1,665 reputation
416
bio website reddit.com/r/…
location meatspace
age
visits member for 2 years, 6 months
seen 22 hours ago

when someone smiles at me, all I see is an ape bearing its teethe


Jun
16
asked Hedging a long position-one period
Jun
10
comment How did we know to invent homological algebra?
I don't know any homological algebra so maybe this isn't what you're looking for. But I know looking at short exact sequences is sometimes a useful way to look at a normal subgroup of a group and the associated quotient group: $N\rightarrow G\rightarrow G/N$.
Jun
9
comment Factor Rings of Polynomial Rings.
Could you explain how you know $\varphi$ is a homomorphism?
Jun
4
accepted Understanding Conditional Expectation
Jun
4
comment Understanding Conditional Expectation
Ok cool yah you're right, although I think you mean $E[X_k]=\frac{1}{2}$, since the coin is biased.
Jun
4
asked Understanding Conditional Expectation
Jun
2
revised Manipulation of probability integrals
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Jun
2
revised Manipulation of probability integrals
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Jun
2
revised Manipulation of probability integrals
added 203 characters in body
Jun
2
revised Manipulation of probability integrals
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Jun
2
asked Manipulation of probability integrals
May
31
revised Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$
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May
31
revised Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$
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May
31
answered Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$
May
30
revised Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$
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May
30
comment Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$
@Shuhao Cao en.wikipedia.org/wiki/Absolute_continuity#Generalizations_2 also in Royden's Real Analysis
May
30
revised Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$
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May
30
revised Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$
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May
30
comment Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$
@srijan this is essentially what I concluded about the function in my post. But I still don't see how there can be a set of measure zero on which the integral of $f'$ is non-zero.
May
30
comment Showing that $f(x)=x\sin (1/x)$ is not absolutely continuous on $[0,1]$
@Shuhao I know it's not of bounded variation, but I can't see how the measure theoretic definition of not being absolutely continuous can possibly be satisfied.