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Solving equations numerically

y. x. Solving equations numerically. The sign - change rule. If the function f(x) is continuous for an interval a  x  b of its domain , if f(a) and f(b) have opposite signs, then there is at least one root of f(x) = 0 between a and b. Consider an equation f(x) = x 2 -5x + 2 = 0.

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Solving equations numerically

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  1. y x Solving equations numerically The sign - change rule If the function f(x) is continuous for an interval a  x  b of its domain , if f(a) and f(b) have opposite signs, then there is at least one root of f(x) = 0 between a and b. Consider an equation f(x) = x2 -5x + 2 = 0 f(0) = 2 There is a sign-change f(1) = -2 This means that there is a solution between x = 0 and x = 1.

  2. f(0) = -3 f(2) = 3 y x Solving equations numerically The sign - change rule Consider an equation f(x) = (2x – 1)(x – 1)(2x – 3) There is a sign-change It is clear from the graph that there are three roots between x = 0 and x = 2

  3. Examples Show that a root of the equation x3 – 5x – 4 = 0 lies in the interval [2, 3]. f(2) = 23 – 5  2 – 4 = 8 – 10 – 4 = - 6 < 0 f(3) = 33 – 5  3 – 4 = 27 – 15 – 4 = 8 > 0 There is a change of sign so the root lies in the interval [2, 3] 4.32 is an approximation to a root of the equation xlnx – 2 – x = 0. Check its accuracy to 2 decimal places. f(4.315) = 4.315  ln4.315 – 2 – 4.315 = - 0.00605 < 0 f(4.325) = 4.325  ln4.325 – 2 – 4.325 = 0.00858 > 0 Change of sign, so the root is accurate to 2 decimal places.

  4. y x Decimal search Consider f(x) = x3 – 5x – 4 ; f(2) = - 6 and f(3) = 8 f(2.5) = 2.53 – 5  2.5 – 4 = -0.875 f(2.6) = 2.63 – 5  2.6 – 4 = 0.576 For 1 decimal place the root is 2.5 or 2.6 f(2.55) = 2.553 – 5  2.55 – 4 = -0.169 x = 2.5 is ignored so the root must be x = 2.6 for one decimal place.

  5. Use decimal search to find each root correct to two decimal places. (i) f(x) = x + (x3 + 1) - 7 x f(x) sign 0 -6.0000 negative 1 -0.5858 negative 2 -2.0000 negative A root lies between 2 and 3 and it is probably closer to 3. 3 0.2915 positive 2.6 -0.0900 negative A root lies between 2.6 and 2.7 and it is probably closer to 2.6. 2.7 0.2479 positive 2.62 -0.0229 negative A root lies between 2.62 and 2.63 and it is probably closer to 2.63. 2.63 0.0180 positive 2.625 -0.0060 negative This confirms that the root is 2.63

  6. Use decimal search to find each root correct to two decimal places. (ii) f(x) = x5 + x3 - 1999 x f(x) sign 2 -1959 negative 3 -1729 negative 4 -911 negative A root lies between 4 and 5. 5 1251 positive 4.5 -62.59 negative A root lies between 4.5 and 4.6. 4.6 158 positive 4.54 23.34 positive 4.53 1.576 positive A root lies between 4.52 and 4.53. 4.52 -20.00 negative 4.525 -9.236 negative This confirms that the root is 4.53.

  7. f(x) x Iteration Consider the equation f(x) = x2 – 5x + 2 = 0 The graph of f(x) shows that one root lies between 0 and 1 and the other root lies between 4 and 5. First step is rearrange x2 – 5x + 2 = 0 in the form x = g(x) Possible rearrangements: x2 – 5x = - 2 x(x – 5) = -2 x = 2/(5 – x) x2 = 5x – 2 x = (5x – 2) x2 = 5x – 2 x = 5 – 2/x 5x = x2 + 2 x = (x2 + 2)/5

  8. Graph of rearranged equations x = 5 – 2/x x = (5x – 2) x = 2/(5-x) x = (x2 + 2)/5

  9. Using xn+1 = g(xn) f(x) = x2 – 5x + 2 = 0  xn+1 = (5xn – 2) Starting value x0 = 4 x8 = 4.5566 x1 = (5  4 – 2) = 4.2426 x9 = 4.5588 x2 = 4.3833 x10 = 4.5600 x3 = 4.4628 x11 = 4.5607 4.56 x12 = 4.5611 x4 = 4.5071 x13 = 4.5613 x5 = 4.5316 x14 = 4.5614 x6 = 4.5451 x15 = 4.5615 x16 = 4.5615 x7 = 4.5525 root = 4.56 (2 d.p.)

  10. f(x) = x2 – 5x + 2 = 0 xn+1 = 5 – 2/xn xn+1 = (xn2 + 2)/5 xn+1 = 2/(5 – x) x0 = 4 x0 = 4 x0 = 4 x1 = 2.0000 x1 = 3.600 x1 = 4.5000 x2 = 1.2000 x2 = 2.9920 x2 = 4.5556 x3 = 0.6880 x3 = 4.5610 x3 = 2.1904 x4 = 0.4947 x4 = 4.5615 x4 = 1.3598 x5 = 0.4489 x5 = 4.5615 x5 = 0.7697 x6 = 0.4388 x6 = 0.5185 x7 = 0.4485 x7 = 0.4538 x8 = 0.4385 x8 = 0.4412 x9 = 0.4389 x10 = 0.4385 x11 = 0.4385

  11. x0 = 4.5 xn+1 = 5 – 2/xn

  12. Example Consider x3 – 5x – 4 = 0 , rearrange in the form xn+1 = f(xn) x3 = 5x + 4  Let x0 = 2 x8 = 2.56154 x1 = 2.41012 x9 = 2.56155 x2 = 2.52250 x10 = 2.56155 x3 = 2.55159 x4 = 2.55902 x5 = 2.56091 x6 = 2.56139 x7 = 2.56151

  13. y x Example Using iteration method formula xn+1 = 2 + ln(xn), find a root of the equation ln(x) – x + 2 = 0, (correct to 3 s.f.) and starting with x0 = 2. x0 = 2 x1 = 2.693 147 181 x2 = 2.290 710 465 x3 = 3.095 510 973 x4 = 3.129 952 989 x5 = 3.141 017 985 x6 = 3.144 546 946 x7 = 3.145 699 825 x8 = 3.146 026 848 So a root is 3.15 (3 s.f.) x9 = 3.146 140 339

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