decimal places as α = 1.353
Example for problems with change of sign method
If the curve touches the x-axis without crossing it, there will be no change of sign, so change of sign methods are doomed to failure. For example, has a repeated rational root, which can be expressed as a recurring decimal i.e = 0.3333……
Sketch graph of f (x) = (3x-1)2(2x+4) is drawn below, you would notice that there is no change of sign on the graph as well as the table. It seems that there is repeated root between the interval 1<x<2 , it touches the x-axis when you looking at the graph, however, the table shows that all the values of f(x) within the interval 1<x<2 are all positive, there is no change of sign. You are unable to do any further by decimal search for a root without a change of sign.
Newton-Raphson Iteration
This method is based on the iteration.
xn+1 = xn - , n = 0,1,2,3,…….with initial approximation x0.
The Newton – Raphson iterative formula is based on evaluating the gradient of the tangent to the curve y = f(x) at
x = x0
The gradient of the tangent at (x0, f (x0) is f ′ (x)
f ′ (x0) =
Rearranging this,
⇒ x1 = x0 –
To solve the equation ex-7x-3, f (x) = ex-7x-3⇒ f ′ (x) = ex-7. This will give rise to the Newton Raphson iteration formula, that is,
xn+1 = xn –
You will then be able to find the upper root α, let the initial approximation x0 = 4
etc
The sequence converge rapidly towards the upper root to give α=3.2478 to 5 s.f.
In this case, Newton-Raphson Iteration gives an extremely rapid rate of convergence. This is the case for examples, even when the first approximation is not particularly good. For manual caluations it is almost always the most efficient method.
You may try to solve the equation using spreadsheet, it gives a
Error Bounds for this root can be established for 5 s.f. by evaluating f(x) for x = 3.2478 ± 0.00005.
f (3.24775) = e3.24775 – 7 (3.24775) –3 = -0.001873116 < 0
f (3.24785) = e3.24785 – 7 (3.24785) – 3 = 0.000000251 > 0
Since there is a change of sign in f(x) over the interval 3.24775<x<3.24785, and the function is continuous, there must be a root in this interval, i.e. α=3.2478 to 5 s.f.
Fixed Point Iteration
The equation f(x) = 0 is rearranged into the form x = g(x). Roots of the equation x = g(x) are therefore roots of the equation f(x) = 0.
This gives rise to the iterative formula x n+1 = g (x n), where n = 0,1,2,3… with initial approximation x0.
The sequence x0, x1, x2, x3… will converge to a root α of the equation x = g(x) provided a suitable starting value x0 is chosen and –1<g′(α)<1.
The equation y = x3-6x+1, f (x)=0 is plotted on the graph below
f (x) = x3-6x+1 = 0 may be re-arranged in a form g(x) as follows:
⇒
⇒
The x-intercept of the graphs of y = x and g (x) =
represent roots of the original equation x3+4x2-3x-7 = 0
The re-arrangement leads to the iteration
To find the middle root α, let initial approximation x0 = 2
x1 = x03+1 = 23 + 1 = 1.5
6 6
x2 = x13+1 = 1.53 + 1 = 0.7291666667
6 6
x3 = x23+1 = 0.72916666673 + 3 = 0.231281045
6 6
x4 = x33+1 = 0.2312810453 + 3 = 0.1687285727
6 6
x5 = x43+1 = 0.1672857273 + 3 = 0.1674672649
6 6
The sequence converges rapidly towards the middle root to give α = 0.16745( to 5 s.f.)
Since g (x) = , g′(x) = x2, hence g′(α) is estimated as g′(0.16745) =
× 0.167452=0.1402 (to 5 s.f.)
The convergence of the iteration is confirmed since –1 <g′(α) < 1
To find the upper root β, let initial approximation xo=3
x1 = x03+1 = 33 + 1 = 4.666666667
6 6
x2 = x13+1 = 4.6666666673 + 1 = 17.10493827
- 6
x3 = x23+1 = 17.104938273 + 1 = 834.2573752
- 6
x4 = x33+1 = 834.25737523 + 1 = 96771821.22
- 6
x5 = x43+1 = 96771821.223 + 1 = 1.510412228E23
- 6
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