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age 25
visits member for 1 year, 8 months
seen Aug 1 at 18:19

Graduate Student in Math at Courant Institute, NYU.


Jul
28
comment A strange trigonometric equation
Setting $u = \cos^2 x$, one obtains an equivalent fourth-order polynomial equation in $u$, for which there is an explicit formula. But this probably isn't "easier".
Jul
21
comment Poincaré Recurrence Theorem (measure theory version)
If $f$ is a function, then the inverse image $f^{-1}$ commutes with unions and intersections.
Jul
21
comment Poincaré Recurrence Theorem (measure theory version)
@math12 If $f : \Omega \to \mathbb{R}$, then every point of $\Omega$ is assigned a value in $\mathbb{R}$, and $\mathbb{R} = \cup_{n \in \mathbb{Z}} (\epsilon n, \epsilon (n + 1)]$.
Jul
16
answered Poincaré Recurrence Theorem (measure theory version)
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
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 $vol_n(\Omega)<vol_n(X)+\varepsilon$