A20 - Approximate solutions by iteration (Higher)

A20 - Approximate solutions by iteration (Higher)

Some equations do not have a convenient exact solution. You will use a starting value and a repeated calculation to get closer to a solution, read suffix notation correctly, and decide how many digits your answer needs. A calculator does the arithmetic; you still choose what to feed back into it.

Feeding an answer back in (Higher)

What number stays unchanged when you square it, subtract twice the number, and get 55? In symbols, solve x2−2x=5x^2-2x=5. A decimal guess of 33 gives 33, while 44 gives 88. The positive solution lies between them. We will build a calculation that homes in on it.

You already know

Substitution replaces a letter with its value. Rearranging preserves an equation when you perform valid inverse operations. Keep the full calculator value during a calculation and round only the final answer.

First try a simpler repeating rule. Start at 22, add 66, then divide by 22. The answer is 44. Put that answer through the same rule to get 55, then 5.55.5. Each output becomes the next input.

Iteration and suffix notation

Iteration repeats a calculation, using the previous output as the next input. In xn+1=xn+62x_{n+1}=\dfrac{x_n+6}{2}, xnx_n is the current value and xn+1x_{n+1} is the next value. The suffix labels the stage: x2x_2 means “the value labelled 22”, not x2x^2 or 2x2x.

A question may start at x0x_0 or x1x_1. Follow its labels. The starting value is supplied; it is not a result you have already calculated. Both examples below are calculator practice, although their arithmetic can be done without one.

Worked example 1: a start labelled zero

Question. Use x0=2x_0=2 and xn+1=xn+62x_{n+1}=\dfrac{x_n+6}{2} to work out x1x_1, x2x_2 and x3x_3.

Plan. Apply the same rule three times, replacing the input by the previous answer.

x1=2+62=4use x0x2=4+62=5use x1x3=5+62=5.5use x2\begin{aligned} x_1&=\frac{2+6}{2}=4 && \text{use }x_0\\ x_2&=\frac{4+6}{2}=5 && \text{use }x_1\\ x_3&=\frac{5+6}{2}=5.5 && \text{use }x_2 \end{aligned}

Why this step? The suffix increases by one after each calculation; the formula uses the previous value, not the suffix number.

Answer. x1=4x_1=4, x2=5x_2=5, x3=5.5x_3=5.5.

Check. Each answer lies halfway between its input and 66, so it moves closer to 66.

Change only the starting value to −2-2. The rule stays the same, so write brackets around the negative input.

Worked example 2: a negative starting value

Question. Use x0=−2x_0=-2 and the same rule to work out x1x_1, x2x_2 and x3x_3.

Plan. Substitute the negative start, then feed each answer back into the rule.

x1=(−2)+62=2use the given startx2=2+62=4use the new answerx3=4+62=5repeat\begin{aligned} x_1&=\frac{(-2)+6}{2}=2 && \text{use the given start}\\ x_2&=\frac{2+6}{2}=4 && \text{use the new answer}\\ x_3&=\frac{4+6}{2}=5 && \text{repeat} \end{aligned}

Why this step? Changing the start changes the values, but not which operation you repeat.

Answer. x1=2x_1=2, x2=4x_2=4, x3=5x_3=5.

Check. The distances from 66 are 88, 44, 22, 11: each step halves the distance. After ten steps the original distance is divided by 10241024. A very simple rule can remove most of a large initial error.

The numerical suffix is a label, not an instruction to substitute that number.

Watch out

Wrong: after finding x1=4x_1=4, putting 22 into the rule to find x2x_2. Right: put 44 into it. On a calculator, an answer-memory key helps preserve all the digits.

The next start changes the first result again. Keep the labels attached to your answers.

Turning an equation into an iteration (Higher)

To solve an equation by iteration, rearrange it so that one xx is alone on the left and an expression involving xx remains on the right. Then use the right side as the repeated calculation.

For example, x3+2x=9x^3+2x=9 becomes x3=9−2xx^3=9-2x, then x=9−2x3x=\sqrt[3]{9-2x}. Replacing the left xx by the next value and the right xx by the current value gives a rule. A solution stays unchanged by that rule.

An iteration from a rearrangement

For an equation rearranged into x=g(x)x=g(x), use

xn+1=g(xn)x_{n+1}=g(x_n)

When the values settle close to a number, substituting it into the original equation checks whether it is an approximate solution. Not every rearrangement or starting value makes the values settle.

Not on the formulae sheet: learn it. A particular iteration rule may be given in the question.

These calculator examples keep the cube-root key round the whole expression. The displayed decimals are for recording; the next calculation uses the unrounded answer.

Worked example 3: isolating a cube

Question. Show that x3+2x=9x^3+2x=9 can be rearranged as x=9−2x3x=\sqrt[3]{9-2x}. Use x0=2x_0=2 to work out x1x_1 and x2x_2.

Plan. Isolate the cube, take a cube root, then apply the resulting rule twice.

x3=9−2xsubtract 2xx=9−2x3take the cube rootx1=9−2(2)3=1.709975947…start at 2x2=9−2(1.709975947…)3use the full answer=1.773696522…\begin{aligned} x^3&=9-2x && \text{subtract }2x\\ x&=\sqrt[3]{9-2x} && \text{take the cube root}\\ x_1&=\sqrt[3]{9-2(2)}=1.709975947\ldots && \text{start at }2\\ x_2&=\sqrt[3]{9-2(1.709975947\ldots)} && \text{use the full answer}\\ &=1.773696522\ldots \end{aligned}

Why this step? Leave the right-hand xx in the rearrangement. It is the input for the repeated calculation.

Answer. x1=1.709975947…x_1=1.709975947\ldots, x2=1.773696522…x_2=1.773696522\ldots.

Check. Cubing the first answer gives 55, which is exactly 9−2(2)9-2(2).

Change only the right-hand constant from 99 to 1010. Make the same change in the iteration; a copied constant would solve a different equation.

Worked example 4: a changed equation

Question. Rearrange x3+2x=10x^3+2x=10 into an iteration rule. Use x0=2x_0=2 to work out x1x_1 and x2x_2.

Plan. Isolate x3x^3, then take a cube root and repeat.

x3=10−2xsubtract 2xxn+1=10−2xn3label next and current valuesx1=63=1.817120593…use x0=2x2=10−2(1.817120593…)3feed back the full value=1.853318496…\begin{aligned} x^3&=10-2x && \text{subtract }2x\\ x_{n+1}&=\sqrt[3]{10-2x_n} && \text{label next and current values}\\ x_1&=\sqrt[3]{6}=1.817120593\ldots && \text{use }x_0=2\\ x_2&=\sqrt[3]{10-2(1.817120593\ldots)} && \text{feed back the full value}\\ &=1.853318496\ldots \end{aligned}

Why this step? The right-hand side must have the current value everywhere, even inside a root.

Answer. xn+1=10−2xn3x_{n+1}=\sqrt[3]{10-2x_n}, with the values shown above.

Check. x23+2x1=10x_2^3+2x_1=10, apart from display rounding. The exact equation need not hold with x2x_2 in both places yet: the iteration is still improving.

An early iterate is usually an approximation, not a solution to quote immediately.

Watch out

Wrong: writing xn+1=103−2xnx_{n+1}=\sqrt[3]{10}-2x_n. The cube root must cover 10−2xn10-2x_n. Right: xn+1=10−2xn3x_{n+1}=\sqrt[3]{10-2x_n}.

Try changing that constant once more. The algebraic rearrangement does not require a calculator.

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