Friday, December 7, 2018

calculus - 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?


Answer



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}$$


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