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 May 25 awarded Popular Question May 19 awarded Notable Question Feb 19 awarded Popular Question Feb 12 accepted Greedy algorithm to make change “getting stuck” Feb 12 asked Greedy algorithm to make change “getting stuck” Feb 8 comment True, false and meaningless statements in math. The language seems to be Georgian. Nkoreli is typing Georgian in the Latin alphabet. Jan 23 accepted How to make recursive computation of $I_n=\int_0^1 \frac{x^n}{x+\alpha}\,dx$ stable? Jan 23 comment How to make recursive computation of $I_n=\int_0^1 \frac{x^n}{x+\alpha}\,dx$ stable? Thanks! That really helped. Jan 23 comment How to make recursive computation of $I_n=\int_0^1 \frac{x^n}{x+\alpha}\,dx$ stable? Umm, we know already what $I_0$ is. We want a formula for $I_n$. I know, for large $n$, $I_0$ is easily calculated, but we are given $I_0$ and asked to find a stable algo for $I_n$. With the recurrence I derived this would involve subtracting something very close to $I_0$ from $I_0$, a classic invite for massive instability. Jan 23 comment How to make recursive computation of $I_n=\int_0^1 \frac{x^n}{x+\alpha}\,dx$ stable? I seem to get $I_0=f(1)-\frac{1}{\alpha}f(2)+\frac{1}{\alpha ^2}f(3)...\frac{1}{\alpha ^ n}I_n$. Is this remotely correct? How on earth is this of any use, especially since I don't know if $n$ is even or odd? This also doesn't look very stable, with the alternating up and down of small quantities... Jan 22 comment How to make recursive computation of $I_n=\int_0^1 \frac{x^n}{x+\alpha}\,dx$ stable? $I_n$ is supposed to be $\int_0^1 \frac{x^n}{x+\alpha}dx$ if it helps. $I_0$ is defined to be $\log \frac{1+\alpha}{\alpha}$ Jan 22 asked How to make recursive computation of $I_n=\int_0^1 \frac{x^n}{x+\alpha}\,dx$ stable? Dec 29 awarded Yearling Dec 17 comment Monty hall problem extended. @Jaydles I don't know, people's intuitions are different. Two years ago, I saw that exact explanation, with a million doors rather than 100 (making it even more "obvious"). I still thought switching made no difference. Nov 20 accepted Proving that $2^n$ is greater than a binomial expression Nov 20 asked Proving that $2^n$ is greater than a binomial expression Nov 16 accepted No idea how to prove this property about symmetric matrices Nov 16 comment No idea how to prove this property about symmetric matrices It is known that symmetric real matrices are orthogonally diagonalizable. Nov 16 asked No idea how to prove this property about symmetric matrices Nov 8 accepted Why does $e$ seem to be an intuitive number?