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age 25
visits member for 1 year, 6 months
seen Jul 9 at 1:26

Graduate Student in Math at Courant Institute, NYU.


Jul
2
awarded  Curious
May
12
reviewed Approve suggested edit on Committee combinatorics
May
6
comment Integrate over different measures
Is this what is called the 'law of the unconscious statistician'?
May
5
answered Closed Graph Theorem
May
5
reviewed Approve suggested edit on Select the approximate values of x that are solutions to $f(x) = 0$, where $f(x) = -7x^2 + 6x + 9$?
May
2
reviewed Approve suggested edit on How to evaluate this integral? $\int \frac {x e^{\arctan(x)}}{{(1+x^2)}^{3/2}} \ dx$
Apr
16
comment If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$
@LAcarguy I don't understand- you can take $a = b = 1/\sqrt{2}$, in which case $2 a b = 1$ holds.
Apr
16
answered If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$
Apr
16
comment If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$
@Lemur This is a good start. You've reduced the problem to showing that the minimum of $a + b$ over the set $a^2 + b^2 = 1, a, b > 0$ is $\sqrt{2}$.
Mar
28
reviewed Reject suggested edit on Isomorphism in cohomology is an isomorphism in homology
Mar
28
reviewed Approve suggested edit on How to find concavity for sine with second derivative $f''(x)= -2(2\sin x-1)(\sin x+1) $?
Mar
17
reviewed Approve suggested edit on modern analysis: limits, integrals, uniform
Mar
17
comment prove that if f is $C^1$ (meaning the derivative $f´$ is continuous) then it can be represented as the sum of an increasing and a decreasing function
By the fundamental theorem of calculus, $f(x) = \int_{a}^x f'(y) dy$. Now decompose $f'$ into its positive and negative parts.
Mar
17
comment There exists an open set such that $vol_n(\Omega)<vol_n(X)+\varepsilon$
@AnaGalois This is because $x \in \Omega = g^{-1}(1 - \epsilon, \infty)$, implies that $1 - \epsilon \leq g(x)$.
Mar
17
comment There exists an open set such that $vol_n(\Omega)<vol_n(X)+\varepsilon$
I apologize for my error. In my head all open sets are '$U$'.
Mar
17
revised There exists an open set such that $vol_n(\Omega)<vol_n(X)+\varepsilon$
Corrected notation
Mar
17
answered There exists an open set such that $vol_n(\Omega)<vol_n(X)+\varepsilon$
Mar
17
comment There exists an open set such that $vol_n(\Omega)<vol_n(X)+\varepsilon$
Then what is the Lebesgue integral to you, and what does it mean for a subset to be Lebesgue measurable for you?
Mar
17
comment There exists an open set such that $vol_n(\Omega)<vol_n(X)+\varepsilon$
I like showing this by showing that the class of all such 'open approximable', Borel measurable subsets is a monotone class, and then using en.wikipedia.org/wiki/Monotone_class_theorem
Mar
3
answered Gaps in the proof