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

AI Thread Summary
For a curve to provide a weak extremum, the Taylor expansion of the integral function I(u) requires u to be small to ensure that higher-order terms (O(u^2)) are negligible compared to the zeroth and first-order terms. While the Taylor series can technically be applied for any size u, its utility diminishes for larger values, as the approximation becomes less accurate. The local nature of differentiation means that the behavior of the function is primarily relevant in a small neighborhood around zero. Thus, using small u allows for a valid linear approximation, which is essential in physics and calculus. Overall, the significance of small u lies in maintaining the validity of the approximation by minimizing error from higher-order terms.
Reuben_Leib
Messages
7
Reaction score
1
TL;DR Summary
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##?
 
Physics news on Phys.org
Reuben_Leib said:
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.
 
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.
 
Reuben_Leib said:
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.
 
  • Like
Likes vanhees71 and Reuben_Leib
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top