Solving Equations Numerically: Newton's Method, Bisection Method

Date: 11/05/2002 at 07:36:30
From: Emily Bardsley
Subject: Solving equations numerically

Hey,

Would you be able to tell me how you can work out equations that can 
only be solved numeraically and not algebraically. Surely the only way 
is not to try out the equation with every algebraic technique, and if 
this is so then do you know how many methods of algebraic solving 
there are?

Thank you very much,
Emily

Date: 11/05/2002 at 08:11:15
From: Doctor Jerry
Subject: Re: Solving equations numerically

Hi Emily,

There are several methods for solving equations numerically.  

If you know calculus, then you can use Newton's method. Suppose you 
want to find a zero of a function f, that is, you want to find a 
number c for which f(c)=0. For Newton's method, you must first find 
an approximation x1 to c. Maybe you can do this graphically.

You calculate the next approximation, call it x2, with the formula

   x2 = x1-f(x1)/f'(x1)

where f' is the derivative of f. Once you have found x2, you find x3 
with a formula having the same form:

   x3 = x2-f(x2)/f'(x2)

In this way you get a sequence x1,x2,x3,... of improving 
approximations to c. Newton's method is usually quite efficient.

Try f(x) = x^2-2, for example. There is a zero, namely c=sqrt(2), near 
x1 = 1.5.  

Okay,

   f(x) = x^2-2
   f'(x) = 2x

   x1 = 1.5
   x2 = 1.41666666667
   x3 = 1.41421568628
etc.

Or you can use the Bisection Method. For this method of a finding a 
zero c of a function f, you begin by figuring out an interval [a1,b1] 
containing c and having the property that f(a1)*f(b1) < 0. (This 
condition means that the graph of f crosses the x-axis somewhere 
between a1 and b1. Where it crosses is a zero of f.) Calculate the 
midpoint m1 of this interval. If f(a1)*f(m1)<0, then a zero of f is in 
[a1,m]. This is the new interval [a2,b2]. Otherwise, a zero of f must 
be in [m1,b1], which is the new interval [a2,b2]. Continue with this 
scheme of bisection an interval containing a zero. The Bisection 
Method is slower than Newton's method, but a bit more certain.

Both of these methods can be programmed on a calculator to speed up 
the calculations.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 11/10/2002 at 11:20:14
From: Emily Bardsley
Subject: Re: Solving equations numerically

Dear Dr Math,

Thank you so much for your help. I was also wondering if you would be 
able to help in suggesting some good equations with which I could try 
out these methods on using graphs. I am looking for equations that 
have about 2 or 3 roots but can't be solved algebraically. I've been 
trying to find some interesting equations such as ones involving 
square roots or ones that have a curve which crosses the x axis at 
one point but only touches the curve (i.e. a parabola) at another. 
I'm unsure of how I could go about deriving these equations that will 
produce good graphs - any ideas?

Thank you for your time so far.
Emily

Date: 11/10/2002 at 15:44:09
From: Doctor Jerry
Subject: Re: Solving equations numerically

Hi Emily,

1. Find the zeros of the function f(x) = 2^x -4*x

2. a. Locate between successive tenths the first four positive zeros 
      of the function f(x) = cos(x)*cosh(x) + 1.
   b. Find the three smallest positive zeros of the equation
      cos(68.617*sqrt(x))*cosh(68.617*sqrt(x)) = -1.

3. Find a zero of f(x) = cos(ln(x)), x>0.

4. Find to within 0.1deg the smallest positive zero of the equation
   tan^3(t) - 4*tan^2(t) + tan(t) - 4 =0.

5. The volume V (in cubic meters) of 1 mole of gas is related to its 
   temperature T (in kelvins) and pressure P (in pascals (Pa)) by the 
   ideal gas law PV=RT.  A more accurate equation is van der Waal's 
   equation

   (P+a/V^2)(V-b) = R*T.

The constant R is 8.314. For carbon dioxide, a=3.592 and b=0.04267.  
Find to two decimal places the volume V of 1 mole of carbon dioxide 
if P=20,000 Pa and T=320K.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 11/10/2002 at 16:33:15
From: Emily Bardsley
Subject: Thank you (Solving equations numerically)

Thanks a lot. Excellent service!

Emily