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?