## Numerical

- Numerical Accuracy
- Polynomials - Horner's rule
- Stirling's Approximation for log N!, also Γ
- Mean & standard deviation
- Integration
- Matrices, product, inverse
- Eigen-Values and Vectors
- ← And other topics (left)

### Solving Equations

A zero, `x`

,
of a real-valued *continuous* function, `f(x)`

,
i.e., such that `f(x)=0`

,
can be found by an application of the
binary search
algorithm.
Values ‘`Lo`

’ and ‘`Hi`

’
are chosen such that
(i) `Lo<Hi`

, and
(ii) `f(Lo)<0`

and `f(Hi)>0`

or v.v..

© L . A l l i s o n |

`Mid=(Lo+Hi)/2`

and `f(Mid)`

is computed.
If `f(Mid)`

has the same sign as `f(Lo)`

then `Lo`

is moved up to `Mid`

.
If it has the same sign as `f(Hi)`

then `Hi`

is moved down to `Mid`

.
The algorithm terminates when `Hi-Lo`

is "small".
It is *not* a good idea to wait until f(Mid)=0.

The HTML FORM below can be used to solve a cubic polynomial
(we'll ignore the fact that there are
better ways to solve cubics).
Change the coefficients of the powers of x
and click on the *solve* button to find a zero:

### Exercises

- Find a solution to x
^{3}+x^{2}+x+1=0. What are the other two solutions? - Multiply out f(x)=(x-1)(x-2)(x-3), by hand, and use the FORM to find one of the solutions to f(x)=0.
- What happens if the coefficient of x
^{3}is set to zero, and the other coefficients are set to 1? Why does this happen? Could the problem be fixed?