457 reputation
518
bio website
location Southern California
age 17
visits member for 2 years, 3 months
seen 23 hours ago

Hi, I'm Cole. Here's my Twitter. I'm the creator of iDecryptIt, and the founder of Hexware.

profile for Cole Johnson on Stack Exchange, a network of free, community-driven Q&A sites" title="profile for Cole Johnson on Stack Exchange, a network of free, community-driven Q&A sites


Aug
12
revised How do I solve $\vert x\vert^{x^2-2x} = 1$?
fixed up english
Aug
12
suggested suggested edit on How do I solve $\vert x\vert^{x^2-2x} = 1$?
Jul
19
suggested suggested edit on Deducing an optimal gambling strategy (using martingales).
Jul
19
suggested suggested edit on Maximal ideals and maximal subspaces of normed algebras
Jul
19
suggested suggested edit on Continuity of a function in two variables
Jul
15
revised Interview riddle
removed smily
Jul
15
suggested suggested edit on Interview riddle
Jul
15
suggested suggested edit on Solving base e equation
Jul
14
revised How do I solve $y' = \sin(x - y)$?
fixed grammar
Jul
14
suggested suggested edit on How do I solve $y' = \sin(x - y)$?
Jul
12
comment Prove that $\sqrt[2012]{2012!}<\sqrt[2013]{2013!}$
Fun fact related to this question: The function $f(x) = \sqrt[x]{x}$ has a maximum at $(e, \sqrt[e]{e})$ and is decreasing on the right of the maximum to infinity. Also, the derivative of $\sqrt[x]{x!}$ is $e^{-1}$.
Jul
12
comment Prove that $\sqrt[2012]{2012!}<\sqrt[2013]{2013!}$
@Lucian Why would that help? $\frac{2013!}{2012!}$ is just $2013$ ($\frac{n!}{(n-1)!} \equiv n$)
Jul
12
revised Why does $1+2+3+\dots = -\frac{1}{12}$?
removed unneeded equation
Jul
12
suggested suggested edit on Why does $1+2+3+\dots = -\frac{1}{12}$?
Jul
12
comment Funny identities
Related: Why does $1+2+3+\dots = -\frac{1}{12}$?
Jul
12
revised Funny identities
Fixing dumped link
Jul
12
comment Funny identities
@ChantryCargill Not when you consider the fact that not many people know (surprisingly) that $\sqrt{x} \equiv x^{\frac{1}{2}}$
Jul
12
revised Funny identities
moving comments into answer
Jul
12
comment How do you find this product?
@DavidRicherby Stuck on a test because you can't find the product? Well, with Amazon's (brand new) Yesterday Shipping™, you'll already have the product! ;)
Jul
12
comment How do you find this product?
Or more roughly put, $P \approx 0$.