# Travis

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bio website location age 32 member for 1 year, 6 months seen 2 days ago profile views 28

I'm earning my PhD in mathematics and have been employed as an analyst since graduating college in 2003.

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 Mar4 revised Differentiable function with bounded derivative, yet not uniformly continuous Since the derivative is not continuous and depends on the point x, I changed f' to f'(x) in two places. Mar4 answered Differentiable function with bounded derivative, yet not uniformly continuous Mar4 comment Differentiable function with bounded derivative, yet not uniformly continuous Your $f'=0$ is nice, but it requires a while to decipher to get there. This is where my mind was headed, but I was trying to get it done with shrinking neighborhoods centered on the peaks of sin waves. I suppose I don't need to constructively define my segments of $X$ though, so my solution below just occurred to me. Mar1 awarded Organizer Mar1 revised Differentiable function with bounded derivative, yet not uniformly continuous added tags Mar1 suggested suggested edit on Differentiable function with bounded derivative, yet not uniformly continuous Mar1 comment Differentiable function with bounded derivative, yet not uniformly continuous My one change would be, I am wondering about the stronger $f'$ bounded on $\mathbb{R}$. Mar1 comment Differentiable function with bounded derivative, yet not uniformly continuous Exactly my question too, beat me to it by a day. ;-) Feb22 answered How do you describe your mathematical research in layman's terms? Nov5 comment Are the following two matrices similar? When OP said we don't know the theory of lambda-matrix, I think they were saying not to use the characteristic polynomial. Sep28 revised Surprising identities / equations Added the parentheses for the cosines to improve readability. Sep28 suggested suggested edit on Surprising identities / equations Sep28 comment Funny identities Hmm, considering that logarithms get at the exponent, and $x$ has a constant exponent ... Since $\ln\left(x^a\right)=a\ln\left(x\right)$ (the log of an expression equals the exponent times the log of the base), then $\ln\left(x^1\right)=1\ln\left(x\right)=x^0\ln\left(x\right)$ might be saying something to the effect that it's more important that your exponent is a constant, than the fact that the log of your base $\ln\left(x\right)$ is growing slowly. Jun6 awarded Teacher Oct22 comment Combinatorial Identity $(n-r) \binom{n+r-1}{r} \binom{n}{r} = n \binom{n+r-1}{2r} \binom{2r}{r}$ Starting with $n+r-1$ & maybe picking someone NOT on one team & using alg identity is overly complicated. Simpler: start with $n+r$, $n$ can be umpire, $r$ cannot. RHS: pick ump, then from $n+r-1$ left, pick $2r$ & divide them into 2 teams of $r$. LHS: pick the ump, then pick $r$ from $n+r-1$ left to be on a team. From the $n$ not on team one, pick $r$ to be on team 2. Whoops, our ump might be on team two. Forget him for a bit (divide by $n$). Pick someone from the $n-r$ left. If he can be ump, good. Otherwise swap him with the orig ump. All combos from RHS can be gotten this way. Sep26 comment Discussion: Differing definitions for the rank of a set @AsafKaragila: I guess I don't see what you mean by there are new sets using the second definition. When I construct $V_\omega$, $\omega$ is not yet a set so I wouldn't be asking what its rank is. When I construct $V_{\omega+1}$ I now have the set $\omega$ and it's a subset of $V_\omega$ so it's rank is $\omega$. So I don't ever have a set whose rank has not been defined yet. It looks like a "labeling" difference to me. I don't see a hierarchical or structural difference. Sep26 accepted Discussion: Differing definitions for the rank of a set Sep26 accepted What is the name of $V_\alpha$? Sep25 awarded Editor Sep21 awarded Custodian