KingOliver
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 Feb10 comment Buy more lottery tickets or more multipliers? Actually I double with chance $1$ in $2$ so isn't it ($1/2 \cdot 2 + 1/3 \cdot 3 + \cdots$)? But thank you this is the way I was thinking about it Jul29 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. Dec9 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 Dec8 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 $|x| < 1$? Dec8 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? Nov8 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 Nov7 comment Squares of differentiable functions Sorry @dfeuer I meant $(f(x))^2$. Thanks Daniel Fischer! Nov5 comment Limit of metric of sequences Ah! I have seen it: $|\rho(x_n,y_n) - \rho(x,y_n)| \leq |\rho(x_n,x) + \rho(x,y_n) - \rho(x,y_n)|$ Nov5 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? Nov5 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 $\mathbb R$ Sep8 comment Showing that $(a_n^2)$ is Cauchy implies that $(a_n)$ is Cauchy Ah! Precisely, thank you André Jun16 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. Jun13 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 Jun13 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 Jun13 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 $p^2 \leq m$? Jun5 comment Trying to prove an identity about a product I will go down that route, thanks for your suggestions @wece Jun5 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? May30 comment Number of possible pairs from $\{1,\dots, n\}$ with $i < j$ My mistake, it is in fact simply ${n \choose 2}$ May23 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 $n \in \Bbb N$? May23 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?