Kopar at Newton is a residential development on Kampong Java Road in the 9th district of Singapore. It is located in an established residential neighborhood and has easy access to Newton MRT and major roads such as Dunearn Road and CTE. A prestigious school such as the Anglo Chinese Junior School is also nearby. It is just minutes from Velocity @ Novena and United Square. Not to mention that the shopping belt Orchard Road is only a few stops away.
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Tag: Newton
acegen – Newton iteration method without multiplier in AceFem
Is there a way in AceFem to use a Newton iteration scheme without using the multiplier λ or the time step t? For example, when trying to calculate a solution for a nonlinear equation, no multiplier is needed.
For the problem I'm facing, I try to use AceFem to compute the geometry of a surface composed of surface patches. Each surface field has a finite element that has three degrees of freedom in each node. These degrees of freedom are twist vectors, and with their variation, the shapes of patches can be changed. To determine the values of twist vectors, I defined the energy potential in the finite element, which must be minimized. This potential is composed of the mean curvature in combination with the Gaussian curvature (2H ^ 22K). Node positions and normal vectors in nodes are always fixed, so no constraints are defined in the analysis (even twist vectors can not be defined as constraints because they are all unknown in each node).
Any help is welcome. Thanks. Tomo
Arithmetic Geometry – Eigenform families mean a lower bound of the Newton polygon of $ U_p $
To let $ p $ be arbitrary and $ k in Z $, and $ f $ be a classic eigenform of the level $ Np $ and weight $ k $, then that's known $ f $ lie in one $ p $adische family of eigen forms.
So the question is why the above result implies that there should be a lower bound on the Newtonian polygon $ U_p $ interact with the space of modular levels $ Np $ and weight $ k $and this barrier should be uniform in $ k $,
Thank you so much!
Pullman residences in Newton
The Pullman Residences is an upcoming residential development on Dunearn Road 18 near Newton Road. This condominium is predominantly surrounded by country houses and highrise condos. Many renowned schools are within a radius of 2 kilometers. Residents will also find it convenient as the Newton MRT Interchange is just 150 meters from the settlement.
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Cheap or expensive brushes / the art suppliers
+ Answer on thread

Cheap or expensive brushes / the art suppliers
In terms of quality and price, the best watercolor brushes range from those in a sealed bargain package of perhaps five hanging on the aisle of your national hobby chain store, to those that can boast a royal seal of approval, such as "On Their Majesty the Queen". As you know, Queen Victoria at Winsor & Newton ordered what would be her favorite size 7 in the Kolinsky Series 7 Squadrar Round.) The previous election could briefly serve to clean a child's boots in your mud room, which could be proud of yours In between, you will find a wide selection of more than just serviceable brushes from recognized fine brands.Some of these brands are: Winsor & Newton Series 7, Isabey, Rafael, Arches, Escoda, Pro Arte and others.Smaller, but always Winsor & Newton Series 666, Princeton and Grumbacher: I recommend buying a good M arke. Avoid these bargains from several; You will do so poorly in practice and in durability that you will be very discouraged. I have mentioned some of each brand and enjoy every brush. And yes, I've bought cheap brushes in the past … and had to throw them away.
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What do you think about most legends were male … Michael Jackson Mike Tyson Albert Einstein Bruce Lee Michael Jordan Isaac Newton.?
For every good male legend in the arts there are also good female legends.
Likewise with the sport, rather with the gymnastics.
And in science, there was probably only one woman whose talents were equal to the two most prized male scientists (Newton and Einstein), and that's Hypatia.
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Algebraic Number Theory – On the Newton Polygon for the Laurent Series
I understand what the Newton polygon should be for a Laurent series. I read & # 39; & # 39; Introducing Dwork's G feature # and devoting only three pages to Newton polygons for the Laurent series: He says what a Laurent series is, as we call it, as the sum of two powers & # 39; can disassemble and treat only the case if we have an annulus of convergence. The purpose of this section is to give a sketch of the proof of the generalized Weierstrass preparation theorem (a version of this sentence, but for Laurent series). I fill in the gaps, complete things and add information. I have everything, even the proposition, whose proof in a sense resembles the case of the power series; But what I do not understand is how the Newton polygon should be. Here is my attempt:
Let us remember that a Laurent series is a series of the form
$$ f (X) = sum_ {n = – infty} ^ { infty} a_ {n} X ^ {n}, quad a_ {n} in mathbb {Q} _ {p} $$
($ v $ is the additive rating on $ mathbb {Q} _ {p} $)
We can write $ f (X) = f ^ {+} (X) + f ^ {} (1 / X) $ from where
$$ f ^ {+} (X) = sum_ {n = 0} ^ { infty} a_ {n} X ^ {n} quad text {and} quad f ^ {} (X) = sum_ {n = 1} ^ { infty} a _ { n} X ^ {n} text {.} $$
We define the Newton polygon of $ f $ as in the case of power series. This is his "definition". So I tried to understand what it means:
So the Newton polygon of $ f $ is the convex hull of the set of points $ (j, v (a_ {j})) $, In the performance series if $ a_ {j} = 0 $ We think the point $ (j, v (a_ {j})) $ as a point on infinity in the upper halfplane, but now the problem is the points, though $ n <0 $, Let's take these examples:
 Consider $ sum_ {n in mathbb {Z}} pX ^ {n} $, we have $ (j, v (a_ {j})) = (j, 0) $ for each $ n in mathbb {Z} $I conclude that the Newton polygon must be all xaxes.
Recall that the number of points in power series is on or above the Newton polygon. This question comes from the next example:
 Consider $ f = sum_ {n = – infty} ^ { infty} p ^ {n} X ^ {2 ^ {n}} $, Here it is more complicated. I tried to get the Newton polygon from $ f $ from the Newton polygons of $ f ^ {+} $ and $ f ^ {} $, The Newton polygon of $ f ^ {+} $ is the positive $ x $Axis. But what happens to the Newton polygon of $ f ^ {} $? The points $ (j, v (a_ {j})) $ With $ j <0 $ are on the negative $ x $Axis and if $ n $ is no power of $ 2 $ the spot $ (j, v (a_ {j})) $ the point seems to be at minus infinity. I think that the Newton polygon of $ f ^ {} $ must be the negative part of $ x $Axis, leaving the Newton polygon of $ f $ has to be everything $ x $Axis.
I start with known power series. These examples are power series in my examples, and the Newton polygon of them has a limited number of pages. In the Power series, I start with an infinite vertical line, then construct the Newton polygon, but in the Laurent series, I do not have a vertical line (we're only dealing with the convergent Laurent series). But what happens when we start from a power series with infinite sides and try to expand $ n in mathbb {Z} $? For example, the Newton polygon of the logarithm is well known, and considering "what" the Newton polygon should be $ f ^ {} $ I have to think the following:
I can define the Newton polygon of a Laurent series $ f $ through the following:
To let $ C (f ^ {+}) $ the convex hull of the point set $ (j, v (a_ {j})) $ from where $ j geq 0 $ and let it go $ C (f ^ {}) $ the convex hull of the point set $ (j, v (a_ {j})) $ With $ j <0 $, Then the Newton polygon of $ f $ is the convex hull of $ C (f ^ {+}) cup C (f ^ {}) $, If $ a_ {j} = 0 $ to the $ j geq 0 $ we ask $ (j, v (a_ {j})) = + infty $ and if $ a_ {j} = 0 $ to the $ j <0 $ we ask $ (j, v (a_ {j})) = – infty $, If $ n <0 $ the amount of points $ (j, v (a_ {j})) $ are on or under $ C (f ^ {}) $, Explicitly for $ C (f ^ {}) $ We start with the positive $ yaxis $ and we turn it clockwise and follow the rules of the process $ f ^ {} $ but in that sense I think so $ C (f ^ {} (X)) = C (f ^ {+} ( X)) $,
Is this "definition" valid? I've searched for books that have this material, but I only have "one course in $ p $– Adic Analysis "by Alain Robert This is the only reference I have in English In French I have" Les nombres padiques "by Yvette Amice, where the Newton polygon of the Laurent series becomes a graph of a particular function and defines then proves that this is equivalent to: the Newton polygon of $ f $ is the limit of the convex upper envelope of the point set $ (j, v (a_ {j})) $, Then there are some examples of Newton polygons, but the examples are polynomials or power series. She never looks at Laurent series.
I really appreciate any reference, discussion, if my definition is valid, anything. As I said, I fill in the gaps and other things, but I also want to have examples of what I say.
Thank you, forgive me also, if I have misprints, I'm from Mexico and I'm not an expert on English. Best regards
Interpolation – How to generate an interpolation polynomial with the Newton formula for the exponential function?
I'm trying to find an interpolating polynomial from $ f (x) = e ^ {3x} $ and interpolation nodes $ x_0 = x_1 = x_2 = 0 and x_3 = x_4 = 1 $ However, with Newton's formula using Wolfram Alpha, I am stuck in the division of zero and try to find split differences:
Do[Do[roots[i,j] = (Roots[i  1, j + 1]  Root[i  1, j]) / (Node[i + j]  knot[j]), {j, 0, n  i}], {i, 1, n}];
Can you tell me what I'm doing wrong? Maybe my formula is wrong or I misunderstand something. I have to admit that I'm new to Numerical Analysis and Wolfram Mathematica.
c ++ – Complex Newton Method
I'm trying to build a complex Newton method. I've submitted a PR here, where you can see the documentation and tests, and get an example to compile. I would be glad if you help to reduce the work of the supervisor.
This uses the idea that if x_{n} converges to x *, then f (x *) = 0 if f * is continuous at x *. When f & # 39; (x_{n}) = 0 At some point she uses Muller's method.
#ifndef BOOST_NO_CXX11_AUTO_DECLARATIONS
/ *
* Why do we set the default maximum number of iterations to the number of digits in the type?
* For double roots, the number of digits increases linearly with the number of iterations.
* Therefore, this default setting should regain full accuracy even in this somewhat pathological case.
* For isolated roots, the problem converges so fast that it does not matter at all.
* /
template
Complex complex_newton (F g, complex conjecture, int max_iterations = std :: numeric_limits:: numbers)
{
typedef typename Complex :: value_type Real;
with std :: norm;
with std :: abs;
with std :: max;
// z0, z1 and z2 can not be the same if we have to use Muller's method immediately:
Complex z0 = conjecture + complex (1,0);
Complex z1 = conjecture + complex (0,1);
Complex z2 = conjecture;
do {
Autopair = g (z2);
if (norm (pair.second) == 0)
{
// Muller's method. Notation follows Numerical Recipes, 9.5.2:
Complex q = (z2  z1) / (z1  z0);
auto P0 = g (z0);
car P1 = g (z1);
Complex qp1 = static_cast(1) + q;
Complex A = q * (pair first  qp1 * P1 first + q * P0 first);
Complex B = (static_cast.)(2) * q + static_cast(1)) * pair.irst  qp1 * qp1 * P1.first + q * q * P0.first;
Complex C = qp1 * pair.
Complex rad = sqrt (B * B  static_cast.)(4) * A * C);
Complex denomination1 = B + rad;
Complex denom2 = B  rad;
Complex correction = (z1z2) * static_cast(2) * C;
if (norm (denom1)> norm (denom2))
{
Correction / = denom1;
}
otherwise
{
Correction / = denom2;
}
z0 = z1;
z1 = z2;
z2 = z2 + correction;
}
otherwise
{
z0 = z1;
z1 = z2;
z2 = z2  (pair first / second pair);
}
// See: https://math.stackexchange.com/questions/3017766/constructingnewtoniterationconvergingtononroot
// If f & # 39; is continuous, then the convergence of x_n > x * f (x *) = 0 means.
// This condition approximates this convergence condition by requiring three consecutive iterations to be grouped.
Real tol = max (abs (z2)) * std :: numeric_limits:: epsilon (), std :: numeric_limits::Epsilon());
bool real_close = abs (z0.real ()  z1.real ()) <tol && abs (z0.real ()  z2.real ()) <tol && abs (z1.real ()  z2.real () ) <tol;
bool imag_close = abs (z0.imag ()  z1.imag ()) <tol && abs (z0.imag ()  z2.imag ()) <tol && abs (z1.imag ()  z2.imag () ) <tol;
if (real_close && imag_close)
{
return z2;
}
} while (max_iterations);
// The idea is, if we can get abs (f) <eps, we should, but if we go through all those iterations
// and abs (f) <sqrt (eps), then the roundoff error just does not allow us to evaluate f to <eps
// This is a bit cumbersome because it is not scale invariant but uses the Daubechies coefficient example code.
// I have found that this condition produces correct roots while the scale invariant condition described here holds:
// https://scicomp.stackexchange.com/questions/30597/definingaconditionnumberandterminationcriteriafornewtonsmethod
// Allowing nonroots as roots.
Autopair = g (z2);
if (abs (pair.first) <sqrt (std :: numeric_limits)::Epsilon()))
{
return z2;
}
return {std :: numeric_limits:: quiet_NaN (),
std :: numeric_limits:: quiet_NaN ()};
}
#endif
Usage:
boost :: math :: tools :: polynomial <std :: complex> p {{1,0}, {0, 0}, {1,0}};
std :: complex esteem {1,1};
boost :: math :: tools :: polynomial <std :: complex> p_prime = p.prime ();
car f = [&](std :: complex z) {return std :: make_pair <std :: complex, std :: complex> (p (z), p_prime (z)); };
std :: complex root = complex_newton (f, rate);
A more intensive example application can be found here
Differential Equations – Draw Gradient and Newton Directions for the Parametric System of Nonlinear ODEs
I have a bucket (system) of chemical kinetic models, a nonlinear dynamical system, given by:
Kf> = 0 and kr> = 0 are the parameters. The initial conditions are A (0) = B (0) = 1 and C (0) = 0. I generated data according to y1 = C (0,5) + noise and y2 = C (2) + noise, where the Noise is normally distributed with mu = 0 and sigma = 0.1 using kf = 0.1 and kr = 2.
odes = {A & # 39;
B & # 39;
C1 & # 39;
C1[0] == 0};
odesData = odes /. {kf > 0,1, kr > 2};
Solution = NDSolve[odesData, {A, B, C1}, {t, 0, 5}][[1]];
seedRandom[10]
y1 = C1[0.5] + RandomVariate[NormalDistribution[0, 0.1]]/. solution
seedRandom[30]
y2 = C1[2] + RandomVariate[NormalDistribution[0, 0.1]]/. solution
Data = {{0.5, y1}, {2, y2}};
I have to visualize the data with an input / output image, a parameter space image and a data space image. For the parameter space and data space images, I also have to plot the gradient and newton directions for several randomly selected parameter values.
With ParametricNDSolve I can get an input / output image.
kfmax = 1;
krmax = 5;
number = 100;
kfrange = range[0, kfmax, kfmax/numsteps];
krrange = range[0, krmax, krmax/numsteps];
soln = ParametricNDSolve[odes, {A, B, C1}, {t, 0, 5}, {kf, kr}];
eqns = Rate[Table[C1[kf, kr]
feweqns = flattening[eqns][[ ;; ;; 1000]];
eqnsplot = plot[feweqns, {t, 0, 2.5}, PlotRange > All]
bestfitplot = plot[model[kf, kr]
AxesLabel > {"t", "C
Using ParametricNDSolveValue and FindFit, I can find the best fitting parameters for the model (line is with fit parameters and points are generated data).
model = ParametricNDSolveValue[odes, C1, {t, 0, 5}, {kf, kr}]
fit = FindFit[data, {model[kf, kr]
C1 & # 39;
bestfitplot = plot[model[kf, kr]
validptplot = ListPlot[{{0.5, y1}, {2, y2}}]; (* the data you are trying to match *)
show[bestfitplot, validptplot]
I can also visualize the parameter space using the log of the cost function:
list1 = eqns /. t > 0.5;
list2 = eqns /. t > 2;
Cost = (list1  y1) ^ 2 + (list2  y2) ^ 2; Costs // MatrixForm; (* kfth row and krth column *)
parspaceplot = ListContourPlot[Log[cost], PlotLegends > Automatic, DataRange > {{0, krmax}, {0, kfmax}}, FrameLabel > {"kr", "kf"}, Contours > 50];
bestfitptplot = ListPlot[{{kr, kf}} /. fit];
parspace = show[parspaceplot, bestfitptplot]
as well as the data space image
max = 2;
dy1 = 0.2; dy2 = dy1;
modely1 = MapThread[model[#1, #2][0.5] &, Table[{i, j}, {i, 0, max, dy1}, {j, 0, max, dy2}][[#]][Transpose]]& /@ Offer[max/dy1];
modely2 = MapThread[model[#1, #2][2] &, Table[{i, j}, {i, 0, max, dy1}, {j, 0, max, dy2}][[#]][Transpose]]& /@ Offer[max/dy1];
plot1 = ListPlot[{Modely1[{Modely1[{modely1[{modely1[[#]]modely2[[#]]}[Transpose] & /@ Offer[max/dy1], Joined > True, PlotStyle > Blue, AxesLabel > {"y1", "y2"}];
plot2 = ListPlot[{Modely1[{Modely1[{modely1[{modely1[Transpose][[#]]modely2 [Transpose][[#]]}[Transpose] & /@ Offer[max/dy1], Joined > True, PlotStyle > Red];
plot3 = ListPlot[{{{y1, y2}}, {{model[kf, kr][0.5] /. fit, model[kf, kr][2] /. fit}}}, PlotStyle > {{Black, dot, point size[0.025]}, {Green}}];
show[plot1, plot2, plot3]
However, if I try to compute the directions of the gradient (j # 39) and Newton ((j # j) .x ==degrees for x), where j === jacobian, j & # 39; j === Fischer information matrix, r === residuals and j & # 39; denotes the transposition of j, I do not understand what I think I should get. That is, the slope direction is not perpendicular to the contour lines.
seedRandom[]
Randmax = 10;
Randoms = table[{RandomReal[{0, krmax}], RandomReal[{0, kfmax}]}, {i, randmax}];
r = {y1, y2}  (model[kf, kr][#] & / @ {0.5, 2});
j =  (degree[model[kf, kr][#]{kf, kr}]{0.5, 2} /0.1;
degree = j [Transpose].r;
Fish = j [Transpose].j /. Random[[#, 1]], kf > Random[[#, 2]]} & /@ Offer[randmax];
grads = grad /. Random[[#, 1]], kf > Random[[#, 2]]} & /@Offer[randmax];
Newts = LinearSolve[Fish[fish[Fisch[fish[[#]]grads[[#]]]& /@ Offer[randmax];
gradarrows = graphics[{Black arrow
Random[{BlackArrow[{Randoms[{BlackArrow[{randoms[[#]]Random[[#]]+ Normalize[grads[grads[grads[grads[[#]]]/ 2.5}]& /@ Offer[randmax]}];
newtarrows = graphics[{Blue arrow
Random[{BlueArrow[{Randoms[{BlueArrow[{randoms[[#]]Random[[#]]+ Normalize[newts[newts[Molche[newts[[#]]]/ 10}]& /@Offer[randmax]}];
show[parspace, gradarrows, newtarrows]
Things still do not look good, even if I only have costs for log[cost],
This is a homework assignment for a degree in Predictive Modeling. A Jupyter notebook is provided for the class, and I know how to numerically calculate the Jacobian (derive sensitivity equations and solve the 9 equations simultaneously with odeint), but I've invested enough in that code to get it to the end to see. I could do the same with Mathematics NDSolve with the 9 equations, but it seemed like I should be able to get the Jacobi with the parameter equation generated by ParametricNDSolve (similar to the parameter sensitivity section in the ParametricNDSolve reference page).
Any suggestions on how I can get the Gradient and Newton directions? (Both math and coding recommendations are welcome).
P.S. This is my first post and for the first time I could not solve things with what I could find online and in this forum, which was very helpful!