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Answer 1

\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}

Answer 2

The integral corresponding to this entry is: \begin{equation} I= \int_{0}^{\infty}\frac{x\: d{x}}{(ax^2+2bx+c)^{n}} \end{equation} 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: \begin{equation} (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{<n+n_1+n_2+n_3>}{\Gamma(n)} \end{equation} Substituting the above expansion in the first expression for I, and employing the definition that $<a> \mapsto \int_{0}^{\infty}x^{a-1}dx $ , we have: \begin{equation} \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><n+n_1+n_2+n_3>}{\Gamma(n)} \end{equation} 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 $(0<ac<b^2)$, 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 $1>Re(\alpha+\beta-\gamma) \geq 0$. Also, for $Re(\alpha+\beta-\gamma) < 0$, the series converges absolutely throughout the entire unit circle. According to the book, the form of the series is as follows: $$F(\alpha,\beta;\gamma;z)=1+\frac{\alpha \cdot \beta}{\gamma \cdot 1}z+\frac{\alpha(\alpha+1)\beta(\beta+1)}{\gamma(\gamma+1) \cdot 1 \cdot 2}z^2+\frac{\alpha(\alpha+1)(\alpha+2)\beta(\beta+1)(\beta+2)}{\gamma(\gamma+1)(\gamma+2) \cdot 1 \cdot 2 \cdot 3}z^3+...$$

  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: $$\begin{array}{c|c|c|c|c|c|c|c|} \text{(a,b,c)} & \text{n} & \text{Int} & \text{RHS} & \text{MoB} & \text{|Int-RHS|} & \text{|Int-MoB|} & \text{|RHS-MoB|} \\ \hline \text{(1,2,1)} & 5 & 0.0045` & 0.0045`,0.0045` & 0.0045` & 0.`,0.` & O(10^{-18}) & O(10^{-18}),O(10^{-18}) \\ \hline \text{(3,7,5)} & 5 & 2.87*10^{-6} & 2.87*10^{-6},2.87*10^{-6} & 2.87*10^{-6} & O(10^{-22}),O(10^{-21}) & O(10^{-21}) & O(10^{-21}),O(10^{-20}) \\ \hline \text{(3,1,5)} & 5 & 0.000031` & 0.000031`,0.000031` & 0.000031` & O(10^{-21}),O(10^{-20}) & O(10^{-21}) & O(10^{-20}),O(10^{-20}) \end{array}$$

  As you could see, there are two comma-separated entries in the column labelled by RHS. The first of these entries correspond to the first of the expressions for $I$ as given in Gradshteyn's book (mentioned at the top), while the second entry corresponds to the second of these expressions. For all set of values of the triad-parameters, the same value of $n=5$ has been maintained, so that $n-\frac{1}{2}=\frac{9}{2}>1$. The curious thing here lies not as much as in the value of $Re(\alpha+\beta-\gamma)$ as much as it lies in the following: for the first
  

  The same table was constructed for

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

Answer 3

We recall that the Taylor series of $(\sin x)/x$ is

$$\frac{\sin{x}}{x}=\sum_{k=0}^\infty \frac{(-x^2)^k}{(2k+1)!},$$ and therefore we may write

$$\frac{\sin(\sqrt{x^2-y})}{\sqrt{x^2-y}}=\sum_{k=0}^\infty \frac{(y-x^2)^k}{(2k+1)!}=\sum_{k=0}^\infty \sum_{n=0}^k \binom{k}{n}\frac{y^n (-x^2)^{k-n}}{(2k+1)!}$$ Interchanging the order of summation then yields $$\sum_{n=0}^\infty \sum_{k=n}^\infty \binom{k}{n}\frac{y^n (-x^2)^{k-n}}{(2k+1)!} =\sum_{n=0}^\infty \frac{y^{n}}{n!}\sum_{k=0}^\infty \frac{(n+k)!}{k!}\frac{(-x^2)^{k}}{(2n+2k+1)!}.$$ Note that, when $n=0$, the sum over $k$ is just the Taylor series of $(\sin x)/x$ again. To sum this series, we take inspiration from the fact that $j_0(x)=(\sin x)/x$ is the zeroth spherical Bessel function of the first kind. Looking up the Taylor series for other $j_n(x)$, we find (see eq. (2) here)

$$j_n(x)=(2x)^n \sum_{k=0}^\infty \frac{(n+k)!}{k!}\frac{(-x^2)^k}{(2n+2k+1)!}$$

We may therefore rewrite our expansion as

$$\frac{\sin(\sqrt{x^2-y})}{\sqrt{x^2-y}}=\sum_{n=0}^\infty \frac{j_n(x)}{n!}\left(\frac{y}{2x}\right)^n.$$

Integrating term-by-term then yields

\begin{align} \int_{-t}^t \frac{\sin(\sqrt{x^2-y})}{\sqrt{x^2-y}}\,dt &=\sum_{n=0}^\infty \frac{j_n(x)}{(2x)^n n!}\int_{-t}^t y^n\,dy\\ &=\sum_{n=0}^\infty \frac{j_n(x)}{(2x)^n n!}\frac{t^{n}-(-t)^{-n}}{(n+1)}\\ &=\sum_{n=0}^\infty \frac{j_n(x)}{(n+1)!}\left(\frac{t}{2x}\right)^n(1-(-1)^n). \end{align}

Cancelling out the even terms yields

$$2\sum_{m=0}^\infty \frac{j_{2m+1}(x)}{(2m+2)!}\left(\frac{t}{2x}\right)^{2m+1}$$

Answer 4

$$\small{\bf F}\!\left(\frac m{2^n}\right)=\sum _{k=0}^n\frac1{\prod_{\mu =1}^k \left(1-2^{-\mu}\right)}\cdot\frac1{\prod_{\nu=1}^{n-k} \left(1-2^{-\nu }\right)}\cdot\frac{\sum_{\ell =0}^{2^k m-1}\,\left(\ell-2^k m+\frac12\right)^{n+k}\,(-1)^{1+\log_2\!\big({\ell+1-\sum _{\lambda =0}^\ell\,(-1)^{\tiny\left(\begin{smallmatrix}\ell\\\lambda\end{smallmatrix}\right)}}\big)}}{2^{n^2+k (k-1)} \, (n+k)!}$$

Answer 5

Available to re-use.${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

Answer 6

Aufgabe

Sei $D$ ein Intervall von $\mathbb{R}$ und

$$f:D\mapsto \mathbb{R}$$

eine reellwertige Funktion mit folgender Eigenschaft:

Zu jedem $x_0$ aus dem Definitionsbereich $D$ existiert ein $\delta_{x_0} > 0$, so dass gilt

$$f(x)>=f(x_0), \;\forall x\in D: \; |x-x_0| < \delta_{x_0}$$

Man zeige, dass der Wertebereich von $f(x)$ höchstens abzählbar ist.

Zu jedem $x_0 \in D$ gibt es also ein $\delta_{x_0} \in \mathbb{R^+}$ und ein Intervall

$$I_{x_0}:=(x_0-\delta_{x_0}, x_0+\delta_{x_0}),$$

sodass

$$f(x_0)\le f(x), \; \forall x\in I_{x_0}.$$

Behauptung

Es gilt für jedes $x_1 \in D$ und für jedes $x_2 \in D:$

$$f(x_1)\ne f(x_2) \implies |x_1-x_2|\ge \min\{\delta_{x_1}, \delta_{x_2}\}$$

Beweis:

Andernfalls wäre $x_2 \in I_{x_1}$ und $x_1 \in I_{x_2}$ und deshalb $f(x_2)\ge f(x_1)$ und $f(x_1)\ge f(x_2)$, also $f(x_1)=f(x_2).$

$\square$

Wir setzen nun

$$ E(N,\varepsilon):=\{x \in D : (|x|<N) \land (\delta_x> \varepsilon) \} $$

Behauptung

$f(E(N,\varepsilon)$ ist endlich.

Beweis

Angenommen $f(E(N,\varepsilon)$ ist nicht endlich, dann gäbe es $$k:=\left\lfloor\frac{2n}{\varepsilon}\right\rfloor+3$$

paarweise verschiedene Werte $y_1,\ldots,y_k$ aus $f(E(N,\varepsilon)$, sodass

$$f(x_1)=y_1\\ f(x_2)=y_2\\ \ldots\\ f(x_k)=y_k $$

für

$$x_1<x_2<\ldots<x_k, \; x_i \in E(N,\varepsilon).$$

Nun ist aber

$$|x_k-x_1| =|x_k-x_{k-1}|+|x_{k-1}-x_{k-2}|+\ldots+|x_2-x_1| \ge \left(\frac{2N}{\varepsilon}+1\right)\varepsilon =2N+\varepsilon $$

Damit kann mindestens eines von $x_1$ und $x_k$ nicht in $(-N,N)$ sein

Also enthält $f(E(N,\varepsilon)$ nur endlich viele Elemente.

$\square$

Für jedes $x \in D$ gilt aber

$$\lvert x \rvert \le \left\lfloor \lvert x \rvert \right\rfloor +1 \\ \delta_{x}\ge \frac{1}{\left\lfloor \frac{1}{\delta_{x}}\right\rfloor+1} $$

und deshalb

$$x \in E(M,\frac 1 M)$$

für

$$M \ge \max\left\{\lfloor|x| \rfloor +1,\left\lfloor \frac{1}{\delta_{x}}\right\rfloor+1\right\}$$

und damit ist $x$ in $E\left(M,\frac {1}{M}\right) .$

Also ist

$$f(D)=\bigcup_{n \in \mathbb{N}} f(E(n,\frac{1}{n}))$$

und das ist damit höchstens abzählbar.

Answer 7

I think it becomes obvious how we try to proceed: we try to use an expression of a partial geometric series of the Binomialmatrix $P$ to get a vector of sums-of-like-powers. An analogy to the same operation on a geometric series on a scalar -say- $|p|<1$ were $$ {1 \over 1 - p} = 1+ p + p^2 + p^3 + \cdots \\ {p^h \over 1 - p} = p^h + p^{h+1} + p^{h+2} + \cdots \\ {1 - p^h \over 1 - p} = 1 + p + p^2 + \cdots + p^{h-1}\\ $$ where the last row gives us the formula for the partial geometric series.
In a matrix-notation -if the inversion were possible- this were formally $$ (I-P^h) \cdot \underset{{\text{impossible}}}{\underbrace{(I - P)^{-1}}} = \sum_{k=0}^{h-1} P^k $$ but it is possible to get the expected result of that impossible matrix-inverse by summing the scalar entries of the powers of $P$ into $\zeta()$-values at zero and negative integer arguments.
Let us denote the use of a $V(x)$-vector as diagonalmatrix by a pre-superfix $^dV(x)$ then a $h$'th power of $P$ is $$ P^h = \,^dV(1/h) \cdot P \cdot \,^dV(h)$$ which can also be written as the Hadamard-product (=elementwise multiplication) of $P$ with a Toeplitzmatrix of $h$ $$ P^h = P \star T(h) $$ where $$T(h) =\small \begin{bmatrix} 1 &h &h^2&h^3& \cdots\\ h^{-1}&1 &h &h^2& \cdots \\ h^{-2}&h^{-1}&1 &h & \cdots \\ \vdots&\vdots&\vdots &\vdots & \ddots \\ \end{bmatrix} $$ After that we rewrite $$\underset{{\text{impossible}}}{\underbrace{(I - P)^{-1}}} \underset{\text{replace}}{=} P \star (T(0)+T(1)+T(2)+ \cdots ) \underset{\text{replace}}{=} P \star Z $$ and define for the infinite sum of the Toeplitzmatrices a $Z$-matrix where the entries contain $\zeta()$ values at integer arguments.
We are already near the solution for the problem of the Bernoulli-polynomials and the Faulhaber-formula, because $\zeta()$-values at negative indexes are simple multiples of the Bernoulli-numbers.

Unfortunately we need special attention at the positions of the $\zeta(1)$. This infinities occur in the first subdiagonal of the $Z$-matrix. By the Hadamard-matrix product with the binomial-matrix $P$ we find there numerically zeros, so we had the indefinite expressions of $ 0 \cdot \infty$. But if we interpret the zeros as evaluation of the binomials $ {k! \over (k+1)! (-1)! } $ then we might take the limit of cancelling $\lim_{\varepsilon \to 0} \zeta(1-\varepsilon) / \Gamma(\varepsilon)=-1 $ we get the full representation of the Hadamard-product, which I called in my matrix-exercises the matrix $ZETA$ such that $$ZETA = P \star_\infty Z$$ where the $\infty$-subscript at the star - operator reminds of the $\zeta(1)/\Gamma(0)$-normalization. $$ ZETA = \small \begin{bmatrix} 1 \zeta(0) & 1 \zeta(-1) & 1 \zeta(-2) & 1 \zeta(-3) & \cdots \\ -1 & 1 \zeta(0) & 2 \zeta(-1) & 3 \zeta(-2) & \cdots \\ . & -1/2 & 1 \zeta(0) & 3 \zeta(-1) & \cdots \\ . & . & -1/3 & 1 \zeta(0) & \cdots \\ \vdots &\vdots &\vdots &\vdots &\ddots \end{bmatrix} \\ = \small \begin{bmatrix} -1/2 & -1/12 & 0 & 1/120 & 0 & -1/252 &\cdots \\ -1 & -1/2 & -1/6 & 0 & 1/30 & 0 &\cdots \\ . & -1/2 & -1/2 & -1/4 & 0 & 1/12 &\cdots \\ . & . & -1/3 & -1/2 & -1/3 & 0 &\cdots \\ \vdots &\vdots &\vdots &\vdots & \vdots & \vdots & \ddots \\ \end{bmatrix} $$

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We can use powers of $P$ : $$\small \begin{array} {} V(x) \cdot P &= V(x+1) & \text{then} & V(x+1) \cdot P &= V(x+2) & \text{ or }\\ (V(x) \cdot P) \cdot P &= V(x+2) &&&& \text { or } \\ V(x) \cdot P^2 &= V(x+2) \\ \end{array}$$ In general $$ V(x) \cdot P^h = V(x+h) $$ Due to the rules of matrix-algebra we can also write $$ V(x) \cdot (I + P + P^2 + P^3) = V(x)+V(x+1)+V(x+2)+V(x+3) $$

where the entries in the result-vector are $$\small \begin{array} {} &V(x)+V(x+1)+V(x+2)+V(x^3) &=& \\ [&1+1+1+1 ,\\ & x+(x+1)+(x+2)+(x+3), \\ & x^2+(x+1)^2+(x+2)^2+(x+3)^2, \\ &\cdots &] \end{array}$$ or $$ V(x) \cdot \sum_{k=0}^{n-1} P^k = \text{vec}[\sum_{k=0}^{n-1}(x+k)^c]_{c=0..\sigma} $$

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I've found some time ago a solution using a toolbox of matrices, which for instance can easily be implemented using Pari/GP.

Ideally all matrices and vectors in the following are meant to be of infinite size, but for the implementation in a software we assume a finite size, defined in a global variable $\sigma$ with for instance $\sigma=16$ or so.

Let us define a type of vector $V_\sigma(x)=[1,x,x^2,x^3,...,]$ up to size of $\sigma$.
Then let us define the upper-triangular Pascal-matrix $P_\sigma$ where $$ P_\sigma=\small \begin{bmatrix} 1&1&1&1&... \\ .&1&2&3&...\\ .&.&1&3&... \\ .&.&.&1&... \\ \vdots&\vdots&\vdots&\vdots&\ddots\\ \end{bmatrix} $$ and truncated to size $\sigma \times \sigma$.

Then we can write the binomial transform on $x$ in a vectorial representation $ [1,x,x^2,x^3,...]_\sigma \cdot P_\sigma =\small{ [1,x+1,(x+1)^2,(x+1)^3,...]_\sigma }$ or better, and for convenience omitting the $\sigma$-parameter $$ V(x) \cdot P = V(x+1)$$

Answer 8

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Answer 9

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Answer 10

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 11

$$ {2^S \over 3^N}\cdot {a_4 \over a_1}= (1+{1 \over 3a_1}) \cdot(1+{1 \over 3a_2})\cdot (1+{1 \over 3a_3}) $$ $$ {2^S \over 3^N}\cdot ({3^N \over 2^S}a_1+{Q(N,S)\over 2^S}){1\over a_1}= (1+{1 \over 3a_1}) \cdot(1+{1 \over 3a_2})\cdot (1+{1 \over 3a_3}) $$ $$ 1+{Q(N,S) \over 3^N}\cdot {1\over a_1}= (1+{1 \over 3a_1}) \cdot(1+{1 \over 3a_2})\cdot (1+{1 \over 3a_3}) $$

Let's use the notation for one step $a_2=U(a_1)$ on odd $a_1$ to odd $a_2$ as $$ U: a_2 = { 3a_1+1 \over 2^{A_1} } \qquad \text{whera $A_1$ is such that $a_2$ is odd}$$ (I use here the letter "U" to avoid the letters "T" and "C" which are used for different definitions of steps and may be remind the reader of the name "Syrac(U)se"-transformation see wikipedia.)
If there are more steps in consideration, let's call the number of steps $N$ and the sum-of-exponents $S$.

For some $N$ , for instance $N=3$ let's look at the following trivial product formula

$$ a_2 \cdot a_3 \cdot a_4 = U(a_1) \cdot U(a_2) \cdot U(a_3) \\ = {3a_1+1 \over 2^{A_1}} \cdot{3a_2+1 \over 2^{A_2}}\cdot {3a_3+1 \over 2^{A_3}} $$ Now, to make the formula smoother, let's prepend $a_1/a_4$ to the equation $$ a_1 \cdot a_2 \cdot a_3 = {a_1 \over a_4} \cdot {3a_1+1 \over 2^{A_1}} \cdot{3a_2+1 \over 2^{A_2}}\cdot {3a_3+1 \over 2^{A_3}} $$ After that, let's rearrange the lhs to the denominators of the rhs and the powers-of-two to the lhs, where also $S=A_1+A_2+A_3$ : $$ 2^S = {a_1 \over a_4} \cdot (3+{1 \over a_1}) \cdot(3+{1 \over a_2})\cdot (3+{1 \over a_3}) $$

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Answer 12

If we reorder the inequality and use $b=n-m$ we can write $$ 2^n-2^m \ge 3^m - 1 \\\ 2^b(2^m-1) \ge 3^m-1 \\ $$ The left expression has the bitstring $$ 2^b(2^m-1)= "\underset{m \text{ ones}} {\underbrace{11..11}}\underset{b \text{ zeros}}{\underbrace{00..00}}"_2 $$ and is defined to be larger than the rhs. Now the rhs has the same length of bitstring and has the form $$ 3^m-1 = "\underset{m \text{ ?}} {\underbrace{1?..??}}\underset{b \text{ ? }}{\underbrace{??..?0}}"_2 $$ So let's write $3^m-1 = (M \cdot 2^b + r) -1 $ and if $2^m-1 > M$ that means we have (at least) one zero in the bitstring of $M$ .
What I've done with this is a set of pictures, what the position of the first zero in the bitstring of $M$ is, or, what the runlength of the leading $1$ in the bitstring of $M$ is depending on $m$ Of course, this is somehow erratic, but using the convergents of the continued fraction of $\log_2(3)$ gives that $m$ for which either the runlength of the leading $1$ in relation to $m$ is maximal

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Answer 13

the first conjecture reflects the open Collatz-problem, whether every number orbits towards the trivial cycle $ s \to \cdots \to 1 \to 1 \to \cdots$

For the analysis of partial problem, such as the properties of $3$-step transformation, I think my notation was a good invention which I propose here too to access your question on the $3$-step transformation.
That means first to just denote the exponents at $2$ in your given summation-formula. That means I propose to reduce the writing of one step on an odd number as $$ b = T(a;A) \qquad \text{ is defined as } b= {3a+1\over2^A} \qquad \text{where } A=\nu_2(3a+1)$$ so we have $T()$ and odd-to-odd expression.
Then a two-step transformation is simply written as $$ c=T(b;B) = T(a;A,B) \qquad \text{or more general} \\ a_{1+N} = T(a_1;A_1,A_2,....,A_N) \qquad \text{for an $N$-step transformation}$$ I'm used to denote $N$ for the number of steps and $S$ as the sum-of-exponents ($=$ number-of steps of divisions-by-$2$) (you used $m$ for what I use $S$ in my own texts) .
To write it out for a three step transformation $$ a_4 = {3^3a_1 + \left(3^2\cdot 2^0 + 3^1 \cdot 2^{A_1} + 3^0 \cdot 2^{A_1+A_2} \right)\over 2^{A_1+A_2+A_3}} = {3^3 a_1\over2^S} +{3^2+ \cdots + 2^{A_1+A_2}\over2^S} $$ which is also $$ a_4\cdot 2^S = 2^S \cdot T(a_1;A_1,A_2,A_3)$$

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Answer 14

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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.