Numerical Analysis and Simulation

The number system now in commonest use worldwide is positional. Consider a number such as 924.37. The position of each digit in the number indicates how significant it is. The digit 9 represents 100s, the 2 10s, the 4 1s, and so on.

The Roman number, or 'numeral' system, is non-positional. The value of a character in a number does not depend on its position only. Rather, the value of a character depends on what the character is and on neighbouring characters. Consider a number such as XLIV. In this case the 'X' means ten but only as a modifier for the 'L' which stands for fifty, so that the combination 'XL' means 40.

There are several that are especially important:

• integers: ..., -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ...
• rational numbers: ie, numbers that can be written as quotients of integers, such as 1/2, 7/8, etc.
• irrational numbers: ie, numbers that cannot be written as quotients of integers, pi, for example
• complex numbers: consisting of a pair of real numbers, and different arithmetic
• numbers 'modulo' a prime, similar to clock arithmetic, where, for instance, 5+9=2 modulo 12.

Some of these are subsets of the reals, others aren't.

There are an infinite series of numbers that are divisible by 228. Two of them are 456 and 684.

On the other hand, if the questioner had meant "what is 228 divisible by?", then the answer is 2, 3, and 19, of which 2 applies twice.
1, 2, 3, 4, 6, 12, 19, 38, 57, 76, 114, 228

I don't have access to maple at the moment so I can't give you the answer as I'm not going to spend 30mins doing an algorithm a computer could do in 1 second, but I will tell you what the algorithm is/how to solve it.

In order to solve a non-linear system numerically by Newton's method, you need to reduce it into a linear system. Recall the newton's method for single equations of one variable, xn+1=xn-f(xn)/f'(xn) the essential theory behind the derivation of this equation is by expressing f(x) as a Taylor polynomial, and assuming that the error (x-x0) is so small that errors of higher terms can be ignored, giving 0≈f(x0)+(x-x0)f'(x0) and solving this equation for x gives us our iterative equation above.

The theory for non-linear systems is similar to this in that instead of having g(x)=x-f(x)/f'(x), we have G(x)=x-A(x)-1F(x) where G(x) and F(x) are vector functions and A(x) is some invertible matrix whose entries are functions from Rn->R i.e. from non-linear to linear. By dividing our non-linear vector function F(x) by the matrix A(x) we are essentially linearising our system. I won't go over the theory of it but as it happens, our matrix A(x) is the Jacobian matrix of our system so we have G(x)=x-J(x)-1F(x)

This gives us an iterative formula x(n+1)=x(n)-J(x(n))-1F(x(n)) however for our algorithm we modify this slightly so that we do not have to calculate the Jacobian and the system at each point. We do this by setting y(n)=-J(x(n))-1F(x(n)) in other words before we start our algorithm we find out what -J(x)F(x) is and set that to y(I am omitting the (x) for simplicity's sake) so that instead of having to calculate the Jacobian and the system separately and then multiply them together, we just calculate y, turning our formula into x(n+1)=x(n)+y(n). This also has the effect of making our formula more accurate by reducing round off errors, as instead of having to do three calculations, finding the value of the Jacobian and the system then multiplying them together, we only have one calculation, finding the value of y.

This means our algorithm is:

INPUT: y(x), initial approximation x=(x1,...,xn), tolerance TOL, maximum number of iterations N.

step 1: set k=1

step 2: if k≤N then do steps 3-6, otherwise do step 7

step 3: calculate y(x)

step 4: set x=x+y(x)

step 5: if y(x)print (k,x)); STOP #comment, this is not meant to be the norm of y but rather the norm between the old x and the x just calculated i.e. we're calculating the difference between approximations as an absolute error value and comparing it against an acceptable degree of error, the tolerance. If it is less than the tolerance the approximation is deemed to be 'good enough' and is given as the solution to the system along with the number of iterations it took to get it.

step 6: set k=k+1, return to step 2.

step 7: OUTPUT ('Procedure unsuccessful, can not find solution within tolerance'); STOP.

And there's how to solve a non-linear system by Newton's method. For your problem this means that F(x)=(x2+y-11, y2+x-7), J(x)=([2x,1],[1,2y]) i.e. 2x and 1 on the top row of the matrix and 1 and 2y on the bottom, J(x)-1=1/(4xy-1)*([2y,-1],[-1,2x]) and y=-J(x)-1F(x)=1/(4xy-1)*(2x2y+y2-22y-x+7, 2xy2+x2-14x-y+11). Note that I have used bold parentheses to indicate a vector, or a matrix with rows separated by [], and that for your initial approximation x=(x1, x2), x1 equates to the coordinate of x, and x2 to the coordinate of y.

Important note: as with most iterative methods, the success of the algorithm is highly dependent on what initial value you choose and, particularly in this case, how close it is to the real solution. If it is too far away it can have a tendency to 'fly away' from the solution and not give you an answer.

The main disadvantage of the bisection method for finding the root of an equation is that, compared to methods like the Newton-Raphson method and the Secant method, it requires a lot of work and a lot of iterations to get an answer with very small error, whilst a quarter of the same amount of work on the N-R method would give an answer with an error just as small.

In other words compared to other methods, the bisection method takes a long time to get to a decent answer and this is it's biggest disadvantage.