Prove that any function z = f(x + y) solves the equation z'x - z'y = 0

[Addez123]

Homework Statement:

Prove that all functions $$z = f(x + y) $$
solves the equation $$z'_x - z'_y = 0$$

Relevant Equations:

$$z = f(x + y) $$
$$z'_x - z'_y = 0$$

How about I prove that to be false instead?
$$z = x$$
$$z'_x(x + y) = 1$$
$$z'_y(x + y) = 0$$
$$z'_x - z'_y = 1$$

 

[Orodruin]

Homework Statement:: Prove that all functions $$z = f(x + y) $$
solves the equation $$z'_x - z'_y = 0$$
Homework Equations:: $$z = f(x + y) $$
$$z'_x - z'_y = 0$$

How about I prove that to be false instead?
$$z = x$$
$$z'_x(x + y) = 1$$
$$z'_y(x + y) = 0$$
$$z'_x - z'_y = 1$$

Incorrect. Your $z$ is not a function of $x+y$, it is just a function of $x$.

For $f(t) = t$, you would obtain
$$z = f(x+y)= x+y$$
The statement is trivially true for this choice of $f$.

 

[Mark44]

Homework Statement:: Prove that all functions $$z = f(x + y) $$
solves the equation $$z'_x - z'_y = 0$$
Homework Equations:: $$z = f(x + y) $$
$$z'_x - z'_y = 0$$

Presumably these are partial derivatives. Primes usually aren't used for partial derivatives.
Instead, you could use $z_x$ to denote the partial of z with respect to x, or $\frac{\partial z}{\partial x}$.
Similar for $z_y$.

How about I prove that to be false instead?

Why?

$$z = x$$

?
It's given that z is some function of x + y, not just x.

$$z'_x(x + y) = 1$$
$$z'_y(x + y) = 0$$
$$z'_x - z'_y = 1$$

I have no idea what you're doing in the three lines above.

You could let $u = x + y$, meaning that u is a function of x and y, and z = f(u). What is $\frac{\partial z}{\partial x}$? Think about the chain rule for partial derivatives.

 

[Orodruin]

Presumably these are partial derivatives. Primes usually aren't used for partial derivatives.

It is actually more common than you might think. I have seen it in at least one textbook. However, I agree that $z_x$ is neater and easier on the eyes ...

 

The question is missing the assumption that the partial derivatives must exist.

 

[Orodruin]

The question is missing the assumption that the partial derivatives must exist.

Technically yes (although stating that $f$ is a differentiable function seems more in line with the problem). However, the context where the question was asked is important. If the question has been asked in the context of something like a mathematical methods for physics/engineering/etc, then typically functions are implicitly assumed to be smooth.

 

[Addez123]

The way I read it is: any function into which I input (x+y) will solve the equation.
Which isn't true obviously but how should I read it?

 

[Orodruin]

The way I read it is: any function into which I input (x+y) will solve the equation.
Which isn't true obviously but how should I read it?

It is true (with the addition that the function needs to be differentiable). Your counter example does not satisfy the premise that it is a function into which you have put $x+y$ as the argument.

 

[Addez123]

It is true (with the addition that the function needs to be differentiable). Your counter example does not satisfy the premise that it is a function into which you have put $x+y$ as the argument.

Ah yes! That DOES make sense!
Thanks :p