- #1
thenewbosco
- 187
- 0
I am supposed to prove using taylor series the following:
[tex]\frac{d^2\Psi}{dx^2} \approx \frac{1}{h^2}[\Psi (x+h) - 2\Psi(x) + \Psi (x-h)][/tex] where x is the point where the derivative is evaluated and h is a small quantity.
what i have done is used:
[itex] f(x+h)= f(x) + f'(x) h + f''(x)\frac{h^2}{2!}+...[/itex]
and solved so that
[tex]f''(x)=\frac{2}{h^2}[f(x+h) - f(x) - f'(x) h][/tex]
i am not sure how to get this into the required form..
I noticed that solving the given equation for [tex]\Psi(x)[/tex] gives a term that looks like [tex]\frac{\Psi(x+h) + \Psi(x-h)}{2}[/tex] i.e. average value on the interval, can this be somehow used to write as a derivative or something?
thanks
[tex]\frac{d^2\Psi}{dx^2} \approx \frac{1}{h^2}[\Psi (x+h) - 2\Psi(x) + \Psi (x-h)][/tex] where x is the point where the derivative is evaluated and h is a small quantity.
what i have done is used:
[itex] f(x+h)= f(x) + f'(x) h + f''(x)\frac{h^2}{2!}+...[/itex]
and solved so that
[tex]f''(x)=\frac{2}{h^2}[f(x+h) - f(x) - f'(x) h][/tex]
i am not sure how to get this into the required form..
I noticed that solving the given equation for [tex]\Psi(x)[/tex] gives a term that looks like [tex]\frac{\Psi(x+h) + \Psi(x-h)}{2}[/tex] i.e. average value on the interval, can this be somehow used to write as a derivative or something?
thanks
Last edited: