Expand the a series of binomial $\left(1-\frac{v^2}{c^2}\right)^{-\frac{1}{2}}$. Enter the first three terms. What is the ratio of third term to second if $\frac{v}{c}=0,1$?
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I assume that "word" you mean "term". In general $$ (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots $$ For your particular case $$ (1-x)^{-1/2} = 1 + \frac{1}{2}x - \frac{3}{8}x^2 + \dots $$ Here $x = v^2/c^2$. Therefore $$ \left ( 1-\frac{v^2}{c^2} \right )^{-1/2} = 1 + \frac{1}{2} \frac{v^2}{c^2} - \frac{3}{8} \frac{v^4}{c^4} + \dots $$ I believe now you should be able to plug in your numerical results and compare the respective terms.
