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Answer 1

Two draws from two urns A and B, without replacement. Urn A contains 5 white 2 red, and urn B contains 3 red 6 black.

For convenience, $$ \text{First draw} \\ \begin{aligned} & \hspace{24pt} A & \hspace{128pt} B \end{aligned} \\ \begin{array}{c} 2 \text{nd} \\ \text{d} \\ \text{r} \\ \text{a} \\ \text{w} \end{array} \begin{array}{l|l|cc|cc} & & w & r & r & b \\ \hline A & w & p\frac57 \cdot p \frac46 & p\frac27 \cdot p \frac16 & (1-p)\frac39 \cdot p \frac57 & (1-p)\frac69 \cdot p \frac57 \\ & r & p\frac57 \cdot p \frac26 & p\frac27 \cdot p \frac16 & (1-p)\frac39 \cdot p \frac27 & (1-p)\frac69 \cdot p \frac27 \\ \hline B & r & p\frac57 \cdot (1-p) \frac39 & p\frac27 \cdot (1-p) \frac39 & (1-p)\frac39 \cdot (1-p) \frac28 & (1-p)\frac69 \cdot (1-p) \frac38 \\ & b & p\frac57 \cdot (1-p) \frac69 & p\frac27 \cdot (1-p) \frac69 & (1-p)\frac39 \cdot (1-p) \frac68 & (1-p)\frac69 \cdot (1-p) \frac58 \end{array}$$

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

Now let's assume $S$ as given, and let's look what does follow for $m$ from this. - what if $S$ is even, say $S=2T$ . Then
$$ (2^{2T} -m^2) \cdot b =(2^T-m)(2^T+m) b= m + 2^B $$ Then we write $$ (2^T-m)\cdot b = { m + 2^B\over m + 2^T} $$ and because the lhs must be positive odd integer equal or larger than 1, we have that $B \le T$ and because $A+B=2=2T$ it follows that $A \le T$.
We have then the other equation $$ (2^T-m)\cdot a = { m + 2^A\over m + 2^T} $$ and now, if $A \lt T$ the rhs were fractional, so as well as $B$ we must have that $A \ge T$ which then forces $A+B \ge 2T$ and by definition of $2T=S$ we can have only equality $A=B$ and from this follows $a=b$ and from this that for even $S$ we have actually detected the 1-step cycle.
Thus for to have a true 2-step cycle we need that $S$ is odd.
There is another form of writing our equation which sheds more light on the relation of $m$ to $S$ . This is, when we equate the product of $a$ and $b$ with the product of their iterates. We write then $$ a \cdot b = {mb+1\over 2^B}{ma+1\over 2^A}$$
Rearranged this gives $$ 2^S = {mb+1\over b}{ma+1\over a} \qquad \text{or} \\ 2^S=(m+\frac1a)(m+\frac1b)$$
If we assume different $a,b$ being odd and positive and $a \lt b$ then the rhs is smaller than $(m+1)^2$ but larger than $(m+\frac1b)^2 \gt (m+\frac1\infty)^2$ and thus $$ (m+0) \lt (m+\frac1b) \lt 2^{S/2} \lt (m+\frac1a) \lt (m+1) $$ (equality in the inner relations can not occur since $2^{S/2}$ is irrational when $S$ is odd)

==========
Algebraic expansion (after inserting) of the two equations above give for $b$: $$ b={m \cdot {m \cdot b+1\over 2^B}+1\over 2^A} = {m^2 \cdot b+m+2^B\over 2^{A+B}} = {m^2 \over 2^{S}}b+{m+2^B\over 2^{S}}$$ This gives $$ (2^S -m^2) \cdot b = m + 2^B $$ and analoguously $$ (2^S -m^2) \cdot a = m + 2^A $$

================ $$ b={m \cdot a+1\over 2^A} \qquad \qquad a={m \cdot b+1\over 2^B} $$ Let's define $N$ the number of odd steps (here: $N=2$) and $S$ the sum of the exponents (here $S=A+B$).

$$ 2^S=(m+\frac1a)(m+\frac1b) $$

Answer 6

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 7

$ \color{silver}{\Tiny \text { I might have found an elegant solution, easily generalizable to more variables.} }$

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Answer 8

You don't actually need to find $\lim_{x \to 0} \frac{\sin x}{x}$ to prove that $f(x) = \sin x$ is continuous.

Define $\sin x$ to be the $x$-coordinate of the point $(\cos x, \sin x)$ on the unit circle. Then we need to prove three things:

$$\lim_{x \to a^+} f(x) = f(a) \tag1$$ $$\lim_{x \to a^-} f(x) = f(a) \tag2$$ $$\lim_{x \to a^+} f(x) = \lim_{x \to a^-} f(a) \tag3$$

Let us first prove $(1)$. To do this, we will need the help of the following diagrams:

Since the large triangle and small triangle are similar triangles, the $x$-coordinate when the angle is $x + \theta$ or $\sin(x + \theta)$ is equal to $\sin(x) - \cos(\theta)/h$, where $h$ is the 'hypotenuse' of the smaller triangle.

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

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Answer 12

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Answer 13

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

Answer 14

Define $P(x)=c_2 x^2+c_1 x+c_0$.

Rewrite the problem as $$w(\eta)\sim\frac1{\Gamma(\alpha)}\int^\eta_0 ds\frac{(1-s/\eta)^{\alpha-1}}{(s+\eta_0)^{\alpha+1}}P\circ w(s)$$

As a shorthand, define the operator $$I=\int^\eta_0 ds\frac{(1-s/\eta)^{\alpha-1}}{(s+\eta_0)^{\alpha+1}}$$

Also define $A$ as the 'asymptotic operator', such that $A[f]$ returns the asymptotics of $f$ up to $o(1)$, according to certain asymptotic scale.

Then, the problem becomes a fixed-point problem: $$w=\frac1{\Gamma(\alpha)}\cdot A\circ I\circ P\circ w\qquad(\star)$$

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It is assumed that $w(\eta)$ grows at most polynomially as $\eta\to\infty$.

First, we investigate $A\circ I$'s action on a monomial $x^q$ ($q\ge0$).

$$\begin{align} I[s^q](\eta) &=\int^\eta_0\frac{(1-s/\eta)^{\alpha-1}}{(s+\eta_0)^{\alpha+1}}s^q ds \\ &=\int^\eta_0\frac{s^q}{(s+\eta_0)^{\alpha+1}}ds+ \underbrace{\int^\eta_0\frac{(1-s/\eta)^{\alpha-1}-1}{(s+\eta_0)^{\alpha+1}}s^q ds}_{I'(q)} \\ \end{align} $$

For $I'(q)$, $$\begin{align} I'(q) &=\int^{\eta}_0 \frac{(1-s/\eta)^{\alpha-1}-1}{(s+\eta_0)^{\alpha+1}}s^q ds \\ &=\sum^\infty_{r=1}\binom{\alpha-1}{r}(-1)^r \eta^{-r}\int^\eta_0\frac{s^{q+r}}{(s+\eta_0)^{\alpha+1}}ds \\ &=\sum^\infty_{r=1}\binom{\alpha-1}{r}(-1)^r \eta^{-r}\left(\frac{\eta^{q+r-\alpha}}{q+r-\alpha}+O(\eta^{q+r-\alpha-1})\right) \\ &=\eta^{q-\alpha}\sum^\infty_{r=1}\binom{\alpha-1}{r}\frac{(-1)^r}{q+r-\alpha}+O(\eta^{q-\alpha-1}) \\ &=\eta^{q-\alpha}\left[\frac{\Gamma(\alpha)\Gamma(q-\alpha)}{\Gamma(q)}-\frac1{q-\alpha}\right]+O(\eta^{q-\alpha-1}) \\ \end{align} $$

Note that $\gamma(q):=\frac{\Gamma(\alpha)\Gamma(q-\alpha)}{\Gamma(q)}-\frac1{q-\alpha}$ converges for $\Re (q)\ge 0$.

Summary $$I[s^q](\eta)=\int^\eta_0\frac{s^q}{(s+\eta_0)^{\alpha+1}}ds+\gamma(q)\eta^{q-\alpha}+O(\eta^{q-\alpha-1})$$

By differentiating under the integral sign, it can be further proved that
$$I[s^q \ln^N s](\eta)= \begin{cases} O(\eta^{q-\alpha}\ln^N \eta), &q>\alpha \\ O\left(\ln^{N+1}\eta\right), &q=\alpha \\ O(1), &\alpha> q\ge 0 \end{cases} $$ (Asymptotic expansions up to $o(1)$ are given in the appendix below.)

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It is obvious that, to match the leading order on both sides of $(\star)$, $w(\eta)\sim C\eta^{\alpha}$ where $C=\frac{\Gamma(2\alpha)}{c_2 \Gamma(\alpha)}$, as the OP obtained.

Now, let's make a slightly different ansatz: $$w(\eta)=C\eta^{\alpha}+\sum^N_{r=0}B_r\ln^r\eta$$

Utilizing asymptotic formulae in the appendix, ($D:=\frac{2\Gamma(2\alpha)}{\Gamma^2(\alpha)}$) $$\begin{align} \frac1{\Gamma(\alpha)}\cdot A\circ I\circ P\circ w &=C\eta^{\alpha}+\frac{DB_N}{N+1}\ln^{N+1}\eta \\ &+D\sum^N_{r=2}\left[\frac{B_{r-1}}{r}+\sum^N_{m=r}\binom{m}{r}\gamma^{(m-r)}(\alpha)B_m\right]\ln^r\eta \\ &+D\left[\frac{c_1\Gamma(\alpha)}{2c_2}+2B_0+\sum^N_{m=1}m\gamma^{(m-r)}(\alpha)B_m\right]\ln\eta \\ &+\text{a messy bunch of constants} \end{align} $$ (The messy bunch of constants can be found using the asymptotic formulae in appendix.)

The mismatch of order, i.e. the presence of $\ln^{N+1}\eta$, is expected. However, in the following calculations, we will give evidence hinting at $$|B_r|\le\frac{K^r_N}{r!}\qquad(\blacksquare)$$

If this is true, under the assumption that $K_N$ does not grow too fast as $N\to\infty$, then:

1. The coefficient of the mismatching order, $\frac{DB_N}{N+1}$, is $\sim D\cdot \frac{(K_N)^N}{(N+1)!}$ and thus vanishes as $N\to\infty$.
2. $w(\eta)=C\eta^{\alpha}+\sum^\infty_{r=0}B_r\ln^r\eta$ converges, and would be the solution to $(\star)$.

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The evidence is based on an inductive argument.

First, equating coefficients on both sides of $(\star)$, $$(1-\gamma(\alpha))B_r=\frac{B_{r-1}}{r}+\sum^N_{m=r+1}\binom{m}{r}\gamma^{(m-r)}(\alpha)B_m$$

Suppose $(\blacksquare)$ is true for $r=i$. Then, neglecting signs, $$\frac{B_{i-1}}{i}+\sum^N_{m=i+1}\binom{m}{i}\gamma^{(m-i)}(\alpha)B_m \le (1-\gamma(\alpha))\cdot\frac{K_N^i}{i!}$$

When $r=i+1$, $$\begin{align} (1-\gamma(\alpha))B_{i+1} &=\frac{B_i}{i+1}+\sum^N_{m=i+2}\binom{m}{i+1}\gamma^{(m-i-1)}(\alpha)B_m \\ &=\frac{B_i}{i+1}+\sum^N_{m=i+2}\binom{m}{i}\cdot\frac{m-i}{i+1}\gamma^{(m-i-1)}(\alpha)B_m \\ &\sim\frac{B_i}{i+1}+\sum^N_{m=i+2}\binom{m}{i}\cdot\frac{m-i}{i+1}\frac{\gamma^{(m-i)}}{m-i}(\alpha)B_m \qquad (1)\\ &=\frac{B_i}{i+1}+\frac{1}{i+1}\sum^N_{m=i+2}\binom{m}{i}\gamma^{(m-i)}(\alpha)B_m \\ &=\frac{B_i}{i+1}+\frac{1}{i+1}\left[(1-\gamma(\alpha))B_i-\frac{B_{i-1}}{i}-\binom{i+1}{i}\gamma^{(1)}(\alpha)B_{i+1}\right] \\ (1-\gamma(\alpha)+\gamma'(\alpha))B_{i+1} &=\frac{2-\gamma(\alpha)}{i+1}B_i-\frac{B_{i-1}}{i(i+1)} \\ (1-\gamma(\alpha)+\gamma'(\alpha))B_{i+1} &\le\frac{2-\gamma(\alpha)}{i+1}\frac{K_N^i}{i!}+\frac{1} {i(i+1)}\frac{K_N^{i-1}}{(i-1)!} \\ B_{i+1} &\le\frac{K_N^{i+1}}{(i+1)!} \\ \end{align} $$ assuming $K_N$ is sufficiently large.

$(1)$: Due to the singularity at $\alpha-1$, the Taylor series of $\gamma(q)$ about $q=\alpha$ has radius of convergence $1$. Hence, $\displaystyle{\left\vert\frac{\gamma^{(j+1)}(\alpha)/(j+1)!}{\gamma^{(j)}(\alpha)/j!}\right\vert\sim 1}$. Signs aside, we get $\displaystyle{\gamma^{(j)}(\alpha)\sim \frac{\gamma^{(j+1)}(\alpha)}{(j+1)}}$.

Calculations above are quite sloppy, but I am sure that a more meticulous analysis would lead to a rigorous proof of $(\blacksquare)$.

Practically, most Volterra integral equations are solved numerically. Therefore, it is feasible to directly plug in the infinte series and solve for the coefficients numerically.

One approach that I haven't tried is substituting $\eta=e^z$ - the infinte series then becomes a Maclaurin series, whose coefficients may be found (probably) by Cauchy's integral formula or simply by differentiation. I may add on this later.

Appendix

For $\alpha<q\le 2\alpha$,

$$I[s^q \ln^N s](\eta)=\int^\eta_0\frac{s^q\ln^N s}{(s+\eta_0)^{\alpha+1}}ds+\eta^{q-\alpha}\sum^N_{r=0}\binom{N}{r}\gamma^{(m-r)}(q)\ln^r\eta+O(\eta^{q-\alpha-1}\ln^N \eta)$$

For $q=\alpha$,

$$\begin{align} I[s^\alpha \ln^N s](\eta) &=\frac{\ln^{N+1}\eta}{N+1}+\eta_0\cdot\frac{\alpha+1}{N+1}\int^\infty_0\frac{s^\alpha\ln^{N+1}s}{(s+\eta_0)^{\alpha+2}}ds \\ &+\sum^N_{r=0}\binom{N}{r}\gamma^{(m-r)}(\alpha)\ln^r\eta+O(\eta^{-1}\ln^N \eta) \end{align} $$

For $0\le q<\alpha$,

$$I[s^q \ln^N s](\eta)=\int^\infty_0\frac{s^q\ln^N s}{(s+\eta_0)^{\alpha+1}}ds+O(\eta^{q-\alpha}\ln^N\eta)$$

Answer 15

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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

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