Bracketing method (Bisection method) used to solve roots of polynomial

[Simonsky]

I'm having a problem with the following Question:

'After n years, Karlie's simple interest account contains £1000(1+0.04n) whilst Tayor's compound interest account contains £1000(1.03)^n.

Use a bracketing method with starting values for n = 8 and n = 40 to find the number of years when both accounts have the same amount of money.'

I decided to start by making a polynomial = 0 : so 1+0.04n = 1.03^n => 1.03^n -0.04n - 1 =0

This can then be converted to 1.03 = nth root of (0.04n +1) or (0.04n + 1)^1/n = 1.03

n=8 gives 1.0353 and n=40 gives 1.0242 (4 pl.) so I assume these are the outer 'brackets' and you have to use iteration to get as close to 0 as possible?

So how does this relate to the number of years the account is the same-will that equate to the number of iterations ? Can't quite grasp this and I'm not sure I'm not off track already.

 

[Simonsky]

n=8 gives 1.0353 and n=40 gives 1.0242 (4 pl.) so I assume these are the outer 'brackets' and you have to use iteration to get as close to 0 as possible?

So how does this relate to the number of years the account is the same-will that equate to the number of iterations ? Can't quite grasp this and I'm not sure I'm not off track already.

I think a light bulb just went on in my head: You just need to find the number for n that produces a number closest to 1.03 when plugged into the iteration? i was making it far too complex.

 

[JeffM]

Before we start worrying about mechanics, let's translate the problem given in English to one in mathematical language.

\(\displaystyle 1000 * (1 + 0.04n) = 1000 * (1 + 0.03)^n \implies 1 + 0.04n - 1.03^n = 0.\)

You were fine to here, but you are not dealing with a polynomial. You have a polynomial and exponential

You must use numerical methods.

\(\displaystyle 1 + 0.04 * 8 - 1.03^8 \approx \text { PLUS } 0.053.\)

\(\displaystyle 1 + 0.04 * 40 - 1.03^{40} \approx \text { MINUS } 0.66\).

So somewhere between 8 and 40 the function must go to zero, right?

True bisection calls for testing next at n = 24, midway between 8 and 40. So you look at the function at n = 24 and decide whether it is a good enough answer. if not you look at the sign to decide what your next bisection will be.

 

[Simonsky]

Before we start worrying about mechanics, let's translate the problem given in English to one in mathematical language.

\(\displaystyle 1000 * (1 + 0.04n) = 1000 * (1 + 0.03)^n \implies 1 + 0.04n - 1.03^n = 0.\)

You were fine to here, but you are not dealing with a polynomial. You have a polynomial and exponential

You must use numerical methods.

\(\displaystyle 1 + 0.04 * 8 - 1.03^8 \approx \text { PLUS } 0.053.\)

\(\displaystyle 1 + 0.04 * 40 - 1.03^{40} \approx \text { MINUS } 0.66\).

So somewhere between 8 and 40 the function must go to zero, right?

True bisection calls for testing next at n = 24, midway between 8 and 40. So you look at the function at n = 24 and decide whether it is a good enough answer. if not you look at the sign to decide what your next bisection will be.

Thanks Jeff. Your explanation is clearer, although I found plugging the numbers into (0.04n + 1)^1/n = 1.03 worked, it was just that it converges to 1.03 instead of 0. Does that matter?

 

[JeffM]

Thanks Jeff. Your explanation is clearer, although I found plugging the numbers into (0.04n + 1)^1/n = 1.03 worked, it was just that it converges to 1.03 instead of 0. Does that matter?

This is a really good question.

As a matter of pure theory, it does NOT make any difference. As tkhunny says, correct answers do not care how you find them.

When you were first taught algebra, you were given problems of the form f(x) = g(x) and told to manipulate it into a form h(x) = some number, where some technique (like the quadratic formula) would let you solve for x. What they did not tell you is that there are an infinite number of functions where there is no known technique for solving for x directly. We are back to guess-and-check. But we want to do that in an efficient manner. (The study of how to do that is called numerical methods.)

Starting from f(x) = g(x), we get to h(x) = f(x) - g(x) = 0 without any complicated algebra, and we have our h(x) = some number instantly. Moreover, when we are dealing with zero, the mere sign of the approximation itself tells us what direction to go in the next iteration without further computation.

So the quickest way to go is with the h(x) = f(x) - g(x) = 0 approach.

EDIT: I should note that bisection is not a particularly efficient method of guess-and-check, but it is frequently used as an initial method to get reasonably close to a useful approximation before switching to a faster method such as Newton's method. The problem with Newton's method is that it only fast if you can start with a good initial approximation.

 

[Denis]

n = 19.6 is closest if rounded (closer than 19.5 or 19.7) ... big deal