calculus – Checking Proof of Claim Integral Minus Quantity is Negative

I just want to clarify a piece of a problem that I’m working on.

Is it fair to say that $int _{0} ^{x}f(y,t) dy – xf(x,t) leq0$?

My reasoning is that for the time slices at each $t$, the integral represents the area under the curve integrated from $0$ to $x$ whereas the multiplication $xf(x,t)$ represents the area of a rectangle with height $f(x,t)$ and length $x$. Looking at it for each fixed $t$ I believe that the area of the rectangle will always be larger than or equal to the integral value. Is this correct? Is there a better way to prove this?

Thank you for your help!

Proof of Evidence

Please can I use a landed property as an evidence but the problem I am having is the document was not written in my name but my aunt has handed the building to me because my uncle who supposed to be the owner of the building is late.

homotopy theory – Elementary proof of the exactness of Čech complex associated to a hypercovering (“Illusie’s Conjecture”)

Let $mathcal{E}$ be a sheaf of abelian groups on a topological space (or a site). For an open covering $mathfrak{U} = (U_i)_i$, it is well known that the augmented Čech complex $0 to mathcal{E} to mathcal{check C}^bullet(mathfrak{U}, mathcal{E})$ is a resolution of $mathcal{E}$. (Depending on the sheaf cohomology of $mathcal{E}$ on the intersections of the members of $mathfrak{U}$, this resolution might compute $mathbb{R}Gamma(mathcal{E})$, but this need not concern us here.)

A simple proof runs as follows: Let $s in mathcal{check C}^{n+1}(mathfrak{U}, mathcal{E})(V)$ such that $ds = 0$. Then $t = (s_{i_text{fix},i_0,ldots,i_n})_{i_0,ldots,i_n} in mathcal{check C}^{n}(mathfrak{U}, mathcal{E})(V cap U_{i_text{fix}})$ is a local preimage of $s$ under $d$, as $(dt)_{i_0, ldots, i_{n+1}} = s_{i_0, ldots, i_{n+1}} – (ds)_{i_text{fix}, i_0, ldots, i_{n+1}} = s_{i_0, ldots, i_{n+1}}$. As $V = bigcup_{i_text{fix}} (V cap U_text{fix})$, that’s all what’s needed.

I’m looking for a similarly elementary proof in the case that $mathfrak{U}$ is a hypercovering. Apparently this fact is sometimes referred to as Illusie’s Conjecture. A proof involving trivial Kan fibrations and an induction reducing to the case of ordinary covers is in Section 01GA of the Stacks Project, but I’d prefer either a more direct proof or some insight as to why a much simpler argument isn’t likely to exist.

fa.functional analysis – Proof of the analytic Fredholm theorem in Borthwick

I’ve stumbled across a proof of the analytic Fredholm theorem given in Theorem 6.1 in Spectral Theory of Infinite-Area Hyperbolic Surfaces by David Borthwick (see below).

Given the notion of being “finitely meromorphic” given in that book, how does the author infer that we can choose a small enough neighborhood $N$ of any point $s_0in N$ so that $N$ contains only finitely many “poles” of $E(s)$?

He didn’t define what a “pole” is, but I guess it’s supposed to mean (in the context of the definition of finite meromorphicity) that a point $ain U$ is a “pole”, if the Laurent series expansion of $A(s)$ given satisfies $m>0$.

If that’s the case, I don’t get how he’s inferring that $N$ contains only finitely many poles. In the ordinary definition of meromorphicity, it is assumed that the set of poles is discrete. Isn’t an assumption of this kind missing here?

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proof of stake – How to generate a Emercoin mainnet/testnet/regtest addresses

I have this settings :

const emercoin: NetworkConfig = {
    mainnet: {
        messagePrefix: 'x18Emercoin Signed Message:n',
        bech32: 'emc',
        bip32: {
            public: 0x0488b21e,
            private: 0x0488ade4,
        },
        pubKeyHash: 0x00,
        scriptHash: 0x05,
        wif: 0x80,
        dustThreshold: 50000,
        timeInTransaction: true,
        maximumFeeRate: 50000,
    },
    testnet: {
        messagePrefix: 'x18Emercoin Signed Message:n',
        bech32: 'emc',
        bip32: {
            public: 0x043587cf,
            private: 0x04358394,
        },
        pubKeyHash: 0x6f,
        scriptHash: 0xc4,
        wif: 0xef,
        dustThreshold: 500,
        timeInTransaction: true,
        maximumFeeRate: 50000,
    },
    regtest: {
        messagePrefix: 'x18Emercoin Signed Message:n',
        bech32: 'emc',
        bip32: {
            public: 0x043587cf,
            private: 0x04358394,
        },
        pubKeyHash: 0x6f,
        scriptHash: 0xc4,
        wif: 0xef,
        dustThreshold: 500,
        timeInTransaction: true,
    },
}

after using this settings for transaction signing the signed transaction can not send over netwotk .
does anyone face this problem?

sorry for my poor english.

algebraic topology – Proof Verification: Only contractible discrete space is the one point space

I am trying to prove that only contractible discrete space is the one point space. The proof of it can be seen here(https://math.stackexchange.com/questions/2218782/show-that-discrete-space-of-x-is-contractible-if-have-one-point) and here(https://math.stackexchange.com/questions/1720201/proving-that-a-three-point-discrete-space-is-not-contractible)

I believe I have a proof, but I do not use connectedness of the unit interval, so I’m asking for verification of my proof.

My proof

Suppose discrete space X was contractible. Any contractible space is path-connected. Indeed, take a homotopy $H colon X times (0,1) to X$ from identity on $X$ to $cost$ on $x_1$. Then for any $x_2 in X$, $H(x_2, -) colon (0,1) to X$ gives a path from $x_1$ to $x_2$ as $H$ being continuous implies $H$ is continuous in each variable.

Now I claim that any discrete space with more than 2 points is not connected. To show this, note that if there is continuous and non-constant map $X to {0,1}$, then $X$ is not connected. As $X$ has more than 2 points, we can find a non-constant set map $X to {0,1}$. And as $X$ is discrete, this non-constant set map must be continuous.

So if $X$ has more than 1 point and contractible, we get that $X$ is path connected but not connected. This is impossible, so $X$ must have one point if it were to be contractilbe.

Thanks!

pr.probability – Pedestrian proof of Gaussian chaos for order-two poynomial?

Let $ell geqslant 1$. Let us consider $(g_n)_{n in mathbb{N}}$ identically distributed idependent real gaussian variables and real number $(a_{n_1,dots n_{ell}})_{(n_1, dots, n_{ell}s)inmathbb{N}^{ell}}$. We consider the random variable
$$G(omega)=sum_{n_1, dots, n_{ell}}a_{n_1, dots, n_{ell}} g_{n_1}(omega) cdots g_{n_{ell}}(omega)$$
where we write $omega in Omega$ the probability space at hand.

It is known as Gaussian polynomial chaos (or Wiener chaos)
$$|G(omega)|_{L^p_{Omega}} lesssim p^{frac{ell}{2}} |G(omega)|_{L^2_{Omega}}.$$

The standard proof of this results uses hypercontractivity of the Ornstein-Uhlenbeck semigroup.

My question is the following: in the case $ell =2$ and $p=2k$ an even integer,

Is it possible to prove “directly by hand” that
$$mathbb{E}(|G(omega)|^{2k}) leqslant C k^{2k}mathbb{E}(|G(omega)|^2)^k.$$
If yes, is there a reference book/article to find it? Or is it possible to finish my attempted proof?

This does not seem impossible. At least, when one expands the left-hand side, we have
$$mathbb{E}(|G(omega)|^{2k}) = sum_{substack{n_1,dots ,n_k\ m_1, dots, m_k}}a_{n_1,m_1} dots a_{n_{2k},m_{2k}}mathbb{E}(g_{n_1}g_{m_1}cdots g_{n_{2k}}g_{m_{2k}}).$$

From there, one can use the fact that $mathbb{E}(g^{2m+1})=0$ for any $m$ to infer that in the above, the expectation is non-vanishing only if the indices ${n_i, m_i}$ appear an even number of time. So in particular, all $n_i, m_i$ are paired.

The right hand side reads
$$k^{2k}mathbb{E}(|G(omega)|^2)^k = k^{2k}sum_{substack{n_1,dots ,n_k\ m_1, dots, m_k}}a_{n_1,m_1} dots a_{n_{2k},m_{2k}}prod_{i=1}^kmathbb{E}(g_{n_i}g_{m_i}g_{n_{k+i}}g_{m_{k+i}}).$$

Also, remark that in the LHS, when all the $g_i$ are equal, then we have $mathbb{E}(g^{4k}) lesssim k^{2k}$, which is the correct magnitude which we look for.

However I am not able to prove that $LHS leqslant RHS$. But I remain convinced that there should be somewhere to find such a proof but I did not locate it in the litterature.

Is This Proof of Pythagorean Theorem?

  1. Given image:
  2. $trianglemathit{BDE}$ is an altitude right triangle as given by the image.
  3. $trianglemathit{BDE}$ and $trianglemathit{FDE}$ both contain 90° given by the image; therefore, $trianglemathit{BDE}$ and $triangle{mathit{FDE}}$ are both right triangles.
  4. $trianglemathit{BDE}$ is an altitude right triangle with $trianglemathit{FDE}$ as given by the image.
  5. If $trianglemathit{BDE}$ is an altitude right triangle with $trianglemathit{FDE}$ then:
    • $overline{mathit{DE}}^{2}=overline{mathit{BE}}overline{mathit{EF}}$
    • $overline{mathit{DF}}^{2}=overline{mathit{EF}}overline{mathit{BF}}$
  6. $overline{mathit{BE}}+overline{mathit{EF}}=overline{mathit{BF}}$ as given by the image
  7. Given these equations algebra can then be performed to solve for $overline{mathit{DF}}$. Algebra:
    • Rearranging:$overline{mathit{DE}}^{2}=overline{mathit{BE}}overline{mathit{EF}}$ to get $frac{overline{mathit{DE}}^{2}}{overline{mathit{EF}}}=overline{mathit{BF}}.$
    • $overline{mathit{DF}}^{2}=overline{mathit{EF}}overline{mathit{BF}}$
    • Substituting:$$overline{mathit{DF}}^{2}=overline{mathit{EF}}(overline{mathit{BE}}+overline{mathit{EF}})$$
    • Substituting Again:$$overline{mathit{DF}}^{2}=overline{mathit{EF}}(frac{overline{mathit{DE}}^{2}}{overline{mathit{EF}}} + overline{mathit{EF}})$$
    • Simplifying:$$overline{mathit{DF}}^{2}=overline{mathit{DE}}^{2}+overline{mathit{EF}}^{2}$$
  • Which is the familiar form of Pythagorean Theorem.
  • Is this a mathematically correct derivation?
    Is this proof general enough to be used as a general proof?

    proof writing – Question about limits at infinity.

    if for example I have fucntions $f$ and $g$, we’ve prove in class that if
    $$lim _{xto a}left(fleft(xright)-gleft(xright)right)=L$$

    $$or$$

    $$lim _{xto a}left(fleft(xright)cdot gleft(xright)right)=L$$

    then both limits of $f$ and $g$ exist, or both don’t exist.

    but what about

    $$lim _{xto ∞/-∞}left(fleft(xright)-gleft(xright)right)=L$$

    $$or$$

    $$lim _{xto ∞/-∞}left(fleft(xright)cdot gleft(xright)right)=L$$

    Does it still hold that both of f,g must exist or both must not exist?
    If so, how do I prove it?

    matrices – Select algorithm proof

    I am reading a paper Fast Monte-Carlo algorithms for matrices I: Approximating matrix multiplication pdf link at author’s site. They provide an algorithm 1 that is used to select elements in pg 137 algorithm 1

    They then give a lemma at p. 136.

    Lemma: Suppose that ${a_1, dots , a_n}, a_i ge 0$, are read in one pass, i.e., one
    sequential read over the data, by the Select algorithm. Then the Select algorithm
    requires $O(1)$, i.e., constant with respect to $n$, additional storage space and returns
    a random i* sampled from the probability distribution probability distribution.

    They then provide proof at p. 136. However I cannot wrap my head around the last part of the proof i.e $i=1,…,ell$ proof. Please help me get the intuition behind it (especially this part
    proof)