## Differential Equation Solution By Power Series

Solve $$(1 + x)y’ = py; y(0) = 1$$, where $$p$$ is an arbitrary constant.

First I plugged in the guess $$y = sum_{n = 0}^infty a_n x^n$$:

$$(1 + x)(sum_{n = 0}^infty a_n x^n)’ = psum_{n = 0}^infty a_n x^n$$

Then I expanded the derivative and multiplication:

$$sum_{n = 0}^infty n a_n x^{n – 1} + sum_{n = 0}^infty n a_n x^n = psum_{n = 0}^infty a_n x^n$$

Then I shifted the left index (the first term yielding $$0$$ allows the lower bound to remain $$0$$) and algebraically combined the summations:

$$sum_{n = 0}^infty (n + 1)a_{n + 1} x^n + (n – p)a_n x^n = 0$$

This leads to the following recurrence relation:

$$a_{n + 1} = frac{p – n}{n + 1}a_n$$

Thus for various values of $$n$$:

$$a_1 = p a_0$$, $$a_2 = frac{p(p – 1)}{2}a_0$$, $$a_3 = frac{p(p – 1)(p – 2)}{6} a_0$$, etc.

So applying definitions for the exponential taylor series and falling factorial, the guessed solution would be:

$$y = sum_{n = 0}^infty frac{p! a_0 x^n}{n! (p – n)!} = sum_{n = 0}^infty a_0 e^x p^{underline n}$$

Solving the initial value problem:

$$1 = sum_{n = 0}^infty a_0 e^0 p^{underline n} implies a_0 = frac{1}{sum_{n = 0}^infty p^{underline n}}$$

My final solution is:

$$y = frac{sum_{n = 0}^infty e^x p^{underline n}}{sum_{n = 0}^infty p^{underline n}}$$

However, the answer is supposed to be $$y = (1 + x)^p$$. Are these identical, or did I make an error somewhere?

## pr.probability – Concentration inequality for norm of solution to nonlinear least-squares problem

Define the piecewise-linear function $$psi(t):=max(t,0)$$ for all $$t in mathbb R$$.

• Let $$n$$, $$d$$, and $$k$$ be large positive integers such that $$n/d,k/d to (0, infty)$$
• Let $$y_1,ldots,y_n in {-1,1}$$ be fixed / determininistic.
• Let $$n$$, $$d$$, and $$k$$ be large positive integers such that $$n/d, k/d to (0,infty)$$.
• Let $$x_1,ldots,x_n$$ be sampled iid uniformly on the $$(d-1)$$-dimensional unit-sphere.
• Let $$w_1,ldots,w_k$$ be sampled iid from $$mathcal N(0,(1/d)I_d)$$, and independently from the $$x_i$$‘s.
• Let $$Z$$ be the $$n times k$$ matrix with $$i$$th row $$z_{i,j} := psi(x_i^top w_j)$$ for all $$(i,j) in (n) times(k)$$.
• Let $$overline{z} := Z^top y = sum_{i=1} y_i z_i in mathbb R^k$$.
• Let $$v in mathbb R^k$$ be the solution to the linear system $$Z v = y$$.

Note that $$Z$$, $$overline{z}$$, and $$v$$ are all random.

Question. What are good concentration inequalities for $$|overline{z}|_2$$ and $$|v|_2$$ ?

Note. I’m really only interested in high-probability lower-bounds for $$|v|_2$$. My interest in $$|Z|_{op}$$ and $$|overline{z}|_2$$ is due to the fact that $$overline{z}^top v = n$$, and so Cauchy-Schwarz gives $$|v| ge frac{n}{|overline{z}|}.$$

Thus, a lower-bound for $$|z|$$ would give a (presumably crude) lower-bound for $$|v|$$.

## Empirical observations

I’ve run some experiments, and it seems $$|overline{z}| = Theta(sqrt{d})$$ and $$|v| = Theta(sqrt{d})$$ w.p $$1-o(1)$$.

## dg.differential geometry – Solution existence in a pde system

If I have a smooth positive scalar function $$f$$ defined on a 2-dimensional manifold $$M$$, then $$f:Mrightarrow (0, infty)$$, where the metric of $$M$$ is $$g=frac{dx^2+dy^2}{y^2}$$, i.e., $$M$$ is Poincare’ half-plane.

$$f$$ must satisfy the following PDEs:

$$begin{cases} Delta f=f/2 \ |nabla f|^2=frac{(f^2+3f)}{2}+1 end{cases}$$

Considering that $$nabla f$$ is the gradient of $$f$$, where $$f$$ is a smooth positive function on manifold $$M$$, where $$M$$ is the Poincaré half plan, so the gradient is referred to the metric: $$g=frac{dx^2+dy^2}{y^2}$$; (The gradient of a smooth function on a manifold is $$nabla f=g^{ij}frac{partial f}{partial x^j} frac{partial}{partial x_i}$$), and $$Delta f$$ is the Laplace-Beltrami operator for $$f$$ on manifold $$M$$ (so referred again to the metric $$g=frac{dx^2+dy^2}{y^2}$$), and the Laplace-Beltrami of a smooth function on a manifold is determined by: $$Delta f=frac{1}{sqrt{|g|}} partial_{i} (sqrt{|g|}g^{ij} partial_{j} f)$$.

QUESTIONS:

Is there a possible solution of that pde system? and if “yes”,

How can it be shown that that pde system admits at least one solution without having to calculate it?

Is there a technique to understand if a solution exists even without calculating it?

## Audio not working in Ubuntu 20.04 after switching over from dual-booted OS (Solution)

Are you having audio issues after switching to Ubuntu from another OS (like windows) on a dual-booted device? Is everything showing as normal, but you just can’t hear anything? Don’t worry, it’s an easy fix.

## graphics – Find point along x axis where yz plane flips to back facing after perspective projection. Have a solution but don’t know why it works

I am trying to find the point along the x axis where a plane with normal pointing along the x axis would flip to back facing after perspective projection. Essentially the red line in this image.

My first attempt to find it I just sampled along the x axis checking if a triangle there was back facing after projection. This only gave me an approximation but I could make the approximation as accurate as I wanted by just sampling more planes so I used this as a baseline to verify other solutions.

I then tried the following:

``````vec4 center = projection.inverse()*vec4(0,0,0,1);
center /= center.w;
float d = dot(center.xyz, vec3(1,0,0));
vec3 point = vec3(1,0,0)*d;
``````

But this did not work, it was close but not right. See this image, the red line is the sampled point and the blue is the one calculated from that.

I then tried changing the calculation for center to `projection.inverse()*vec4(0,0,-1,1)` to see what that would look like. With the the sampled point in red, the blue line being the previous version and the green the new version. I got the following image which as you can see is still not correct.

But I noticed that the green was always directly between the blue and the red so I tried the following.

``````vec4 c1 = projection.inverse()*vec4(0,0,0,1);
c1 /= c1.w;
float d1 = dot(c1.xyz, vec3(1,0,0));
vec4 c2 = projection.inverse()*vec4(0,0,-1,1);
c2 /= c2.w;
float d2 = dot(c2.xyz, vec3(1,0,0));
float d = (d2-d1)*2 + d1;
vec3 point = vec3(1,0,0)*d;
``````

Which seems to be correct but I have no idea why that works. It is always extremely close to the sampled point so I assume it is correct.

My question is why does this work?

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.

## linear programming – estimating optimal solution for LP with strict inequalities

I have an LP problem with strict inequalities that cannot be relaxed. I understand that most LP solvers require the problem to not have any strict inequalities as it is impossible to find an optimum for all problems.

However, is there precedent (ex. a paper or package) in allowing strict inequalities and then “stepping” towards the solution via something like gradient decent.

This would allow an estimation of the optimum (especially given the restriction of a convex polytope) with a fairly high degree of accuracy (which can be increased with more steps), which seems like it would be useful in many situations.

## Show that the analytical solution of a PDE is equal to a given Wave Equation

I am only stuck on how to do the first part of the question, in the photo.

If someone can help me break-through this, I really appreciate it.

## sql server – solution? an expression for non-boolean type specified in a context where a condition is expected, near ‘order’

``````SELECT   Kelas.idkelas           AS id_kelas,
Matakuliah. Mata kuliah AS nama_matakuliah,
Dosen. Nama             AS nama _dosen,
COUNT(krs.idkelas)      AS jumlah_peserta
FROM     kelas,
matakuliah,
dosen,
krs
WHERE    kelas. Dikelas = krs.idkelas
AND kelas.nip = dosen.nip
AND kelas.kode = matakuliah.kode
AND krs. nilai IS NOT NULL
GROUP BY kelas.idkelas,
matakuliah.Matakuliah,
dosen. Nama
HAVING   COUNT(krs.idkelas)
ORDER BY kelas.Id kelas
``````

## TAPE%ANY%( 91 =99 14 7032 22 ) Divorce Problem Solution baba ji Bahadurgarh

TAPE%ANY%( 91 =99 14 7032 22 ) Divorce Problem Solution baba ji Bahadurgarh… | Read the rest of https://www.webhostingtalk.com/showthread.php?t=1833949&goto=newpost