1,677 reputation
1817
bio website nicolasessisbreton.com
location Montreal, Canada
age 30
visits member for 3 years, 2 months
seen 2 days ago

I'm a graduate math student at Concordia University, Montreal.


Oct
13
revised Show $\left\lvert\sum_{k=-n}^n \frac{\sin k t }{k}\right\rvert \le \pi + 2$ for all $n$ and $t$, $t \in [0,2\pi]$
no -2 in the integral
Oct
13
revised Show $\left\lvert\sum_{k=-n}^n \frac{\sin k t }{k}\right\rvert \le \pi + 2$ for all $n$ and $t$, $t \in [0,2\pi]$
a factor of 2 missing in the fourier series
Oct
13
revised Show $\left\lvert\sum_{k=-n}^n \frac{\sin k t }{k}\right\rvert \le \pi + 2$ for all $n$ and $t$, $t \in [0,2\pi]$
there was a minus missing in the fourier series
Oct
13
accepted Show $\left\lvert\sum_{k=-n}^n \frac{\sin k t }{k}\right\rvert \le \pi + 2$ for all $n$ and $t$, $t \in [0,2\pi]$
Oct
12
comment Show $\left\lvert\sum_{k=-n}^n \frac{\sin k t }{k}\right\rvert \le \pi + 2$ for all $n$ and $t$, $t \in [0,2\pi]$
@Pragabhava I don't see the two functions to which I should apply Parseval theorem.
Oct
12
revised Show $\left\lvert\sum_{k=-n}^n \frac{\sin k t }{k}\right\rvert \le \pi + 2$ for all $n$ and $t$, $t \in [0,2\pi]$
there was an i in the sum, shouldn't be one
Oct
12
comment Show $\left\lvert\sum_{k=-n}^n \frac{\sin k t }{k}\right\rvert \le \pi + 2$ for all $n$ and $t$, $t \in [0,2\pi]$
@johnmangual You are right. Thank you.
Oct
12
revised Show $\left\lvert\sum_{k=-n}^n \frac{\sin k t }{k}\right\rvert \le \pi + 2$ for all $n$ and $t$, $t \in [0,2\pi]$
typo detected in comment
Oct
12
revised Show $\left\lvert\sum_{k=-n}^n \frac{\sin k t }{k}\right\rvert \le \pi + 2$ for all $n$ and $t$, $t \in [0,2\pi]$
typos in title
Oct
12
asked Show $\left\lvert\sum_{k=-n}^n \frac{\sin k t }{k}\right\rvert \le \pi + 2$ for all $n$ and $t$, $t \in [0,2\pi]$
Oct
9
revised Show $\exists f \in C \ni \|f\|_\infty \le 1$, but it's Fourier series diverges
error in dirichlet summation per davide comment
Oct
9
accepted Show $\exists f \in C \ni \|f\|_\infty \le 1$, but it's Fourier series diverges
Oct
9
asked Show $\exists f \in C \ni \|f\|_\infty \le 1$, but it's Fourier series diverges
Oct
9
accepted Brownian motion: Show $\lim \sum W_{i} (W_{i+1}-W_{i})=\frac12 W^2_t-\frac12 t$ in probability.
Oct
9
revised Brownian motion: Show $\lim \sum W_{i} (W_{i+1}-W_{i})=\frac12 W^2_t-\frac12 t$ in probability.
1/2 factor added per davide answer
Oct
9
comment Brownian motion: Show $\lim \sum W_{i} (W_{i+1}-W_{i})=\frac12 W^2_t-\frac12 t$ in probability.
@Sasha your comment mad me realize I didn't grasp the book argument. I added the complete argument plus the edits ou suggest. Thank you.
Oct
9
revised Brownian motion: Show $\lim \sum W_{i} (W_{i+1}-W_{i})=\frac12 W^2_t-\frac12 t$ in probability.
added more detail from the book, per sash comment
Oct
9
asked Brownian motion: Show $\lim \sum W_{i} (W_{i+1}-W_{i})=\frac12 W^2_t-\frac12 t$ in probability.
Sep
25
comment Show with DCT: $\lim_{R \to \infty} \int_0^{2\pi} \frac{id\theta}{\sqrt{1+R^{-2}e^{-2i\theta}}}=\int_0^{2\pi} id\theta$
@did I believe you are pretty strong, finding this lower bound on the spot. Any idea for the original question?
Sep
25
comment Show with DCT: $\lim_{R \to \infty} \int_0^{2\pi} \frac{id\theta}{\sqrt{1+R^{-2}e^{-2i\theta}}}=\int_0^{2\pi} id\theta$
@did We showed $\sqrt{1+|R|^{-2}-2\Im(R^{-1}e^{-i\theta})} \ge 1-|R|^{-1}$. But I need a lower bound for $\sqrt{1+R^{-2}e^{-2i\theta}}$ to apply the DCT. What is the link between these?