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### 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
```
Associated Topics:
College Calculus
High School Calculus

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