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visits member for 1 year, 11 months
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Kids in rectangles irritating sick urchins rattling foxes, directory.kirisurf.org lol


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?
Nov
8
comment Finding dual of incredibly complex LP; any trick?
Perhaps... I don't think our prof allows us to request full solutions anyway.
Nov
7
awarded  Custodian
Nov
7
reviewed Approve Why does $e$ seem to be an intuitive number?
Nov
7
comment Why does $e$ seem to be an intuitive number?
Umm, what does the root of 10 have to do with logarithms?
Nov
7
asked Why does $e$ seem to be an intuitive number?
Nov
7
asked Finding dual of incredibly complex LP; any trick?
Nov
5
comment Is 10 closer to infinity than 1?
@LarsH: No, just that most languages have less shitty clauses for comparisons. In Japanese the equivalent would translate as "WRT infinity, is 1 close or 10 close".
Nov
4
comment Is 10 closer to infinity than 1?
@LarsH: English just sucks. Move along and use Lojban, or just almost any other language on the Earth that does not have two ways to parse almost all sentences.