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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.


Apr
21
comment Find $t$ which minimizes $\|A(x+ty)-b\|^2_2$
Why did you assume $f(t)=0$?
Apr
19
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
18
revised Show that the iteration $x_{n+1} = x_n - 2\frac{f(x_n)}{f'(x_n)}$ converges quadratically to $x_*$ provided $x_0$ is sufficiently close to $x_*$
added 41 characters in body
Apr
18
answered Show that the iteration $x_{n+1} = x_n - 2\frac{f(x_n)}{f'(x_n)}$ converges quadratically to $x_*$ provided $x_0$ is sufficiently close to $x_*$
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
answered How would Intermediate Value Theorem be used in this question?
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
revised Laurents Series Expansion Complex Analysis
added 205 characters in body
Apr
17
answered Laurents Series Expansion Complex Analysis
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
revised Number of zeros of $ z^7+4z^4+z^3+1$
added 30 characters in body
Apr
17
revised Number of zeros of $ z^7+4z^4+z^3+1$
added 30 characters in body
Apr
17
answered Number of zeros of $ z^7+4z^4+z^3+1$