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Mathematics Stack Exchange

2h ago

Define the complex function ( f(z) ) as follows:
[ f(z) = \frac{1}{\sin z} - \frac{1}{z} ]
For each ( n \in \mathbb{N} ), let ( a_n = \left(n + \frac{1}{2}\right)\pi ), and consider the contour ( C_n ) that traces the square with vertices at ( a_n + i a_n ), ( -a_n + i a_n ), ( -a_n - i a_n ), and ( a_n - i a_n ) in a counterclockwise direction. Then, answer the following questions:
(1) For any ( n \in \mathbb{N} ), show that if ( z \neq \pm n\pi ) and ( z \notin C_n ), then
[ \lim_{n \to \infty} \oint_{C_n} \frac{f(w)}{w(w - z)} , dw = 0 ..read more

Mathematics Stack Exchange

2h ago

I have initial condition $?_1=2, ?_1=1$, and the given recurrence relations: $?_{?+1}=2?_?,$ $?_{?+1}=2?_?+\frac{1}{2} ?_?.$
I need to show that that, $v_i=\Theta(n_i\log n_i).$
I observe that $?_?=2^?$, how to proceed from here ..read more

Mathematics Stack Exchange

2h ago

Let $\mathcal{A}$ be abelian category, and $X\in\operatorname{ob}(\mathcal{A}).$ $\operatorname{Hom}(X,-)$ is left exact functor from $\mathcal{A}$ to $\mathsf{Ab}$, which means that when $0 \to A \xrightarrow[]{f} B \xrightarrow[]{g} C$ is exact sequence in $\mathcal{A}$, $0 \to \operatorname{Hom}(X,A) \xrightarrow[]{f'} \operatorname{Hom}(X,B) \xrightarrow[]{g'} \operatorname{Hom}(X,C)$ is exact in $\mathsf{Ab}$, given $f'$ by $h\mapsto f\circ h$ for all $h\in \operatorname{Hom}(X,A)$ and $g'$ by $i\mapsto g\circ i$ for all $i\in \operatorname{Hom}(X,B)$.
I understand that it is sufficient t ..read more

Mathematics Stack Exchange

2h ago

I have been running around in my head all afternoon trying to figure out how to create an acceleration function that accumulates into a desired position function. It has led me to this:
Where each curve accumulates to the area of its containing curve. If anyone has seen something like this, a name or some reference to its related field of study would be great.
I would like to figure out the equation for these curves; the best I've found is sin(2*asin(sin(x))).
Any info helps, thank you ..read more

Mathematics Stack Exchange

2h ago

Prove: Two non-empty sets $ A $ and $ B $ satisfy $ A \times B = B \times A $ if and only if $ A = B $.
Let $ j \in A \wedge n \in B $, then $ (j,n) \in A \times B $ and $ (n,j) \in B \times A $.
We assume that $ A \neq B $, so $ \exists u, v \mid u \neq v $ such that $ u \in A \wedge v \in B $. From $ u,v $, the ordered pair $ (u,v) $ exists, which is a subset of $(n,j) $, and also the ordered pair $ (v,u) $, which is a subset of $ (j,n) $.
Given that $ u \neq v $, it is demonstrated by the theorem of ordered pairs that $ (u,v) \neq (v,u) $.
If we suppose that $ A \times B = B \times A $, thi ..read more

Mathematics Stack Exchange

2h ago

I am trying to do the following exercise:
Imagine rolling two dice and summing the results up. (So we can get values between $2$ and $12$). What sum has the highest probability? What if we have $n \in \mathbb{N}$ dice?
My approach: In case of the two dice, the most simple approach would be to consider every single case. But this gets way too time-consuming for more than 2 dice. So here is the other approach I have been thinking about:
Let $A_k$ be the event that the sum equals $k$, where we know that $2 \leq k \leq 12$. Thus a successful outcome $\omega$ would be of the form $(i,k-i)$ for $i ..read more

Mathematics Stack Exchange

2h ago

Is there an article of proving Lagrange theorem on cyclic group using graph theory? this is my thesis idea, please help ..read more

Mathematics Stack Exchange

2h ago

Given 2 events A and B, it's common to see the concept of independence being defined as several combinations of:
$P(A|B)=P(A)$
$P(B|A)=P(B)$
$P(A\cap B)=P(A)P(B)$
What is correct to say? Condition 1 is necessary and sufficient to stablish what is called independence (because it would imply 2)? Condition 1 and 2 are the necessary ones? Condition 3 implies 1 and 2 and is the only one necessary? Some of them are a specific case of others ..read more

Mathematics Stack Exchange

2h ago

Modules are known to generalize abelian groups since all abelian groups are $\mathbb{Z}$-module (scalar multiplication : $n\cdot x=x+x+\dots +x$ (n times)) However, all modules are abelian group, when not considering scalar multiplication. Therefore it's hard to understand that module is more generalized concept than abelian group. What I mean is, definition of module is much more specific than that of abelian group(it fully contains it). What is the real difference between abelian groups and modules? Can you provide some intuition in understanding ..read more

Mathematics Stack Exchange

2h ago

Suppose $R$ is a relation on $A$ and let $S$ be the symmetric closure of $R$. Prove that $$\text{Dom}(S) = \text{Ran}(S) = \text{Dom} (R)\cup \text{Ran}(R).$$ Is the following deduction correct? We have $S = R \cup R^{-1}$, then
\begin{align*} \text{Dom}(S) &= \text{Dom}(R \cup R^{-1})\\ &= \text{Dom}(R)\cup \text{Dom}(R^{-1})\\ &= \text{Dom}(R) \cup \text{Ran}(R) \end{align ..read more