The algorithm was presented without proof to avoid others claiming the idea as their own. The method is clearly the same as the formula method of solution. He showed the solution of particular equations, for example: x2 - 10x = -9. He called the numbers by the names:
1 is "the square"
-10 is "the unknown"
-9 is "the absolute number"
Brahmagupta's work yields only one solution for each quadratic equation. However, by the 12th century, scholars were identifying both solutions. Problems were now being written to teach mental skills rather than to solve problems from astronomy.
x - 2.3) = x2- 3.8x + 3.45 the quadratic function can be form
This is a quadratic function.
When x is 3 this quadratic function is zero.
x2 - 4x + 3
x = 3
32 - 4 × 3 + 3
9 - 12 + 3
The Factor Theorem states that if a function of xis zero when x = 3, then (x - 3) is a factor of the function.
f(3) = 0 (x - 3) is a factor of f(x) = x2 - 4x + 3
The other factor can then be found to give the completely factorised form of the quadratic.
By multiplication it is found that the other factor is (x -1) and f(x) = (x - 3)(x -1)
The Factor Theorem states that if f(a) = 0 then (x - a) is a factor of the function f(x).
(x - a) is a factor of f(x) because f(a) = 0
The process can be used in reverse to find a quadratic function with particular properties. Find the function with factors (x - 3) and (x + 5)
The function with factors (x -3) and (x + 5) is
(x - 3)(x + 5)
x2 + 2x - 15
Sometimes functions with higher powers of x can be factorised by this method .
The process of obtaining the factors becomes longer and sometimes involves long-division.
f(x) = x3 - 12x2 + 47x - 60 f(5) = 0 (x - 5) is a factor.
Long division takes practise. Here is an example:
Divide: f(x) = x3 - 12x2 + 47x - 60 by the factor (x -5) to find the other factors
x cubed divided by x gives x squared multiply the factor (x -5) by x2 to give x3 - 5x2 subtract this from the first two terms and then divide the answer by x This gives -7x and after multiplication 7x2 + 47x Subtract to leave 12x which when divided by x gives 12
Long division is not essential. Other factors can be found using the Factor Theorem again. Factorising continues with the quadratic factor.
However, here h tends to 0. All terms here contain an h value, apart from 3x�. So therefore it simplifies to... 3x� + (3 x 0 x x) + 0� = 3x�. Therefore, this is the gradient function. This also corresponds to the increment methods in the tables.
Second order of convergence: In order to find the first order of convergence Errors: To begin with I will look at a graph of the approximation of the root compared to real value of the root. If we look at this graph we can be certain that our answer is merely an approximation.
the point of intersection between y = x and y = g(x) = (-4x2+2)1/5 is between x = 0 and x = 1. Like before, I will take my starting value of x (x1) to be 0.5. The same process of substituting values of x into the iterative formula, as in the last Rearrangement is carried out here.
root is found then I apply 3 to then something strange happen as shown: This sequence is not converging to the target root but instead it diverges to infinity. To prove this furthermore a spread sheet is set up to give more evidence n xn g(xn)
This is a lot quicker in excel compared to using pen, paper and calculator. It also allows the recording of all calculations. A spreadsheet is instantly responsive to changed input values which enables exploration of the effect of variables within a process (Mathematical Association, 2002, p39)
x ? ? For y = tan x, domain is ?/2 ? x ? ?/2 . The inverse function of y = sin x is y = arcsin x The inverse function of y = cos x is y = arccos x The inverse function of y = tan x is y = arctan x Trig Identities sin(x y)
3rd 18 From the table above, it shows there are 36 calculations for change of signs, 5 calculations for Newton Raphson and 18 calculations for x=g(x). To conclude, Newton Raphson method is the quickest in find the root, followed by x=g(x) method and the slowest is change of signs method.
With Mandelbrot?s set, we can determine the fate of the orbits. For example, when we substitute the constant, z0, as 1, and keep the seed of the iteration as 0: z1 = 02 + 1 = 1 z2 = 12 + 1 = 2 z3 = 22 + 1 =
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