P3 - Numerical Solution of Equations
Trapping Roots You Can't Solve Exactly 🎯
Introduction
1. Introduction
Some equations, like or , cannot be solved exactly with algebra. This chapter gives you two tools to pin their roots down to any accuracy you like. First, the sign-change method shows that a root lies in a given interval. Then, the iterative method homes in on the root step by step. The crucial skill is knowing when to stop: two iterates that agree to 3 decimal places are not always enough, and a sign-change check is how you prove your answer. Let's get started!
2. Locating Roots by Sign Change
Alright, let's talk about finding roots. A 'root' of an equation like is just the fancy name for the x-value where the graph of crosses the x-axis. Think of it like the ground level in a video game. Sometimes your character is above it, sometimes you're in a secret cavern below it. To get from one to the other, you have to pass through ground level, right?
That's the whole vibe of the sign change rule. If you have a continuous function (basically, a graph you can draw without lifting your pen, no weird teleports or gaps like a character glitching across the map), and you find a point where the function is positive (like ) and another point where it's negative (like ), then you've just proven that the graph must cross the x-axis somewhere between and .
That's the whole vibe of the sign change rule. If you have a continuous function (basically, a graph you can draw without lifting your pen, no weird teleports or gaps like a character glitching across the map), and you find a point where the function is positive (like ) and another point where it's negative (like ), then you've just proven that the graph must cross the x-axis somewhere between and .
It's like your bank account balance after getting a part-time job. If you were at -£20 one day and +£50 the next after getting paid, you know at some point your balance hit exactly £0. That's our root! The key conditions are: 1) The function must be continuous on the interval , and 2) There must be a change of sign between and . This method is awesome for trapping a root in a specific interval, often between two consecutive integers like 1 and 2.
Worked example
Worked Example: Applying the Sign Change Rule
Let's Hunt for a Root! 🎯
The equation has a single root, . Show that lies in the interval .
- 1First up, we need to define our function based on the equation given. The goal is to show that when we plug in the interval's endpoints (2 and 3), the function's value changes sign from positive to negative, or vice versa.
- 2Now, let's test the lower bound of the interval, . We'll substitute this into our function and see what we get. Keep a few decimal places for accuracy.
- 3Okay, that one's negative. Now let's do the same for the upper bound, . We're hoping this gives a positive result to confirm the sign change.
- 4Success! We have a negative value and a positive value. We just need to write a concluding sentence that mentions the two key conditions: the sign change and the continuity of the function.
Answer
3. The Iterative Method: Using xₙ₊₁ = F(xₙ)
Alright, so you know how some equations are just impossible to solve perfectly? Like, you can't get on its own, no matter how hard you try. It's the math equivalent of trying to get a perfect screenshot of a specific moment on a Snapchat video. Frustrating, right? That's where iterative methods come in. Think of it as a 'get closer' machine. The core idea is the formula . It looks a bit wild, but it's actually simple. You start with an initial guess for the root, which we call . You plug this guess into the function , and it spits out a new, better guess, . Then you take , plug it back in, and get an even better guess, . You keep repeating this process, or iterating, and your values get closer and closer to the actual root. It's like when your GPS keeps recalculating to get you to your destination; each new route is a better approximation.
This process of getting closer is called convergence. If you were to graph it, you'd see the values either spiralling inwards (a 'cobweb' diagram) or stepping closer and closer (a 'staircase' diagram) to the point where the line crosses the curve .
This process of getting closer is called convergence. If you were to graph it, you'd see the values either spiralling inwards (a 'cobweb' diagram) or stepping closer and closer (a 'staircase' diagram) to the point where the line crosses the curve .
. A root of is also a root of the original equation, because is just that equation rearranged; if the iteration converges, its limit satisfies .
So, when do you stop? The question will tell you the accuracy, like 'to 3 decimal places'. Keep full accuracy on the calculator (use Ans) and write each iterate to 2 more decimal places than the answer. Then look at the pattern:
• Cobweb (values jump over and under the root): the root is trapped between consecutive iterates, so two in a row agreeing to 3 d.p. is enough.
• Staircase (values creep steadily one way): two iterates agreeing to 3 d.p. is not enough, because the root lies beyond both. Keep going until they agree to one more decimal place (4 d.p.), then round.
To prove the answer, use a sign change: the root is to 3 d.p. if and have opposite signs.
So, when do you stop? The question will tell you the accuracy, like 'to 3 decimal places'. Keep full accuracy on the calculator (use Ans) and write each iterate to 2 more decimal places than the answer. Then look at the pattern:
• Cobweb (values jump over and under the root): the root is trapped between consecutive iterates, so two in a row agreeing to 3 d.p. is enough.
• Staircase (values creep steadily one way): two iterates agreeing to 3 d.p. is not enough, because the root lies beyond both. Keep going until they agree to one more decimal place (4 d.p.), then round.
To prove the answer, use a sign change: the root is to 3 d.p. if and have opposite signs.
Worked example
Worked Example: Applying an Iterative Formula
Let's Get This Bread (and this Root) 🍞
The equation has a root between and . Use the iterative formula with a starting value of to find the root correct to 3 decimal places.
- 1First, let's set up the calculator. This is a pro-gamer move to make life easier. Type your starting value, , into your calculator and press the '=' key. This stores as the most recent 'Ans'.
- 2Now, type the iterative formula into your calculator, but instead of using 'x', use the 'Ans' button. This will let you repeat the calculation super fast.
- 3Time to iterate! Press '=' repeatedly and write down each value to 5 decimal places, keeping full accuracy in the calculator.
- 4Look at the pattern: the values increase steadily, so this is a staircase. Watch out: and both round to , and stopping there would give the wrong answer, because the sequence is still climbing and the root is above both. So keep going until two iterates agree to 4 decimal places.So the root is to 3 d.p. (not ).
- 5Finally, prove it with a sign change either side of , using . If the sign changes between and , the root rounds to .The root lies between and , so the root is (correct to 3 d.p.).
Answer
The root lies between and , so the root is (correct to 3 d.p.).
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