10.2 The Interval Bisection Method
Not every equation can be solved using algebra!
The Interval Bisection Method is one technique that can be used to solve an equation for which there is no algebraic way of solving it. To find the solution, we go through an iterative process, coming up with better and better estimates of the solution.
We've developed a lot of different techniques that we can use to solve a lot of different types of equations. However, there are many equations that we'd like to solve for which algebra can't give us the answer. In fact, we just saw that if an equation involves a polynomial is of degree 5 or higher, then there it is not soluble by radicals. Further, because so many relationships in the physical world are so complex, many, many equations don't have a clean way of solving them. Therefore, an entire branch of mathematics, known as numerical analysis has been developed to study so-called "numerical methods" of solving equations.
Just to give you a taste of one method, let's look at the interval bisection method through an example. Suppose that you have $10,000 to invest and need $80,000 to pay for college. If you can earn interest on a well-producing stock at 7%, then how long will it take for you to come up with the $80,000?
Here we can use the compound interest formula that was discussed earlier and we can set up the equation:
80000=10000(1+.07)^t
Then, using the tools that we already have, we can at least divide both sides by 10000 to get that
8=1.07^t
But how we're stuck! Unfortunately, this type of equation doesn't fit into any of the techniques that we've learned so far. Then, we can either try to develop a new algebraic method of solving the equation or try to solve it using a numerical method.
The interval bisection method works by first rewriting the equation and setting it equal to 0:
8-1.07^t=0
Then, let's first find a value of t that makes the left-hand side bigger than 0; for example, if t = 0, then 8-1.070= 7, which is bigger than 0.
Now let's find a value of t that will make the left-hand side less than 0: for example, if t = 40, then 8-1.0740 = -6.97, which is less than 0.
What does this tell us? Well, we now know that our solution is between 0 and 40 since the left-hand side is negative when t is 40 and positive when t is 0 (it's got to cross over from positive to negative somewhere in-between!). Let's now break our interval between 0 and 40 in half. In other words, let's test out t = 20.
Then, we get that 8-1.0720= (to 3 decimal places) which is telling us that the solution has to be between
See the full discussion below.
Since we haven't crossed 0 yet when t = 20 (for t=0 and t=20 our left-hand side is positve), the correct value of t has to be bigger than 20 still! We can keep playing this game, halving the interval each time and evaluating at the midpoint. Now that we know that our answer is in the interval (20, 40), checking the midpoint at t = 30, we see that 8-1.0730 = .3877 which is still bigger than 0, so we know that our solution is between (30, 40).
Try to work out several more iterations, using the same process and then compare your work to the table below.
| Iteration | Interval to Check | Midpoint (4 decimal places) |
Value of 8-1.07t at the midpoint |
|---|---|---|---|
| 1 | (0, 40) | 20.0000 | 4.1303 |
| 2 | (20, 40) | 30.0000 | 0.3877 |
| 3 | (30, 40) | 35.0000 | -2.6766 |
| 4 | (30, 35) | 32.5000 | -1.0151 |
| 5 | (30, 32.5) | 31.2500 | -0.2841 |
| 6 | (30, 31.25) | 30.6250 | 0.0589 |
| 7 | (30.625, 31.25) | 30.9375 | -0.1107 |
| 8 | (30.625, 30.9375) | 30.7813 | -0.0255 |
| 9 | (30.625, 30.78125) | 30.7031 | 0.0169 |
| 10 | (30.703125, 30.78125) | 30.7422 | -0.0043 |
Notice that our final column is getting closer and closer to 0. Testing out the last row, we see that the difference between 1.0730.7421875 and 8 is really small — we're getting closer and closer to the exact answer. In fact, if we plug 30.74 into the right side of our original equation, we get: 10000(1.07^{30.74})=80,030.83
This tells us that you will either need to invest more money, find a higher interest rate, or hope that you won't go to college for another 30 years!
It should be noted that for this method to work, you actually have to be able to start with "reasonable" guesses: in other words, this process won't work if you can't find a value that's bigger than 0 on one side and less than 0 on the other.
Explore!
Let's use the interval bisection method on an equation that we can actually solve using algebraic techniques. If we're trying to solve x^2+8x+7=0 then, by factoring, we see that (x+7)(x+1)=0 so that x=-1 or x=-7. If we didn't know how to factor could we have used the interval bisection technique?
If we plug in x = -3 into the equation we get -8, which is negative; if we plug in x = 0 into the equation, we get out 7 which is positive. Try a few iterations before looking at the table below:
| Iteration | Interval to Check (4 decimal places) |
Midpoint | Value of x2+8x+7 at the midpoint |
|---|---|---|---|
| 1 | (-3.0000, 0.0000) | -1.5000 | -2.7500 | 2 | (-1.5000, 0.0000) | -0.7500 | 1.5625 | 3 | (-1.5000, -0.7500) | -1.1250 | -0.7344 | 4 | (-1.1250, -0.7500) | -0.9375 | 0.3789 | 5 | (-1.1250, -0.9375) | -1.0313 | -0.1865 | 6 | (-1.0313, -0.9375) | -0.9844 | 0.0940 | 7 | (-1.0313, -0.9844) | -1.0078 | -0.0468 | 8 | (-1.0078, -0.9844) | -0.9961 | 0.0235 | 9 | (-1.0078, -0.9961) | -1.0020 | -0.0117 | 10 | (-1.0020, -0.9961) | -0.9990 | 0.0059 | 11 | (-1.0020, -0.9990) | -1.0005 | -0.0029 | 12 | (-1.0005, -0.9990) | -0.9998 | 0.0015 | 13 | (-1.0005, -0.9998) | -1.0001 | -0.0007 | 14 | (-1.0001, -0.9998) | -0.9999 | 0.0004 | 15 | (-1.0001, -0.9999) | -1.0000 | -0.0002 | 16 | (-1.0000, -0.9999) | -1.0000 | 0.0001 | 17 | (-1.0000, -1.0000) | -1.0000 | -0.0000 |
It took us several iterations, but we did eventually get one of the answers!
Hopefully by the two examples above you can see that sometimes it's easier to use algebra and sometimes it's better to use a numerical method. Also, numerical analysis is a very broad and deep topic. And though this section is presented to give you just a taste of it, you should realize that you could study the topic of numerical analysis for several years in order to develop a deep understanding of the different numerical methods.
In the next section, we're going to open the door to another branch of mathematical study: Calculus. For those of you who go on to take Calculus, it will be a great way to be introduced to the topic. And, if you don't plan on taking Calculus, this will be your chance to become acquainted with one of the major mathematical accomplishments in human history.
