Expert: Scott A Wilson - 1/30/2009

Question
QUESTION: I understand this topic a bit only & i'm lack of resources on this topic. I'm wondering if you can provide me some more examples on this topic & some websites providing examples with questions on this. A)Use the iterative formula x_(n+1)=-(4-e^(x_n))^.5 & x0=-2 to find in turn x1,x2 & x3 & hence write down an approximation to the negative root of the equation,giving your answer to 3 decimal places. An attempt to evaluate the positive root of the equation is made using the iterative formula x_(n+1)=(4-e^(x_n)) with x0=1.3. B)describe the result of such an attempt. Thanks a million.

ANSWER: A) Use x0 as -2.  Evaluate the function (4 - e^x)^0.5 with x=x0.
In other words, calculate e^x, subtract that from 4, and
then take the squareroot.  This will give you some value that
is called x1.

With x1, do the same thing that was done with x0.
Using x=x1, calculate e^x, subtract if from 4, and take the
squareroot.  This will give you a value for x2.

With x2, find x3.

B) Do the same thing, but start with x0 = 1.3.
Using the formula, if you haven't guessed, that's use x=x0,
calculate e^x, subtract the result from 4, and take the root.
This will give you x1.

Using x=x1, do the formuula (4 - e^x)^0.5 again to get x2.
Do it again to get x3.

Now back to the fact you need more examples, here we go.

Suppose you have the function f(x) = x - √3.
We know that f'(x) = 1.

At this time, I'll give you a better way.
It's called Newton's method.

x[n+1] = x[n] - f(x[n])/f'(x[n]).

Let's take a wild answer that's not even close, like x[0] = 70.
By Newton's method, x[1] = 1.732, and that's close enough.

Let's try f(x) = x² - 7.
It can be seen that f'(x) = 2x.

Taking x[n+1] = x[n] - f(x[n])/f'(x[n]), we get
-1   -6   -2
-4   9   -8
-2.875   1.23   -5.75
-2.655   0.05   -5.309782609
-2.646   0, so there's no need to go and farther.  It can be stated
that x² - 7 will give √7, which (to three decimals) we have seen
is -2.646.  Note that if we started with a different root, we
might have gotten to 2.646.  Now if you look at the functional
error, you get around 0.001316.

The main website to be used is
http://en.wikipedia.org/wiki/Root-finding_algorithm,
for that the website with definitions to many math functions.

Another website is http://www.efunda.com/math/num_rootfinding/num_rootfinding.cfm.
This will show how the Bisection, Secant, false position, and
Newton-Raphson methods are done.


---------- FOLLOW-UP ----------

QUESTION: (a) i got this answer x=1.965 is it correct? (b) can i said it is divergent & no real root. Thanks

Answer
(a) Here are my results:
  xn   4   exp(x)    -4+EXP(x)
  --   -   ------   -----------
x0   -2   4   0.13534   -3.864664717    =x1
x1   -3.8647   4   0.02097   -3.979030048    =x2
x2   -3.9790   4   0.01870 -3.981296228    =x3
x3   -3.9813   4   0.01866   -3.981338566   

(b) I got a root, and there is only 1, for the second derivative
will give you any extreme points, and there are none since it
is never 0, which means (since the function is continuous)
it can only cross the axis once.