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Lesson goal: The while-loop: solving equations

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$\sqrt{n}-\sqrt{n-1}<0.01$

In a past lesson, a while loop was used to guess a number you were thinking of by bracketing in your guess, with tighter and tighter bounds, until your number was found. This was meant as an example of a while loop, and has a much more useful purpose: to solve equations in math.

Suppose you have a function

$f(x)$ and you want to know for what

$x$ is

$f(x)=0$ . For example, what about

$f(x)=x^2+7x+10$ , for what

$x$ is

$f(x)=0$ ? This is called "finding the roots" or "solving" the equation. Yes, we know, for

$x^2+7x+10=0$ you could use the quadratic formula, but what would you do to solve

$x+cos(5x^3)=0$ ?

In this lesson, we'll show you how to solve math equations using a while loop.

In order for this to work, you need to tell the computer two numbers, between which is the actual solution. Also, the sign of

$f(x)$ must be different at

$a$ and

$b$ . This is based on the premise that if a function changes sign between two numbers, it must pass through zero at some point in between. In other words, if the sign of

$f(a)$ is opposite to the sign of

$f(b)$ , then we expect

$f(x)$ to be zero at some point

$x$ for

$a\le x\le b$ . If you can deliver the

$a$ and

$b$ , then the following code will go and find your solution. It is known as the "bisection method" (ref).

Now you try. Try to put in an equation for $f(x)$ in the `function f(x)` `return` statement, and see if you can direct the program with your choices of $a$ and $b$ to find the roots of the equation.

Type your code here:

 
function f(x)
 return(x^2+7*x+10)
end
​
try=0
a=-10
b=-3
​
if f(a)*f(b)>0 then
 print("f(x) must have different signs at a and b.")
 os.exit()
end
​
stop = false
while stop == false do
 x=(a+b)/2
 if math.abs(f(x)) < 0.001 then
  stop = true
 end
 
 if f(a)*f(x) > 0 then
   a=x
 else
   b=x
 end
x=(a+b)/2
print(x)
​
try=try+1
if try > 100 then
   stop = true
 end
​
end
print("solution is x=",x)

See your results here:

The code will run just fine..it's hard enough to understand this algorithm, so we didn't break it anywhere.

As far as learning about while-loops goes, see if you can understand the following:

The loop keeps going as long as variable stop is false.
The loop is told to stop if after bisections (i.e. $(a+b)/2$ , no solution is found.
If $|f(x)|$ at some $x$ is $\le 0.001$ we assume this is close to zero and we've found a solution, so stop.

Lesson goal: The while-loop: solving equations

Now you try. Try to put in an equation for f(x)f(x) in the function f(x) return statement, and see if you can direct the program with your choices of aa and bb to find the roots of the equation.

Share your code

Now you try. Try to put in an equation for $f(x)$ in the `function f(x)` `return` statement, and see if you can direct the program with your choices of $a$ and $b$ to find the roots of the equation.