1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding a left inverse

  1. Sep 28, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##S ## be a cylinder defined by ##x^2 + y^2 = 1##, and given a parametrization ##f(x,y) = \left( \frac{x}{ \sqrt{x^2 + y^2}}, \frac{y}{ \sqrt{x^2 + y^2} },\ln \left(x^2+y^2\right) \right)## , where ##f: U \subset \mathbb R^2 \rightarrow \mathbb R^3 ## and ## U = \mathbb R ^2 /{(0,0)}##

    1. Find a left inverse of ##f##
    2. Show that ##f## is injective.

    2. Relevant equations


    3. The attempt at a solution
    I'm not even sure where to begin with this, My professor has done a very poor job of explaining how he wants us to go about these problems, and the book doesn't help much at all. The first thing that confuses me is that ##S## isn't a cylinder, it's a circle.

    In class, we did inverses of simple functions which you could fairly easily solve, such as ##f(x,y) = (\sin xy, y)##, where we would just set it equal like this: ##y' = y## and ##\arcsin x' = xy##, then we would have ##f^{-1} (x',y') = (\frac{\arcsin (x'y')}{y'},y')##.

    Also regarding the second question, I'm not aware of anyway to show that a function is injective unless its derivate is a square matrix, in which case you can take the determinant of the derivative and see where it's nonzero.
     
  2. jcsd
  3. Sep 28, 2016 #2

    Mark44

    Staff: Mentor

    In R2, you are correct: the equation ##x^2 + y^2 = 1## represents a circle. In R3, though, the same equation represents a cylinder whose central axis lies along the z-axis. Your function f is a map from R2 to R3.
    What's the definition of a left inverse that you are using?
     
  4. Sep 28, 2016 #3
    The standard one: http://mathworld.wolfram.com/LeftInverse.html

    I see what you're saying about ##S##, I didn't realize that at first.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Finding a left inverse
Loading...