Find the coefficient of x ^ k in (x + a) (x + b) … (x + n), where a, b and n are integers.

I am unable to tackle this problem

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# Tag: coefficient

## expansion coefficient

## statistics – Weighted correlation coefficient

## Reference Request – Local Coefficient Systems in Cohomology

## Complexity of the calculation of $ f ^ {(n)} (0) $ / Extracting a coefficient of a function series

## non-linear – Non-constant coefficient in the heat equation

## st.statistics – Maximize a regression coefficient: $ frac {a ^ dagger x} {x ^ dagger Mx} $ for $ x in mathbb {Z} ^ n $

## Break with binomial coefficient – math stack exchange

## ordinary differential equations – remove expressions from the coefficient of the first derivative term of an ODE

## real analysis – uniqueness for autonomous ODE with discontinuous coefficient

## Performance Optimization – Generate a coefficient array from the general formula of the linear equation system

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Find the coefficient of x ^ k in (x + a) (x + b) … (x + n), where a, b and n are integers.

I am unable to tackle this problem

Because you only compare the candidates to the average sample of each class, you lose information about the distribution of each class. You try to compensate for this by assigning a different meaning to different aspects of the signal. However, a good model should take this uncertainty directly into account.

As a very simple improvement, you can compare a candidate with multiple samples of the class and then combine the correlations. The more samples you take, the more accurate the correlation estimate will be, but the longer the calculation will take. With such a technique, you can even determine the uncertainty of your correlation estimate by bootstrapping.

You can also use a different metric to show the class affiliation, and this metric can use more information. Your classification problem can be interpreted as a model selection problem. Each class represents a model and you want to choose the model that best explains the candidate sample. With the Bayesian approach, you would choose the model that offers the best *proofs*,

While you are currently describing the model by the mean of the distribution of your training samples, you can expand the model, such as: to include the variance as well. Unimportant features have a high variance within a class and are therefore automatically classified as less important. This is pretty easy if you can assume that the features have a normal / Gaussian distribution. If you manually assign weights to features, this can be useful to speed up the algorithm. However, this is not necessary or fundamental.

Before you start creating complicated models, consider whether a learning approach could work that does not require a model to be abstracted from the data. For example, a KNN algorithm could perform well: select the option *k* Next is the training pattern (with a distance metric in the feature space that could be your correlation metric), and then assign the candidate to the class to which most of its neighbors belong.

However, this is more of a statistical and data science problem than a software development problem. From the point of view of software development, I found it useful to create an interface that all algorithms have to implement in order to easily exchange algorithms for testing and comparison. This also makes it pretty easy to create a reinforced classifier that combines the output of several simpler classifiers.

To let $ mathcal F $ to be a locally constant sheaf with values in it $ mathbb $ Say on a sufficiently beautiful place a compact distributor. The etale room of $ mathcal F $ defines a cover $ p: tilde X to X $,

Is there a connection between (Cech) cohomology? $ H ^ * (X, mathcal F) $ and the (singular) cohomology $ H ^ * ( tilde X, mathbb C) $?

Many combinatorial problems can be solved using generation functions.

In such a case, we get a function $ f (x) $that (usually) has a Taylor expansion:

$$

f (x) = sum_ {n ge 0} a_n x ^ n

$$

Thus the coefficients $ a_n $ are exactly the elements of the sequence we are looking for.

Furthermore, we can assume that the function $ f (x) $ is given in closed form, d. H. as a finite function constructed using only elementary functions.

Unlike the determination of Taylor expansion, we now want a single coefficient $ a_k $fixed for some $ k $,

What is the most efficient way to do this?

addendum:

I've already asked the question on math.stackexchange, where I got an answer to which I am bound $ mathcal O (log_2 (n) ^ alpha) $ is possible. However, I can not figure out from the answer how this limit is established.

I have to solve the following heat equation over a cylindrical domain.

In cylindrical coordinates, the PDE equation is:

```
pde = 1 / r D[k*r*D[T[t, r, z], {r, 1}], {r, 1}]+
D[k*D[T[t, r, z], {z, 1}], {z, 1}]- rho * c * D[T[t, r, z], {t}]
```

The problem is that the coefficients k (conductivity), rho (material density) and c (heat capacity) are generally all temperature-dependent quantities.

After the FEM tutorial, it should be possible to solve such equations if the PDE is specified in the coefficient form

In addition, some articles mention that the inactive form of the PDE is needed to make the work easier for the FEM solver …

My question is: What is the appropriate form of the PDE mentioned above?

Do I have to formulate it first in Cartesian coordinates?

Incidentally, I use Mathematica version 12.

Request for help and thanks in advance. Greetings from Germany!

Are there any nice accelerations / domain cleanups for the following maximization problem? To the $ mathbf {M} in mathbb {R} ^ {d times d} $ positive-definite and $ mathbf {a} in mathbb {R} ^ d $:

begin {align}

text {maximize} & frac { mathbf {a} ^ dagger mathbf {x}} { mathbf {x} ^ dagger mathbf {m} mathbf {x}} \

text {subject to} & mathbf {x} in mathbb {Z} ^ d, \

& mathbf {x} neq 0

end

Since then I am satisfied with brute force in practice $ d about $ 5 and $ mathbf {a}, mathbf {M} $ are well conditioned.

Ideally, I would like to refer to an already existing problem with the next grid vector, but I can not figure out the details.

I want to prove that $ binom {a / m} {k} = frac {b} {m ^ n} $ With $ b in mathbb {Z}, n in mathbb {N} $, I thought of proving it by induction, but could not transform myself $ binom {a / m} {k + 1} $, Another idea was to use the binomial series as $ sum_ {k = 0} ^ infty binom {a / m} {k} x ^ k = left ( sum_ {k = 0} ^ infty binom {a} {k} x ^ k right) ^ frac {1} {m} $ but I am not sure how to proceed at this point.

In a second-order ordinary differential equation governing a function $ S ( theta) $containing the term $ S & # 39; ( theta) $ is

$ cot theta left (1 tan theta ~ frac {f # ( theta)} {f ( theta)} right) S & # 39; ( theta) $.

from where $ f (θ) = 1-c ^ 2 cos ^ 2 theta $. $ c in mathbb {C} $,

Q.: How to find a transformation? $ S rightarrow Q $ so that the above expression becomes

$ cot theta ~ Q # ( theta) + V ( theta) Q ( theta) $.

from where $ V ( theta) $ is a function of the independent variable $ theta $?

Consider the following scalar ODE: for all initial data $ x (0) =: x_0 in (a, b) subset mathbb {R} _ + $.

begin {equation}

dot {x} (t) =

begin {cases}

1-x, & text {if} x

end {cases} \

end {equation}

from where $ c in (a, b) $, We search for $ textbf {Caratheodorys Solution} $ for all $ t geq 0 $,

For the existence:

i): If $ x (0) in (a, c) $: I found a solution

$ x (t) = 1- (1-x_0) e ^ {- t} $ to the $ t < tilde {t} $ and $ x (t) = c $ to the $ t geq tilde {t} $ from where $ tilde {t} $ is the time $ x (t) $ reached $ c $ and notice that $ x (t) $ is strictly increasing $ (a, c) $;

ii): If $ x (0) = 0 $is a solution $ X (t) = 0 $ for all $ t geq 0 $;

iii): if $ x (0) in (c, b) $is a solution

$ x (t) = x_0e ^ {- t} $ to the $ t leq has {t} $ and $ x (t) = c $ for all $ t geq has {t} $, from where $ has {t} $ is the time $ x (t) $ reachese $ c $, Note that $ x (t) $ is strictly decreasing while it is in $ (c, b) $,

My main concern, however, is the $ textbf {uniqueness} $ the Caratheodory solution for this ODE. Are these solutions unique? If not, can we impose mild conditions to restore uniqueness?

This is a problem that occurs when implementing the finite difference method (FDM).

Here is a toy example. Suppose we want to solve the boundary value problem (BVP).

$$ y & # 39; & # 39; (x) = sin (x), y (0) = 0, y (1) = 0 $$

```
eq = y & # 39; & # 39;[x] == Sin[x];
bc = {y[0] == 0, y[1] == 0};
```

With FDM we can then program as follows. First write down the corresponding difference formula:

```
Clear @ dx;
generalformula = (y[x + dx] - 2 years[x] + y[x - dx]) / dx ^ 2 == sin[x];
```

Then define the grid size and generate difference equations at each grid point:

```
Points = 25; domain = {0, 1};
dx = @@ domain / subtract (1 - points);
var = table[y[y[y[y[x], {x, ##, dx}]& @@ domain;
differenceeq = table[Generalformula{xDomäne[[Generalformula{xdomain[[allgemeineFormel{xDomäne[[generalformula{xdomain[[1]]+ dx, domain[[-1]]- dx, dx}]~ Join ~ bc;
```

Finally, solve the difference equations with z. `NSolve`

:

```
Solrule = NSolve[differenceeq, var][[[[[1]]; // Absolute timing
(* {0.00462427, zero} *)
```

Incidentally, a comparison with the analytical solution:

```
asol = DSolveValue[{eq, bc}, y, x]
plot[asol[asol[asol[asol[x], {x, ##}, PlotRange -> All]~ Show ~
ListPlot[solrule[[All, -1]], DataRange -> {##}]& @@ domain
```

Alternatively, we can extract the coefficient array `Differenzäq`

and solve it `LinearSolve`

as in the document of `CoefficientArrays`

, This approach is faster than the one used `NSolve`

:

```
{barray, marray} = CoefficientArrays[differenceeq, var]; // Absolute timing
(* {0,000614324, zero} *)
sol = LinearSolve[marray // N, -barray]; // Absolute timing
(* {0,000488251, zero} *)
```

However, this is not the end. As already mentioned, we have generated a difference equation **at every grid point** and then **extract** the coefficient array from the symbol system in the code above. This procedure can be quite memory intensive and slow if `Points`

is big. Is it possible to generate `Barray`

and `Marray`

directly from `general formula`

and `bc`

**programmatically**in a memorable and not so slow way?

If the above toy example is too simple to illustrate the underlying problem, the following is more complicated to solve the problem mentioned here:

```
r = 0.8; Glass = (1.45) 2; Air = 1; k = (2π) / 1.55;
n2 = (function[{x, y}, #] & @ (Simplify`PWToUnitStep @
PiecewiseExpand @ if[X^2+y^2[X^2+y^2[x^2+y^2[x^2+y^2<= r^2, glass, air]));
ClearAll[fw, bw, delta]
SetAttributes[#, HoldAll] & /@ {fw, bw};
fw@D[expr_, x_] :=
With[{Δ = delta@ToString@x},
Subtract @@ (expr /. {{x -> x + Δ}, {x -> x}}) / Δ]bw @ D[expr_, x_] : =
With[{Δ = delta@ToString@x},
Subtract @@ (expr /. {{x -> x}, {x -> x - Δ}})/Δ]
delta[a_ + b_] : = Delta @ a + Delta @ b
delta[k_. delta[_]]: = 0
degree = function[expr, fw@D[expr, #] & / @ {x, y}];
div = function[expr, Total@MapThread[bw@D@## &, {expr, {x, y}}]];
elst = e[#][x, y] & /@ Offer[2];
discretelhs =
With[{N2=n2[{N2=n2[{n2=n2[{n2=n2[x, y]},
div @ grad @ elst + (n2 k ^ 2) elst - grad[div@elst - 1/n2 div[n2 elst]]];
Points = 100; L = 2; domain = {-L, L}; grid = array[# &, points, domain];
del = #[[2 ;; -2]]&;
delta["x"] = Delta["y"] = -Subtract @@ domain / (points - 1);
vare = Outer[e[e[e[e[#][#2, #3] &, Area @ 2, del @ grid, del @ grid, 1]// flattening;
discretelhslst =
Flattening when transposing[
Block[{e}, e[i_][L | -L, _] = e[i_][_, L | -L] = 0;
table[discretelhs, {x, del@grid}, {y, del@grid}]], {2, 3, 1}]; // Absolute timing
{barray2, marray2} = CoefficientArrays[discretelhslst // N, vare // N]; // Absolute timing
{val, vec} =
Eigensystem[marray2, -6,
Method -> {"Arnoldi", "Shift" -> k^2 glass}]; // Absolute timing
mat = ArrayReshape[#, {2, points - 2, points - 2}] & / @ vec;
MapThread[ArrayPlot[#[[ArrayPlot[#[[ArrayPlot[#[[ArrayPlot[#[[1]], PlotLabel -> Sqrt @ # 2, PlotRange -> All,
ColorFunction -> "AvocadoColors"]&, {mat, val}]
```

```
ByteCount[marray2]/ 1024 ^ 2. "MB"
(* 2.76122 "MB" *)
ByteCount[discretelhslst]/ 1024 ^ 2. "MB"
(* 180,695 "MB" *)
```

The corresponding problem is how to arrive `marray2`

from `discreet`

and `e[i_][L | -L, _] = e[i_][_, L | -L] = 0;`

without generating `discreet`

efficient enough?

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