# Differentials in the argument of a function

Does the following equation make mathematical sense?

$$2 f(t + dt, x) = f(t, x - dx) + f(t, x + dx)$$

Its form appears to resemble a PDE, but I cannot find a way to manipulate the differentials inside the function arguments to demonstrate that idea.

My only thought is to relate $f(x + dx)$ with the definition of the derivative. That is,

$$\frac{df}{dx} = \lim_{h \to \infty} \frac{f(x + h) - f(x)}{h}.$$

However, $h$ is not a differential, and I have run out of ideas. Is my fundamental understanding of a differential variable incorrect, or can the equation above be revived?

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It might have a sensible meaning in the language of differential forms, but I don't know much about it. Although with one variable, $f(t+dt)=f(t)+df$ I think. – anon Jun 1 '12 at 4:07
That would be an extremely useful equation, but it is my intuition that it would fail for most non-linear functions of $f$. I could be wrong though, as I can't think of a proof for either case. – Vortico Jun 1 '12 at 4:18
In light of your comment, let me point out that $dt$ does not stand for "$d$ times $t$" (same with $df$); rather it is the differential. – anon Jun 1 '12 at 4:20
I believe I understand that distinction, but does $h = dx$ in the above definition? If so, $f(x + dx) = f(x) + dx f'(x)$. – Vortico Jun 1 '12 at 4:30
You should replace the terms with their Taylor expansions about $f(t,x)$ and see what comes out. – Rahul Jun 1 '12 at 5:21

It could make sense with a change: Subtract $2f(t,x)$ from both sides of the equation, to get $$2\big(f(t+dt,x)-f(t,x)\big)=f(t,x-dx)-2f(t,x)+f(t,x+dx).$$ If, instead of this, it were $$2\frac{f(t+dt,x)-f(t,x)}{dt} = \frac{f(t,x-dx)-2f(t,x)+f(t,x+dx)}{dx^2}$$ then you would say it is an approximation to the heat equation $$2f_t = f_{xx}$$
Very clear answer! I think that if I look closely at the steps I followed to obtain the first equation, I might find that there should in fact be a factor of $dt$ and $dx^2$. – Vortico Jun 6 '12 at 2:21