\Big)^k \\ &= \sqrt{\pi}\Big(\frac{1}{\Gamma(n)}\sum_{k=0}^{\infty}\frac{\Gamma(k+n-1)}{\Gamma(k+\frac{1}{2})}\Big(\frac{b^2}{ac}\Big)^k\Big)_{n=1} \\ &= \sqrt{\pi}\sum_{k=0}^{\infty}\frac{\Gamma(k)}{\Gamma(k+\frac{1}{2})}\Big(\frac{b^2}{ac}\Big)^k \end{align} which blows up at k=0.

For sake of numerical consistency check, we set n$$=$$5, a$$=$$3, b$$=$$2, c$$=$$2. The differences between the numerical values obtained via employing different expressions is of the minuscule order of $$10^{-16}$$.

Now, let's look at the case $$b^2=ac$$. Since we see that for n$$=$$5, $$I_2$$ yields a nice solution as long as $$|\frac{b^2}{ac}|<1$$, we can take a dig at $$|\frac{b^2}{ac}|=1$$, using Gauss's hypergeometric theorem, which states the following: $$_2{F}_1(a,b;c;1)=\frac{(c-b)_{-a}}{(c)_{-a}}=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\qquad [Re(c-a-b)>0]$$ Hence, we may write the following:

\begin{align} _2{F}_1(1,n-1;\frac{1}{2};1) &= \frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{2}-n)}{\Gamma(\frac{1}{2}-1)\Gamma(\frac{1}{2}-n+1)} \\ &= \frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{2}-n)}{\Gamma(-\frac{1}{2})\Gamma(\frac{1}{2}-n+1)} \\ &= \frac{\Gamma(-\frac{1}{2}+1)\Gamma(\frac{1}{2}-n)}{\Gamma(-\frac{1}{2})(\frac{1}{2}-n)\Gamma(\frac{1}{2}-n)} \\ &= \frac{-\frac{1}{2}\Gamma(-\frac{1}{2})}{(\frac{1}{2}-n)\Gamma(-\frac{1}{2})} \\ &= \frac{1}{2n-1} \end{align} For $$b^2=ac$$, the second term in $$I_2$$ blows up, while the first one evaluates to a finite value, for given values of a, c, and n. The blowing up of the second term is actually due to one of the roots of b in $$b^2=ac$$, namely, $$b=-\sqrt{ac}$$.

That $$b=\sqrt{ac}$$ is indeed responsible for the finite part, and that $$b=-\sqrt{ac}$$ is responsible for the divergence, could be ascertained as follows:

• $$b=\sqrt{ac}$$:$$\int_{0}^{\infty}\frac{x\: d{x}}{(ax^2+2bx+c)^{n}}=\int_{0}^{\infty}\frac{x\: d{x}}{(ax^2+2\sqrt{ac}x+c)^{n}}=\int_{0}^{\infty}\frac{x\: d{x}}{(\sqrt{a}x+\sqrt{c})^{2n}}\qquad$$ which yields a finite result for given values of a, c, and n. It is seen that this result is in exact numerical agreement with that of the first term in $$I_2$$, for any given set of parameters (a,c,n).
• $$b=-\sqrt{ac}$$:$$\int_{0}^{\infty}\frac{x\: d{x}}{(ax^2+2bx+c)^{n}}=\int_{0}^{\infty}\frac{x\: d{x}}{(ax^2-2\sqrt{ac}x+c)^{n}}=\int_{0}^{\infty}\frac{x\: d{x}}{(\sqrt{a}x-\sqrt{c})^{2n}}\qquad$$ which rapidly diverges for any given set of parameters (a,c,n).

Now that I am done with all the details of the calculation (please, feel free to ask me to show some step explicitly if the corresponding calculation seems blurry), let me put forth my queries:

• In the book by Gradshteyn and Ryzhik (7th ed., pg. 1005, section 9.10), it is said that for non-troublesome values of the triad of parameters $$(\alpha,\beta,\gamma)$$, a hypergeometric series $$F(\alpha,\beta;\gamma;z)$$ diverges on the entire unit circle, if $$Re(\alpha+\beta-\gamma) \geq 1$$, and converges throughout the entire unit circle (except at $$z=1$$), for $$1>Re(\alpha+\beta-\gamma) \geq 0$$.

Now looking at $$I_1$$, it could be seen that $$Re(\alpha+\beta-\gamma)=n-\frac{1}{2}$$, which is $$<1$$ for $$n=1$$, but $$>1$$ $$\forall$$ n $$\in$$ $$\mathbb{N}\setminus\{1\}$$.

Putting aside the calculations for awhile, let's now take a look at the following table: $$\begin{array}{c|c|c|c|c|c|c|c|} \text{(a,b,c)} & \text{n} & \text{Int} & \text{RHS} & \text{MoB} & \text{|Int-RHS|} & \text{|Int-MoB|} & \text{|RHS-MoB|} \\ \hline \text{(1,2,1)} & 5 & 0.0045 & 0.0045,0.0045 & 0.0045 & 0.,0. & O(10^{-18}) & O(10^{-18}),O(10^{-18}) \\ \hline \text{(3,7,5)} & 5 & 2.87*10^{-6} & 2.87*10^{-6},2.87*10^{-6} & 2.87*10^{-6} & O(10^{-22}),O(10^{-21}) & O(10^{-21}) & O(10^{-21}),O(10^{-20}) \\ \hline \text{(3,1,5)} & 5 & 0.000031 & 0.000031,0.000031 & 0.000031 & O(10^{-21}),O(10^{-20}) & O(10^{-21}) & O(10^{-20}),O(10^{-20}) \end{array}$$

(this question is not complete yet---I am currently editing it, as it is too long to write at one go)

My post needs more work, and I don't have the time for it today. Leaving this to be used by anyone.

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This slot is open for anyone to use.

My first ansatz, which gives at best an asymptotic power series (with convergence radius zero!), seemed to be rough and likely much suboptimal. But for completeness I'll give the method here, also because it provides that nice initial guess $$a_\text{init}$$ which is very near at the true value.

The ansatz employs essentially the Bernoulli-numbers, but not as in the classic Bernoulli-polynomials but in some "transposed" manner.

The individual $$\log(1+1/a_k)$$ - epressions have a mercator series, such that we can reformulate the original sum $$S_d(a,N)$$ as a double sum $$\begin{array}{rll} S_d(a,N) &=& 1/a - 1/2/a^2 + 1/3/ a^3 - 1/4/ a^4 + \cdots - \cdots \\ &+& 1/(a+d) - 1/2/(a+d)^2 + 1/3/ (a+d)^3 - 1/4/ (a+d)^4 + \cdots - \cdots \\ &+& 1/(a+2d) - 1/2/(a+2d)^2 + 1/3/ (a+2d)^3 - 1/4/ (a+2d)^4 + \cdots - \cdots \\ &...& \\ &+& 1/(a+(N-1)d) - 1/2/(a+(N-1)d)^2 + 1/3/ (a+(N-1)d)^3 - 1/4/ (a+(N-1)d)^4 + \cdots - \cdots \\ \end{array}$$ Summing this vertically gives the column-sums $$\begin{array}{rll} C_{d,1}(a,N) &=& 1/a + 1/(a+d)+ 1/(a+2d) + ... + 1/(a+(N-1)d) \\ C_{d,2}(a,N) &=&\frac12 ( 1/a^2 + 1/(a+d)^2+ 1/(a+2d)^2 + ... + 1/(a+(N-1)d)^2 )\\ C_{d,3}(a,N) &=&\frac13 ( 1/a^3 + 1/(a+d)^3+ 1/(a+2d)^3 + ... + 1/(a+(N-1)d)^3 )\\ &...& \end{array}$$ Such sums are essentially Hurwitz-Zeta sums, and can be expressed as power-series with zeta at negative integers as coefficients.
For easiness of handling and memorizing I've the "ZETA"-matrix in my toolbox, which provides that coefficients for the application to power series.
The ZETA-matrix looks like

and has the composition of Zeta-values
Reading it horizontally one observes the coefficients for the integrals of the Bernoulli-polynomials which are configured to allow the summing-of-like-powers.
However, reading vertically, they give the coeffients for power series to determine the sums-of-reciprocals-of-like-powers. However, that power series have convergence-radius zero, but because of alternating signs we can use them as asymptotic power series, truncated at some meaningful index.

For instance, the power series taken from the second,third, fourth and so on column, and calculated as $$f_1(a) = a z_{0,1} + a^2 z_{1,1} + a^3 z_{2,1} + ... \to \zeta(2,a) \\ f_2(a) = a z_{0,2} + a^2 z_{1,2} + a^3 z_{2,2} + ... \to \zeta(3,a) \\ ...$$ where $$z_{r,c}$$ indicates the element of the ZETA-matrix from row $$r$$ and column $$c$$ (index based at $$0$$).
That allows immediately to determine the sums of reciprocals from $$a$$ to $$a+(N-1)$$ by $$s_1(a) = f_1(a) - f_1(a+N) = 1/a^2 + 1/(a+1)^2+... + 1/(a+(N-1))^2 \\ s_2(a) = f_2(a) - f_2(a+N) = 1/a^3 + 1/(a+1)^3+... + 1/(a+(N-1))^3 \\ ...$$

For the first column we need a correction factor which comes out to be $$f_0(a) = a z_{0,0} + a^2 z_{1,0} + a^3 z_{2,0} + ... \to \\ \log(1/a)+\left(\sum_{k=1}^{a-1} \frac1k \right) -\gamma$$
such that $$s_0(a) = f_0(a) - f_0(a+N) = 1/a + 1/(a+1)+... + 1/(a+(N-1)) \\ + \log(1/a) - \log(1/(a+(N-1))$$

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Consider the following initial value problem (IVP) to the first order ODE:

$$\tag{1} \dot x = f(t, x), \ \ \ x(t_0) = x_0.$$

The Existence and Uniqueness Theorem describes when one has exactly one solutions. This is true (e.g.) when $$f$$ is continuously differentiable.

There are examples where uniqueness fails and existence fails (here or here):

In all of the examples where I am aware of, whenever one has more than one solutions, one actually has infinitely many. Hence my question:

Can an IVP has more than one solutions, but only finitely many solutions?

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The claim is made that $$\frac NS \ge \frac {\ln 2}{ \ln3}$$ which also means $$\text{conj: } \liminf _{k \to \infty} N^*(k) \ln3 - k \ln2 \ge 0$$ This would be satisfied if in our construction of infinitely many partial trajectories on $$\{x_1,x_2,...,x_{N_x},y_1,...\}$$ only finitely many exceptions for the inequality $$N_x \ln3 - S_x \ln2 \ge 0 \qquad \qquad \qquad \text{or} \qquad \qquad \qquad \qquad \qquad \qquad \\ {2^{S_x} \over 3^{N_x}} \le 1 \qquad \text{ for some partial trajectory x_1,...,x_{N_x} }$$ Now we rewrite our formula to adapt $${2^S\over 3^N} = \frac{a_1}{b_1} \cdot (1+{1\over 3a_1 })(1+{1\over 3a_2 })\cdots(1+{1\over 3a_N}) \le 1$$ That means also $$a_1 \cdot (3a_1+1)(3a_2+1)\cdots(3a_N +1) \le b_1 \cdot 3a_1 \cdot 3a_2 \cdots 3a_N \\ 3^N a_1 a_2 \cdots a_N + ... + (3a_1+3a_2+\cdots +3a_N) + N \le b_1/a_1 \cdot 3^N a_1 a_2 \cdots a_N$$

Let's shorten the formula by the first parenthese on $$a_1$$: $$2^{S-1} = 3^{N-1} \cdot \frac{1}{b_1} \cdot ({3a_1+1\over 2 })(1+{1\over 3a_2 })\cdots(1+{1\over 3a_N})$$

A well known tool developed for the discussion of cycles is useful also here: the productformula of all elements of a (partial) trajectory equalling the product of its iterates. Of course we have equality $$a_2 a_3 \cdots a_N a_{N+1} = {3a_1+1\over 2^{A_1}}{3a_2+1\over 2^{A_N}}\cdots{3a_N+1\over 2^{A_N}}$$ Writing $$S$$ for the sum of all exponents $$A_k$$, rearranging and cancelling $$a_2,...a_{N+1}$$ gives $$2^S = ({3a_1+1\over 1 })(3+{1\over a_2 })\cdots(3+{1\over a_N}) \cdot \frac1{a_{N+1}}$$ A smoothing occurs if we extract $$a_1$$ from the first parenthese and rewrite $$2^S = \frac{a_1}{a_{N+1}} \cdot (3+{1\over a_1 })(3+{1\over a_2 })\cdots(3+{1\over a_N})$$ We finally extract the $$N$$ fold occuring $$3$$ in the parentheses and also introduce that $$a_{N+1} = b_1$$ by our definitions to write the basic formula $$2^S = 3^N \cdot \frac{a_1}{b_1} \cdot (1+{1\over 3a_1 })(1+{1\over 3a_2 })\cdots(1+{1\over 3a_N})$$ From our construction of the sequence of partial trajectories we have the following properties: $$a_2= {3a_1+1\over2 } \qquad \text{ because a_1 \to a_2 must be increasing } \\ a_1< b_1 < a_2 = {3a_1+1\over2 } \qquad \text{ and even } \\ a_1< b_1 < \min(a_2,a_3,...,a_N)\qquad \text{ by construction of partial trajectories }$$

In the following I use the "syracuse" (or "compacted") notation for the Collatz-transformation $$Y:= a_{k+1} = {3a_k+1\over 2^{A_k}}$$ with $$a_k$$ odd and the exponent $$A_k\ge1$$ such that also $$a_{k+1}$$ is odd
I use $$N$$ for the number of subsequent iterations and $$S(N)$$ or simply $$S$$ for the sum of exponents of the subsequent iterations and always capital letters for exponents and small letters for the elements (also using indexes) or simple temporary variables.

Let's define an assumed divergent trajectory from its minimal element $$a_1$$ on $$\{a_1,a_2,...,a_{N_a}, b_1,b_2,...,b_{N_b},c_1,... \}$$ where we assume $$a_1$$ the minimal element of the whole trajectory, $$b_1$$ the minimal element of the remaining trajectory after $$a_1$$, then $$c_1$$ the minimal element of the remaining trajectory after $$b_1$$ and so on. We construct with this a monotonic increasing sequence of elements, separating the full trajectory into infinitely many partial trajectories. There cannot be equal elements because then we would have a cycle whose trajectory obviously cannot diverge infinitely and which is thus not in the scope of this discussion.

We have always $$a_1 and $$b_1 and so on except if $$b_1=a_2$$ but then we would reduce the first partial trajectory to the element $$a_1$$ only and $$N_a=1$$ and use then $$a_2$$ only to keep index-notations in formulae consistent/easy.

Function $$f$$ is defined as $$f(x)=\frac{1}{(x-1)^3(x^3+2x^2+x+1)}\tag 1$$ Note that $$x^3+2x^2+x+1=(x+1)(x^2+x+1)\tag 2$$ and so $$f(x)=\frac{1}{(x-1)^3(x+1)(x^2+x+1)}\tag 3$$ and also $$f(x)=\frac{1}{(x-1)}\cdot\frac{1}{(x^2-1)}\cdot\frac{1}{(x^3-1)}=\sum_{i=0}^\infty x^i \cdot\sum_{k=0}^\infty x^{2k} \cdot\sum_{j=0}^\infty x^{3j}$$

$$\frac{1}{(x-1)^3(x+1)(x^2+x+1}=\frac{A_1}{(x-1)}+\frac{A_2}{(x-1)^1}+\frac{A_3}{(x-1)^3}+\frac{B}{(x+1)}+\frac{Cx+D}{(x^2+x+1)}$$ $$1=A_1(x-1)^2(x+1)(x^2+x+1)+A_2(x-1)(x+1)(x^2+x+1)+A_3(x+1)(x^2+x+1)+B(x-1)^3(x^2+x+1)+(Cx+D)(x-1)^3(x+1)$$

• Oh god, I still remember having tests that required computing these king of generating functions by hand. If result was correct, the method was verified, otherwise no points. – enedil Feb 11 at 21:53

Edit: This answer got more popular than I expected, so let me add some details about what the structure here actually looks like.

Before talking about manifolds, which we'll get to in a bit, let's discuss some algebra. If you're not comfortable with algebra you can skip down to where I start talking about topology again. There are lots of examples in algebra of times where we are talking about the same underlying sets, but considering different structures, and the best way to think about those structures is through their morphisms.

For example, if we have the category of $$R$$-modules, over some ring $$R$$, we are considering the (homo)morphisms which preserve the structure of being a module over $$R$$. What does that mean, precisely? Given a pair of modules $$M,M' \in \textbf{R-Mod}$$ we say that a map $$f:M \to M'$$ is a homomorphism iff $$f$$ is a homomorphism of abelian groups and $$f$$ preserves the $$R$$ action (i.e. $$r \cdot f(a) = f(r\cdot a)$$).

We have a whole rich set of other possible morphisms we could care about in this context, though, depending on what we are doing. The classical example is that if there is some subring $$S \subset R$$, then we can consider morphisms that only preserve the action by $$S$$. This actually gives us a new category, $$S-\textbf{Mod}$$, the category of $$S$$ modules. Another example is that we could just drop the $$R$$ action all together and just think about homomorphisms of abelian groups. We could also take the approach where we don't care about the group structure at all and instead think only about $$f$$ preserving the group action by $$R$$. These all lead to different and important areas of math.

My point in this diversion is simply to say that the richness of mathematical objects actually doesn't come from their inherit structure, like their topology or their geometry, instead it comes from studying those structures paired with the morphisms that preserve certain aspects of them. It is from this pairing that we start to see the bigger picture about what is and isn't true for a whole class of objects, and even how those classes relate to each other. So that is the approach I'll take in describing some more detail.

Alright, lets get to manifolds now. Before introducing anything about Riemannian or Smooth manifolds let me start out by contrasting a topological space and a topological manifold. The reason I want to briefly focus on this contrast is that the morphisms of a topological space behaves a lot like those of other objects you learn about in math, but the morphisms of a topological manifold, although preserving a very similar structure, behave much more like those of other kinds of manifolds than anything.

A pair of topological spaces $$X, Y$$ with topology $$\mathscr{T,T'}$$ has a morphism $$f: X \to Y$$, called a continuous map, iff $$\forall A \in \mathscr{T'}, f^{-1}(A) \in \mathscr{T}$$. This is an isomorphism, called a homeomorphism, iff $$\forall A \in \mathscr{T}, f(A) \in \mathscr{T'}$$. This is what we see all over math, so it should look familiar.

A topological manifold is a topological space $$X$$ with the local property of being locally homeomorphic to $$\mathbb R^n$$. Here we have introduced two things that will make all manifolds behave "differently" than topological spaces: a local property, and a relationship to $$\mathbb R^n$$.

It's been a long time since I last did this stuff, but nobody better seems to have turned up, so I'll have a go.

I'll give the box $$\square \phi$$ its usual meaning here: "when I translate $$\phi$$ into the formal system using Gödel numbers, then there is a collection of mechanical transformations inside the formal system that together form an internal-system-proof of $$\phi$$'s quotation". I will also refer to the formal system in question in the first person, with the pronoun "I".

Then the formula $$\square (\square \phi \to \phi) \to \square \phi$$ should be read as:

If there is a number which encodes a proof that "having a number which encodes a proof of $$\phi$$ means I can actually prove $$\phi$$", then there is a number which encodes a proof of $$\phi$$.

Or, a bit more loosely,

If I, internally, can prove that my proof of $$\phi$$ respects the logic in which I operate (so my proof of $$\phi$$ means that in my surrounding logic, $$\phi$$ is actually true), then in fact there is a proof of $$\phi$$ that I will be satisfied with.

This is more concretely understood through its universal quantification, Löb's theorem (which is true in any system rich enough to support Peano arithmetic):

If I can prove within myself that my own proof-checking process respects the semantics of the logic in which I operate, then the logic realises that actually I can satisfy myself of any statement at all.

Specifically, $$\square (\square \phi \to \phi)$$ means "I contain a proof that my proofs are semantically valid in the surrounding logic", while $$\square \square \phi \to \square \phi$$ means "it's true in the surrounding logic that if I want to prove $$\phi$$, it's enough to prove provability-of-$$\phi$$".

The two statements were very unlikely to be the same: one is talking about proof within the system, and one is talking about facts of the surrounding logic.