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location Albany, CA
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visits member for 2 years, 8 months
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If I see further than most, it is because I have stood on the toes of giants and they kicked me high into the air...

However, there is nothing to suggest that I see further than most.

I am (inasmuch as one 'is' an occupation) an engineer. My mathematical skills are rather pedestrian, with a rare insight every now and then. I have been lucky enough to sit in the same room with some famous mathematicians & engineers. My Erdős number is 5, which undoubtedly makes me as unique as an Irishman in a pub. And I have been spotted in The Pub from time to time.

In professional circumstances, my value add has usually been the injection of common sense and completion of grunt work when perspectives and energies are unfocused.

My current age is the smallest number composed of the first two primes that would allow me to have watch on RTÉ the slightly contradictory small step for man and giant leap for mankind.

My real name is Joe Higgins. Apparently, my last name means 'of the Viking' in Irish.

My alma mater is University College, Cork in Ireland, and I had the additional privilege of obtaining my Phd. from the University of California at Berkeley under the enlightened tutelage of Lucien Polak.

I can be reached at joe dot higgins at gmail dot com.


Jun
5
comment All unit vector has bounded components?
$V$ is not necessarily square.
Jun
5
answered All unit vector has bounded components?
Jun
5
answered Integration by substitution help please
Jun
5
comment An approximation of the sign function
@riem: Correct, I was a little sloppy. Since ${1 \over k} T_k$ agrees with $\operatorname{sgn}$ on $|x|>k$, the derivative must be zero there.
Jun
5
comment An approximation of the sign function
@riem: The limit is infinity, not zero. See my comment above.
Jun
5
comment An approximation of the sign function
You can write ${1 \over k} T_k(x) = \max(-1, \min({x \over k},1))$, so the only region it differs from $\operatorname{sgn}$ is $|x| \le k$. The $\operatorname{sgn}$ is a function, albeit not-continuous. The functions $T_k$ are continuous, and differentiable at zero, but $T_k'(0) = {1 \over k}$ which is unbounded.
Jun
5
revised Suppose $f$ is analytic on a connected open set $D$ and $f'(z)= 0$ for all $z \in D$
edited title
Jun
5
comment Suppose $f$ is analytic on a connected open set $D$ and $f'(z)= 0$ for all $z \in D$
You have not used connectedness in your answer, so you cannot conclude that $f$ is constant.
Jun
5
answered Suppose $f$ is analytic on a connected open set $D$ and $f'(z)= 0$ for all $z \in D$
Jun
4
comment Permutations of a string with duplicate characters
...and you left out 'import itertools'...
Jun
3
revised Prove that the continuous function in $[0,1]$ with $f(x)=x+\frac{\sin(f(x))}{2}$ exist and is an unique function.
added 936 characters in body
Jun
3
revised Prove that the continuous function in $[0,1]$ with $f(x)=x+\frac{\sin(f(x))}{2}$ exist and is an unique function.
added 1172 characters in body
Jun
3
answered Prove that the continuous function in $[0,1]$ with $f(x)=x+\frac{\sin(f(x))}{2}$ exist and is an unique function.
Jun
3
comment how to derive the canonical form of a transfer second order equation?
This follows directly from the relevant differential equation.
Jun
3
awarded  continuity
Jun
3
comment Proving u-substitution the hard way — use only definition of integration with partitions
Well, for example, $(\sup_{x \in [a,b]} f(x))(b-a) = (\sup_{x \in [\phi^{-1}(a),\phi^{-1}(b)]} f(\phi(t)))({b-a \over \phi^{-1}(b)-\phi^{-1}(a)} )(\phi^{-1}(b)-\phi^{-1}(a))$. For 'small enough' partitions, ${b-a \over \phi^{-1}(b)-\phi^{-1}(a)} \approx {1 \over (\phi^{-1})'(a)} = \phi'(a)$.
Jun
3
comment Proving u-substitution the hard way — use only definition of integration with partitions
Can you use the implicit function theorem (to show that ${ 1 \over (\phi^{-1})'(t)} = \phi'(\phi^{-1}(t))$).
Jun
3
comment Proving u-substitution the hard way — use only definition of integration with partitions
Did you mean $\phi'(x) >0$?
Jun
2
comment Question about Angle-Preserving Operators
@mSSM: Note that if $T^T T = \sigma^2 I$, then (since ${1 \over \sigma}$ is orthogonal) we have $|\lambda_k|^2 = \sigma^2$ for all eigenvalues of $T$. Does this answer your question?
Jun
2
comment Question about Angle-Preserving Operators
@mSSM: It basically means that $T$ is a scaled rotation.