Strange invocation of Taylor series

[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?

 

[SteamKing]

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?

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).

 

[paisiello2]

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

 

[noahcharris]

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

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 $\frac{dV(x + \epsilon)}{dx}$ ??

 

[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).

 

[paisiello2]

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 $\frac{dV(x + \epsilon)}{dx}$ ??

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

 

[noahcharris]

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