I Why does ##u## need to be small to represent the Taylor expansion

[Reuben_Leib]

TL;DR

Taylor series

Necessary condition for a curve to provide a weak extremum.
Let $x(t)$ be the extremum curve.
Let $x=x(t,u) = x(t) + u\eta(t)$ be the curve with variation in the neighbourhood of $(\varepsilon,\varepsilon')$.
Let $$I(u) = \int^b_aL(t,x(t,u),\dot{x}(t,u))dt = \int^b_aL(t,x(t) + u\eta(t),\dot{x}(t) + u\dot{\eta}(t))dt$$
"Taylor’s theorem indicates that, for $u$ sufficiently small, $I(u)$ can be represented by"
$$I(u) = I(0) + u \left(\frac{\textrm{d}I}{\textrm{d}u}\right)_{u=0} + O(u^2)$$
My question is: Why does $u$ need to be small to represent the Taylor expansion of $I(u)$? doesn't the Taylor series work for any size $u$?

 

[kuruman]

My question is: Why does $u$ need to be small to represent the Taylor expansion of $I(u)$? doesn't the Taylor series work for any size $u$?

It does, but then you will not be able to ignore "$+O(u^2)$" because the higher order terms will become significant relative to the zeroth and first order terms.

 

[Dale]

If you take the series expansion to infinite order then yes, it will work for any size $u$. But in this case you are taking only the first term in the expansion, so this will deviate from the original function with errors of order $O(u^2)$. Those will only be small if $u$ is small.

 

[fresh_42]

My question is: Why does $u$ need to be small to represent the Taylor expansion of $I(u)$? doesn't the Taylor series work for any size $u$?

Yes, it does.

Where did you read it? Small values of $u$ are the second nature of physicists if they read, write, or hear Taylor. It is usually applied to discover linear approximations, i.e. you need small, neglectable values for $u^n \;(n>1).$

There is simply not much use of the series for $u\gg 0.$ Furthermore, note that differentiation is a local quality. Differentiable in a neighborhood of $0$ means in a very small radius around $0.$ So we are already limited to a local phenomenon. Why does it make sense to have local behavior in the derivative terms and global in the stretching terms? But yes, formally, there is no restriction for calling the series a Taylor series. However, there is something "small" if we write down the theorem:

Let $f:I\longrightarrow \mathbb{R}$ be a $n$ times continuously differentiable function, and $p\in I,$ Then we have for all $x\in I$
$$
f(x)=\sum_{k=0}^n \dfrac{f^{(k)}(p)}{k!}\,(x-p)^k + \eta(x)(x-p)^n
$$
where $\displaystyle{\lim_{x\to p}\eta(x)=0.}$

The estimation for the remainder term converges to zero.