"More generally, an $n × n$ complex matrix $M$ is said to be positive definite if $z^*Mz$ is real and positive for all non-zero complex vectors $z$; where $z^*$ denotes the conjugate transpose of $z$. This property implies that $M$ is an Hermitian matrix."
@Peter Tamaroff Yeah I saw that in wiki, but then it mentions this a few lines lower: "Some authors use more general definitions of "positive definite" that include some non-symmetric real matrices, or non-Hermitian complex ones. However, some of those extended definitions are incompatible,"
@skullpatrol Guess what? If I remember the correct "said person" who ignores you, I noticed yesterday that said person ignored me. Blatantly. I'm good otherwise, but I get hurt easily, like when ignored.
But why? I don't have a clue. Unless I was found to be rude and interrupting, or maybe because I said "his loss, and not yours!" Oh well...I guess I'm not alone!! :D
@Ethan You could try...there's also a site (beta, stackexchange) called "Academia"
@Ethan Look at practicing researchers/profs that are doing what you'd love to do, doing work you admire...see where they're located...look into the programs...You really need a high quality mentor! (anyone does, but I think you would do well to have one or two, personally speaking - they've made all the difference in my life!)
@Ethan Someone who will nurture your talent, both professionally speaking, but also like an advisor. Someone who takes a personal interest in your progress and work.
Some of the best work I've done, across disciplines, is learning from and work with mentors.
@Ethan It doesn't happen overnight...but those who mentor get satisfaction from doing so, when it's someone eager to learn, and someone who is as talented as you are.
@amWhy Bill duke, a proffesor at ucla, asked me to type up some of my work and send it to him, but I don't know how to do that, what type of program do you type up papers like that in
Most of the greatest have had mentors, across history...
@Ethan Do your best with what you know. Check out TeX.SE for getting latex installed...or manage with word, if you have to, Bill Duke will be forgiving...he won't expect you to give him publishable-quality formatted work!! Just don't miss the opportunity trying to be perfect... :D
Bill Duke wouldn't invite you to send him some of your work unless he meant it...unless he was serious.
@GaryJamesMakinson You are missing the closing quotation mark: “Your work is going to fill a large part of your life, and the only way to be truly satisfied is to do what you believe is great work.*"
In the proof of theorem 3.1 they put : $\langle X,\nabla f \rangle =X(f)$ after that they say that:
for a curve $c$ on $M$ then $\left\langle\dfrac{\mathrm{d}c}{\mathrm{d}t},\nabla f\right\rangle=\dfrac{\mathrm{d}(f\circ c)}{\mathrm{d}t}$
i don't understand why $<X,\nabla f > =1$ ?
please
...
@Gustavo I agree with @shobon that calculus and real analysis (mostly the latter) are a very good place to start. They can provide a gentle introduction to mathematical thinking and proof-writing, augmented with sufficient amounts of intuition that can help you get along. I first learned to properly write proofs in my analysis courses.
The less appealing part of it may be that there's no way around just doing it. You have to write down proofs and have other people read them. The arguments can be perfect in your mind, but if you can't write them down it will result in unnecessarily low marks.
@Charlie he messed up with skew symmetric and hermitian. He said the standard scalarproduct on $\mathbb{C}$ is skew symmetric. I told him in the break that it is called hermetian cause skew symmetric is that $\langle a,b\rangle = - \langle b,a\rangle$. He said that hermetian is only for matrices not for the scalarproduct -.-
and he calls an endomorphism an operator, and every operator is an endomorphism, so why not just saying endomorphism ?
@Dominic However, other books by Conway (notably A Course in Functional Analysis) that I have studied indicate that he does like to leave significant parts of proofs to the reader.
Some cancellation can occur if we calculate
$$\int_{1/n}^n \frac{\log x}{1+x^2} dx$$
Namely, substitute $y = 1/x$, then we get:
$$\int_{1/n}^n \frac{\log x}{1+x^2} dx = \int_n^{1/n} \frac{\log y^{-1}}{1+y^{-2}} \cdot \frac{-1}{y^2} dy = \int_n^{1/n} \frac{\log y}{y^2+1}dy$$
Swapping the integ...
However, I'd split things from $0$ to $1$ and then from $1$ to $\infty$. Integrals involving Catalan's constant, only the sign makes the difference. If you add them up you have $C-C=0$
Split at $1$. On $[0,1]$ it converges since $\int_0^1 \log x=-1$ while it converges on $[1,\infty)$ since $\int_1^\infty \frac{dx}{1+x^2}=\frac{\pi}4$ and the denom dominates the $\log$.
In the proof of theorem 3.1 they put : $\langle X,\nabla f \rangle =X(f)$ after that they say that:
for a curve $c$ on $M$ then $\left\langle\dfrac{\mathrm{d}c}{\mathrm{d}t},\nabla f\right\rangle=\dfrac{\mathrm{d}(f\circ c)}{\mathrm{d}t}$
i don't understand why $<X,\nabla f > =1$ ?
please
...
Some cancellation can occur if we calculate
$$\int_{1/n}^n \frac{\log x}{1+x^2} dx$$
Namely, substitute $y = 1/x$, then we get:
$$\int_{1/n}^n \frac{\log x}{1+x^2} dx = \int_n^{1/n} \frac{\log y^{-1}}{1+y^{-2}} \cdot \frac{-1}{y^2} dy = \int_n^{1/n} \frac{\log y}{y^2+1}dy$$
Swapping the integ...
