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I'm an Emeritus Associate Professor of Mathematics at University of British Columbia and an Optimization Algorithms Researcher at D-Wave Systems in Burnaby BC


4h
answered Finding a matrix with the following property
4h
comment Radius of convergence and complex power series
Once you have it for first derivative, mathematical induction gives it to you for all orders.
4h
comment $2^{q-1}\equiv 1\pmod{q}.$
No, it can't be asking that. Look at the second sentence. You need not only to consider $2^{q-1} \mod q$, but also $2^k \mod q$ for $1 < k < q-1$.
4h
comment Easy way to check for a valid solution in this triple equality?
You can, but you still need two equations, otherwise you lose some information. So you can use equation 1 and equation 2, or you can use equation 1 and (equation 1 - equation 2), or ...
4h
answered Basic question on the transformation of Exponential distribution.
1d
answered About compact operator
1d
comment Is't true that for a linear transformation $T:\mathbb R^n\to\mathbb R^n,~T$ is positive definite $\iff\langle Tx,x\rangle>0~\forall~x\ne 0$
$x^t T x$ and $\langle Tx, x\rangle$ are different notations for the same thing (at least when the scalars are real).
1d
comment funcitonal series convergence… SOS…
This is not a duplicate. The other question just asks about the radius of convergence: the real question here is about the limit of $f(x)$ as $x \to \infty$.
2d
comment funcitonal series convergence… SOS…
Or are you asking whether $f(x)$ has a finite limit as $x \to +\infty$?
2d
comment funcitonal series convergence… SOS…
What do you mean by "even $x \to$"? For every given $x$, the series converges. It doesn't converge uniformly, though, if that's what you're asking, because for every $k$ there are $x$ that make the $k$'th term as large as you please.
May
21
answered Prove that if $p(A)=0$ where A is a matrix of a linear operator ($A \in L(V)$), $p(\lambda)=0$ if $\lambda \in \sigma(A)$
May
21
answered Incomplete space
May
20
comment Results following from Analyticity on a domain
Yes, that's right.
May
20
comment Easy way to check for a valid solution in this triple equality?
If the $x$ equations were $$\eqalign{x_1 + 2 x_2 + 3 x_3 &= 0\cr 4 x_1 + 5 x_2 + 6 x_3 &= 0\cr}$$ then the $y$ inequalities would be $$\eqalign{y_1 + 4 y_2 &< 0\cr 2 y_1 + 5 y_2 &<0\cr 3 y_1 + 6 y_2 &<0\cr}$$
May
20
comment Easy way to check for a valid solution in this triple equality?
Each inequality for $y$'s corresponds to an $x$ variable, and each $y$ variable corresponds to an equation for the $x$'s. The coefficients are the same. Thus the coefficient of $x_2$ in the first equation $(-1)$ is the coefficient of $y_1$ in the second inequality.
May
20
revised Why the root of this tree has to be “1”?
added 1 characters in body
May
20
answered Why the root of this tree has to be “1”?
May
20
answered Results following from Analyticity on a domain
May
20
revised Easy way to check for a valid solution in this triple equality?
added 578 characters in body
May
20
answered Easy way to check for a valid solution in this triple equality?