85,804 reputation
781201
bio website
location Minneapolis, MN
age
visits member for 3 years, 3 months
seen 4 hours ago

I have a Ph.D. with a minor in mathematics and a major in statistics.


1d
revised prove the following equivalence $\ln(-\frac{l-x+\sqrt{r^2+(l-x)}}{l+x-\sqrt{r^2+(l+x)}})={}$ArcSinh$(\frac{l+x}{r})+{}$ArcSinh$(\frac{l-x}{r})$
added 135 characters in body; edited title
1d
revised continuity of bilinear
added 1 character in body
1d
comment Normal Vector Affecting The Divergence Theorem
@guyuzB : In the expression $T =\{(x,y,z): x^2+y^2=z^2, 0\leq z\leq3\}$, you put the {curly braces} outside of MathJax. I edited so now they're inside. That assures proper spacing and alignment and matching fonts, and is standard usage. And of course they need backslashes: T =\{(x,y,z): x^2+y^2=z^2, 0\leq z\leq3\} (I also created a new command making Div and operator name.) ${}\qquad{}$
1d
revised Normal Vector Affecting The Divergence Theorem
added 54 characters in body
1d
comment How to show that an infinite decimal is equal to a unique real number?
@mauna : I added an example. Does that help?
1d
revised Graph of the function $f(x)=$the 1st number in the decimal expansion of $x$
added 6 characters in body; edited title
1d
revised Graph of f(x) = the first number in the decimal expansion of x
added 21 characters in body
2d
comment every Abelian group is a converse lagrange theorem group
$<x>\ \mapsto\ \langle x\rangle$. ${}\qquad{}$
2d
revised every Abelian group is a converse lagrange theorem group
added 65 characters in body
2d
comment Computing $\sum_{i=0}^{\infty}\frac{i}{2^{i+1}}$
I've posted a second answer. It's intuitive rather than logically rigorous.
2d
answered Computing $\sum_{i=0}^{\infty}\frac{i}{2^{i+1}}$
2d
revised Computing $\sum_{i=0}^{\infty}\frac{i}{2^{i+1}}$
edited body
2d
revised geometric interpretation of the norm: $\|\vec x\|={(|x_1|+|x_2|)\over 3}+{2\max(|x_1|,|x_2|)\over 3}$
deleted 28 characters in body
2d
revised What is the limit regarding $a$
edited body
2d
comment Indefinite integral of $\frac{\ln(x)}{(x-3)^2}$
Why do you use \displaystyle when actually displaying the line would clearly work better? I've taken the liberty of changing two lines to dipslayed lines. Lots of posters here do that; I've never known why. ${}\qquad{}$
2d
revised Indefinite integral of $\frac{\ln(x)}{(x-3)^2}$
added 1 character in body
2d
comment Computing $\sum_{i=0}^{\infty}\frac{i}{2^{i+1}}$
@SpamIAm : Done. I actually wrote that fraction before I noticed that I had neglected changing $i=0$ to $i=1$. ${}\qquad{}$
2d
revised Computing $\sum_{i=0}^{\infty}\frac{i}{2^{i+1}}$
added 457 characters in body
2d
answered Computing $\sum_{i=0}^{\infty}\frac{i}{2^{i+1}}$
2d
revised Exponential Form Conversion
added 1 character in body