Error range for the Taylor polynomial Lorentz factor $γ$

Question

Consider the Lorentz factor (in special theory of relativity) as the function

$$γ(x)=\frac{c}{\sqrt{c^2-x^2}},\;x\in[ 0 , c \rangle$$

Where $x=$ is the velocity of an object moving relative to another at rest. $C=$ speed of light in the vacuum.

Wikipedia says that if γ (x) approaches with the Taylor polynomial centered on $ x = 0 $ of second degree $ P_2 (x) $, then the approximation error is as follows:

The approximation $γ ≈ 1 + (1/2)β^2$ may be used to calculate relativistic effects at low speeds. It holds to within 1% error for $x < 0.4 c$, donde $β=x/c$.

I suppose that refers to the error that is obtained with the rest of Lagrange. According to my own calculations this is:

$$R_3(α,x)=\frac{{{\gamma ^{(3)}}(\alpha )}}{{3!}} \cdot {x^3}=\frac{{c\!\cdot\!\alpha\!\cdot\!\left( {{\alpha ^2} + \frac{3}{2}{c^2}} \right)}}{{{{\left( {{c^2} - {\alpha ^2}} \right)}^{7/2}}}} \cdot {x^3},\quad 0<α<x$$

Then the question is:

How can I prove that

$$\color{blue}{R(α,x)<1\%, \;\,\textrm{if}\; x < 0.4 c\,?}$$

Answer 1

I am not sure that I properly understood the problem.

Taylor expansion is $$\gamma(x)=\frac{c}{\sqrt{c^2-x^2}}=1+\frac{x^2}{2 c^2}+\frac{3 x^4}{8 c^4}+O\left(x^6\right)$$ So, you want $$\frac{3 x^4}{8 c^4} \lt \frac 1 {100}\implies x^4<\frac {8c^4} {300}\implies |x| <\sqrt[4]{\frac 8 {300}}\,c\approx 0.404\,c$$ If you consider the next term of the expansion, it is $$\frac{5 x^6}{16 c^6}$$ and for the above value of $x$, it would be $\frac{1}{300 \sqrt{6}}\approx 0.00136$ which is one oreder lower.

Answer 2

The value is normally found as follows:

Given $\gamma=(1-\beta^2)^{-1/2}$, when is the approximation $1+\beta^2/2$ good?

$$\gamma=1+\frac{\beta^2}{2}+\cdots$$

It's clear from the Taylor series that as $\beta\rightarrow 0$ the approximation becomes exact so that the relative error strictly improves below the $\beta$ where the relative error is $1\%$: (if you do not believe this, take the derivative of the relative error and confirm it is positive when $\beta\in(0,1)$ - this is the proof)

$$\frac{\gamma-(1+\beta^2/2)}{\gamma}=0.01$$

Solving this you will find the relevant solution is approximately $\beta=0.398$ which is usually quoted as $0.4$.

In your question you are calculating the absolute error. Since $\gamma\approx 1$ it's amusing that the absolute error $=\gamma-(1+\beta^2/2)$ is also approximately $0.01$ when $\beta=0.4$, and the same proof strategy works for this error measure.

(n.b. this means your question, interpreted strictly, is false - the error is just barely above $1\%$ at $0.4c$)