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Jul
29
comment Flipping heads 10 times in a row
Alex, could you provide your closed form solution? I have recently asked this question myself and your question here was most helpful.
Dec
9
comment Proving that $\sum_j x^j$ is differentiable $(-1,1)$
Ah, so it converges on every interval $[-\beta, \beta]$ for $|\beta| < 1$? Many thanks for your help @EricAuld
Dec
8
comment Proving that $\sum_j x^j$ is differentiable $(-1,1)$
Yes but I am not sure how to show that $\sum_j (j+1)x^j$ converges to a finite number for $|x| < 1$?
Dec
8
comment Showing that $\sum_j e^{-jx}x^j$ converges uniformly
Hmm, thanks Mhenni! Any idea as to how I would actually compute the sum?
Nov
8
comment How to prove Riemann integrable with partitions
I understand, however, I am trying to prove this Riemann integrability from first principles, and so would like to find a way that involves a choice of partition rather than appeal to another theorem
Nov
7
comment Squares of differentiable functions
Sorry @dfeuer I meant $(f(x))^2$. Thanks Daniel Fischer!
Nov
5
comment Limit of metric of sequences
Ah! I have seen it: $|\rho(x_n,y_n) - \rho(x,y_n)| \leq |\rho(x_n,x) + \rho(x,y_n) - \rho(x,y_n)|$
Nov
5
comment Limit of metric of sequences
Hmm, in that case I suppose I'm just not quite sure how to show that something like $|\rho(x_n,y_n) - \rho(x, y_n)|$ can be made small?
Nov
5
comment Limit of metric of sequences
Ah yes I was unsure here whether I could use the absolute value metric $|\cdot|$ to prove convergence or whether I had to use a general $\rho$ to prove convergence. And $\rho$ is indeed a metric on $\mathbb R$
Sep
8
comment Showing that $(a_n^2)$ is Cauchy implies that $(a_n)$ is Cauchy
Ah! Precisely, thank you André
Jun
16
comment Checking an inductive proof on a combinatorial product
@coffeemath when you say "yes" are you confirming that I do indeed have this right? I have added the $k \geq 2$ condition thanks for this input. Regards.
Jun
13
comment Binomial coefficients equal to a prime squared
Never mind, I now see that you would need $2p \cdot (2p - 1) \cdots (p)(p-1)\cdots$. Thanks again
Jun
13
comment Binomial coefficients equal to a prime squared
I accepted Jyrki's answer, but I want to thank you for the referral to the paper, this is useful for the more general aspects of my project
Jun
13
comment Binomial coefficients equal to a prime squared
many thanks for this response. A neat proof! However I wonder if you could elaborate on why it is necessary that $2p \leq m$, and not the more trivial bound that $p^2 \leq m$?
Jun
5
comment Trying to prove an identity about a product
I will go down that route, thanks for your suggestions @wece
Jun
5
comment Trying to prove an identity about a product
@wece I thought that I should induct on $p$ and then perhaps have a "nested" induction on $n$ inside the proof. How can I induct on $n$ but still establish that it is only those above tuples that give me the desired dimension?
May
30
comment Number of possible pairs from $\{1,\dots, n\}$ with $i < j$
My mistake, it is in fact simply ${n \choose 2}$
May
23
comment How to find instances when $d(a,b) = p^2$ for $p$ a prime.
@MarkBennet it appears after looking at your answer and the original problem more closely that there are no such $(a,b) \in \Bbb N$ that satisfy this for $p$ a prime. Do you know how I might find the integral points on the surface $F(a,b,n) = 0 = d(a,b) - n$ for $n \in \Bbb N$?
May
23
comment How to find instances when $d(a,b) = p^2$ for $p$ a prime.
@AmireBendjeddou $a, b \in \Bbb N$, sorry I should have mentioned these constraints. @ Mark Bennet: I'm sorry I think I may have misunderstood you. So your comment is that there are no pairs of the form $(1, b)$ satisfying the above equality?
Apr
27
comment Computing eigenvalues for $\mathrm{Sym}^2(\mathrm{Sym}^3 V))$ for $V = \Bbb C^2$
Thanks Jack, this works perfectly. I too noticed that the elements of $\mathfrak{sl}_2$ were acting like derivatives in terms of reduction of powers (it was clear that they would act via the typical derivation action on tensor products). I appreciate you taking the time to work this out for me in a very motivated manner.