123,126 reputation
478206
bio website math.ubc.ca/~israel
location Richmond, Canada
age
visits member for 3 years, 4 months
seen 21 hours ago

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


10h
awarded  analysis
21h
answered solving system of equations(nonlinear)
21h
answered Prove two solutions of differential equation are the same
2d
answered Definition and analyticity of $T^z$ where $T$ is a positive operator
2d
comment Why does $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$?
Maple gave me a parametric solution which simplifies to $$\left\{ a={{\it \_Z1}}^{4}+2\,{{\it \_Z1}}^{3}+4\,{{\it \_Z1}}^{2}+3 \,{\it \_Z1}+3,b={{\it \_Z1}}^{2}+{\it \_Z1}+2,c=1+{\it \_Z1},d={\it \_Z1} \right\} $$
Jul
23
awarded  Nice Answer
Jul
23
answered Mathematical Intuition Behind Schizophrenic Numbers?
Jul
23
answered a question about sequence and series. prove $ \lim_{n \to \infty}( n\ln n)a_{n}=0$?
Jul
22
revised Optimum set partitioning with constraint
added 194 characters in body
Jul
22
revised Optimum set partitioning with constraint
added 194 characters in body
Jul
22
comment Optimum set partitioning with constraint
The question is nontrivial if $\sum_{i \in A} i \ge m$. For example with $m=5$ and $A = \{1,2,3,4\}$, an optimal solution is $\{1,3\},\{2\},\{4\}$.
Jul
22
answered Optimum set partitioning with constraint
Jul
22
answered Factoring in Maple
Jul
22
comment Choosing random marbles until one is divisible by $X$
The way you stated the problem, $X$ is always $1$: choose one marble and its number will certainly be divisible by $1$. If that's not what you mean, please explain what you do mean.
Jul
22
comment Solution of the Legendre's ODE using Frobenius Method
No. You can take $a_1$ to be anything. The recurrence then tells you want $a_3$ is in terms of $a_1$, what $a_5$ is in terms of $a_3$, etc.
Jul
21
answered Solution of the Legendre's ODE using Frobenius Method
Jul
21
comment Radius of convergence of power series
@Libertron That is indeed what it means. The radius of convergence $\ge r$ because the function is analytic in $\{z: |z - z_0| < r\}$, and $\le r$ because $|f(z)| \to \infty$ as $z$ approaches the pole, so you conclude it is exactly $r$.
Jul
19
awarded  discrete-mathematics
Jul
19
answered Matrix Algebra, Signs of solution
Jul
19
comment Find all positive integers $n$ such that sum of digits of $2^n$ is equal to $n$.
Yes, but the effect on most of the digits is very small.