Infinite series

http://www.physics.miami.edu/~nearing/mathmethods/series.pdf

Series worth knowing

Geometric Series
\sum\limits_0^\infty x^n = 1 + x + x^2 + x^3 + \ldots = \frac{1}{1-x}
Consider: (1 + x + x^2 + \ldots + x^N)(1-x) = 1 - x^{N+1}
so \sum\limits_0^N x^N = \frac{1-x^{N+1}}{1-x}

Power Series

  • e^x = 1 + x + \frac{x^2}{2!} + \ldots = \sum\limits_0^\infty \frac{x^k}{k!}
  • \sin x = x - \frac{x^3}{3!} + \ldots = \sum\limits_0^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}
  • \cos x = 1 - \frac{x^2}{2!} + \ldots = \sum\limits_0^\infty (-1)^k \frac{x^{2k}}{(2k)!}
  • \ln (1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots = \sum\limits_1^\infty (-1)^{k+1} \frac{x^k}{k} for |x| < 1
  • (1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha - 1)x^2}{2!} + \ldots = \sum\limits_0^\infty \frac{\alpha(\alpha-1) \ldots (\alpha - k + 1)}{k!} x^k for |x| < 1 (Binomial series)

Taylor Series
\displaystyle \begin{array}{rl}  f(x) &= f(0) + xf'(0) + \frac{x^2}{2!} f''(0) + \frac{x^3}{3!} f'''(0) + \ldots  \\ &= f(t_0) + (t - t_0)f'(t_0) + \frac{(t - t_0)^2}{2!} f''(t_0) + \ldots \end{array}

Convergence Tests

Comparison test
Let u_k, v_k be sequences of positive reals and \exists k_0 < \infty s.t. \forall_{k < k_0} u_k < v_k and v_k converges, then u_k converges

Ratio test
If for large k, \frac{u_{k+1}}{u_k} \leq x \le 1, then \sum u_k converges. Derived from comparison test with geometric series.

Integral test
If f(x) is decreasing positive and \int_0^\infty f(x) dx converges, \sum\limits_0^\infty f(x) converges

Stirling’s approximation

\displaystyle n! = \Gamma (n + 1) = \int_0^\infty t^n e^{-t} dt = \int_0^\infty e^{-t + n\ln t}

Using newton’s method to approximate this to the 2nd order:

\displaystyle \Gamma (n + 1) \sim n^n e^{-n} \int_{-\infty}^\infty e^{-(t-n)^2 / 2n} dt = n^n e^{-n} \sqrt{2\pi n} (for moderately large n)

The last step in the above makes use of the Gaussian integral:

\displaystyle \int_{-\infty}^\infty e^{- \alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}

using parametric differentiation:

\displaystyle \int_{-\infty}^\infty x^2 e^{-\alpha x^2} dx = -\frac{d}{d\alpha} \sqrt{\frac{\pi}{\alpha}} = \frac{1}{2} \left( \frac{\sqrt{\pi}}{\alpha^{3/2}} \right)

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