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 $mathbb R^d$
nabla cdot a(x) nabla f (x)=g text{ in } D \
f(x) = 0 text{ in } partial D,

where $g$ for some $g$.
Consider now $x_0in D$ and $delta < d(x,partial D)$ and the function $f_{x_0}$ which solves
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).

I was wondering whether it is possible to bound quantities such as

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!?

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