13,373 reputation
31232
bio website
location
age
visits member for 3 years, 3 months
seen 1 hour ago

Hopeful mathematics student with some competition level problem solving experience. Studying algebraic topology and algebraic geometry at the moment, but I feel like anyone who know anything know more than I do.


23h
comment If $X$ is Hausdorff, then so is $E$
You have to divide into cases depending on whether $q(x)=q(y)$.
1d
comment What is the discrete log used for?
Many cryptography schemes rely on the fact that discrete logarithm is difficult to calculate.
1d
comment Estimated wait time for a library book
A quick estimate is the following: twenty books each being out twenty-one days means that on average, (a little less than) one book is delivered back each day. So, a little over forty days.
1d
revised Proving $P$ by proving $\neg Q$ and knowing $P\lor Q$
Added math environment, retagged
1d
comment how to add supremums
First off, be careful with the word "group". It has a specific meaning in mathematics, and I don't thing it's the meaning you intend. You probably meant "set" instead. Second, the set $S+T$ is the set of all possible results of adding a member of $S$ with a member of $T$.
1d
comment How can I find the degree of the extension?
Now you know that it's either $1,2,3$ or $6$. I'll admit that as I wrote the answer I thought the degree of $\omega_7$ was seven. But it isn't too difficult to prove that no second or third degree rational polynomial evaluates to zero at $\omega_7+\omega_7^5$.
1d
comment Define a bijection function
Hint: Add $10$.
1d
comment Why before $e^{x}$,the solution was not possible?
Using $e^{mx}$ instead of $10^{nx}$ is mostly done to avoid all those pesky $\ln(10)$ coefficients you keep getting.
1d
comment Determine $1^{x}+2^{x}+3^{x}+4^{x}+5^{x}+6^{x}+7^{x}+8^{x}+9^{x}=10^{x}$
@Peter WolframAlpha backs that up.
1d
comment Determine $1^{x}+2^{x}+3^{x}+4^{x}+5^{x}+6^{x}+7^{x}+8^{x}+9^{x}=10^{x}$
Is it enough to say "Between $5$ and $6$", or do you need an exact answer?
1d
revised Determine $1^{x}+2^{x}+3^{x}+4^{x}+5^{x}+6^{x}+7^{x}+8^{x}+9^{x}=10^{x}$
deleted 39 characters in body; edited title
2d
comment How write down PMF when random variable follows conditionally discrete uniform distributions with different support.
Yes, but you need to be careful with $P(F=i)$. Let's say we pick a random kid from another town, where half the families have one kid and half the families have two kids. Then $P(F=2)=2/3$. I believe it's easier to just count the number of first, second and third borne children.
2d
comment What does $\mathbb{R}^2$ domain mean?
$\Bbb R^2$ is the common $xy$-plane. That is probably what $R^2$ means as well.
Dec
16
answered trying to understand what a polynomial ring is
Dec
16
comment Proving a surjective function by given property
@AlanWang You can work to make the contradiction argument less visible, but I don't think that this result is even provable without using some form of argument by contradiction. Here are some more details on the proof I had in mind: If $f$ is not surjective, then that means that there is at least one $y\in F$ such that no $x\in E$ gives $f(x) = y$. Then by definition of $f^{-1}$, we have $f^{-1}(\{y\}) = \emptyset$, the empty set. That means that $f(f^{-1}(\{y\})) = f(\emptyset) = \emptyset \neq \{y\}$. This contradicts the condition $A\subseteq F, f(f^{-1}(A)) = A$.
Dec
16
answered Proving a surjective function by given property
Dec
16
comment Find all $a,b,c$ such that $\binom{a}{b} \binom{b}{c}=2\binom{a}{c}$
Use the definition of $\binom{m}{n}$, and simplify.
Dec
16
comment Is there an error in my matrix proofs (Also: potato quality jpeg errors present)
$A^3$ is also symmetric: $$(A^3)^T=(A^2A)^T=A^T(A^2)^T=AA^2=A^3$$ The same goes for any other power of $A$ (this shows that the product of two commuting, symmetric matrices is symmetric).
Dec
16
comment Divisors of sequence $1!+2!+\ldots+n!$
Your question is equivalent to asking whether my division condition holds for an infinite number of primes. I do not know the answer. By the way, how does $n!$ behave mod prime $p$ for $1<n<p$? Perhaps we can get somewhere from there.
Dec
15
comment Divisors of sequence $1!+2!+\ldots+n!$
For any single prime $p$, what must happen for $p$ to divide $1! + \cdots + (p-1)!$?