Could someone please explain this (simple) fact about Taylor expansions?

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.
 

HallsofIvy

Science Advisor
41,618
821
Let [itex]u= x+ \Delta x[/itex] and expand L(u) in a Taylor's series about the point u= x (NOT [itex]\Delta x[/itex]). Then, in that Taylor's series, replace u by [itex]x+ \Delta x[/itex] so that [itex]u- x= \Delta x[/itex].
 
Let [itex]u= x+ \Delta x[/itex] and expand L(u) in a Taylor's series about the point u= x (NOT [itex]\Delta x[/itex]). Then, in that Taylor's series, replace u by [itex]x+ \Delta x[/itex] so that [itex]u- x= \Delta x[/itex].
Thanks - that's very helpful.

Is it correct to argue that when you evaluate [tex]dL/du[/tex] at [itex]u=x[/itex], you can just replace the "u" in the "du" with an "x" and thereby turn [tex]dL/du[/tex] into [tex]dL/dx[/tex] (and so on and so forth, for higher derivatives of [itex]L[/itex])?
 
1,699
5
no you can't argue like that. you have to argue through the chain rule. i.e.

[tex]\frac{dL}{dx}=\frac{dL}{du}\frac{du}{dx}[/tex]

now calculate [itex]\frac{du}{dx}[/itex] and discover why dL/du = dL/dx
 

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