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age 21
visits member for 2 years, 5 months
seen Oct 22 at 19:01

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Dec
8
asked Proving that $\sum_j x^j$ is differentiable $(-1,1)$
Dec
8
accepted Showing that $\sum_j e^{-jx}x^j$ converges uniformly
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?
Dec
8
asked Showing that $\sum_j e^{-jx}x^j$ converges uniformly
Dec
3
accepted How to prove Riemann integrable with partitions
Nov
18
accepted How to show that $\sum {2^j + j \over 3^j - j}$ converges
Nov
16
asked How to show that $\sum {2^j + j \over 3^j - j}$ converges
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
asked How to prove Riemann integrable with partitions
Nov
7
comment Squares of differentiable functions
Sorry @dfeuer I meant $(f(x))^2$. Thanks Daniel Fischer!
Nov
7
asked Squares of differentiable functions
Nov
5
accepted Limit of metric of sequences
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 $\mathbb R$
Nov
5
asked Limit of metric of sequences
Nov
4
awarded  Notable Question
Oct
31
asked Finding a nonzero continuous function that satisfies this integral equation, but not unique?
Sep
14
accepted Showing that $(a_n^2)$ is Cauchy implies that $(a_n)$ is Cauchy
Sep
8
comment Showing that $(a_n^2)$ is Cauchy implies that $(a_n)$ is Cauchy
Ah! Precisely, thank you André