– 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) = x^{3} -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 x^{3} -8x – 10 = 0 can be rearranged as
x = \(\sqrt{8x + 10}\)
Let x_{0} = 3 and use the iteration x_{n+1 = \(\sqr{8x + 10}\) to gi}
x_{1} = 3.2396
x_{2} = 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 x_{0} , then
x_{1} = x_{0}= \(\frac{f(x)}{f'(x)}\)
And the process is repeated
\(x_n - \frac{f(x)}{f'(x)}\)– Numerical Integration
The Trapezium rule
The area under the curve y = f(x)is given by
A = \(\int f(x) \) ≈ \(\frac{1}{2}\)(y_{0}+ y_{n} + 2(y_{1}+ y_{2} + y_{3}+…………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.
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