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

[AxiomOfChoice]

My professor just told me that if \Delta x is small, then we can expand L(x+\Delta x) as follows:

<br /> 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,<br />

where each of the derivatives above is evaluated at x. Could someone please explain why this is true? I just can't see it. Thanks.

 

[HallsofIvy]

Let u= x+ \Delta x and expand L(u) in a Taylor's series about the point u= x (NOT \Delta x). Then, in that Taylor's series, replace u by x+ \Delta x so that u- x= \Delta x.

 

[AxiomOfChoice]

Let u= x+ \Delta x and expand L(u) in a Taylor's series about the point u= x (NOT \Delta x). Then, in that Taylor's series, replace u by x+ \Delta x so that u- x= \Delta x.

Thanks - that's very helpful.

Is it correct to argue that when you evaluate dL/du at u=x, you can just replace the "u" in the "du" with an "x" and thereby turn dL/du into dL/dx (and so on and so forth, for higher derivatives of L)?

 

[ice109]

no you can't argue like that. you have to argue through the chain rule. i.e.

\frac{dL}{dx}=\frac{dL}{du}\frac{du}{dx}

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