# 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:

$$L(x + \Delta x) = L(x) + \frac{d L}{d x} \Delta x + \frac{1}{2!} \frac{d^2 L}{d x^2} (\Delta x)^2 + \ldots,$$

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

Science Advisor
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

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