## Total convergence of a function series

Let us consider the following function series:

$$\sum_{n=1}^{\infty} \frac{1}{n^x + n^{-x}}.$$

It seems that there is total convergence in $$(- \infty, -1-\delta] \cup [1+ \delta, + \infty)$$ for $$\delta > 0$$. Thank-you in advance for any help!

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## convergence rate of newton’s method

So, I’m currently studying Newton method used for finding the 0’s of a function, however my professor has only announced that the speed of this algorithm can be more than quadratic, however I’m wondering when this happens, since in the demonstration used by him to explain the quadratic case, there is no evidence about the "more than quadratic"

Using Taylor and Lagrange
$$f(\xi)=f(x_n)+f'(x_n)(\xi-x_n)+\frac{f”(z_n)}{2}(\xi-x_n)^2$$

$$-\frac{f(x_n)}{f'(x_n)}=\xi-x_n+\frac{f”(z_n)}{2f'(x_n)}(\xi-x_n)^2$$

$$x_{n+1}-x_n=\xi-x_n+\frac{f”(z_n)}{2f'(x_n)}(\xi-x_n)^2$$

$$e_{n+1}=\left|x_{n+1}-\xi\right|=c_ne_n^2 \quad \text{with} \quad c_n=\frac{1}{2}\frac{\left|f”(z_n)\right|}{\left|f'(x_n)\right|}$$

Can someone please tell me when (example) the order of the convergence is cubic, and why?=

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## Improving mesh and NDSolve solution convergence

I have developed the code below to solve two PDEs; first mu[x,y] is solved for, then the results of mu are used to solve for phi[x,y]. The code works and converges on a solution as is, however, I would like to decrease the size of a, b, and d even further. To accurately represent the physical process I am trying to simulate, a, b, and d would need to be ~100-1000x smaller. If I make them smaller, I don’t believe the solution has actually converged because the values for phi along the right boundary change significantly with a change in mesh size (i.e. if I make them smaller and the code below produces a value of phi=-0.764 at the midpoint between y2 and y3 along the right boundary, a change in size1 to 10^-17 and size2 to 10^-15, changes that value of phi to -0.763, and a change in size2 to 10^-16 changes that value again to -0.860), but I cannot make the mesh size any smaller without Mathematica crashing.

Are there any better ways to create the mesh that would be less computationally taxing and allow it to be more refined in the regions of interest? Or are there any ways to make the code in general less computationally expensive so that I can further refine the mesh?

ClearAll["Global*"] Needs["NDSolveFEM"] (* 1) Define Constants*) e = 1.60217662*10^-19; F = 96485; kb = 1.381*10^-23; sigi = 18; sigini = 0; sigeni = 2*10^6; T = 1000; n = -0.02; c = 1;  pH2 = 0.2; pH2O = 1 - pH2; pO2 = 1.52*^-19; l = 10*10^-6; a = 100*10^-7; b = 50*10^-7; d = 300*10^-7; y1 = 0.01; y2 = 0.5*y1; y3 = y2 + a; y4 = y3 + d; y5 = y4 + b; mu1 = 0; mu2 = -5.98392*^-19; phi1 = 0;  (* 2) Create mesh*) m = 0.1*l; size1 = 10^-16; size2 = 10^-15; size3 = 10^-7; mrf = With[{rmf =       RegionMember[       Region@RegionUnion[Disk[{l, y2}, m], Disk[{l, y3}, m],          Disk[{l, y4}, m], Disk[{l, y5}, m]]]},     Function[{vertices, area}, Block[{x, y}, {x, y} = Mean[vertices];      Which[rmf[{x, y}],        area > size1, (0 <= x <= l && y2 - l <= y <= y2 + l),        area > size2, (0 <= x <= l && y3 - l <= y <= y3 + l),        area > size2, (0 <= x <= l && y4 - l <= y <= y4 + l),        area > size2, (0 <= x <= l && y5 - l <= y <= y5 + l),        area > size2, True, area > size3]]]]; mesh = DiscretizeRegion[Rectangle[{0, 0}, {l, y1}],     MeshRefinementFunction -> mrf];  (* 3) Solve for mu*) bcmu = {DirichletCondition[mu[x, y] == mu1, (x == 0 && 0 < y < y1)],    DirichletCondition[     mu[x, y] ==       mu2, (x == l && y2 <=  y <=  y3) || (x == l && y4 <= y <= y5)]}; solmu = NDSolve[{Laplacian[mu[x, y], {x, y}] ==       0 + NeumannValue[0, y == 0 || y == y1 ||         (x == l && 0 <= y < y2) || (x == l &&            y3 < y < y4) || (x == l && y5 < y < y1)], bcmu},     mu, {x, y} \[Element] mesh, WorkingPrecision -> 50];  (* 4) Solve for electronic conductivity everywhere*) pO2data = Exp[(mu[x, y] /. solmu)/kb/T]; sige0 = 2.77*10^-7; sigedata = Piecewise[{{sige0*pO2data^(-1/4), 0 <= x <= l - m},     {sige0*pO2data^(-1/4), (l - m < x <= l && 0 <= y < y2)},     {(sigeni - sige0*(pO2data /. x -> l - m)^(-1/4))/m*(x - (l - m)) +        sige0*(pO2data /. x -> l - m)^(-1/4), (l - m < x <= l &&         y2 <=  y <= y3)},     {sige0*pO2data^(-1/4), (l - m < x <= l && y3 < y < y4)},     {(sigeni - sige0*(pO2data /. x -> l - m)^(-1/4))/m*(x - (l - m)) +        sige0*(pO2data /. x -> l - m)^(-1/4), (l - m < x <= l &&         y4 <= y <= y5)},     {sige0*pO2data^(-1/4), (l - m < x <= l && y5 < y <= y1)}}];  (* 5) Solve for phi*) Irxn = -(2*F)*(c*pO2^n ); A = (Irxn - sigi/(4*e)*(D[mu[x, y] /. solmu, x] /. x -> l))/(-sigi); B = sigi/(4*e)*(D[mu[x, y] /. solmu, x] /.        x -> l)/(sigi + sigedata /. x -> l - m); bcphi = DirichletCondition[phi[x, y] == phi1, (x == 0 && 0 < y < y1)]; solphi = NDSolve[{Laplacian[phi[x, y], {x, y}] ==       0 + NeumannValue[0,         y == 0 ||          y == y1 || (x == l && 0 <= y < y2) || (x == l &&            y3 < y < y4) || (x == l && y5 < y < y1)] +        NeumannValue[-A[], (x == l && y2 <= y <= y3)] +        NeumannValue[-B[], (x == l && y4 <= y <= y5)], bcphi},     phi, {x, y} \[Element] mesh, WorkingPrecision -> 50];  (* 6) Print values to check for convergence*) P[x_, y_] := phi[x, y] /. solphi; P[l, (y3 - y2)/2 + y2] P[l, (y5 - y4)/2 + y4] 
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## Why doesn’t Mathematica provide an answer while Wolfram|Alpha does, concerning a series convergence?

Among other series I’ve been working on, I was asked to find whether $$\sum_n 1-\cos(\frac{\pi}{n})$$ converged, and Mathematica’s output to SumConvergence[1 - Cos[Pi/n], n] simply was repeating the input, without further information. Wolfram|Alpha, though, at least told me which test were or not conclusive.

I’m new to Mathematica, and even though I’ve looked both on Google and into Wolfram’s documentation, I haven’t found information that could help me figure out how to get, from Mathematica, the conditions for the convergence of a series involving something else than powers of a variable.

I would appreciate if you could give me some clues on the typical procedure to make Mathematica correctly evaluate the convergence of a series, or/and to return the conditions for convergence. Thank you in advance.

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## Convergence of Conjugate gradient method

I have implemented my own matrix library in Java to solve fluid simulations. So I have also implemented the conjugate gradient method and I got a little bit confused.

What I have done to test my CG-method is the following:

1. Generate M, a $$n\times n$$ matrix, initialized with values between 0 and 1 (exclusive)
2. Calculate $$A = M \cdot M^T$$ (which must be symmetric positive definite)
3. Calculate the conjugate gradient and display the residuum and the actual distance to the target (I used cholesky to solve the equation and get an exact solution).

My results have been the following (n = 10):

residuum:                               actual distance: 0.9199326513658944                      72.7313310733776 0.7300075150837726                      72.24654586114522 1.2209457036576696                      69.71996519380937 2.062353336085214                       50.76028746131138 0.5685046122325719                      44.90515825964432 0.6994645262755821                      43.72541529795878 0.6572329365491988                      33.37442078957705 1.0192100131027082                      18.410526779358293 1.0808876324653045                      4.531060524696579 0.19130828691828886                     4.436357203678213 3.423483345725764E-10                   1.8117326798071072E-10 `
1. Why does the actual distance always decrease but the residuum does not?
2. Why is the distance so big to begin with? Is there a way to scale the Matrix or the vector to reduce A) the amount of steps required B) reduces the distance?
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## When does this convergence to zero?

For which complex values of $$s$$ is this true? $$\lim_{x \rightarrow \infty} \overline{\zeta(s,x)}\ \left(x^{1-s} – (x-1)^{1-s} \right) = 0$$ with $$\zeta(s,1) = \zeta(s) =$$ Riemann zeta

## Weak convergence for generic periodic points that obtains from specification

Let $$X$$ be a compact metric space and let $$f:X\rightarrow X$$ be a Anosov and transitive map.

Let $$x$$ be a generic point so that $$\mu_{n_{i}}:=\frac{1}{n_{i}}\sum_{j=0}^{n_{i}-1} \delta_{T^{j}(x)} \rightarrow \mu$$ for some invariant measure $$\mu$$.

It is well known that $$T$$ satisfies the specification property. Let $$p\in X$$ be a periodic point associated to $$\delta, S$$ and $$\{ x, f(x), . . . , f^{n_{i}}(x)\}$$ by the specification property. In particular, $$f^{n_{i}+S}(p) = p$$ and orbit of $$p$$ cover orbit of $$f^{n_{i}}(x)$$.

My first question is as follows:

$$\mu_{n_{i},p}:=\frac{1}{n_{i}+S}\sum_{j=0}^{n_{i}+S-1} \delta_{T^{i}(p)} \rightarrow \mu$$?

If yes. For $$\mu\in\mathcal M(X)$$ we write $$\text{supp} \mu$$ for the support of $$\mu$$. It is easy to see that if a sequence of measures $$(\mu_n)_{n=1}^\infty$$ weak$$^*$$ converges to some $$\mu\in\mathcal M(X)$$, then $$\text{supp}\, \mu\subset \liminf_{n\to\infty}(\text{supp}\,\mu_n).$$ (By $$\liminf$$ above I mean the Kuratowski limit inferior, for a definition see wikipedia entry.)

According to above, Can we say $$supp \mu$$ is periodic point? Because $$\mu_{n_{i},p}$$ has periodic support.

## Overconcentration of Poles on the Circle of Convergence of a Power Series with Bounded Coefficients

Let $$V$$ be an arbitrary set of infinitely many positive integers, and let: $$\varsigma_{V}\left(z\right)\overset{\textrm{def}}{=}\sum_{v\in V}z^{v}$$ Let $$T_{V}$$ denote the set of all $$t\in\left[0,1\right)$$ for which the limit:

$$c_{V}\left(t\right)\overset{\textrm{def}}{=}\lim_{x\uparrow1}\left(1-x\right)\varsigma_{V}\left(e^{2\pi it}x\right)$$ exists and is non-zero. Is $$T_{V}$$ necessarily finite (that is, would an “overconcentration” of radii on which $$\varsigma_{V}\left(z\right)$$ grows like $$\frac{1}{1-\left|z\right|}$$ as $$z$$ tends radially to the unit circle result in a contradiction against the boundedness of $$\varsigma_{V}\left(z\right)$$‘s power series coefficients)? Or do there exist $$V$$ for which $$T_{V}$$ is infinite?

Part of the problem is that there is so much literature about the boundary behavior and singularities of power series that trying to find information about something this specific is like looking for the needle in the proverbial haystack. Any assistance would be much appreciated.

## Question of convergence of an infinite number of series with special conditions

I am trying to think about the following type of problems, and, actually, if we could get some results, they could be applied inside of mathematics, even to some unsolved problems.

I will not now write about potential applications, but proceed to a direct presentation of a problem.

i am studying (although, with no conclusions yet) the sum $$\sum_{3 \leq a_{1,1},a_{1,2}< + \infty} \frac {1}{a_{1,1}a_{1,2}}+\sum_{3 \leq a_{2,1},a_{2,2},a_{2,3}< + \infty} \frac {1}{a_{2,1}a_{2,2}a_{2,3}}+…+ \sum_{3 \leq a_{n,1},a_{n,2},…,a_{n,n+1}< + \infty} \frac {1}{a_{n,1}a_{n,2},…a_{n,n+1}}+…$$.

where $$\{a_{1,1}a_{1,2}\}$$ and $$\{a_{2,1}a_{2,2}a_{2,3}\}$$ and … and $$\{a_{n,1}a_{n,2},…a_{n,n+1}\}$$ and… are all of density zero in $$\mathbb N$$ and every term of every sequence is odd.

Does all of this implies convergence of this infinite sum of infinite sums?

## Conclude $X^n_t\to X_t$ weakly from knowing that $X^n_0\to X_0$ weakly and convergence of the corresponding generators

Let

• $$E$$ be a locally compact separable metric space
• $$(T(t))_{t\ge0}$$ be a strongly continuous contraction semigroup on $$C_0(E)$$ with generator $$(\mathcal D(A),A)$$
• $$(T_n(t))_{t\ge0}$$ be a uniformly continuous contraction semigroup on $$B(E)$$ (bounded Borel measurable functions $$E\to\mathbb R$$ equipped with the supremum norm) for $$n\in\mathbb N$$
• $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space
• $$(X_t)_{t\ge0}$$ be an $$E$$-valued càdlàg process on $$(\Omega,\mathcal A,\operatorname P)$$ with $$\operatorname E\left[f(X_t)\mid(X_r)_{r\le s}\right]=(T(t-s)f)(X_s)\;\;\;\text{almost surely}\tag1$$ for all $$f\in C_0(E)$$ and $$t\ge s\ge0$$
• $$(X^n_t)_{t\ge0}$$ be an $$E$$-valued càdlàg process on $$(\Omega,\mathcal A,\operatorname P)$$ with $$\operatorname E\left[f(X^n_t)\mid(X^n_r)_{r\le s}\right]=(T_n(t-s)f)(X_s)\;\;\;\text{almost surely}\tag2$$ for all $$f\in C_b(E)$$ and $$t\ge s\ge 0$$

Assume $$X^n_0\xrightarrow{n\to\infty}X_0\tag3$$ weakly and that for all $$f\in\mathcal D(A)$$ and $$t\ge0$$, there is a $$(B_n)_{n\in\mathbb N}\subseteq\mathcal B(E)$$ and a $$(f_n)_{n\in\mathbb N}\subseteq B(E)$$ with $$\operatorname P\left[\forall s\in[0,t]:X^n_s\in B_n\right]\xrightarrow{n\to\infty}1\tag4,$$ $$\sup_{n\in\mathbb N}\left\|f_n\right\|_\infty<\infty\tag5$$ and $$\left\|f_n-f\right\|_{B_n}+\left\|A_nf_n-Af\right\|_{B_n}\xrightarrow{n\to\infty}0\tag6$$ (where $$\left\|g\right\|_{B_n}:=\sup_{x\in B_n}|g(x)|$$).

Fix $$f\in\mathcal D(A)$$ and $$t\ge 0$$. Are we able to conclude $$\operatorname E\left[f(X^n_t)\right]\xrightarrow{n\to\infty}\operatorname E\left[f(X_t)\right]$$?

The claim would follow from $$\operatorname E\left[f(X^n_t)-(T(t)f)(X^n_0)\right]\xrightarrow{n\to\infty}0.\tag7$$ We know that there are $$B_n$$ and $$f_n$$ as above. Let $$U^n:=f_n(X^n)$$ and $$V^n:=(A_nf_n)(X^n)$$. Then we may somehow use that $$U^n-\int_0^{\;\cdot\;}V^n_s\:{\rm d}s$$ is a martingale. Maybe we need to stop this martingale at $$\tau_n:=\inf\left\{t>0:\left(\int_0^t\left|(A_nf_n)(X^n_s)\right|^2\:{\rm d}s\right)^2>\sqrt t(\left\|Af\right\|_\infty+1)\right\},$$ noting that $$\operatorname P\left[\tau_n\le t\right]\xrightarrow{n\to\infty}0$$. The idea is that we may insert this martingale in $$(7)$$ and somehow show the convergence.

Remark: There is a superior result in section 8 of chapter 4 in the book of Ethier and Kurtz, but I would like to find a direct proof in this special situation.