155 reputation
8
bio website
location United States
age 26
visits member for 3 years, 3 months
seen Dec 15 at 6:21

Oct
31
comment Probability (X >Y) when X and Y have the same distribution?
"Let Y be the next day after X" - i.e. Y = X + 1 (mod 7). The only way that X + 1 > Y is if X = 7. Basically, you're reasoning isn't wrong, it's just that you misread the question. I do this all the time.
Sep
25
awarded  Commentator
Sep
20
answered For what value of K is the following identity
Aug
1
comment Confused by definition of an open set in “All the Mathematics You Missed”
That's what I thought, but I am not nearly confident enough of a reader to say "that's wrong". Thanks.
Aug
1
asked Confused by definition of an open set in “All the Mathematics You Missed”
Apr
16
awarded  Critic
Mar
26
awarded  Popular Question
Jan
9
accepted Average proportion for proportions with different denominators
Jan
9
accepted Limit of $1/x^2$ - Apostol 3.2, Example 4
Oct
29
awarded  Tumbleweed
Oct
22
asked Average proportion for proportions with different denominators
Apr
23
awarded  Scholar
Apr
23
accepted Proof that $\int_1^x \frac{1}{t} dt$ is $\ln(x)$
Feb
15
comment Proof that $\int_1^x \frac{1}{t} dt$ is $\ln(x)$
@BrianM.Scott Yeah, sorry. I looked through and I couldn't find it, but once gingerjin gave a proof I recognized it. I was too impatient, is all. Thanks anyways!
Feb
15
comment Proof that $\int_1^x \frac{1}{t} dt$ is $\ln(x)$
Thank you. This is a really clear explanation!
Feb
15
comment Proof that $\int_1^x \frac{1}{t} dt$ is $\ln(x)$
@JonasMeyer I'll see if I can find something about $e^u$. Maybe that will help me understand this.
Feb
15
comment Proof that $\int_1^x \frac{1}{t} dt$ is $\ln(x)$
@HenningMakholm Apostol's "One Variable Calculus". If there is a proof, I'm missing it.
Feb
15
revised Proof that $\int_1^x \frac{1}{t} dt$ is $\ln(x)$
edited body
Feb
15
asked Proof that $\int_1^x \frac{1}{t} dt$ is $\ln(x)$
Nov
29
comment Limit of $1/x^2$ - Apostol 3.2, Example 4
Oh! I think I get it. Any neighborhood N(0) will contain points such that 0<x<{1\over A+2}, and for those points f(x) > A+2. You can't get around that, no matter what δ you choose. What makes me sure that I get it is that now I don't understand why I didn't see that in the first place.