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location Albany, CA
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visits member for 2 years, 3 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 divides the current year.

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.


17h
comment Improper double integral evaluation by changing the order of integration
It is good that you pay attention to the detail, many people are sloppy with such things.
1d
comment Computing a contour integral over curve not centered at origin
And its not defined if the circle passes through zero.
1d
comment Proof of subspace, finding a basis in polynomials
Suppose $p \in P_2$, then $p(t) = p_0+p_1 t + p_2 t^2$, and so $p'(t)+tp(0) = p_1+(p_0+2 p_2)t$. The representation of $p$ in the basis is $(p_0,p_1,p_2)^T$ and the representation of $t \mapsto p'(t)+tp(0)$ is $(p_1, p_0+2 p_2, 0)^T$.
1d
comment Prove this function is absolute continuous and Lipschitz of order $\alpha$
+1: Nice! ${}{}{}{}$
1d
revised Proof of subspace, finding a basis in polynomials
added 109 characters in body
1d
answered Proof of subspace, finding a basis in polynomials
1d
comment Linear Transformation $T(A)=A^t$
Michelle, you need to distinguish between $I$ the identity matrix and $I$ the identity mapping on the space of matrices. $T$ is a linear operator on the space of matrices, it is not a matrix.
1d
comment Linear Transformation $T(A)=A^t$
@GitGud: I think I understand what the confusion is. If $T(A) = A^T$, then $T(T(A)) = A$. So, we write $T^2 =I$, where $T^2$ means $T$ composed with itself, and $I$ is the identity mapping on the space of matrices, that is $I(A) = A$.
1d
comment Linear Transformation $T(A)=A^t$
What has $T(A) = A^T$ got to do with anything?
1d
comment Linear Transformation $T(A)=A^t$
What is the question?
1d
revised The Lipschitz property of an upper envelope
added 58 characters in body
1d
comment The Lipschitz property of an upper envelope
How do you conclude that ${{\sup }_{\alpha }}\left| f\left( {{x}_{1}} \right)-{{f}_{\alpha }}\left( {{x}_{1}} \right) \right|={{\sup }_{\alpha }}\left| {{f}_{a}}\left( {{x}_{2}} \right)-f\left( {{x}_{2}} \right) \right|=0$? In general this can't be true. Take $f_\alpha(x) = \alpha$ with $\alpha \in [0,1]$. Then $f(x) = 1$, of course, but $\sup_\alpha |f(x)-f_\alpha(x)| = 1$ for all $x$.
1d
answered The Lipschitz property of an upper envelope
1d
comment Fast way to inverse B'CB+D
Any matrix $B: \mathbb{R}^n \to \mathbb{R}^m$, with $n>m$ must have a non-trivial kernel (that is, there is some $x\neq 0$ such that $Bx=0$. Hence $B^T CB x = 0$ also). This follows from the reduced row echelon form.
1d
comment If $x^T\!Ay=0$ for all $x,y $ in $ \mathbb{R}^n$ then $A=0$
Your comment is correct, this is another way of looking at it.
1d
answered If $x^T\!Ay=0$ for all $x,y $ in $ \mathbb{R}^n$ then $A=0$
1d
comment Fast way to inverse B'CB+D
If $n>m$ then it is impossible for $A$ to be invertible.
1d
comment 3D reconstruction from 2 images with baseline and single camera calibration
You need to express this as a mathematical problem. As is, it depends on knowledge of what you mean by Q, F, H1,H2,...
1d
comment find the minimum value of $x^2-6x+9+ \dfrac{64}{x^2}$
The function is convex on $(-\infty,0)$ and $(0, \infty)$ separately.
1d
answered Linear mapping in an inner product space