NUMERICAL METHODS FOR SOLVING EQUATIONS

Transcription

1 Mathematics Revision Guides Numerical Methods for Solving Equations Page of M.K. HOME TUITION Mathematics Revision Guides Level: AS / A Level AQA : C3 Edecel: C3 OCR: C3 NUMERICAL METHODS FOR SOLVING EQUATIONS Version : 4.3 Date: 0--04

2 Mathematics Revision Guides Numerical Methods for Solving Equations Page of NUMERICAL METHODS. Overview. Most equations cannot be solved algebraically to give an eact answer, and therefore we have to resort to numerical approimations. There are four methods discussed at A-level, but only the two in bold are widely used. The decimal search (All) Formula iteration (All, ecept OCR MEI) The bisection method The Newton-Raphson method The first two methods involve trapping the root of f () = 0 between two values a and b such that f (a) and f (b) have different signs, and that f is continuous between a and b. The second two methods rely on choosing a single starter approimation to the root, with the intention of improving the accuracy with each successive step. Continuous Functions change of sign. If the graph of a function has no breaks within an interval = a to = b, then the function is said to be continuous on the interval. For eample, all polynomial functions are continuous everywhere, but the reciprocal function / is not continuous on any interval which contains the value = 0. If a function f () is continuous between = a and = b, and if f (a) and f (b) have different signs, then the interval from a to b will contain a root of f () = 0. Eample (): Show that f () = = 0 has a root between and. Substituting for gives f () = 3 and f () = - ; there is a change of sign between = and =.. The function is continuous, therefore there is a root between and. Eample (a): The function f ( takes ) a value of when = 0, and - when =. There is a sign change in the interval 0 < <, but it does not signify a root here! This is because the interval in question includes a discontinuity at =. In fact the equation f () = 0 has no solution.

3 Mathematics Revision Guides Numerical Methods for Solving Equations Page 3 of The Decimal Search. This is a 'trial and improvement' method used to trap the root of an equation, familiar from GCSE. Eample (): Find the particular root of the equation from Eample (), correct to decimal places. Using f () = = 0 in Eample (), we know that f () takes a zero value between and. We therefore work out f (.), f (.)... up to f (.9) and find where the sign changes. (Spreadsheet programs and graphic calculators are a great help here!) f () The change of sign is now known to occur between when is between.5 and.6. (Values in italics need not have been calculated we could have stopped at the change of sign) We therefore repeat the previous step, calculating f (.5), f (.5)... up to f (.59) f () The equation changes sign between.57 and.58, but we still need to find which of those two values is closer to the root, correct to two decimal places. We therefore take the value of the function at the midpoint of those two values, namely f (.575). This value is 0.06, which means that the change of sign occurs between =.57 and =.575. The correct value of the root to decimal places is therefore.57 and not.58.

4 Mathematics Revision Guides Numerical Methods for Solving Equations Page 4 of Formula iteration. With this method, we start with an approimation to a root and improve its accuracy by substituting into a formula. We can then repeat this process until we have the desired level of accuracy. To create a simple iterative formula, we rearrange the equation so that is epressed as a function of itself. From there, we can turn it into an iterative formula. The eample below will be given in considerable illustrated detail, but there is no need to memorise it for eamination questions. Eample (3): Solve the equation - - = 0 by creating an iteration formula, and use and 0.5 as the starting values for. Stop the iteration when the result has converged to 4 decimal places. (Although this is a quadratic, which can be solved using the general formula, we are choosing this eample to illustrate the method. Its solutions are.68 and 0.68 to 3 decimal places.) The equation can be rearranged in several ways. One way is to re-epress it (dividing by ) as - - = 0 = +. This can be turned into an iterative formula by subscripting on each side as follows: n (Formula 3a) n The roots of the original equation correspond to the - coordinates of the points of intersection between the line y = and the curve y = +. Another rearrangement is = +. The corresponding iterative formula is n n (Formula 3b) The positive root of the original equation corresponds to the - coordinate of the point of intersection between the line y = and the curve y. A third one is to re-epress it as = -. The derived iterative formula is (Formula 3c) n n The roots of the original equation correspond to the - coordinates of the points of intersection between the line y = and the curve y = -.

5 Mathematics Revision Guides Numerical Methods for Solving Equations Page 5 of Using formula (3a): n n (Calculator hint for most makes: Key in, =, and then, +, (,,, Ans, ), = to produce each term in the sequence). n n n n n n This converges to the positive root when choosing a starting value of. This can be shown diagrammatically: Start from point (, 0) on the -ais (corresponding to the starting ). Move up to the curve, and then across to the line y =. This gives the net value of, i.e.. Move down to the curve, and across to the line y = again. This gives the net -value, i.e..5. Repeating this process creates a cobweb diagram as each -value oscillates around the root (.680 to 4 dp) in ever-smaller intervals. When trying to find the negative root, choosing a starting value of 0 or - causes the iterative process to fall over one of several things that can go wrong with a formula iteration. n n 0-0 Falls over - division by zero! Different starting values also lead to wildly varying results: n n n n n n Falls over! A starting value of -0.6 causes the iteration to fall over, and a starting value of -0.7 ends up converging to the positive root of the equation.

6 Mathematics Revision Guides Numerical Methods for Solving Equations Page 6 of Using formula (3b): n n (Calculator hint for most makes: Key in, =, and then ( Ans + ) = to produce each term in the sequence). n n n n This also converges to the positive root when choosing a starting value of. Start from point (, 0) on the -ais (corresponding to the starting ). Move up to the curve, and then across to the line y = to give the net value of, or.44 to 4 d.p. Move up to the curve, and across to the line y = again. This gives the net -value,.5538 to 4 d.p. Repeating this process creates a staircase diagram as each -value approaches the root (.680 to 4 dp) more closely. This is a different pattern from the cobweb from formula (3a). If we tried to find the other root (between 0 and ), there would be a problem: n n This looks to be converging to the positive root as well.

7 Mathematics Revision Guides Numerical Methods for Solving Equations Page 7 of n n Using formula (3c): This formula will be found to be unsatisfactory for iteration, but will highlight several of the problems that might be encountered when choosing an inappropriate formula. We will try and find the positive root of the equation by starting with 0 =. (Calculator hint for most makes: Key in, =, and then Ans,, -, keys., followed by repeated presses of the = key to produce each term in the sequence easily and rapidly). n n By choosing 0 =, the formula appears to end up oscillating between 0 and - without giving a root. Start from point (, 0) on the -ais, we are already on the curve, so we move across to the line y = to give the net value of, namely or 0. Moving down to the curve, and across to the line y = again, gives the net value of, or. Moving up to the curve, and across to the line y = again, gives the net value of, 3 or 0. Moving down to the curve, and across to the line y = again, gives the net value of, 4 or. The iteration has become stuck in an infinite loop!

8 Mathematics Revision Guides Numerical Methods for Solving Equations Page 8 of Trying to find the positive root using 0 = is also no help! n n The iteration blows up by diverging further and further from the solution! We start from point (, 0) on the -ais, so we move up to the curve and then across to the line y = to give the net value of, namely or 3. Moving up to the curve, and across to the line y = again, gives the net value of, or 8. It can clearly be seen that we have a diverging staircase scenario, where the subsequent values of become indefinitely large.

9 Mathematics Revision Guides Numerical Methods for Solving Equations Page 9 of Finally we try and find the negative root using a value other than 0 or, say 0 = We already know that using a starting value of 0 or would lead us into an infinite loop, so we try a different starting value. n n n n Start from point (-0.5, 0) on the - ais (corresponding to 0 ). Move down to the curve, and then across to the line y =. This gives, or Move up to the curve, and across to the line y = again. This gives, i.e Move down to the curve, and then across to the line y =. This gives 3, or We can see that, although we have a cobweb diagram here, successive iterations etend outwards away from the root, so that the process will not converge. (In fact, it will end up oscillating between 0 and ).

10 Mathematics Revision Guides Numerical Methods for Solving Equations Page 0 of Some eamination questions give a suggested formula for you. Eample (4): Verify that the formula n n n is a rearrangement of - - = 0. Substitute 0 = and -, and hence find the roots of the equation to 6 decimal places. What do you notice about the convergence? (-) = + - = = = 0. Removing the subscripts and rearranging the iteration formula we have Key in = and then ( Ans + ) ( ( Ans) ) = = = n n n n This rearrangement converges far more rapidly than the other two, giving 6-decimal place values of and Eample (5): Find the roots of e = 0 by choosing suitable iteration formulae, to 4 decimal places. Use 3 and as the starting values. Rearrange as e = 5 + 4, and by taking logs of both sides, we have = ln (5 + 4) and hence n+ = ln (5 n + 4) n n n n This rearrangement falls over at once when starting with = -, as the first step would attempt to find ln ( ). We can also rearrange the original equation as 5 = e 4 e 4 5 and thus n e n 4 5 The first root was found with a starting value of 3, so try finding the root using a start value of -: n n n n the roots of e = 0 are.943 and to 4 decimal places.

11 Mathematics Revision Guides Numerical Methods for Solving Equations Page of Eample (6) : i) Show that the equation 9 tan - = 0 where is in radians and non-zero, produces the iteration formula n tan n 3. ii) Use a starting value of = 0. to solve the equation 9 tan - = 0 correct to 3 decimal places. i) We rearrange the original equation as 9 = tan -, and taking square roots of both sides, 3 tan tan The final iteration formula is The iteration results are : 3 n tan n 3. n n n n The non-zero solution of 9 tan - = 0 is 0. to 3 decimal places.

12 Mathematics Revision Guides Numerical Methods for Solving Equations Page of Eample (7): Two game shows have a similar format where the prize money increases on a daily basis and remains unclaimed until the champions are beaten by the challengers, who then win the prize money. On Smartheads, the prize money starts at 000 on day and increases by 000 per day. In other words, it forms an arithmetic series with starting value 000 and common difference 000. The prize money sequence (in ) goes 000, 000, and the general term is 000t. On Brainboes, the prize money starts at 000 on day and increases by 0% per day. This time, it forms a geometric series with starting value 000 and common ratio.. The prize money sequence (in ) goes 000, 400, and the general term is 000(. t- ) After 3 days, the payout on Smartheads is higher, but the payout on Brainboes regains the lead several days later. i) Show that the time in days, t, when the prize money payouts on Smartheads and Brainboes are equal again, can be represented by the iterative formula t ln(0.6t n) ln. n. ii) Find the value of t using t 0 = 5, and hence state on which day the payout on Brainboes eceeds that on Smartheads again. t i) The payouts on both shows are equal when 000(.) 000t t (.) t (.). t t (multiplying top and bottom by. to get rid of the t inde) t (.) t ln( 0.6t) t (.) 0.6t t ln. ln(0.6t ) t, and hence the 0.6 ln. ln(0.6t n) required iterative formula is tn. ln. Iteration results : n t n n t n Limit 9.6 The prize payout in Brainboes equals that in Smartheads on Day 9.6, so therefore it eceeds it by Day 0. (Check: Smartheads payout for Day 0: 0,000; Brainboes payout for Day 0: ( ) = 0,30 to nearest ).

13 Mathematics Revision Guides Numerical Methods for Solving Equations Page 3 of Checking the suitability of an iteration formula. Sometimes, an iterative formula might be unsatisfactory because a) convergence is too slow, b) the iteration becomes stuck in an infinite loop, c) the iteration blows up or d) the iteration falls over. The full study of iterative formulae requires difficult analysis beyond the scope of A-level, and therefore an eamination question would include a ready-tweaked formula, as in the net eamples. Eample (8): Use iterative formulae to find the two positive roots of f () = = 0. (One of them has already been found to decimal places using decimal searching, in Eample ()). Find both to 4 decimal places, and use 0 = in each case. One arrangement is = n. n 6 6 = 0 = 4 -, to give an iterative formula of n n n n n n This iteration formula has converged to the root between 3 and 4, and its value is 3.54 to 4 decimal places. Another arrangement is = 0 4 = to give an iterative formula of n n n n n n n n This iteration formula converges to the root between and, and its value is.570 to 4 decimal places, but the convergence is a bit slow. Also, neither formula is of use when trying to find the negative root. The net two eamples show how a tweaked formula can be used for better convergence.

14 Mathematics Revision Guides Numerical Methods for Solving Equations Page 4 of Eample(9): i) Show that the iterative formula rearranging the equation = 0. 6 n n n can be obtained by ii) Hence, find the negative root of = 0, using a starting value of -. i) We begin by writing the formula as ii) The results of the formula iteration are as follows: n n The negative root of = 0 = to 4 decimal places.

15 Mathematics Revision Guides Numerical Methods for Solving Equations Page 5 of Eample(0): i) Show that the iterative formula rearranging the equation = 0. 6 n 3 4 n n can be obtained by ii) Hence, find the root of = 0 lying between and, using a starting value of. How does this compare to the result from Eample (8)? In Eample (8), we used a different iterative formula and obtained a convergence to.570, but the process was rather slow, taking 0 iterations to achieve accuracy to 4 decimal places. i) We begin with ii) The results of the formula iteration are: n n This iterative formula converges to the root of.570 much more rapidly than the iterative formula of 3 n n 6 from Eample (8), which took over 0 iterations.

16 Mathematics Revision Guides Numerical Methods for Solving Equations Page 6 of Eample (): (Omnibus eam-style question) i) A function is defined as f () = - sin. (Note: is measured in radians!) Show that the equation f () = 0 has a solution in the range.9 < <.0. ii) Use a decimal search method to find the solution of f () = 0 to two decimal places. iii) Form an iteration formula from the equation f () = 0. iv) Taking the result from part ii) as the starting value 0 for the iteration, use this formula to find the values of, and 3. v) Hence find the solution of the equation f () = 0 to four decimal places. i) Substituting =.9 into the equation gives.9 sin(.9 c ) = Substituting =.0 into the equation gives.0 sin(.0 c ) = Given that f () = - sin is a continuous function, the change of sign in the value of the function between =.9 and =.0 shows that there is a root in the interval, probably closer to.9. ii) We calculate f (.9), f (.9)... until the sign of f () changes f () no need no need no need no need no need The equation f () = 0 changes sign between =.93 and =.94. The midpoint of those two values is.935, and f (.935) = The change of sign occurs between =.93 and =.935. The correct value of the root to decimal places is therefore.93. iii) The epression - sin can be rewritten as = sin +, from which the iteration formula of n+ = sin n + can be derived. iv) Using the iteration formula of n+ = sin n +, and a starting value of 0 =.93, the net three iterations give: =.936 = = v) Further iterations give: 4 = = = The iteration converges to the root of f () = - sin = 0, or.9346 to four decimal places.

17 Mathematics Revision Guides Numerical Methods for Solving Equations Page 7 of The Bisection (Midpoint) method. (Skip this if not on your syllabus) This method is less hit-miss than formula iteration, but again is fairly slow, though slightly faster than the decimal search. This time we start with two values on each side of the root and keep halving the desired interval by chopping out the irrelevant half. Eample () : The equation f () = = 0 has a root between and. Use the bisection method to find its value correct to decimal place. We begin by working out f () for the end values of. Here f () = -7 and f () = 4, so f () changes sign from negative to positive between those two values. We therefore proceed to halve the interval by finding out f (.5), that being the mean of the two end values. The value of f (.5) = -.65, and so the function changes sign between.5 and, and we can discard the interval < <.5, and keep the interval.5 < <. After one bisection, we have narrowed down the possible range of values for the root it is now in the interval.5 < <. We can continue the process by taking the midpoint of this reduced interval, namely.75, and then finding out f (.75). This works out at 0.359, so f () changes sign in the interval.5 < <.75. We therefore chop out the irrelevant interval.75 < <. After two bisections, the root has been found to lie in the interval.5 < <.75. y = y 0 y After bisection: Root is between.5 and Midpoint y After bisections: Root is between.5 and.75 Midpoint y After 3 bisections: Root is between.65 and.75 Midpoint y After 4 bisections: Root is between.6875 and.75 Midpoint We then continue the process until there is no uncertainty about the first decimal place. (Third bisection) The midpoint of the interval.5 < <.75 is.65. Now f(.65) = -.0, and so f() changes sign in the interval.65 < <.75.

18 Mathematics Revision Guides Numerical Methods for Solving Equations Page 8 of (Fourth bisection) The midpoint of the interval.65 < <.75 is Now f(.6875) = -0444, and so f() changes sign in the interval.6875 < <.75. Both end values in the interval are.7 to one decimal place, so we have obtained the required accuracy after four bisections. The root of f () = = 0 =.7 to one decimal place. The working can be seen more clearly when the results are tabulated: Interval f(lower f(upper Midpoint f(midpoint) Sign change in interval bound) bound) of interval < < < <.5 < < < <.75.5 < < < < < < < <.75 Eample (a) : Continue the previous eample to find the root of the equation f () = = 0 correct to decimal places. We left off the last eample with the root in the range.6875 < <.75, thus obtaining a value of.7 to decimal place. We continue taking the midpoints as before until the estimate is correct to decimal places. (Previous workings in italics). Interval f (lower bound) f (upper bound) Midpoint of interval f (midpoint) Sign change in interval < < < <.5 < < < <.75.5 < < < < < < < < < < < < < < < < < < < < < < < <.77 The root lies in the interval.788 < <.77, and since each end value =.7 to decimal places, the solution of f () = = 0 is =.7 to d.p. The bisection method is a slight improvement on the decimal search, but is still slow, and eventually the fractions become tedious to input in full (hence the approimations to 4 decimal places).

19 Mathematics Revision Guides Numerical Methods for Solving Equations Page 9 of The Newton-Raphson method. (Skip this if not on your syllabus) This is another method of finding roots of an equation. Take the equation f() = 0 with an unknown root, in other words f() = 0. If 0 is a reasonable approimation to, then an improved one is f ( n ) f ( ) n n. n The method usually converges quickly to the required root, but it might fail in eceptional cases, such as when f() is close to zero or if there are discontinuities in the curve. (On the occasions when the method fails, a better first approimation will usually give convergence.) Eample (3): Find the root of e - -3 = 0, using a starting value, 0, of. Use three iterations of the Newton-Raphson method and give the result to 4 decimal places. This equation has a root fairly close to, as the graph shows. We need to differentiate the function to be able to find the root. f () = e 3 f() = e

20 Mathematics Revision Guides Numerical Methods for Solving Equations Page 0 of Substituting for 0 =, we have f () f () e e The diagram illustrates the method geometrically. By continuing the tangent to the curve until it meets the -ais, we will have an improved estimate of the actual root here it is =.885 to 4 d.p. If we draw another tangent to the curve, this time at (, f ( )), then that tangent will meet the -ais even closer to the root. Repeated iterations give 3 f (.8850) f (.8850) f (.8734) f (.8734) The root of the equation is.873 to 4 decimal places.. (Since f(.87305) = and f(.8733) = , there is no danger of the result creeping down to.8730).

21 Mathematics Revision Guides Numerical Methods for Solving Equations Page of Eample (4): Use the Newton-Raphson method with two iterations to find to 4 decimal places the three roots of the equation = 0, used in earlier eamples. Use starting values of = -.,.6 and 3.5. Differentiating f() = , we have f() = = -. f (.) f (.).43 f (.08643) f (.08643).97 0 =.6 f (.6) f (.6) 5... f (.57875) f (.57875) = 3.5 f (3.5) f (3.5) f (3.5486) f (3.5486) The roots of = 0 are -.086,.570 and 3.54 to 4 decimal places. In actual fact, the values are also correct to si places!

22 Mathematics Revision Guides Numerical Methods for Solving Equations Page of Eample (5): The equation of f () = has a root between and 3. i) Use three iterations of the Newton-Raphson method and a starting value, 0, of.5. Give the result to 4 decimal places. ii) Use a suitable lower bounding value of to show that the result in part i) is indeed correct to 4 decimal places i) Differentiating f() = , we have f() = 3 -. f (.5) f (.5) f (.4) f (.4) f (.39487) f (.39487) The root of f () = = 0 is thus.3949 to 4 decimal places. ii) Since f (.39485) = and f (.39487) = , there is a change of sign in f () between = and = There is therefore no danger of the result converging down to a value that rounds down to.3948.