Numerical Analysis and Simulation

I'm assuming you're trying to write f(x)=exp(x)-3x2=0 as your function. This has 3 solutions where f(x)=0 between (-0.5,-0.4), (0.85, 0.95) and (3.7, 3.8) I won't find all of the roots, only the one between 3.7 and 3.8 in order to show you the procedure/algorithm.

What we do is choose two points x=a, b on the function where it takes on opposite values. We then find the mid point (a+b)/2 between these values and find the value of the functions at f((a+b)/2) we then choose the smallest interval where the points take on opposite values and repeat until we have a small enough value for x.

We will do our iteration until we get within 0.01 of the solution (i.e. the first number to give an answer between -0.01 and 0.01) and i will be using c instead of writing out (a+b)/2 all the time.

iteration 1: a=3.7 b=3.8 c=3.75 f(c)=0.3335 +ve

iteration 2: a=3.7 b=3.75 c=3.725 f(c)=-0.155 -ve

iteration 3: a=3.725 b=3.75 c=3.7375 f(c)=0.086 +ve

iteration 4: a=3.725 b=3.7375 c=3.73125 f(c)=-0.035 -ve

iteration 5: a=3.73125 b=3.7375 c=3.734375 f(c)=0.025 +ve

iteration 6: a=3.73125 b=3.734375 c=3.732813 f(c)=-0.005 between -0.01 and 0.01

ONE solution for f(x)=exp(x)-3x2=0 is x=3.732813 (actual value is x=3.733079029 so we have an absolute error of 0.000266029, a relative error of 0.00007126262207 and a percentage error of 0.007126262207%)

If this is a homework assignment, please consider trying to answer it yourself first, otherwise the value of the reinforcement of the lesson offered by the assignment will be lost on you.

The possible ways of arranging a line of eleven people is equivalent to the number of permutations of eleven things taken eleven at a time. That is eleven factorial, or 39,916,800.

The rate of convergence of an iterative method is represented by mu (μ) and is defined as such:

Suppose the sequence{xn} (generated by an iterative method to find an approximation to a fixed point) converges to a point x, then

limn->[infinity]=|xn+1-x|/|xn-x|[alpha]=μ,where μ≥0 and α(alpha)=order of convergence.

In cases where α=2 or 3 the sequence is said to have quadratic and cubic convergence respectively. However in linear cases i.e. when α=1, for the sequence to converge μ must be in the interval (0,1). The theory behind this is that for En+1≤μEn to converge the absolute errors must decrease with each approximation, and to guarantee this, we have to set 0<μ<1.

In cases where α=1 and μ=1 and you know it converges (since μ=1 does not tell us if it converges or diverges) the sequence {xn} is said to converge sublinearly i.e. the order of convergence is less than one. If μ>1 then the sequence diverges. If μ=0 then it is said to converge superlinearly i.e. it's order of convergence is higher than 1, in these cases you change α to a higher value to find what the order of convergence is.

In cases where μ is negative, the iteration diverges.

it is unstable as an approximation method due to round-off errors involved and the fact that it involves taking a limit of h->0. Doing this for an iteration method is a really bad idea because as h gets very close to 0, say h=1x10-10, f(x0+h) and f(x0) are virtually the same number and subtracting near equal numbers in an iteration formula has the effect of producing huge errors and as such would not give a very reliable answer.

In fact if they are to close the computer can't recognise a difference between the two when doing the algorithm and you end up with 0/h so you don't get an answer at all! It is because of these reasons that taking the formula for numerical differentiation as it is, with the limit of h-> 0, is unstable and not suitable as an approximation method.

It is however, a good place to start. What we do instead is take the formula, but get rid of the limit and make it so we take h at very small values but not so small as to create huge round errors, so values such as h=0.1 or 0.001 or something like that. Then to get the answer we obtain from the formula more accurate, we apply a truncation error to it. i.e. we subtract a small amount from it to get it closer to the answer. This error correction term is f''([xi])|h|/2 where x0<[xi]

f'(x0)=(f(x0+h)-f(x0))/h - (h/2)*f''([xi])

This is known as the forward-difference formula if h>0and the backward difference formula if h<0

The real number system can be sliced and diced in an infinite number of ways. In other words, you can't list all the different sets of numbers in the real number system. However, here are some of the commonest:

• the real number system itself (on the grounds that every set is a subset of itself)
• the integers
• the whole numbers: 1, 2, 3, ...
• the even number: ..., -6, -4, -2, 0, 2, 4, 6, ...
• the rational numbers