1,653 reputation
1817
bio website nicolasessisbreton.com
location Montreal, Canada
age 30
visits member for 3 years
seen Aug 17 at 16:10

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


Sep
24
comment Show translation is not continuous in $\text{Lip}_\alpha(T)$
@DavideGiraudo Yes, $T$ is the unit circle.
Sep
24
asked Show translation is not continuous in $\text{Lip}_\alpha(T)$
Sep
23
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 see your point. How do I get a lower bound for this radius?
Sep
23
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 So $\|1+iR^{-1}e{-i\theta}\|^2=1+R^{-2}e^{-2i\theta} \ge 1$, and DCT follows. Is it correct?
Sep
23
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 This is the radius of the complex number $1+iR^{-1}e^{-i\theta}$.
Sep
23
asked Show with DCT: $\lim_{R \to \infty} \int_0^{2\pi} \frac{id\theta}{\sqrt{1+R^{-2}e^{-2i\theta}}}=\int_0^{2\pi} id\theta$
Sep
16
accepted Does $\int_a^b \text{Im}(f(t))e^{int}dt=0$ implies $f=0$ on $[a,b]$
Sep
16
revised Does $\int_a^b \text{Im}(f(t))e^{int}dt=0$ implies $f=0$ on $[a,b]$
corrected question per Alex Becker comment
Sep
16
comment Does $\int_a^b \text{Im}(f(t))e^{int}dt=0$ implies $f=0$ on $[a,b]$
@AlexBecker Good eyes, I meant $\text{Im}(f(t))$ is identically zero. Thank you.
Sep
16
asked Does $\int_a^b \text{Im}(f(t))e^{int}dt=0$ implies $f=0$ on $[a,b]$
Sep
16
accepted Does the complex conjugate of an integral equal the integral of the conjugate?
Sep
16
asked Does the complex conjugate of an integral equal the integral of the conjugate?
Sep
10
awarded  Civic Duty
Aug
29
accepted Find $\displaystyle{\int_C \left(1+ \frac{2}{z}\right) dz}$ , where $C(\theta)= e^{i\theta}, 0 \le \theta \le \pi$
Aug
28
asked Find $\displaystyle{\int_C \left(1+ \frac{2}{z}\right) dz}$ , where $C(\theta)= e^{i\theta}, 0 \le \theta \le \pi$
Aug
1
awarded  Yearling
Jun
28
comment Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
Our answers are complementary. But You were of precious help and I think you deserve the points. I learn a lot. Thank you.
Jun
28
accepted Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
Jun
28
comment Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
@CameronBuie I improved the first paragraph of the proof. You are of great help. Thank you for sharing your insight.
Jun
28
revised Show $g(a)= \operatorname*{arg}_{b \in B} \, f(a,b)=0$ is continuous when $g(a)$ is single-valued and $f$ continuous
First paragraph of the proof improved per Cameron comment