# Strange invocation of Taylor series

1. Jul 28, 2015

### noahcharris

Hi all,

I was working through a chapter on Lagrangians when I cam across this:

"Using a Taylor expansion, the potential can be approximated as
$V(x+ \epsilon) \approx V(x)+\epsilon \frac{dV}{dx}$"

Now this looks nothing like any taylor expansion i've seen before. I'm used to

$f(x) = \sum\frac{f^{(i)}(a)(x-a)^i}{i!}$

What am I missing here?

2. Jul 28, 2015

### SteamKing

Staff Emeritus
You just need to expand the definition of the Taylor series over the first couple of terms. Remember to account for x + ε as the argument of V(x).

3. Jul 28, 2015

### paisiello2

What are the first two terms with a = x + ξ ?

4. Jul 28, 2015

### noahcharris

Looks like it's $f(x) \approx f(x + \epsilon) - f^1(x + \epsilon)\epsilon$

Oh, ok, I see now. So the $\frac{dV}{dx}$ is really a $\frac{dV(x + \epsilon)}{dx}$ ??

5. Jul 28, 2015

### TeethWhitener

You don't even need to use a Taylor series. Just use the definition of a derivative:
$$\frac{dV}{dx}= \lim_{\epsilon \to 0} \frac{V(x+\epsilon)-V(x)}{\epsilon}$$
Therefore, for smaller and smaller $\epsilon$, the above becomes a better and better approximation. The expression from your first post just comes from taking the above equation, dropping the limit, and isolating $V(x+\epsilon)$.

6. Jul 29, 2015

### paisiello2

Sorry, do what SteamKing suggested but also take a=x as where the approximation is taken from. That should give the same answer.

7. Jul 29, 2015

### noahcharris

Ok I finally figured it out. I was wrong in my previous comment. $x \mapsto x + \epsilon$ and $a = x$
Thanks everyone.