S. Verona Lišková
Table of Contents
Equations
The simplest way to solve any equation - except linear or quadratic equations - is by trial and error.
Example: Interpolation
\(\frac{1}{1 + x^2} = 2x\)
\[\begin{array}{|l|l|l|l|} \mathbf{x} & \textbf{LHS} & \textbf{RHS} & \textbf{Difference} \newline 0 & 1 & 0 & 1 \newline 1 & .5 & 2 & -1.5 \newline .4 & .862 & .8 & .062 \ . \end{array}\]
Example: Iteration
Solve the equation for \(x\).
\(x = \frac12(1/1 + x^2)\)
\[\begin{array}{rl} \textbf{trial } \mathbf{x} & \textbf{result } \mathbf{x} \newline 0 \to & .5 \newline .5 \swarrow & .4 \newline .4 \swarrow & .431 \newline \dots \swarrow & \dots \end{array}\]
If the equation is solved in the form \(x = \sqrt{\frac1{2x} - 1}\), however, the results diverge. If the results do diverge, solving the equation in the inverse mannerThat is to say, inverting the side that is not \(x.\) will make them converge.
Example: Newton's method
Solve the equation in the form \(f(x) = 0\).
Let \(x_1\) be the first trial value; then the second trial value is given by the relation \(x_2 = x_1 - f(x_1)/f’(x_1)\). This procedure is equivalent to extrapolating back along the tangent. If the curve is full of wiggles things may diverge.
\[\begin{gathered} f(x) = 1/(1 + x^2) - 2x = 0\newline f’(x) = -2x/(1 + x^2)^2 - 2 \end{gathered}\]
\[\begin{array}{|l|l|l|} \mathbf{x} & \mathbf{f(x)} & \mathbf{f’(x)} \newline 0 & 1 & 2 \newline .5 & 2 & \dots \newline \dots & \dots & \dots \end{array}\]
Problems
- Find the first positive root of \(e^{-x} = \cos x\) by all three methods to 2 decimals. Estimate higher roots.
- Prove \(1 + x + x^2 + x^3 + \dots = 1/(1 - x)\).
- Sum \(1 + \alpha \cos \theta + \alpha^2 \cos 2\theta + \alpha^3 \cos 3\theta + \dots.\)
Solution
Assume \(\vert \alpha\vert < 1\). Then we have:\[\begin{aligned} \sum_{n = 0}^\infty \alpha^n \cos n\theta &= \Re\left[\sum_{n = 0}^\infty \alpha^n (\cos n\theta + i \sin n\theta)\right] \newline &= \Re\left[\sum_{n=0}^\infty \alpha^n e^{in \theta}\right] \newline &= \Re\left[\sum_{n=0}^\infty (\alpha e^{i\theta})^n \right] \newline &= \Re\left[\frac1{1 - \alpha e^{i\theta}}\right]\newline &= \Re\left[\frac1{1 - \alpha \cos\theta - \alpha i\sin\theta}\right]\newline &= \Re\left[\frac{1 - \alpha \cos\theta + \alpha i\sin\theta}{(1 - \alpha \cos \theta)^2 - (\alpha i \sin \theta)^2} \right]\newline &= \frac{1 - \alpha \cos \theta}{(1 - \alpha \cos \theta)^2 + (\alpha \sin \theta)^2} \newline &= \frac{1 - \alpha \cos \theta}{1 - 2\alpha\cos\theta + \alpha^2}. \end{aligned}\]
Power Series
Summation by Integration and Differentiation
\[\begin{aligned} 1(x) - \frac12 x^2 + \frac13 x^3 - \frac14 x^4 + \dots &= S(x) \newline 1 - x + x^2 - x^3 + x^4 + \dots = S’(x) \newline S’(x) = \frac1{1+x}\newline S(x) = \ln(1 + x) + C \end{aligned}\]
Evaluate constant for \(x = 0\):
\[\begin{aligned} S(0) &= \ln 1 + C \newline 0 &= 0 + C \newline S(1) &= \ln 2 \approx .69315 \end{aligned}\]
Convergence: Ratio Test
If the \(\lim_{m \to \infty} x_{m + 1}/x_m < 1\) the series converges.