Reputation
21,898
Next tag badge:
95/100 score
110/20 answers
Badges
2 5 36
Impact
~102k people reached

9h
comment Deriving the inverse Fourier transform without knowledge of the form it will take
@Ryan I started with your assumption in the yellow box, multiplied both sides by $e^{-ixl}$, and integrated over $[-R,R]$ with respect to $x$. The order of integration was then swapped on the right side.
10h
comment How does $u^Tv = p \cdot \|u\|$ follow from the projection onto line?
The inner product of a vector with itself is the square of the norm of the vector: $(w,w)=\|w\|^{2}$. That allows you to write $\frac{p\cdot w}{w\cdot w}w=\left[p\cdot(\frac{1}{\|w\|}w)\right](\frac{1}{\|w\|}w)$, which may look more familiar to you, due to the fact that $\frac{1}{\|w\|}w$ is a unit vector.
12h
comment How does $u^Tv = p \cdot \|u\|$ follow from the projection onto line?
@DaBuj : Your formula at the top, in your problem statement, is the same as mine. I put the vector $\vec{w}$ on the right because the fraction is a scalar. Dot product for a real space is transpose of column vector times a column vector, which is the same as Euclidean dot problem, and the same as an inner product. I'm using $\cdot$ as Euclidean dot product here, but that's optionally a matrix operation. Is there something I'm still missing in this explanation?
13h
answered Does the set of all piecewise constant functions form a subspace of the vector space $\mathbb{R}^\mathbb{R}$ over $\mathbb{R}$?
13h
answered How does $u^Tv = p \cdot \|u\|$ follow from the projection onto line?
15h
comment why is $\overline{span\{e_n\mid n\in\mathbb{Z}\}}=L^2(T)$?
I've asked the moderators to consider deleting your previous version of this question.
15h
awarded  Citizen Patrol
16h
comment A Fourier cosine series of type $f(x)=\frac{a_0}{2}+\sum_{k=1}^\infty a_k \cos(2^k x)$.
@Blezej : $\{ 1,\cos(x),\cos(2x),\cdots\}$ is a complete orthogonal basis of $L^{2}[0,\pi]$. Similarly, $\{ \sin(x),\sin(2x),\sin(3x),\cdots\}$ is a complete orthogonal basis of $L^{2}[0,\pi]$. Pointwise convergence at the endpoints is a separate issue, and does not affect the $L^{2}$ convergence, or the interior pointwise convergence.
16h
answered Deriving the inverse Fourier transform without knowledge of the form it will take
17h
comment Orthonormal basis of $L^2(T)$
For future reference, note that your notation is non-standard because $\int_{T}f(z)dz=\int_{0}^{1}f(e^{2\pi i t})e^{2\pi i t}2\pi i dt$.
18h
revised How numerical radius help us to conclude an operator is normal and partial isometry?
deleted 1 character in body
18h
revised How numerical radius help us to conclude an operator is normal and partial isometry?
added 151 characters in body
1d
revised How numerical radius help us to conclude an operator is normal and partial isometry?
deleted 22 characters in body
1d
revised How numerical radius help us to conclude an operator is normal and partial isometry?
deleted 22 characters in body
1d
revised How numerical radius help us to conclude an operator is normal and partial isometry?
deleted 22 characters in body
1d
answered How numerical radius help us to conclude an operator is normal and partial isometry?
1d
comment Does the identity ${|\cosh z|}^2={\cos}^2x+{\sinh}^2y$ given in my text hold?
@Obinoscopy : Both cannot be right. One expression is bounded for all real $x$ and the other is bounded for all real $y$. It cannot be bounded in both.
1d
revised Orthonomal bases and cross products
added 298 characters in body
1d
comment Correctness of proof that weak convergence implies pointwise convergence in C([0,1])
Yes. $|\int fd\mu| \le \|f\|_{\mathscr{C}[0,1]}\|\mu\|$, where $\|\mu\|$ is the total variation of the complex Borel measure. In fact, $\Phi_{\mu}(f)=\int fd\mu$ has norm exactly equal to $\|\Phi_{\mu} \|=\|\mu\|$.
1d
comment Correctness of proof that weak convergence implies pointwise convergence in C([0,1])
Looks like David and I were posting at the same time.