## fa.functional analysis – Reference request: Is if possible to estimate the local behaviour of the solution of \$nabla cdot a(x) nabla f=g\$ via constant coefficients?

Consider the divergence form uniformly elliptic operator $$nabla cdot a(x) nabla$$
where the coefficient $$a$$ are smooth and bounded and $$D$$ is a bounded
and smooth domain of $$mathbb R^d$$
$$begin{cases} nabla cdot a(x) nabla f (x)=g text{ in } D \ f(x) = 0 text{ in } partial D, end{cases}$$
where $$g$$ for some $$g$$.
Consider now $$x_0in D$$ and $$delta < d(x,partial D)$$ and the function $$f_{x_0}$$ which solves
$$begin{cases} nabla cdot a(x_0) nabla f^delta_{x_0} (x)=g text{ in } B(x_0,delta)\ f^delta_{x_0}(x) = f(x) text{ in } partial B(x_0,delta). end{cases}$$

I was wondering whether it is possible to bound quantities such as
$$M(x_0,delta,r,p):=r^{-d}|f-f^delta_{x_0}|_{L^p(B(x_0,r))}$$
for $$r < delta$$ and for some $$p in (1,infty)$$. In particular, I was wondering about the case asymptotic behaviour for $$r to 0^+$$. That is, can I show that
$$M(p, gamma):= sup_{x_0 in D} sup_{delta < d(x,partial D)wedge c_a} sup_{r le delta} frac{M(x_0,delta,r,p)}{r^gamma},$$
is finite for some $$gamma>0$$ and some constant $$c_a >0$$? If so, does that bound depends on the smoothness of $$g$$?

The idea being that if $$delta$$ is sufficiently small, $$a(x)approx a(x_0)$$ in the ball $$B(x_0,delta)$$ and therefore the two equations should behave similarly. I am not sure if this is indeed enough or if I would need to ask $$delta$$ to vanish as well.

I would appreciate any references or even what are the keywords to find such type of estimates.

## improper integrals – Convergence of \$ I=int_{0}^{+infty} frac{1}{sqrt{t}} cdot sin left(t+frac{1}{t}right) dt \$

improper integrals – Convergence of \$ I=int_{0}^{+infty} frac{1}{sqrt{t}} cdot sin left(t+frac{1}{t}right) dt \$ – Mathematics Stack Exchange

## probability – In a multinomial distribution, compute \$mathbb E[I_i cdot I_j]\$

Suppose there are $$n$$ independent trials each of which can take the values : $$n_1,n_2, cdots, n_r$$ with probabilities : $$p_1,p_2,cdots,p_r$$.

Let the Indicator variable $$I_j = 1$$ when the event $$j$$ never occurs in the $$n$$ trials and $$0$$ otherwise.

I am trying to prove that $$mathbb E(I_i cdot I_j) = ( 1 – p_i – p_j)^n$$

Attempt: From a previous problem, I know that $$I_i, I_j$$ are not independent.

Now $$mathbb E(I_i cdot I_j) = 1 cdot P(I_i cdot I_j = 1)$$

$$P(I_i cdot I_j = 1) = P(I_i = 1 ~cap~I_j = 1) = P(I_i = 1 ~cup~I_j = 1) – P(I_i = 1) – P(I_j = 1)$$

How can I evaluate this further. Help is much appreciated!

## Demonstrate that \$1^{2}+2^{2}+3^{2}+cdot cdot cdot +n^{2}=frac{n(n+1)2(n+1)}{6}\$

$$1^{2}+2^{2}+3^{2}+cdot cdot cdot +n^{2}=frac{n(n+1)2(n+1)}{6}$$

Excuse me, I am new to demonstrations. I don’t know how to demonstrate this.. Any help would be very much appreciated! Or tips!

Kind wishes!

## Sufficient condition for uniform convergence of \$f_{epsilon}(x, cdot)\$ to \$f(x, cdot)\$ implies that \$f_{epsilon} rightarrow f\$ uniformly.

Consider functions $$f_{epsilon}: X times U rightarrow mathbb{R}$$ for all $$epsilon >0$$ and $$f: X times U rightarrow mathbb{R}$$. $$X subset mathbb{R}^n$$ and $$U subset mathbb{R}^m$$ are compact spaces.
Suppose that given $$x in X$$, $$f_{epsilon}(x, cdot)$$ and $$f(x, cdot)$$ are continuous (on $$U$$), and $$f_{epsilon}(cdot, u)$$ and $$f(cdot, u)$$ are continuous (on $$X$$).

If $$f_{epsilon}(x, cdot)$$ converges uniformly to $$f(x, cdot)$$, then does it imply that $$f_{epsilon} rightarrow f$$ uniformly?
I guess that from the uniform convergence of $$f_{epsilon}(x, cdot)$$ to $$f(x, cdot)$$,
we may choose $$overline{epsilon}(x) > 0$$ for any $$epsilon’ >0$$ such that $$|f_{epsilon(x)}(x, u) – f(x, u)| < epsilon’, forall epsilon(x) leq overline{epsilon}(x), forall (x, u) in X times U$$. Then, $$epsilon := min_{x in X}overline{epsilon}(x)$$ implies the desired result.

Actually, I’m not sure that such $$epsilon$$ exists. Is it correct? If not, are there any sufficient conditions to derive similar conclusions?

## Determine : \$lim _{n rightarrow infty} left[displaystylefrac{1 cdot 1 !+2 cdot 2 !+ldots+n cdot n !}{(n+1) !}right]^{(n+1)!}\$

What is the limit of the following :
$$lim _{n rightarrow infty} left(displaystylefrac{1 cdot 1 !+2 cdot 2 !+ldots+n cdot n !}{(n+1) !}right)^{(n+1)!}$$
I think it is clear that the numerator approaches infinity faster than the denominator so the result should be $$+infty$$?

## combinatorics – A combinatorial formula for different sums \$acdot k_1+a cdot k_2+…+k_n\$

I am looking for a combinatorial rule for the following: let $$n in mathbb{N}$$ and $$k_1,k_2,…,k_n in mathbb{N}$$ (these are known). Since we know what the sum $$sum_{j=1}^{n}k_i$$ is, let’s say that $$sum_{j=1}^{n}k_i=B$$.

Now, let us define a new combinatorial arrangement of $$k_i$$‘s, let us label this as $$k_{x_1},k_{x_2},…,k_{x_n}$$. Also, let $$a in mathbb{N}$$ be known. Now, we want to determine how many different values may the following sum take:

$$acdot k_{x_1}+acdot k_{x_2}+…+k_{x_n}=acdot(k_{x_1}+…+k_{x_{n-1}})+k_{x_n}$$?

Here is an example: let $$k_1=2$$, $$k_2=3$$ and $$k_3=1$$ (i.e. $$n=3$$ and $$k_1+k_2+k_3=6$$). Now we would have the following arrangements (let $$A_i$$ stand for arrangement):

$${k_{x_1},k_{x_2},k_{x_3}}:$$ $$A_1={2,3,1}$$, $$A_2={3,2,1}$$, $$A_3={2,1,3}$$, $$A_4={3,1,2}$$, $$A_5={1,2,3}$$ and $$A_6={1,3,2}$$.

Now, if we want to determine different described sums for these arrangements, we may see the following:

For $$A_1$$ we have $$acdot(2+3)+1=5a+1$$.

For $$A_2$$ we have $$acdot (3+2)+1=5a+1$$.

For $$A_3$$ we have $$acdot (2+1)+3=3a+3$$.

For $$A_4$$ we have $$acdot (3+1)+2=4a+2$$.

For $$A_5$$ we have $$acdot (1+2)+3=3a+3$$.

For $$A_6$$ we have $$acdot (1+3)+2=4a+2$$.

Thus, we have the total of $$3$$ different formulas for $$acdot (k_{x_1}+…+k_{x_n})+k_{x_n}$$. However, we do not know the exact value of $$ain mathbb{N}$$. For example, if we have $$a=1$$ then we would have to have $$3a+3=4a+2=5a+1$$, thus reducing the number of different sums to exactly $$1$$. However, if $$a=2$$ then $$5a+1=11$$, $$3a+3=9$$ and $$4a+2=10$$ and we would have $$3$$ different sums. Would there be a combinatorial formula that takes into consideration all different possibilities and would determine the exact number of different possible sums given the original $$k_i$$‘s (in the example $$k_1=2$$, $$k_2=3$$ and $$k_3=1$$) and an arbitrary choice of $$a in mathbb{N}$$?

## algorithm analysis – Solve the recurrence \$ T(n) = sqrt{n} cdot T(sqrt{n}) + 1 \$

$$T(n) = sqrt{n} cdot T(sqrt{n}) + 1$$

I’ve found so many similar questions but I couldn’t understand any of the answers explanations. When I try to draw a recurrence tree, I see that each ‘level’ has as many operations as nodes (because of only $$1$$ operation in each node) so in the first level it has $$1$$ node then $$sqrt{n}$$ nodes then $$sqrt{sqrt{n}}$$ nodes and so forth to $$n^{frac{1}{2^k}}$$ on the lowest level on the tree.

I get the same answer when unrolling it:

$$T(n) = n^frac{1}{2} T(n^{frac{1}{2}}) + 1 = n^{frac{1}{2}}(n^{frac{1}{4}} T(n^{frac{1}{4}}) + 1) + 1= n^{frac{1}{2}} ( n^{frac{1}{4}}(n^{frac{1}{8}} T(n^{frac{1}{8}}) + 1 ) + 1) + 1 = …$$

But it is really inconvenient working with this form.

Also tried using substitution as mentioned here and then applying master theorem but with I can’t understand how to make the transition back. Also a similar question here but no further explanations in the answers. I would rather use tree recurrence to solve it but substitution and master theorem also good.

## Proving that \$sum_{k=0}^{k=n} binom{2n}{k} cdot k = 2^{2n -1} cdot n\$

I have to show that $$sum_{k=0}^{k=n} binom{2n}{k} cdot k = 2^{2n -1} cdot n$$.

What I know is that $$sum_{k=0}^{k=n} binom{n}{k} cdot k = 2^{n -1} cdot n$$.

How do I proceed from there?

## linear algebra – In any vector space, \$(-1+1) ; cdot ; vec{v} ; = ; 0 ; cdot ; vec{v} ; = ; vec{0} \$?

My work:

Let us consider “$$+$$” to be “$$#$$“, and “$$cdot$$” to be “$$*$$“, so that I can use you them to represent the normal addition and multiplication.

$$a ; # ; b = 2cdot a ; + ;b$$

$$a ; * ; b = a^2 ; + ;b^2$$

According to the book (Linear algebra by Jim Hefferon
Third edition), item $$(2)$$:

$$(-1 cdot vec{v}) + vec{v} = color{red}{(-1+1)} cdot vec{v} = 0 cdot vec{v} = vec{0}$$

But $$(-1 ; # ; 1) = 2cdot (-1) ; + ;1 = -2 ; + ;1 ; = ; -1$$. So it is not zero!?