– Locate roots of f(x)=0 by the change in sign

If y = f(x) changes sign between x = a and x = b  and f (x) is continuous in this region then a root of f (x) = 0 lies between x = a and x = b.

Example : The equation f(x) = x³ -8x – 10 has a root between x=3 and x=4 as f(3)=27-24-10=-7 and f(4)=64-32-10=22
f(x) changes sign and f(x) is continuous, therefore there is a root between 3 and 4.

– Solve equations by simple iterative methods

Many equations of the form f (x) = 0 can be re–arranged as x = g (x) which leads to the iteration x_n+1 = g(x_n)

If an equation is rearranged as x = g(x) and if there is a root x = á

then the iteration

(i) will converge if -1 < g ‘(α) < 1
(a) will converge without oscillating if 0 <g'(α)<1
(b) Will oscillate and converge if -1 < g'(α)<0
(ii) will diverge if g'(α)<-1 or if g'(α)>1

Example : The equation x³ -8x – 10 = 0 can be rearranged as

x = \(\sqrt{8x + 10}\)

Let x₀ = 3 and use the iteration x_{n+1 = \(\sqr{8x + 10}\) to gi}

x₁ = 3.2396

x₂ = 3.23994………..

And so on

– Newton- Rhapson ‘s Method

The Newton-Rhapson’s method uses an iterative method to calculate the roots of the equation f(x)=0
Suppose the there is a root x₀ , then

x₁ =  x₀= \(\frac{f(x)}{f'(x)}\)

And the process is repeated

– Numerical Integration

The Trapezium rule
The area under the curve y = f(x)is given by
A = \(\int  f(x) \) ≈ \(\frac{1}{2}\)(y₀+ y_n + 2(y₁+ y₂ + y₃+…………y_n-1))

– Numerical Integration

The Trapezium rule

The area under the curve y = f(x) is given by

\(\int_a^{f}f(x)dx\)  ≈    \(\frac{1}{2}(y_0 + y_n + 2(y_1 + y_2 + y_3............y_{n-1))}\)

– Numerical methods

Evaluate \(\int^1_0{\sqrt{x + 2dx}}\) using the values at x=0, x=0.25, x=0.50, x=0.75, and x=1
Then calculate the areas of the trapezium for the overestimate and the underestimate.