Solving Equations by Iteration

The question gives you x n plus one equals the cube root of the whole of five x n plus eight, and x nought equals two. Find x three. There is no x three anywhere in that formula, so what goes in, and how do you get to x three?

This is video seven of eight in Quadratic and Simultaneous Equations. It is Higher tier, and not on every board's paper, so check your own.

It is one move, made three times, and the first move uses the two. Doing the same step over and over, with each answer fed into the next go, is called iteration. A formula built to be used that way is an iterative formula. Read this one from the inside out. Take a number, times it by five, add eight, then take the cube root. The small number after each x is its suffix, and it counts the steps. x nought is the start. x one is what comes out when x nought goes in. And x n plus one means the number after x n. First pass. x nought is two, so two goes in where the formula has x n. Five times two is ten. Ten add eight is eighteen. The cube root of eighteen is two point six two one, to three decimal places. That is x one. Next, which number goes in to make x two: two, two point six two one, or eighteen? Two point six two one, the answer that came out. The two has done its job, and it never goes back in. Eighteen was only a stop along the way. On a calculator, the answer key, marked A N S, does the feeding back for you. Type two and press equals. Then key in the cube root of five times the answer key, add eight, and press equals. The display shows two point six two nought seven and more digits: that is x one again, unrounded. Press equals again, and the formula runs once more with that full value going in: the cube root of five times two point six two nought seven, add eight. The display starts two point seven six three, so x two is two point seven six three. Retyping the rounded x one instead can knock x two out by one in the last place. Keep the full value in the calculator, and round only when you write a value down. Here's another. Is x three bigger or smaller than x two, and by more or less than before? Bigger, but by less. One more press of equals gives two point seven nine four, and the steps are shrinking. Two point seven nine four is the x three the question asked for. Three passes of the formula, starting from two. Every pass is the same three moves. Put the value into the formula, write the answer down rounded, and feed the full value back in. Your handle for this video: work it, write it, feed it back. That is the loop, said in one breath.

Now keep pressing equals and watch the values. After two point seven nine four come two point eight nought one, two point eight nought two, then two point eight nought three, and there they stay. The formula comes from the equation x cubed minus five x minus eight equals zero. Rearranged, that says x equals the cube root of five x plus eight, and the question does the rearranging for you. Why would a solution of that equation go in and come out unchanged? Because a solution makes both sides match. Put it in on the right, and what you get is that same number on the left. That fixed number is what the list closes in on, about two point eight nought three. This equation has no whole-number solution to factorise out, which is why the loop is the way in. A shortcut can look tempting at this point. x three is three steps along, so one student writes x three equals three times x one, which is seven point eight six three. What is wrong with multiplying x one by three? The three in x three counts passes. It is not a multiplier. Each step is one full run of the formula, so x three needs three runs. One examiner's report puts it this way. It is clear that many didn't really understand iterative processes or how to use an iterative formula. That question was about growth over a number of years. Most students multiplied by the number of years, instead of using the formula once for each year. The fix is to count passes, never to multiply by the count. x three means three passes of the formula, each one starting from the answer before it.

Now a formula you have not met yet. x n plus one equals the square root of three x n plus seven, and x nought equals three. The job is to find x one, then feed it back in to find x two. Pause and work out both values, each to two decimal places. x nought is three, so three goes in. Three times three is nine. Nine add seven is sixteen. The square root of sixteen is four, so x one is four point nought nought. Now four goes in, not three. Three times four is twelve. Twelve add seven is nineteen. The square root of nineteen is four point three six, to two decimal places, and that is x two. One student works through the same steps, writes x two equals root nineteen, and stops there. What is wrong with leaving x two as root nineteen? It is a surd, a root left unworked. The question asked for two decimal places, and a surd is not a decimal. A second report, on a Higher paper, records this. Some students wrote down unprocessed surds as their answers which scored zero. In plain words, those students set the calculation up and never pressed equals to finish it. The fix is the write-it move. Press equals, and write the decimal to the accuracy the question asks for.

Some questions ask something different: show that a solution is there at all. Take x cubed plus four x equals thirty, and show it has a solution between x equals two and x equals three. First, move the thirty across so one side is zero. Call the other side f of x, which is a name for its value at any x. A solution is an x that makes f of x zero. Put in two. Two cubed is eight, and four times two is eight. Eight add eight is sixteen, and sixteen take away thirty is minus fourteen. That is one end of the interval, the stretch from two to three. The other end is yours. What is f of three, the value at the other end? Nine. Three cubed is twenty-seven, and four times three is twelve. Twenty-seven add twelve is thirty-nine, and thirty-nine take away thirty is nine. Now a harder one. Why must f of x be zero somewhere between two and three? Because the graph of f of x is one unbroken curve. It starts below zero at minus fourteen and ends above zero at nine, so it has to cross zero on the way. Picture climbing the stairs from a cellar to an upstairs room. Somewhere on the way, you pass ground level. Three students each write a last line under those two values. The first writes, x is about two point five. The second writes, f of two is minus fourteen and f of three is nine. The third writes, the sign changes, so there is a solution between two and three. Which line finishes the proof: the first, the second or the third? The third. Only that line turns the two numbers into a reason. The second stops at the arithmetic, and the first is a guess nobody asked for. Write that reason in full, every time. f of two is negative and f of three is positive. There is a change of sign, so a solution lies between x equals two and x equals three.

The whole lesson comes down to four questions, and the first one uses a formula you have not seen. x n plus one equals the square root of x n plus six, and x nought is ten. What is x one? Four. Ten add six is sixteen, and four times four is sixteen. Now try another. What goes in to make x two: ten, or four? Four, the answer that came out. Four add six is ten, and the square root of ten is three point one six two, to three decimal places. Here's a different one. f of one is minus two and f of two is five. What do you write? There is a change of sign, so a solution lies between x equals one and x equals two. One more, then we're done. Work it, write it, and then what? Feed it back. The full value goes back in for the next pass. And that is the question from the start. x nought is two, three passes of the formula give x three, and x three is two point seven nine four.

If this one is in the bag, a thumbs-up marks it done and you can leave it behind. If it has not landed yet, save the playlist, because three formulas in, the loop starts to feel normal.

The next video is Solving Equations with Algebraic Fractions.

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