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location Grinnell, IA
age 19
visits member for 1 year, 5 months
seen 2 hours ago

Currently enrolled in Grinnell College

Majoring in math


1d
revised Proving that this function is negligible
added 66 characters in body
1d
answered Proving that this function is negligible
1d
comment Proving that this function is negligible
How about using the fact that $e^{x/2}<2^x$ and Taylor expansion for getting some nice inequality for $2^{\sqrt{n}}$?
2d
comment Putnam definite integral evaluation $\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$
Substitution $y=\tan^2 x$.
2d
comment How to select the right books?
I always choose the books required in the corresponding course of MITOCW, especially the ones with the solution available online or at the end of the book. For the level beyond that, I just pick the book most highly recommended by the experts in Math Overflow.
2d
comment Motivation for Putnam (soft question)
Thanks for your link!
2d
asked Motivation for Putnam (soft question)
2d
comment $ \lim_{n \to \infty} \int_0^{\frac{\pi}{2}} \sum_{k=1}^{n} \left( \frac {\sin kx}{k} \right)^2 \, \mathrm{d}x $
You can express $\sin^2 kx$ in terms of $\cos 2kx$ and a constant term. Separate them, and use integration by parts.
Dec
17
comment Sketch the graph of $y=(g-f)(x)$ given the graphs of $g(x)$ and $f(x)$
The above graph says $g(0)=1$ and $f(0)=-4$. So, $(g-f)(0)=5$.
Dec
17
comment Sketch the graph of $y=(g-f)(x)$ given the graphs of $g(x)$ and $f(x)$
I mean you made mistakes at the points of $x=-1,0,1$, since these points should have positive y-coordinate. But other part is fine.
Dec
17
comment find the possible values of z
Let $y=iw$. The conditions can be restated as $|z|\leq 1$, $|y|\leq 1$, $|z+y|=2$. The equality holds if and only if $z=y$ and $|z|=1$. Therefore, $z$ can be any complex number with absolute value 1.
Dec
17
comment Sketch the graph of $y=(g-f)(x)$ given the graphs of $g(x)$ and $f(x)$
The points of $-1\leq x\leq 1$ should be on the upper half plane, since in that region $g$ is non-negative, and $f$ is negative and therefore $g-f$ should be positive.
Dec
17
comment How do I create a transformation matrix from a polynomial transformation?
Let $p(x)=\sum_{k=0}^n a_kx^k$. Then $\int_0^1p(x)dx=\int_0^1\sum_{k=0}^n a_kx^k dx=\sum_{k=0}^n \frac{a_k}{k+1}$. So, $N(T)$ is the set of polynomials such that the above sum becomes 0. Since the set clearly consists of more than one element, T is not one-to-one. Since the sum can take an arbitrary real number if you take an arbitrary polynomial, T is onto.
Dec
15
awarded  Caucus
Dec
15
comment Magnitude of a complex expression
The magnitude should be real, but that's not real. So, the problem is not yet solved.
Dec
15
comment Magnitude of a complex expression
Yes. I just factored the denominator $3+2i\lambda\sin\theta$ to the two factors.
Dec
15
comment Magnitude of a complex expression
Does this help? $ \frac{2 + (1-2ia\lambda \sin \theta)^{1/2}}{3 + 2ia\lambda\sin\theta}=\frac{2 + (1-2ia\lambda \sin \theta)^{1/2}}{(2 + (1-2ia\lambda \sin \theta)^{1/2})(2 - (1-2ia\lambda \sin \theta)^{1/2})}=\frac{1}{2 - (1-2ia\lambda \sin \theta)^{1/2}}$
Dec
14
comment Differential Geometry of Curves and Surfaces
Thanks for all the responses to my question. I will try to grasp the essence of this topic, especially Gauss-Bonet theorem, as quickly as possible by reading Do Carmo's corresponding book, which is fortunately not that time-consuming due to its mathematical simplicity.
Dec
14
accepted Differential Geometry of Curves and Surfaces
Dec
14
asked Differential Geometry of Curves and Surfaces