8.2009 Approximate Solution To Equations

Approximate solutions to equations When a mathematical model is applied to a reallife situation the resulting equations are very different from those we have seen so far: there are fewer whole numbers and the solutions tend to be the best approximations from the available information. A different approach is necessary and a vast array of techniques has evolved together with the formidable assistance of computers. In this chapter we shall see how we can find solutions to certain types of equations. Given y = f(x). To find the root/roots of the equation y = f(x), we draw the curve y = f(x) and find the value/values of x at which the curve y = f(x) cuts the x–axis. Another method is by finding the value/values of x at which two graphs intersect. Finding the number of roots Consider the equation x 3 − x − 2 = 0 . To find the number of roots to this equation i) Draw the curve. If this curve is drawn accurately, the curve cuts the xaxis only once and therefore there is only one real root to this equation. ii) Rearrange the equation x 3 − x − 2 = 0 into x 3 = x + 2 . Draw the two graphs i.e. y = x 3 and y = x + 2 . Then find the intersection of these two graphs to obtain the root of the equation x 3 − x − 2 = 0 . In both cases however, it is difficult to obtain the exact root of the equation, but we can easily determine the number of real roots of the equation y = f(x). In general, roots of the equation f(x) = 0 can be located by considering the sign change of the function f(x). Consider the function f(x) = x 3 − x − 2 . f(1) = 13 − 1 − 2 = − 2 (negative value)

f(2) = 23 – 2 – 2 = 4 (positive value) The sign of f(x) changes from x = 1 to x = 2. Therefore, we say that the root of the equation f(x) = x3 – x – 2 lies between x = 1 and x = 2. Or we write 1 < α < 2. This is the first stage in finding an approximate numerical value to the root. 1. Show that the following equations have a root in the given interval: a. e x = 5x between x = 0 and x = 1 b. x = 3 sin x between x = 2.2 and x = 2.3 c. x + x + 1 + x + 2 = 5 between x = 1.8 and x = 2 d. tan x = 2x between x = 1 and x = 1.2 − 1 e. sinx = cos x between x = 0.7 and x = 0.8 2. Determine the number of roots of the following functions. State the interval between which the roots of the equation f(x) = 0 lies.

1

π π <x < 3 2

a. f(x) = ex – 10 cos x b. f(x) = 3 ln x + x – 8 3 < x <4

c. f(x) = tan x – 2x

1 < x <1.2

We have already seen how we can find an approximate value for a root of the equation f(x) = 0 by looking for a change in sign of the function f(x). Iteration The next stage is to improve the accuracy of the first approximation. This is done by generating a sequence of approximation which approach more and more closely to the value of the root that we’re looking for. We use a recurrence relation of the form xn+1 = F(xn) so that when we substitute xo for xn the expression gives us the value for x1. We then substitute x1 for xn and the value x2 emerges and so on. We keep going until the approximation is correct to the required degree of accuracy. This method of using the last value to estimate the next is called iteration, and the recurrence relation is called the iterative formula. Suppose for example, that we wanted to find 2 and we are given the iterative formula xn +1 =

xn2 + 2 , together with the starting point xo = 1. 2xn

The

formula with n = 0 tells us that 12 + 2 x1 = = 1.5 2(1) 1.5 2 + 2 x2 = = 1.417 = 1.42 2(1.5) 2 1.417 + 2 x = = 1.4142 = 1.41 3 2(1.417) 2 1.412 + 2 x = = 1.4142 = 1.41 4 2(1.412)

Since x3 and x4 have the same rounded off value, therefore the root, α = 1.41, correct to 3 significant figures.

Iterative Formulae Each equation has many possible rearrangements and these might lead to different roots, or possibly none at all. As an example, let’s take the

2

equation ex = 5x. If we put f(x) = ex – 5x, we find that f (0) = 1, f (1) = – 2.3, f (2) = – 2.6, f (3) = 5.1 and so we can say that there is a root of the equation, α, between 0 and 1 and another, β, between 2 and 3. 1 5

If we rearrange ex = 5x ⇒ x = e x ⇒ xn +1 =

1 xn e . 5

Alternatively, by taking ln’s of both sides, ln e x = ln 5x ⇒ x = ln 5x ⇒ xn +1 = ln 5xn

Lets take xo = 2 1 xn e 5 x0 = 2 x1 = 1.47 x2 = 0.88 x3 = 0.48 x4 = 0.32 x5 = 0.276 x6 = 0.264 x7 = 0.260 x8 = 0.259

xn +1 = ln 5xn

xn +1 =

x0 = 2 x1 = 2.303

. . . . . . x8 = 2.543

The first iterative formula shows that the iteration converges to the root between 0 and 1, while the second iterative formula converges to 2.543, which is the root between 2 and 3. So, different formula can converge to different roots from the same starting point. The first approximation that we use is also important. If we start just a little bit higher, say at x0 = 2 . 6, the sequence gets bigger and bigger very quickly and so we say that the approximation diverges from the root. Derivation of the iterative formula A formula can be found by rearranging the original equation until we end up with something like x = f(x) and then putting x = xn+1 into the LHS and x = xn into the RHS to give xn+1 = f(xn). For example, we could rearrange the equation x 3 − x − 2 = 0 into x3 = x – 2 ⇒ x = 3 x - 2 . From this we could derive the iterative formula xn +1 = 3 xn - 2 Rearrange the following equations into the given form: a

a) 2x3 +3x – 5 = 0

into

xn +1 =

b) x2 +7x – 2 = 0

into

xn +1 =

c) e2x = 2 – x

into xn+1 = a ln (b + cxn)

b + cxn2 a +b xn

3

d) 3x + sin x – 4 = 0

into xn+1 = sin-1(a + bxn)

e) x2 – 5x + 1 = 0

into

xn +1 = axn + b

f) x2 – 5x + 1 = 0

into

xn +1 = a +

g) 4 ln x = x

into xn+1 = e axn

h) 3ex + x – 9 = 0

into xn+1 = ln (a + bxn)

i) x ln x = 5

into

j) 2 cos x + 3x – 1 = 0

into

xn +1

b xn

a xn =e

xn+1 = cos

-1

(a + bxn)

Answer a) a = 5, b = 3, c = 2 (one root ⇒ α = 1) b) a = 2, b = –7 (only for negative approx.) c) a = ½ , b =2, c = –1; α = 0.4 (1 d.p) d) a = 4, b =–3 In each case find an equation in its simplest form possible, from which the iterative formula could have been derived: a) xn +1 = (2xn − 5) c) xn +1 = e) xn +1 =

1

3

4 xn2

−2

5xn − 2 xn

b)

xn +1 =

2 xn + xn 5

d)

xn +1 =

1 3 (xn − 3) 2

f)

xn +1 = 3 +

2 xn2

Iterative Process The general iterative formula is xn+1 = F(xn). If the iteration xn+1 = F(xn) is convergent, it will produce a sequence of approximations to a root of the equation x = F(x). In general xn+1 = F(xn) converges to root α if i) xo = is a fairly good approximation to α and ii) –1 < f '(α) < 1. (Perkins & Perkins pg 481) a) Pg 484 ex/23.4 Q2 - Q10

4

b) Show that the equation x =

1

has a root α between 0 and 1. 2+ x By using an iterative formula of the form xn+1 = F(x n), find α correct to two decimal places. You should show clearly your sequence of approximations. x+a c) Rearrange the equation 2x2 – x – 20 = 0 into the form x = , stating b the values of the constants a and b. Starting with x0 = 3 use the xn + a iteration formula x n +1 = with the values of a and b that you have b found to find the positive root of the equation. d) The iterative formula xn+1 = 1 – sin xn can be used to solve the equation sin x = 1 – x. Use this formula with x0 = 0.5 to find the root of this equation correct to 2 decimal places. Answer b)

α = 0.38

c)

a = 20, b = 2

α = 3.422

d)

α = 0.51

Approximate Solutions To Equations 1. Solve the following equations, using the iterative formula and starting at the given point correct to 3 significant figures. 3 ; x0 = 0 (a) x 2 + 5 x − 3 = 0 using x n +1 = xn + 5 (b) x 3 − 4 x − 2 = 0 using x n +1 = 3 4 x n + 2 ; x0 = 2 (c) ex = 3− x using x n +1 = ln(3 − x n ) ; x0 = 0.8 xn + 1 (d) x ln x = 1 using x n +1 = ;x = 2 1 + ln x n 0 5

(e) 2 x = tan x using x n +1 = tan −1 (2 x n ) ; x0 = 1.2 (f) sin x = cos-1x using xn +1 = cos (sin xn ) ; x0 = (g) x = e −

x 5

using x = e n +1

−

xn 5

π 4

; x0 = 1

2. The equation f(x) = 0, where f(x) is a polynomial, can be solved using the iterative formula

1

x n +1 = ( x n + 10) 4 a. Find the polynomial f(x). b. Starting with x0 = 1, find x1 , x 2 , x3 and x 4 to 4 decimal places. Hence give the value of the positive root of the equation correct to 3 decimal places. 1 c 3. (a) Show that the iterative formula x n +1 = ( x n + 2 ) can be used to find 2 xn the cube root of c. (b) Starting with x0 = 1.6, find 3 4 to 3 significant figures. 4. f(x) = 5x – 4 sin x – 2. a) Evaluate to 2 significant figures, f(1.1) and f(1.15). b) State why the equation f(x) = 0 has a root in the interval [1.1, 1.15]. An iterative formula of the form xn +1 = p sin xn + q is applied to find an approximation to the root of the equation f(x) = 0 in the interval [1.1, 1.15]. c) Stating the values of p and q, use this iteration formula with x0 = 1.1, to find x4 to 3 decimal places. Show the intermediate results in your working.

5. (a) By sketching the curves with equations y = 4 − x 2 and y = ex , show that the equation x 2 + ex − 4 = 0 has one negative root and one positive root. (b)Use the iteration formula x = (4 − ex n ) , x0 = −2, to find n +1 x1 , x 2 , x3 and x 4 and hence write down an approximation to the negative root of the equation giving your answer to 4 decimal places. An attempt to evaluate the positive root of the equation is made using the iteration formula x = (4 − ex n ) with x0 = 1.3. n +1 (c) Describe the result of such an attempt.

6

6. f(x) = e 0.8x −

1 3 − 2x

;x ≠

3 2

(a) Show that the equation f(x) = 0 can be written as x = 1.5− 0.5e-0.8x (b) Use the iteration formula x n +1 = 1.5− 0.5e-0.8xn with x0 = 1.3, to obtain x1 , x 2 and x3 . Give the value of x3 , an approximation to a root of f(x) = 0, to 3 decimal places. (c) Show that the equation f(x) = 0 can be written in the form x = p ln(3 – 2x), stating the value of p. (d) Use the iteration xn +1 = p ln(3 - 2xn ) with x0 = −2.6 and the value of p found in part (c), to obtain x1 , x 2 and x3 . Give the value of x3 , an approximation to the second root of f(x) = 0, to 3 decmal places. 7. (a) Show that the equation 21− x = 4 x + 1 can be arranged in the form 1 x = (2 − x ) + q, stating the value of the constant q. 2 1 −x (b) Using the iterative formula x n +1 = ( 2 n ) + q, x0 = 0.2 , with the value 2 of q found in part (a), find x1 , x 2 , x3 and x 4 . Give the value of x4 , to 4 decimal places.

8.

7

B

r

θ O

r

A

The diagram shows a sector OAB of a circle of radius r. The chord AB subtends an angle θ radians at the center O of the circle. Given that AB divides the sector into two regions of equal area, show that θ = 2 sin θ. Use theiterative formula θn +1 = 2 sin θn with θ1 = 2, to find the value of θ correct to 1 decmal place. xn , with x1 = 1, 9. The sequence given by the iteration formula x n +1 = tan x n converges to the number α, where 0 < α < 1. (a) Find x 2 and x3 correct to 3 decimal places. (b) State the equation of which α is a root and hence find the exact value of α 10.The equation x 3 + 3x − 10 = 0 has exactly one real root, α. An iteration for finding α is

1

Use this iteration, with a first x n +1 = (10 − 3 x n ) 3 . approximation x1 = 1.6 , to find α correct to 3 decimal places. 1

11.Show, by means of a graphical argument, that the equation x = 10e − 4 x has exactly one real root (denoted by α), and determine the pair of consecutive integers between which α lies. The iterative formula 1

x n +1 = 10e − 4

xn

can be used to find α. Starting with the nearest integer

above α, carry out four applications of this iteration, and state the number of significant figures to which α has been determined as a result of these calculations.

8

12.Use the iterative formula x n +1 =

2( x n3 + 1) , where x0 = 1, to calculate x1 , x 2 3 x n2

and x3 . Give your answers correct to 6 decimal places.(1.333333; 1.263889; 1.259933) Given that the iteration converges, find and simplify an equation in x whose real root is approximated by this approximation. x3 – 2 = 0 Hence write down the exact value of this root. exact root = 3 2 Find the absolute error e1 , e2 , e3 in the first three approximations x1 , x 2 and x3 , giving your answers to 6 decimal places. 0.073412; 0.003968; 0.000012 13.By means of sketches of curves y = ln(x – 1) and y = ¼ x, show that the equation ln (x – 1) = ¼ x has exactly two real roots, α and β, (α < β). (a) Verify that 3< α < 4 and 7 < β < 8. (b) Use the iterative formula x n +1 = 4 ln( x n − 1), with x0 = 7.2, to find the value of β, correct to one decimal place. (c) Using the same iterative formula as in part (b), a student tried to find an estimate forα. He started with x0 = 4, find the values of x1 and x 2 . What comment would you make about these results? 14.Show that the equation x 3 + 3 x 2 − 7 = 0 may be arranged into the form a , and state the values of a and b. Hence, using the iteration x= x+b a formula x n +1 = with x0 = 2 , together with your values of a and b, xn + b find the approximate solution x4 of the equation, giving your answer to an appropriate degree of accuracy. 15.The solution to

3x ex = 1

is to be found using simple iteration.

The

1 − xn .

equation is rearranged to give xn+1 = e 3

1 3

(a) Carefully sketch the graphs of y = x and y = e− x on the same axes. (b) Use the iteration with x1 = 1, to find x 2 , x3 and x 4 . (c) Use these four values and a cobweb diagram on your graph to illustrate how this method works.

9

16.(a) Show that the equation x sin x – 1 = 0 has two roots in the interval [0, π]. (b) Verify that the smaller root lies between 1.0 and 1.2 and use the 1 , to estimate this root, giving your iterative formula x n +1 = sin x n answer correct to 2 decimal places. (c) The curve with equation y = x sin x has one maximum point in the interval [0, π]. Show that the x-coordinate of this maximum point satisfies the equation x + tan x = 0. Answer 1(a) α = 0.541 (b) α = 2.21 (c) α =0.792 d) α = 1.76 (e) α = 1.17 (f) α = 0.77 (g) α = 0.845 4 2(a) x − x − 10 = 0 b) x1 = 1.821, x2 = 1.854, x3 = 1.856 and x 4 = 1.856; α = 1.856 3 b) α = 1.59 4 a) f(1.1) = - 0.065; f(1.15) = 0.099 c) p = 0.8 ; q = 0.4 ; x1 = 1.1, x2 = 1.1130, x3 = 1.1176 and x 4 = 1.119 5 b) x 0 = - 2. x1 = 1.9659, x2 = 1.9647 , x3 = -1.9646 and x 4 = −1.9646; root = - 1.9646 6 b) d)

x 0 = 1.3, x1 = 1.323, x2 = 1.327 and x3 = 1.327

p=−

5 4

x 0 = −2.6, x1 = −2.630, x2 = −2.639 and x3 = −2.642

7 a) q = − 8.

c)

1 4

b)

x1 = 0.1853, x2 = 0.189, x3 = 0.1884 and x 4 = 0.1888

θ = 1.9

9 a)

x1 = 1

x2 = 0.801 x3 = 0.789

b)

tan x = 1 ; α =

π 4

10. 1.699 11. α = 4; 1 significant figure 12. 1.333333; 1.263889; 1.259933; exact root = 0.073412; 0.003968; 0.000012

3

2

10