471 reputation
26
bio website
location North Carolina
age 19
visits member for 2 years, 5 months
seen Dec 11 at 16:20

I am a high school student with a passion for math. I'm trying to get some exposure to mathematics aside from what I can get at school (although I am fortunate enough to attend a school with a number of courses beyond basic calculus).


Sep
24
awarded  Autobiographer
Jul
4
awarded  Yearling
Jul
4
awarded  Yearling
Jul
19
comment Why do we say the harmonic series is divergent?
I certainly intended no offense, and I regret that my wording offended you. As @did says, I was simply checking your understanding, as your second sentence seems to imply (to me) that any constantly increasing series is divergent regardless of whether it tends to infinity at the limit. We can answer most usefully and clearly if we know whether to focus on the terminology or the proof. I hope there are no hard feelings.
Jul
18
answered Why do these two methods of calculating the probability of winning a best-of-7 series give the same answer?
Jul
18
awarded  Commentator
Jul
18
awarded  Critic
Jul
18
comment Determine if it is possible to fit 2 circles in a rectangle
Unless I misunderstand, shouldn't it be $d[(r_1,r_1), (l-r_2,h-r_2)] \geq r_1+r_2$?
Jul
18
revised Swatting flies with a sledgehammer
Fixed typo in title
Jul
18
suggested approved edit on Swatting flies with a sledgehammer
Jul
18
comment Why do we say the harmonic series is divergent?
Do you know what the word "divergent" means here?
Jul
18
awarded  Editor
Jul
18
revised How many rolls until probability of a 5 is at least 1/2?
improved clarity
Jul
18
comment How many rolls until probability of a 5 is at least 1/2?
That would be the probability of rolling a $5$ every time. An event of probability $P$ occurring $n$ times in a row has probability $P^n$, and the probability of the complement (opposite) of $P$ occurring during $n$ trials is $1-P^n$.
Jul
18
answered How many rolls until probability of a 5 is at least 1/2?
Jul
16
comment Solve for $a$: $V=2(ab+bc+ca)$
Not quite; perhaps I've not explained my hint clearly. Try it this way: I see you refer to "take out the a". Do this from your (correct) expression $\left(\frac{V}{2}\right)-bc=ab+ca$.
Jul
16
comment Solve for $a$: $V=2(ab+bc+ca)$
Since you did that one step forwards and then backwards and got something different, you know that you did something wrong there. If you distribute your $2a$ in what you just wrote, you get $\left(\frac{V}{2}\right)-bc=2ab+2ca$. This is very close to what it should be, but the right is a bit different - there's an extra factor of $2$ on each term. That should tell you what you did wrong.
Jul
16
comment Solve for $a$: $V=2(ab+bc+ca)$
By backwards, I mean take your expression $\dfrac{\left(\frac{V}{2}\right)-bc}{b+c}=2a$ and multiply it by $b+c$. Do you get the expression you had before you divided by $b+c$?
Jul
16
answered Solve for $a$: $V=2(ab+bc+ca)$
Jul
14
answered Does $\sum_{n\ge1} \sin (\pi \sqrt{n^2+1}) $ converge/diverge?