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So I've got a function here: $$m(v)= \frac {M}{\sqrt{1- \frac{v^2}{c^2}}}$$ which basically states that the mass $m$ of an object with rest mass $M$ (a positive constant) changes with its velocity $v$. In the given equation above, $c$ is the speed of light, and is another constant aside from $M$.

The kinetic energy, as a function of velocity, is then given by: $$K(v) = m(v)c^2 - Mc^2$$

And so I am trying to find the degree 2 Taylor Polynomial $T_2(v)$ of $K(v)$ near $v=0$.

I started off by calculating for $K(0)$, which is $0$. Then I went off to calculate for $K'(v)$ and $K''(v)$ and input $v = 0$ to both functions, respectively. The weird thing was that both of them return $0$. So when I tried substituting all the values to the Taylor Polynomial formula: $$K(v)+K'(v)(x-v)+ \frac {K''(v)}{2}(x-v)^2$$ $$0 + 0(x-0) + \frac{0}{2}{x-0}^2 = 0$$

And an answer of $0$ does not make sense. So I probably made a mistake while calculating for the first and second order derivatives of $K(v)$, which was honestly pretty confusing as there are a lot of constant-variables.

I would appreciate any help here!

Edit:

To answer the comments, $$K(v) = \frac{Mc^2}{\sqrt{1- \frac{v^2}{c^2}}} - Mc^2$$ $$K(0) = Mc^2 - Mc^2 = 0$$ Since $M$ and $c$ are both constant, their derivatives are $0$. $$K'(v) = -\frac{\frac{Mc^2v}{\sqrt{1-\frac{v^2}{c^2}}}}{1-\frac{v^2}{c^2}}$$ $$K'(0) = 0$$

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  • $\begingroup$ Can you show us your workings for $K''(v)$, i.e. the double derivative with respect to $v$, and then for $K''(0)$? It should not be $0$ $\endgroup$
    – Henry
    Dec 5, 2021 at 1:56
  • $\begingroup$ you should definitely recheck your differentiation. Make sure you applied the quotient/product and chain rules correctly. You should be getting $K(v)=\frac{Mv^2}{2}+\cdots$ (i.e the elementary formula for kinetic energy). $\endgroup$
    – peek-a-boo
    Dec 5, 2021 at 2:07
  • $\begingroup$ If $v\ne 0$ then $K(v)=K(v)-K(0)=vK'(0)+(v^2/2)K''(0)+(v^3/6)K'''(x)$ for some $x$ strictly between $v$ and $0$.... BTW, let $y=-v^2/c^2$ and $r=-1/2.$ You can use Newton's Generalized Binomial Theorem $(1+y)^r=\sum_{n=0}^{\infty}y^nC(n,r)$ where $C(0,r)=1$ and $C(n+1,r)=C(n,r)\cdot\frac {r-n}{n+1}$ for each $n$. This theorem holds for all real $r$ and for all real $y$ such that $|y|<1.$ $\endgroup$
    – DanielWainfleet
    Dec 5, 2021 at 4:25

1 Answer 1

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I get $K''(0)=Mc^3\left[\frac{1}{(c^2-v^2)^{3/2}}+\frac{3v^2}{(c^2-v^2)^{5/2}}\right]_{v=0}=M$.

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