Applications of Taylor Series

Final exam: Wednesday, March 22, 7-10pm in PCYNH 109. Bring ID!
Last Quiz 4: Today (last one)
Today: 11.12 Applications of Taylor Polynomials
Next; Differential Equations

This section is about an example in the theory of relativity. Let $ m$ be the (relativistic) mass of an object and $ m_0$ be the mass at rest (rest mass) of the object. Let $ v$ be the velocity of the object relative to the observer, and let $ c$ be the speed of light. These three quantities are related as follows:

$\displaystyle m = \frac{m_0}{\displaystyle \sqrt{1-\frac{v^2}{c^2}}}$   (relativistic) mass

The total energy of the object is $ mc^2$:
$ E = mc^2.
In relativity we define the kinetic energy to be

$\displaystyle K = mc^2 - m_0 c^2.$ (6.4)

What? Isn't the kinetic energy $ \frac{1}{2} m_0 v^2$?

Notice that

$\displaystyle mc^2 - m_0 c^2 = \frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}} - m_0 c^2
= m_0 c^2 \left[ \left(1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}} - 1\right].


$\displaystyle f\left(x\right) =
\left(1 - x\right)^{-\frac{1}{2}} - 1

Let's compute the Taylor series of $ f$. We have

$\displaystyle f(x)$ $\displaystyle = (1 - x)^{-\frac{1}{2}} - 1$    
$\displaystyle f'(x)$ $\displaystyle = \frac{1}{2}(1 - x)^{-\frac{3}{2}}$    
$\displaystyle f''(x)$ $\displaystyle = \frac{1}{2}\cdot \frac{3}{2} (1 - x)^{-\frac{5}{2}}$    
$\displaystyle f^{(n)}(x)$ $\displaystyle = \frac{1\cdot 3 \cdot 5 \cdots (2n-1)}{2^n} (1-x)^{-\frac{2n+1}{2}}.$    


$\displaystyle f^{(n)}(0) = \frac{1\cdot 3 \cdot 5 \cdots (2n-1)}{2^n}.


$\displaystyle f(x)$ $\displaystyle = \sum_{n=1}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$    
  $\displaystyle = \sum_{n=1}^{\infty} \frac{1\cdot 3 \cdot 5 \cdots (2n-1)}{2^n \cdot n!} x^n$    
  $\displaystyle = \frac{1}{2}x + \frac{3}{8}x^2 + \frac{5}{16}x^3 + \frac{35}{128}x^4 +\cdots$    

We now use this to analyze the kinetic energy ([*]):

$\displaystyle mc^2 - m_0 c^2$ $\displaystyle = m_0 c^2 \cdot f\left(\frac{v^2}{c^2}\right)$    
  $\displaystyle = m_0 c^2 \cdot \left(\frac{1}{2} \cdot \frac{v^2}{c^2} + \frac{3}{8}\cdot \frac{v^2}{c^2} + \cdots\right)$    
  $\displaystyle = \frac{1}{2} m_0 v^2 + m_0 c^2 \cdot \left(\frac{3}{8} \frac{v^2}{c^2} + \cdots\right)$    

And we can ignore the higher order terms if $ \frac{v^2}{c^2}$ is small. But how small is ``small'' enough, given that $ \frac{v^2}{c^2}$ appears in an infinite sum?

William Stein 2006-03-15