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2d
answered Boundedness theorem question proof check
May
25
revised Euler's Equation
added 4 characters in body
May
24
comment Counting poles that are shared between $f$ and $g$
@pndev : There's still the problem of finding a "nice" $h$ so that your contour integral is actually defined. And then there's the fact that the bottom sum can be $0$ when the second summand is the opposite of the first, even if they are both not $0$.
May
24
comment Counting poles that are shared between $f$ and $g$
How about something like $$z\mapsto\cfrac{1}{\cfrac{1}{h(f(z))}+\cfrac{1}{h(g(z))}}$$ assuming you can find $h$ so that $|z|\to \infty \iff h(z) \to \infty$
May
24
comment Counting poles that are shared between $f$ and $g$
@Adayah : Whatever the OP meant by $D$. The interior of $\gamma$ I guess.
May
24
comment Counting poles that are shared between $f$ and $g$
@Adayah : I didn't notice the denominator was $0$. But yes, making a constant function was intended. Fixed trivial example : $$P(z)=\frac{\sum_{f_i=g_i\in D}1}{2\pi i}\frac{1}{z}$$
May
7
comment Derivatives and integrals of polynomials of two variables
The partial derivative doesn't give you that much about $p$. If $r(x,y) = p(x,y) + q(y)$, then $\partial r / \partial x = \partial p / \partial x$. Probably not what you're searching for but $R[X,Y] \equiv R[X][Y]$ as rings and $R[X] \equiv R[X] \times R$ as vector spaces via $p \mapsto (p', p(a))$ (where $a$ is a constant). If you combine those two facts, you get $R[X,Y]\equiv R[X,Y]\times R[X]$ as vector spaces via $p\mapsto (\partial p / \partial y, (x\mapsto p(x,a))$.
May
2
awarded  Yearling
Apr
29
comment Zeroes of sin(x)
Might help: math.stackexchange.com/questions/349143/…
Apr
25
comment When can I divide both sides of an equation if one side is zero
For equality, you can multiply both sides by the same number (and therefore also divide by any non-zero number since that's just multiplication by the inverse of the number). Whether the number is positive or negative doesn't matter. It does however when you have an inequality because if you multiply by a negative number, you have the inverse the inequality.
Apr
17
comment Power set equinumerosity. Is this proof correct?
$H(B)$ isn't defined. $B\in \mathcal P(B)$ which is the codomain of $H$. And to prove that a function is the inverse of another, you just prove that when you compose them (on both sides), you get the identity.
Apr
17
comment Power set equinumerosity. Is this proof correct?
You didn't define $H$, you only gave its domain and codomain. $H:X\mapsto \{f(a) / a \in X\}$. And I think it'd be easier to build its inverse and prove that it is indeed its inverse instead of proving that it's bijective.
Jan
23
comment $x=yx$. Can this statement be true when we don't know that $y=1$?
$yx=x \implies yx-x = 0 \implies x(y-1)=0 \implies x= 0 \text{ or } y = 1$
Jan
21
comment Finding $|a|$, a complex number, given a system of equations
Have you tried using $z\overline{z}=|z|^2$?
Jan
18
comment [Verification]$G$ is a group whereby $(a\cdot b)^{i} =a^i\cdot b^i$ for three consecutive integers $i$ for all $a, b \in G$, show $G$ is abelian.
a^n or a^{stuff}${}{}{}{}{}{}$
Dec
27
comment Is there any 100% sure numerical method to find all roots in a polynomial equation of degree n without fail?
The idea is that the roots are the eigenvalues of the companion matrix, and that we know how to computer the eigenvalues of a matrix numerically.
Dec
27
comment Is there any 100% sure numerical method to find all roots in a polynomial equation of degree n without fail?
See mathworld.wolfram.com/PolynomialRoots.html
Dec
20
comment Derivative of a quadratic form
Have you tried looking at an example? Like $X=\begin{pmatrix}0&i\\-i&0\end{pmatrix}?$
Dec
14
comment Basic Axiomatic Definitions in Categories and Allegories (Freyd and Scedrov)
For example $x\mapsto \cfrac{1}{x}$ isn't defined at $0$ so it's not total (as a function of $\mathbb R \to \mathbb R$). And a binary partial operation is to a binary operation what a partial function is to a function.
Dec
14
comment Basic Axiomatic Definitions in Categories and Allegories (Freyd and Scedrov)
And a binary operation is a function $f:A\times A \to A$. You can also represent a function $f:X\to Y$ by a subset $F$ of $X \times Y$ so that $\forall x\in X,\forall y_1,y_2 \in Y, (x,y_1)\in F \land (x,y_2)\in F \implies y_1 = y_2$ ([functional] given an input, you have at most one output) and $\forall x\in X, \exists y \in Y, (x,y)\in F$ ([total] given an input, you have at least one output). A partial function is like a function in the sense that it is functional but it doesn't have to be total. In other words, a partial function is a function that can't give an output on some inputs.