- #1
AxiomOfChoice
- 533
- 1
My professor just told me that if [tex]\Delta x[/tex] is small, then we can expand [tex]L(x+\Delta x)[/tex] as follows:
[tex]
L(x + \Delta x) = L(x) + \frac{d L}{d x} \Delta x + \frac{1}{2!} \frac{d^2 L}{d x^2} (\Delta x)^2 + \ldots,
[/tex]
where each of the derivatives above is evaluated at [tex]x[/tex]. Could someone please explain why this is true? I just can't see it. Thanks.
[tex]
L(x + \Delta x) = L(x) + \frac{d L}{d x} \Delta x + \frac{1}{2!} \frac{d^2 L}{d x^2} (\Delta x)^2 + \ldots,
[/tex]
where each of the derivatives above is evaluated at [tex]x[/tex]. Could someone please explain why this is true? I just can't see it. Thanks.