I don't understand part two of the following question:
a) Let $x\in[0;1)$. Prove by using Taylor series that $(1-x)^{-\frac{1}{2}}=1+\frac{x}{2}+\frac{3}{8}(1-\xi)^{-\frac{5}{2}}x^2$.
This was straight forward. I calculated the Taylor series to the first degree to the point $a=0$:
$P_{1,a=0}(x)=f(a)+f'(a)(x-a)=1+\frac{x}{2}$
Afterwards I calculated the remainder:
$$R_{1,a=0}(x)=\frac{f''(\xi)}{(n+1)!}(x-a)^{n+1}=\frac{3x^2}{8(1-\xi)^{\frac{5}{2}}}$$
Adding these two together, I get:
$$(1-x)^{-\frac{1}{2}}=P_{1,a=0}(x)+R_{1,a=0}(x)=1+\frac{x}{2}+\frac{3x^2}{8(1-\xi)^{\frac{5}{2}}}$$
The second part is:
b) According to Einstein, a particle's kinetic energy is given by:
$$E_{kin}(v)=m_0c^2\left(\frac{1}{\sqrt{1-(\frac{v}{c})^2}}-1\right), 0≤v<c$$
Where $m_0$ is the invariant mass, $c$ is speed of light $(3*10^5\frac{km}{s})$ and v is the velocity of the particle.
The classic kinetic energy is defined as:
$$T(v)=\frac{1}{2}m_0v^2$$
The relative error made by replacing $E_{kin}(v)$ with $T(v)$ can be calculated by:
$$E=\frac{E_{kin}(v)-T(v)}{E_{kin}(v)}$$
Show by using the approximating polynomial from the point $a=0$ out of $(1-x)^{-\frac{1}{2}}$ that
$$E<\frac{3(\frac{v}{c})^2}{4(1-(\frac{v}{c})^2)^{\frac{5}{2}}}$$
I feel completely lost on this part. I'd appreciate your help.
You should apply the Taylor expansion with $x=(v/c)^2$ (this you probably guessed?). There is a small catch as you divide by $E_{\rm kin}$ so you also need to produce a lower bound for that. This is, however, not difficult as you may use the very same Taylor expansion and note that the last correction term is positive, so $(1-x)^{-\frac12}-1\geq \frac{x}{2} $.