Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding a formula from a single variable differentiable function

  1. Aug 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose that f is a differentiable function of a single variable and F(x,y) is defined by F(x,y) = f(x^2 - y)

    Problem: Given that F(0,y) = sin y for all y, find a formula for F(x,y)


    2. Relevant equations



    3. The attempt at a solution

    This is what the tutor had put up on the board, but I couldn't make any sense of it:

    F(0,y) = sin y
    F(0,y) = f(-y) = sin y = sin (-(-y))

    F(x,y) = sin (-(x^2-y)) = sin(y-x^2)


    I understand that he got f(-y), and given the equation at the start that it results in sin y, but I do not understand how he got sin (-(-y)) and onwards.
     
  2. jcsd
  3. Aug 8, 2011 #2
    [itex]F(0,y)=\sin(y)[/itex]
    but also, since [itex] F(x,y)=f(x^2-y)\Rightarrow F(0,y)=f(x^2-y)|_{x=0}=f(-y)\,\,\mathrm{hence}\,\, F(0,y)=\sin(y)=f(-y)\Rightarrow\\ f(y)=\sin(-y)\Rightarrow f(\xi)=\sin(-\xi)\Rightarrow F(x,y)=f(x^2-y)=\sin(-\xi)|_{\xi=x^2-y}=\sin(y-x^2)[/itex]

    Got it?
     
  4. Aug 8, 2011 #3

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Just making it more readable.
     
  5. Aug 8, 2011 #4
    Sorry, sometimes mathematic formalism takes over me :D

    [itex]F(0,y)=\sin(y)[/itex](1)
    but also, since
    [itex]F(x,y)=f(x^2-y)[/itex]if you evaluate [itex]F(0,y)[/itex] you get [itex]f(x^2-y)|_{x=0}=f(-y)[/itex] hence, from (1), you get [itex]F(0,y)=\sin(y)=f(-y)[/itex]. From here, we can deduce the function of a single variable [itex]f(\xi)=\sin(-\xi)[/itex] (we call it [itex]f(\xi)[/itex] for clarity). So [itex]F(x,y)[/itex] is [itex]f(\xi)[/itex] evaluated for [itex]\xi=x^2-y[/itex]: [itex]F(x,y)=f(\xi)|_{\xi=x^2-y}=\sin(-\xi)|_{\xi=x^2-y}=\sin(y-x^2)[/itex]

    Is it clearer?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook