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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.


13h
comment Find t which minimizes ||A(x+ty)-b||$^2_2$
Why did you assume $f(t)=0$?
1d
comment A question about a continuous function that satisfies the property $\forall x\in\mathbb{R},\exists x<y\in\mathbb{R},f(x)<f(y)$
What is the question?
Apr
17
comment Formal definition of convexity for multivariate function?
$f(X) = \operatorname{tr} (I - X_1 X_2)$. $f(\lambda X + (1-\lambda)Y) = \operatorname{tr} (I - (\lambda X_1 + (1-\lambda)Y_1) (\lambda X_2 + (1-\lambda)Y_2))$. Etc, etc.
Apr
17
comment Formal definition of convexity for multivariate function?
The formal definition is the same regardless of space. Look at the first comment above. Think of $f(X) = F(X_1,X_2)$ (where $X = (X_1,X_2)$). en.wikipedia.org/wiki/Convex_function
Apr
17
comment Formal definition of convexity for multivariate function?
Well, just try the scalar version to see that it is not convex (the Hessian is not positive semi-definite).
Apr
17
comment Formal definition of convexity for multivariate function?
Did you mean $f: M \times M \to \mathbb{R}$? In any case, think of it as $f((1-\lambda)M_1+\lambda M_2) \le (1-\lambda) f(M_1)+\lambda f(M_2)$. This way the definition is the same regardless of underlying space.
Apr
17
comment A doubt in the rigorous definition of limits.
However, with different topologies (not an issue here) open sets are used to express limits, so becoming familiar with the idea may be worthwhile.
Apr
17
comment A doubt in the rigorous definition of limits.
You can express the limits using $\le$ instead of $<$ as long as the $\epsilon,\delta >0$.
Apr
17
comment Prove with Lebesgue’s Criterion for integrablility that the composition $f\circ g$ is integrable
You need to show that the composition is bounded.
Apr
17
comment How would Intermediate Value Theorem be used in this question?
$f \to h$. ${}{}{}{}$
Apr
17
comment Laurents Series Expansion Complex Analysis
Yes. I wanted to make the value of $f_{-1}$ clear since there is overlap.
Apr
17
comment Laurents Series Expansion Complex Analysis
Partial fractions and expansion of ${1 \over z-a }$ are useful tools/tricks. ${1 \over z (z-1)(z-2) } = {1 \over 2z} - {1 \over z-1} + {1 \over 2(z-2) }$.
Apr
17
comment Number of zeros of $ z^7+4z^4+z^3+1$
The answers are 4 and 7 respectively.
Apr
16
comment Proving that (0,1) and [0,1] are numerically equivalent.
What does $[0,1] \in A$ mean?
Apr
16
comment Find the coordinates of all points that satisfy certain conditions.
Draw a picture.
Apr
16
comment Prove that Orthogonal Set Is Linearly Independent
Depending on your definition, the inner product is linear (or conjugate linear) in one argument with the other fixed. Hence if either argument is zero the inner product will be zero.
Apr
16
comment Prove that Orthogonal Set Is Linearly Independent
For any inner product, you will have $\langle 0, x \rangle = 0$. The missing part above is that $\langle \sum_i a_i x_i , x_k \rangle = a_k \langle x_k, x_k \rangle= a_k \|x_k\|^2$ if the $x_j$ are orthogonal.
Apr
16
comment Why optimization problems cannot be solved by simple derivative?
If a point is an unconstrained maximum and also satisfies the constraints then it is a solution. But it is 'unlikely' that the unconstrained maximum, if it exists, will satisfy the constraints.
Apr
16
comment Why optimization problems cannot be solved by simple derivative?
Your question is a bit vague. The derivative may be used as part of an algorithm that solves the problem. Even if you have a convex quadratic cost, the solution must depend on the constraints in some way, so it is unrealistic to expect that the derivative of the cost will provide a general, constraint-independent solution. (I may be misunderstanding your questions.)
Apr
16
comment Why optimization problems cannot be solved by simple derivative?
If $f$ is linear, the derivative is constant, so it tells you little about the solution.