33,049 reputation
270126
bio website sos440.tistory.com
location Los Angeles, CA
age 27
visits member for 3 years, 5 months
seen 5 hours ago

I am currently a UCLA student, particularly interested in probability theory.

I love mathematics, especially calculating some sort of integrals and summations with inner aesthetic appealing.

Language fluency: (G:green, A:amber, R:red)

  • Korean: Native
  • English: (Reading: G, Listening : A, Speaking : R, Writing : A)
  • Japanese: (Reading : A, Listening : A, Speaking : A, Writing : R)

6h
comment How to compute $\int_{-\infty}^\infty\exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx$
+1, The second lemma is impressive!
19h
comment How to compute $\int_{-\infty}^\infty\exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx$
+1 for its potent to be generalized (to a higher degree, maybe?).
19h
comment How to compute $\int_{-\infty}^\infty\exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx$
+1, This is also what came to my mind first.
19h
awarded  Explainer
Sep
28
answered Computing $\lim_{N\to\infty}\sum_{i = 1}^N \sum_{j = 1}^N \frac{(-1)^{i+j}}{i+j}.$
Sep
19
awarded  Enlightened
Sep
19
awarded  Nice Answer
Sep
16
awarded  Good Answer
Sep
15
answered An integral of Wolstenholme:$\int_0^{+\infty}\frac{\sum_1^n A_k\cos{a_k x}}{x}\mathrm {d} x$ where $\sum A_k=0$ and $a_k>0$
Sep
11
awarded  Nice Answer
Sep
2
awarded  Nice Answer
Aug
24
comment How can I prove that $a^n + b$ is composite?
Hint: $2^{33} + 1 = 2^{33} - (-1)^{33}$. Do you remember any formula that factors $a^n - b^n$?
Aug
24
awarded  Nice Answer
Aug
23
awarded  Good Answer
Aug
19
revised How prove this limit $\left(\frac{1}{2}+\sum_{k=1}^{n-1}(-1)^{\lfloor\frac{mk}{n}\rfloor}\{\frac{mk}{n}\} \right)^n=\frac{1}{\sqrt{e}}$
added 612 characters in body
Aug
19
answered How prove this limit $\left(\frac{1}{2}+\sum_{k=1}^{n-1}(-1)^{\lfloor\frac{mk}{n}\rfloor}\{\frac{mk}{n}\} \right)^n=\frac{1}{\sqrt{e}}$
Aug
19
comment How prove this limit $\left(\frac{1}{2}+\sum_{k=1}^{n-1}(-1)^{\lfloor\frac{mk}{n}\rfloor}\{\frac{mk}{n}\} \right)^n=\frac{1}{\sqrt{e}}$
Observations suggest that the sum is equal to $(n-1)/2n$, but I am still seeking for a proof.
Aug
19
revised Closed form of a complex series sum
added 213 characters in body
Aug
19
answered Closed form of a complex series sum
Aug
17
awarded  Nice Answer