Sharp EL-5230, EL-5250 Equations that are difficult to solve, Solving y = sin x for y =

Equations that are difficult to solve

Newton’s method has problems in solving certain types of equations, either because the tangential lines it uses to approximate the solutions iterate only slowly toward the correct answer, or because they do not iterate there at all. Examples of such equations include equations of which steep slopes are a feature (e.g. y = 10x–5), periodic functions (e.g. y = sin x), functions featuring an inflection (e.g. y = x3–3x2+ x + 5) and functions where the unknown variable appears as a denominator (e.g. y = 8/x + 1).

Many of those equations may become soluble if a range of expected values is defined that corresponds closely to the real solution.

•For periodic functions such as sin x and cos x, the gradient near peaks or troughs is very shallow. If the initial expected value falls too close to a peak or trough, the calculator may iterate to a totally different cycle of the function and will not obtain an accurate solution. Make sure the initial expected value is an appropriate distance between a peak and a trough.

•Where appropriate, you can try rearranging the equation so that the unknown variable is no longer a denominator.

Appendix

y

x

Solving y =10x – 5 for y = 0. Because of the steep slope, it takes a long time to iterate to the correct solution. Set limits a and b as close as possible either side of your expected solution.

Solving y = sin x for y = 0.

If the Initial expected value is too close to a peak, the calculator will iterate away from the correct solution.

y

x

Solving y = x3 – 3x2 + x + 5 for y = 0.

If the initial expected value is x = 3, no solution is obtained. However, setting x to –3 gives the correct solution of –1.

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