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age 28
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Hi.


Jul
28
comment $f(r) \leq \int_r^{r+1} f(t)dt$
Good call. Thanks for your input!
Jul
28
comment $f(r) \leq \int_r^{r+1} f(t)dt$
Nice! Would the statement be true if, instead of $r+1$ as the upper limit of integration, we had $+\infty$?
Jul
28
asked $f(r) \leq \int_r^{r+1} f(t)dt$
Jul
2
awarded  Curious
May
26
accepted Some computation with a power series
May
26
comment Some computation with a power series
Excellent. Thanks again!
May
26
comment Some computation with a power series
Great. In the "easy" case, the supremum is infinite, right?
May
26
comment Some computation with a power series
Thanks for the response! One question: shouldn't the remainder of our Taylor polynomial be $F^{(k)}(\xi)x^k/k!$ rather than what you have written (without the factorial)?
May
26
comment Some computation with a power series
Haha oops! It would've helped to have more quantifiers. I edited.
May
26
asked Some computation with a power series
Apr
18
comment Maximum among $1, 2^{1/2}, 3^{1/3}, 4^{1/4},…$
It might just be me, but this argument is not satisfactory. Yes, I can look at the graph and say 'it looks bigger at 3', but that is not at all rigorous! If I didn't care about rigor, I could get my calculator to give me decimal approximations to these numbers and call it a day.
Apr
18
comment Maximum among $1, 2^{1/2}, 3^{1/3}, 4^{1/4},…$
Why is it "clear" that $|n^{1/n}-e^{1/e}|$ is minimized when $n=3$? Again, it needs justification.
Apr
18
comment Maximum among $1, 2^{1/2}, 3^{1/3}, 4^{1/4},…$
"The closest number in your list to $e^{1/e}$ is $3^{1/3}$" needs justification.
Apr
18
comment Maximum among $1, 2^{1/2}, 3^{1/3}, 4^{1/4},…$
The above link shows that the maximum of $f(x)$ occurs at $x=e$, so you need to figure out which of $2^{1/2}$ and $3^{1/3}$ is larger.
Apr
18
revised Help in Stats, Joint p.d.f
Formatting
Apr
18
comment Joint p.d.f stats help
FYI - you should add the homework tag if you're asking homework questions.
Apr
18
comment Help in Stats, Joint p.d.f
Cov$(X,Y)=E(XY)-E(X)E(Y)$.
Apr
18
suggested suggested edit on Help in Stats, Joint p.d.f
Mar
30
comment Limit at infinity for $e^{-\sin x}$
Okay. I must be misinterpreting your answer then. The way I read your answer is 'this thing is bounded, hence it oscillates and has no limit'.
Mar
30
comment Limit at infinity for $e^{-\sin x}$
@Hey123 Just because something is bounded doesn't mean the function oscillates. Also, just because something oscillates, doesn't mean it has no limit. For example $\sin(1/x)$ is bounded, oscillates, and has a limit as $x$ approaches infinity.