How to choose the correct function to use for a Taylor expansion?

[zenterix]

Homework Statement

I have a question regarding a mathematical tool used a lot in physics apparently: Taylor expansions.

Relevant Equations

Here is the context.

I calculated the electric field at a particular point in 3D space.

$$E_p=\begin{cases}
\frac{2k_eQ}{R^2}(1-\frac{z}{(z^2+R^2)^{1/2}}) & \text{for } z \geq 0\\
\frac{2k_eQ}{R^2}(-1-\frac{z}{(z^2+R^2)^{1/2}}) & \text{for } z < 0\\
\end{cases}$$

And I am interested in knowing what happens when $z>>R$

First, to isolate just the mathematical operation of Taylor expansion, let's consider

$$\frac{z}{(z^2+R^2)^{1/2}}$$
$$=\frac{1}{(1+\frac{R^2}{z^2})^{1/2}}$$

When $z>>R$ we also have $\frac{R}{z}$ and $\frac{R^2}{z^2}$ both approach zero.

Consider two different Taylor expansions.

First, let $f_1(s)=(1+s)^{1/2}$

$$f_1'(s)=-\frac{1}{2(1+s^{3/2})}$$

Near $s=0$, we have the first order Taylor expansion
$$f_1(s) \approx 1 - \frac{s}{2}$$

Now consider a different choice for $f(s)$

$$f_2(s)=(1+s^2)^{1/2}$$
$$f_2'(s)=-\frac{s}{(1+s)^{3/2}}$$
$$f_2''(s)=-\frac{1}{(1+s)^{1/2}}+\frac{3s^2}{(1+s^2)^{5/2}}$$

Near $s=0$, we have the first order Taylor expansion
$$f_2(s)\approx 1$$

and the second order Taylor expansion
$$f_2(s) \approx 1 - \frac{s^2}{2}$$

My question is simply: how do we choose which approximation to use?

Considering the original context of what happens to the electric field if $z>>R$, it appears that the choice of Taylor expansion doesn't affect the result. I guess I am trying to wrap my head around the intuition of why this is.

If I use $f_1$, then

$$E_p=\begin{cases}
\frac{2k_eQ}{R^2}(1-1+\frac{R^2}{2z^2}) & \text{for } z \geq 0\\
\frac{2k_eQ}{R^2}(-1-1+\frac{R^2}{2z^2}) & \text{for } z < 0\\
\end{cases}$$

$$E_p=\begin{cases}
\frac{k_eQ}{z^2} & \text{for } z \geq 0\\
\frac{k_eQ}{z^2}-\frac{4k_eQ}{R^2} & \text{for } z < 0\\
\end{cases}$$

If I use $f_2$, then

$$E_p=\begin{cases}
\frac{2K_eQ}{R^2}(1-1+\frac{R^4}{2z^4}) & \text{for } z \geq 0\\
\frac{2K_eQ}{R^2}(-1-1+\frac{R^4}{2z^4}) & \text{for } z < 0\\
\end{cases}$$

$$E_p=\begin{cases}
\frac{k_eQ}{z^2} & \text{for } z \geq 0\\
\frac{k_eQ}{z^2}-\frac{4k_eQ}{R^2} & \text{for } z < 0\\
\end{cases}$$

ie, same result in both cases.Here are some plots

 

[PeroK]

My question is simply: how do we choose which approximation to use?

It depends how precise an answer you want.

 

[zenterix]

It depends how precise an answer you want.

Inoticed that both calculations give the same result ultimately. The result differs in the variable $s$ but it is the same in $R$ and $z$.

 

[PeroK]

Inoticed that both calculations give the same result ultimately. The result differs in the variable $s$ but it is the same in $R$ and $z$.

I'm not sure I know what that means.

 

[zenterix]

What I mean is that if you just consider the functions $f_1$ and $f_2$ as I defined them, they are in fact two different functions of $s$. The Taylor expansions will be different for each function.

However, we actually defined $s$ differently in each case. For $f_1$ we had $s=\frac{R}{z}$ and for $f_2$ we had $s=\frac{R^2}{z^2}$. When we substitute these $s$ expressions into the respective Taylor expansions, the Taylor expansions are the same.

 

[PeroK]

What I mean is that if you just consider the functions $f_1$ and $f_2$ as I defined them, they are in fact two different functions of $s$. The Taylor expansions will be different for each function.

However, we actually defined $s$ differently in each case. For $f_1$ we had $s=\frac{R}{z}$ and for $f_2$ we had $s=\frac{R^2}{z^2}$. When we substitute these $s$ expressions into the respective Taylor expansions, the Taylor expansions are the same.

That makes sense. They must be the same.

 

[zenterix]

I think I've gotten some practice some with Taylor expansions :)

 

[kuruman]

Homework Statement:: I have a question regarding a mathematical tool used a lot in physics apparently: Taylor expansions.
Relevant Equations:: Here is the context.

I calculated the electric field at a particular point in 3D space.

$$E_p=\begin{cases}
\frac{2k_eQ}{R^2}(1-\frac{z}{(z^2+R^2)^{1/2}}) & \text{for } z \geq 0\\
\frac{2k_eQ}{R^2}(-1-\frac{z}{(z^2+R^2)^{1/2}}) & \text{for } z < 0\\
\end{cases}$$

And I am interested in knowing what happens when $z>>R$

I would assume that you are interested in knowing what happens to the electric field, not part of the electric field, when $z>>R$. In that case, your function should be the entire expression for the electric field, not just a piece of it. That is always safe. You have $$E_p=\begin{cases}
\frac{2k_eQ}{R^2}(1-\frac{z}{(z^2+R^2)^{1/2}}) & \text{for } z \geq 0\\
\frac{2k_eQ}{R^2}(-1-\frac{z}{(z^2+R^2)^{1/2}}) & \text{for } z < 0\\
\end{cases}.$$ You can expand the top function in Taylor series and then note that if you subtract $2\times \dfrac{2k_eQ}{R^2}$ from the top function, you get the bottom function. So if you expand the top function to any order, you immediately have the bottom function to the same order. That is also safe.