Axler Linear Algebra What does this notation mean?

Saladsamurai
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In chapter 1 of Axler's LA Done Right, he defines a polynomial as such:

"Our next example of a vector space involves polynomials. A function p:\mathbf{F}\rightarrow\mathbf{F} is called a polynomial with coefficients in
F ..."

Can someone translate this "p:\mathbf{F}\rightarrow\mathbf{F}" into words? I have never seen that notation before.
 
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"p: A->B" means p is a function with domain A and codomain B. Or in other words, p takes elements from the set A and outputs members of the set B.
 
Thanks! How would one know that btw? That is, what subject does one learn that in?
 
Normally, one learns what "F:A-> B" means in basic algebra or pre-calculus.
 
Saladsamurai said:
Thanks! How would one know that btw? That is, what subject does one learn that in?

Normally, one uses textbooks in conjunction with an instructor, and the instructor answers such questions. I guess this board is a (slow) substitute for having an instructor...
 
g_edgar said:
Normally, one uses textbooks in conjunction with an instructor, and the instructor answers such questions. I guess this board is a (slow) substitute for having an instructor...

Yeah. That's the route I am going. My maths seem to be lacking when it comes to the fundamentals. I have completed all of the math in my engineering requirement and have performed very well, but a lot of the base details seem to be missing. Not sure why. But I am reteaching myself all of the math I have learned (and more) from a more 'pure' perspective.

As a result, PF will have to deal with the onslaught of stupid questions that I usually harass my teachers with :redface:

GO PF!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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