Sandbox for drafts of long, complex posts

Question

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.

When you are happy with your draft here, you may simply copy the code and paste it to the desired location.

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Answer 1

(this answer slot is free for use)

Answer 2

This slot is as vacant as the expression of someone who's been staring at the same line of a proof for three hours.

Answer 3

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

Answer 4

This entry is available for anyone to commandeer since Oct. 17th, 2018.

Answer 5

This space free for whoever wants it

Answer 6

This answer is available for anyone to use.

Answer 7

This slot is open for anyone to use.

Answer 8

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 $$ s_0(a) = f_0(a) - f_0(a+N) = 1/a + 1/(a+1)+... + 1/(a+(N-1)) \\ + \log(1/a) - \log(1/(a+(N-1)) $$

Answer 9

This answer is free for anyone to use.

Answer 10

Unused, available for anyone.$ $

Answer 11

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?

Answer 12

This entry is available for anyone to commandeer since Oct. 13th, 2018.

Answer 13

This entry is free for anyone to commandeer since Oct. 27th, 2018

Answer 14

This answer is being edited right now.

The Fabius function is a continuous function $F:\mathbb R\to[0, 1]$ uniquely determined by the following conditions:

• $F(0)=0$,
• $F(1)=1$, and
• $F'(x)=2\,F(2{\tiny\text{ }}x)\,$ for all $x\in\mathbb R.$

It is well known that the Fabius function assumes rational values at dyadic rational arguments, and several efficient algorithms to compute those values have been published (usually they use some auxiliary rational sequences defined by bulky recurrence relations).

I have been looking for a non-recursive, self-contained formula for them for quite a long time. After years of experimentation and looking for patterns in values of the Fabius function I came up with the following empirical formula: $$F\!\left(\frac m{2^n}\right) = \frac{2^{-n^2}}{\left(\frac12;{\tiny\text{ }}\frac12\right)_n}\,\sum _{k=0}^n \frac{2^{-k{\tiny\text{ }}(k-1)}\binom n k_{1/2}}{(n+k)!}\,\sum _{r=0}^{2^k{\tiny\text{ }}m-1}\,(-1)^{s_2\left(r\right)}\left(r-2^km+\tfrac12\right)^{n+k}{\small,}$$ where $m,\,n$ are non-negative integers, $\left(a;{\tiny\text{ }}q\right)_n$ is the q-Pochhammer symbol, ${\binom n k}_q$ is the q-binomial coefficient, and $s_2(r)$ is the sum of binary digits of $r$.

Do not ask me how exactly I found it — it was a long and peculiar way; at some point I had a formula with four nested summations, and then I gradually simplified it. It agrees with other algorithms for all values I tried, but, unfortunately, so far I could not find a way to prove it.

A more general formula that works not only for dyadic rational arguments is $$F\!\left(x\right) = \lim_{n\to\infty} \frac{2^{-n^2}}{\left(\frac12;{\tiny\text{ }}\frac12\right)_n}\,\sum _{k=0}^n \frac{2^{-k{\tiny\text{ }}(k-1)}\binom n k_{1/2}}{(n+k)!}\,\sum _{r=0}^{\left\lfloor2^{n+k}{\tiny\text{ }}x-\frac12\right\rfloor}\,(-1)^{s_2\left(r\right)}\left(r-2^{n+k}x+\tfrac12\right)^{n+k}{\small.}$$ We can observe that for a dyadic rational $x$ the limit in this formula is trivial, because the values under the limit stabilize (become constant) for large enough $n$.

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Let's define $t_n$ by the recurrence $$t_0 = 1, \quad t_n = (-1)^n \, t_{\lfloor n/2\rfloor}\tag1$$ It is easy to see that $|t_n|=1$, and the signs follow the same pattern as the Thue–Morse sequence. The same sequence is given by a non-recurrent, but rather clumsy formula $$t_n = 1-\frac83 \, \sin^2\left(\frac\pi3 \cdot \sum_{k=1}^n \left[2 + (-1)^{\binom n k}\right]\right)\tag2$$

Let's define $c_n$ by the recurrence $$c_0 = 1, \quad c_n = \frac1{(4^n - 1)(2n+1)} \, \sum_{k\ge1} \binom{2n+1}{2k+1} \, c_{n-k},\tag3$$ so we have $$c_1 = \frac19, \quad c_2 = \frac{19}{675}, \quad c_3 = \frac{583}{59535}, \quad c_4 = \frac{132809}{32531625}, \quad \text{etc.}\tag4$$ Note that only finite number of terms in the sum are non-zero. I am not aware of any non-recurrent formula for this sequence, but it would be very nice to find one.

The values of the Fabius function at dyadic rationals are given by $$F\!\left(\tfrac{m}{2^n}\right) = \frac1{n! \, 2^{\binom{n+1}2}} \, \sum_{k\ge0}\binom n {2k} \, c_k \sum_{1 \le r \le m}(2r-1)^{n-2k} \, t_{m-r}.\tag5$$ Again, note that only finite number of terms in each sum are non-zero.

Answer 15

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 $$f(x)=\frac{1}{(x-1)}\cdot\frac{1}{(x^2-1)}\cdot\frac{1}{(x^3-1)}=\sum_{i=0}^\infty x^i \cdot\sum_{k=0}^\infty x^{2k} \cdot\sum_{j=0}^\infty x^{3j}$$

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

Answer 16

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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 $\mathbb R^n$.

Answer 17

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.