Finding a formula from a single variable differentiable function

[NewtonianAlch]

Homework Statement

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)

Homework Equations

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.

 

[DiracRules]

F(0,y)=\sin(y)
but also, since 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)

Got it?

 

[SammyS]

F(0,y)=\sin(y)

but also, since 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)

Got it?

Just making it more readable.

 

[DiracRules]

Sorry, sometimes mathematic formalism takes over me :D

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

Is it clearer?