P3.6 - Numerical methods

If a function is continuous on an interval and its values at the two endpoints have opposite signs, its graph must cross the x-axis somewhere between them. Therefore the equation f(x)=0 has at least one root in that interval.

State a suitable function f, calculate f(a) and f(b), show that one is negative and the other positive, then state both the sign change and continuity before concluding that a root lies between a and b.

For $f(x)=x^3-x-1$, $f(1.3)=-0.103$ and $f(1.4)=0.344$. Their product is negative and this polynomial is continuous, so $f(x)=0$ has a root in $(1.3,1.4)$. To confirm a value rounds to 3 decimal places, apply the same test at the two rounding boundaries, such as $1.1335$ and $1.1345$ for $1.134$.

An iteration repeatedly substitutes the latest approximation into a given recurrence. For $x_{n+1}=g(x_n)$, start from the stated $x_1$ and use each calculated value to produce the next one; a settled value is a fixed point satisfying $x=g(x)$.

First rearrange the original equation into the required fixed-point form if asked. Enter the given starting value, evaluate the recurrence with the correct brackets, and carry the unrounded calculator value into the next step. Continue for the requested number of iterations or until successive values agree to the requested accuracy.

To approximate the positive solution of $x^2+x-5=0$, use $x_{n+1}=\sqrt{5-x_n}$ with $x_1=2$. Then $$x_2=\sqrt3=1.732050\ldots,$$ $$x_3=\sqrt{5-1.732050\ldots}=1.807747\ldots.$$ Repeating gives values that settle near $1.7913$; substituting this value into the original equation provides a check.