Calculation and analysis: why does Mathica lose some solutions for this integral?

When calculating the integral of two orthogonal sine functions

sin (n Pi x) sin (m Pi x), n, m integers> 0

Mathematica seems to lose the solution where n = m. I don't understand exactly why this is lost. It is possible to find the solution by taking the limit of the expression it gives; In the following example, this is found by taking the limit as n_> 5.

Have I coded something incorrectly?

Thank you.

Sample Code

(* compute the integral of two orthogonal sine functions  *)
a = 
 Integrate(Sin(5 Pi x) Sin(n Pi x), {x, 0, 1}, 
   Assumptions -> {n  (Element) Integers && x (Element) Reals && 
      m  (Element) Integers  && n > 0 && m > 0})
(* Simplify the result *)
B(n_) = FullSimplify(a, 
  Assumptions -> {n  (Element) Integers && x (Element) Reals && 
     m  (Element) Integers  && n > 0 && m > 0})

Out(31)= (5 Sin(n (Pi)))/(25 (Pi) - n^2 (Pi))

Out(32)= 0

(* Apparently, FullSimplify misses the case where n=5.  Zero is correct only when n≠5 *)

(*Taking the appropriate limit of the function before FullSimplify yields the correct result *)

In(30):= Limit((10 Sin(n (Pi)))/(25 (Pi) - n^2 (Pi)), n -> 5)

Out(30)= 1/2

pdes analysis – Dirichlet problem extension solutions

I got caught in the following problem.

Leave $ Delta = left {| z | leq 1 right } subset mathbb {C} $ be the disk drive, and leave $ r $ be a holomorphic function in $ Delta $ which is soft in $ bar { Delta} $.

We are looking for a holomorphic function $ f $ in $ Delta $ for which
$$
1 + | r | ^ 2 = | f | ^ 2 quad text {in the $ partial Delta $} limit.
$$

For this Dirichlet problem, there is a unique solution $ f $ that has no zeros in $ Delta $.

Now suppose that $ r $ it is a rational function in $ mathbb {C} $ (meromorphic function in the sphere $ mathbb {CP} ^ 1 $) We can extend $ f $ be a rational function in $ mathbb {C} $? What if $ r $ Is it a polynomial?

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reference request – Limited heat equation solutions

Consider the solution of the heat equation in $ Omega times (0, T) $ :
$$
partial_tu- Delta u + c (x) u = 0,
u (x, 0) = f $$

With zero Dirichlet B.C. Where $ T> 0 $ fixed and $ Omega $ It is a limited domain of $ mathbb {R} ^ n $. $ f $ It is mild initial condition limited s.t. $ f ge0, ; f en L ^ infty ( Omega) $Y $ c = c (x) in L ^ infty ( Omega) $.

When $ c (x) = 0 $, the solution is limited: $ u in L ^ infty ( Omega times (0, T)) $.

The simple test I know is based on the properties of the fundamental solution and particularly on the initial conditions.

My question is: do we have the same result in the case of $ c (x) in L ^ infty ( Omega) $? Can we get a result like
$$ | u (x, t) | le C, para ; a.e ; (x, t) in mathbb {R} ^ n times (0, T), $$
where $ C $ it is a positive constant depending on $ T, | c | _ infty $ Y $ | f | _ infty $?

Any suggestion or reference that includes some ideas to demonstrate the limitation of the solution of this equation would be useful.

complexity theory – Post-correspondence problem: find the total number of solutions

Leave $ A = {a, b } $ be an alphabet

$ P = A ^ * times A ^ * $

A PCP instance is a non-empty List $ D = (d_1, d_2, …, d_n) in P ^ n $ of word pairs

To a PCP instance $ D in P ^ n $ an index sequence $ (i_1, i_2, …, i_m) in {1, …, n } ^ m, m in mathbb {N ^ +}, $ It is a solution for D, if it has the following property:

for $ (t, b) = d_ {i_1} diamond d_ {i_2} diamond … diamond d_ {i_m}: t = b $

C) $ n = 3: d_1 = (a, aba), d_2 = (ba, a), d_3 = (aba, b) $. Find a solution for this problem.

My answer: for $ (t, b) = d_1 diamond d_2 diamond d_1 diamond d_3 diamond d_2 Rightarrow t = b $

$ a.ba.a.aba.ba \ aba.a.aba.b.a $

re) Now: $ d_1 = (ab, a), d_2 = (ab, ba), d_3 = (ba, b). $ Explain why this PCP instance does not have a solution.

My answer: the first part of the pair always has two letters, while the second only has one letter, except one pair. That means that the first part would always get bigger than the second part, except if only $ d_2 $ since both the first and the second part have two letters each. But you can't start with $ d_2 $ because the first letters don't match, then you should start with $ d_1 $ or $ d_3 $ which means that the first part would be much larger than the second part. Then you would never reach a solution.

Would this answer be enough? Or is there a more formal way to answer the question?

me) What are the possibilities for the total number of solutions for any random PCP instance?

My answer: I tried to think about the different numbers of letters that each domino has for each part. But it doesn't seem right.

F) Find couples $ d_1 = (t_1, b_1), d_2 = (t_2, b_2) $ Y $ d_3 = (t_3, b_3) $(all in P), so that $ t_1, t_2, t_3, b_1, b_2, b_3 $ they are different words $ ($ that means: $ | {t_1, t_2, t_3, b_1, b_2, b_3 } | = 6) $ and so that there is a solution for an instance $ D = (d_1, d_2, d_3) $ of the PCP. Each domino $ d_i $ must be used at least once. Find a solution for $ D $.

My answer: ???

RESIDENTIAL STORAGE SOLUTIONS REQUIRED – Everything else

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I own a beautiful house of more than 2,500 square feet of area near Leo Baeck Day School in Vaughan Road (Toronto). My father bought the house about a decade ago in 2009. Every home improvement I made to this old house has scared me of doing another one. In addition to the usual wiring and plumbing replacements, I plan to repair the floor and make some structural changes in the house. I have rented a small apartment for this short time. I have many things to change and the new place to stay is very congested. I am planning to contact any self-storage provider in Burlington or in a nearby area. What are the procedures to follow regarding this? Let me know if you have any ideas about residential and residential storage solutions.

plot – Miminize with regions for solutions and criteria for other ways to solve equations

I noticed that NMinimize is very, very slow if one has conditions. For example, here is my code:

F(x_, y_, l_, m_, n_) = 
  x*Log(x) + y*Log(y) + (1 - x - y)*Log(1 - x - y) - l*x*y - m*x^2 + 
   n*(x + y)^4;
Mux(x_, y_, l_, m_, n_) = D(F(x, y, l, m, n), x);
Muy(x_, y_, l_, m_, n_) = D(F(x, y, l, m, n), y);
p(x_, y_, l_, m_, n_) = 
  F(x, y, l, m, n) - x*Mux(x, y, l, m, n) - y*Muy(x, y, l, m, n);

sol(x1_, l_, m_, n_) := 
  FindMinimum({(p(x1, y1, l, m, n) - 
        p(x2, y2, l, m, n))^2 + (Mux(x1, y1, l, m, n) - 
        Mux(x2, y2, l, m, n))^2 + (Muy(x1, y1, l, m, n) - 
        Muy(x2, y2, l, m, n))^2, Element(x2, Reals), 
    Element(y2, Reals), 
    Element(y1, Reals)}, {{x2, 0.4}, {y2, 0.4}, {y1, 0.05}});

n0 = 0;
m0 = 10;
l0 = 20;
{Plot({y1 /. sol(x, l0, m0, n0)((2)), x2 /. sol(x, l0, m0, n0)((2)), 
   y2 /. sol(x, l0, m0, n0)((2))}, {x, 0, 0.1}, 
  PlotLegends -> "Expressions"), 
 Plot({sol(x, l0, m0, n0)((1))}, {x, 0, 0.1}, 
  PlotLegends -> "Expressions")}

Then I wondered how to do it faster.

In addition, I wanted to know when it is better to use NMinimize or NSolve or FindRoot. Here I have 3 equations $ p (x_1, y_1) = p (x_2, y_2), mu_x (x_1, y_1) = mu_x (x_2, y_2), mu_y (x_1, y_1) = mu_y (x_2, y_2) $. I find NMinimize quite convenient and efficient in the way I used it. What do you think about it ? How would you approach this system of equations?

Also, I'm looking to make many graphics like that:

Clear(l0)

Table({Plot({y1 /. sol(x, l0, m0, n0)((2)), 
    x2 /. sol(x, l0, m0, n0)((2)), y2 /. sol(x, l0, m0, n0)((2))}, {x,
     0, 0.1}, PlotLegends -> "Expressions"), 
  Plot({sol(x, l0, m0, n0)((1))}, {x, 0, 0.1}, 
   PlotLegends -> "Expressions")}, {l0, {5, 10, 20}})

Do you think it's a good strategy with NMinimize and Table?

Analysis and classic falls – Solutions to holonomic $ D $ modules: when are they squared?

I want to apply the theory of $ D $-modules to solve operator equations of several variables in the Bargmann space
$$ mathcal H: = bigg { psi in mathcal O ^ text {an} _ { mathbb {C} ^ n} , , bigg | , , , int_ { mathbb {C}} d ^ {2n} x , | psi (x) | ^ 2e ^ {- | x | ^ 2} < infty bigg }. $$
I can find many references in $ D $-modules that give good algorithms that solve this type of equations. However, I cannot find good references that consider the convergence properties of these solutions.

I had a naive idea about how to quickly calculate the dimension of the solution space $ text {Ann} (I) $ associated with a given holonomic ideal, and this is based on the correspondence between holonomic $ D $-modules and local systems in $ mathbb {C} ^ n $. In fact, to calculate the behavior of this local system, we would analyze the limiting behavior of the connection form in general $ | x | to infty $:
$$ A_j (x) dx ^ j underset {| x | to infty} { sim} sum_ {| α | = m} x ^ α A_j ^ {(α)} dx ^ j. $$
Here, $ m: = deg A $ Gives the main order contribution to the connection form. When analyzing the spectral properties of the $ A (?) _ J $, this would determine how many solutions are normalizable or not (suppose there are no singular points, for simplicity). However, this analysis fails for some simple examples. My question is, when this analysis fails, why does it fail:

Test case: Kummer equation (WORKS)

Leave $ n = 1 $and consider the main ideal in Weyl algebra generated by $$ l: = x partial ^ 2- (x-b) partial -a. $$
We can always solve the ideal $ I $ When building a local system, in this case we get the following flat connection:
begin {align}
A (x) dx & = frac {1} {x} begin {pmatrix} 0 & a \ a & x + b end {pmatrix} dx underset {| x | to infty} { sim} begin {pmatrix} 0 and 0 \ 0 and 1 end {pmatrix} dx
end {align}

In this case, $ A (?) It has two spaces of its own, one with its own zero value and one with its own value. Then we would expect the following asymptotic:
begin {align}
psi_1 (x) & underset {| x | to infty} { sim} e ^ x cdot ( text {lower order terms}) \
psi_2 (x) & underset {| x | to infty} { sim} e ^ 0 cdot ( text {lower order terms})
end {align}

In any case, both solutions (if they are analytical) would be in the Bargmann space $ mathcal H $. In fact, the solutions to the holonomic ideal are explicitly
begin {align}
psi_1 (x) = , _ 1F_1 (a; b; x), ~~~~~~~ psi_2 (x) = , U (a; b; x)
end {align}

where here $ U $ of note TricomiThe confluent hypergeometric function. In general $ | x | to infty $, Kummer's hypergeometric function actually grows as $ x ^ a-b} e ^ x $, while the Tricomi function grows as $ x ^ – a $ (which is in fact in Bargmann's space, when it is analytical), confirming our intuition from the perspective of local systems.

Test case: Hermite equation (DOES NOT WORK)

Leave $ n = 1 $and consider the main ideal in Weyl algebra generated by $$ partial ^ 2 + x partial + a. $$
In this case we get the following flat connection:
begin {align}
A (x) dx & = begin {pmatrix} 0 & -a \ 1 & -x end {pmatrix} dx underset {| x | to infty} { sim} -x bigg ( begin {pmatrix} 0 & 0 \ 0 & 1 end {pmatrix} + O bigg ( frac {1} {x} bigg) bigg ) dx
end {align}

In this case, $ A (?) It has two spaces of its own, one with zero own value and one with its own value $ x $. Then we would expect the following asymptotic:
begin {align}
psi_1 (x) & underset {| x | to infty} { sim} e ^ frac {-x ^ 2} {2} cdot ( text {lower order terms}) \
psi_2 (x) & underset {| x | to infty} { sim} e ^ 0 cdot ( text {lower order terms})
end {align}

In this case, one of the solutions is not normalizable, while the other should be normalizable. Then the solution space must be one-dimensional. In short, based on the theory of $ D $-modules, we have two analytical solutions (such as $ text {Sing} (I) = 0 $), and we must always lose a solution when moving into Bargmann's space:
$$ dim text {Ann} (I) = 2, ~~~~~~~ dim text {Ann} (I) cap mathcal H = 1. $$

However, the solutions to the holonomic ideal are explicitly
begin {align}
psi_1 (x) = , _ 1F_1 (a / 4; 1/2; x ^ 2), ~~~~~~~ psi_2 (x) = x , _ 1F_1 (a / 4 + 1/2 ; 3/2; x ^ 2)
end {align}

In fact, both functions have the asymptotic type $ O (e x ^ 2) $ and they are not normalizable, violating our intuition that was based on mapping to a local system. So, in fact, we lose both of them solutions:
$$ dim text {Ann} (I) = 2, ~~~~~~~ dim text {Ann} (I) cap mathcal H = 0. $$

Is this problem well known and what solutions exist?

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Entry:

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  • Total source and sump flow match

  • A concave cost function that provides a cost per edge in terms of flow capacity

Expected Performance:

  • The minimum cost graph (not necessarily completely connected) that connects all sources to sinks and allows the required flow, without creating additional nodes.

Therefore, the network topology is not specified and should be optimized.

Thank you.

Discrete mathematics: total number of whole solutions with restrictions

Find the number of ways in which 5 dice can be rolled to get a sum of 25.

In solving this question, the way we solve it is $ x_1 + x_2 + x_3 + x_4 + x_5 $ $ = $ 25 where $ 1 <= x_i <= 6 $

Then we replace $ x_i $ by $ y_i = 6-x_i $ , which $ x_i = 6-y_i $

replacing $ x_i $ In the previous equation we obtain it as → $ (6 * 5) – (y_1 + y_2 + y_3 + y_4 + y_5) $ $ = $ 25

$ (y_1 + y_2 + y_3 + y_4 + y_5) $ = $ 5 $

After solving this equation by the formula of whole solutions $ (n-r + 1)! / (n! * (r-1)!) $ we get the ans as → $ 126 $


Now consider this problem,

The number of non-negative integer solutions such that $ x_1 + x_2 + x_3 = $ 17 where $ x_1> 1, x_2> 2, x_3> 3 $ is ___________________

By solving this, we are solving it as → $ y_1 = x_1-2 $ , $ y_2 = x_2 -3 $ , $ y_3 = x_3-4 $

so, $ x_1 = y_1 + 2 $ , $ x_2 = y_2 + 3 $ , $ x_3 = y_3 + 4 $

Now we substitute this in our original equation to get →

$ y_1 + 2 + y_2 + 3 + y_3 + 4 = $ 17

$ y_1 + y_2 + y_3 = 8 $

and after solving this we get the ans as $ 45 $


Now i have a $ DOUBT $ here, in the second problem since when $ x_1> 1 $ we do it like $ x_1 = y_1 + 2 $ , but in the first problem all the dice must have value $> 0 $ So why in that case we haven't done $ x_i = y_i + 1 $ for all cases?


And besides, what would happen if the question were

$ x_1 + x_2 + x_3 = 12 $ , $ 2 <= x <= $ 5 so how to solve this using an entire solution and applying the formula $ (n-r + 1)! / (n! * (r-1)!) $ ?