Quod Erat Demonstrandum | 數里無人到，山黃始知秋。岩間一覺醒，忘卻百年憂。



Applied Mathematics (II) Revision Notes and Exercises

Differential Equations

A. First Order

| |Variable Separable |Homogeneous |Linear First Order |

|Pattern |[pic] |[pic] |[pic] |

|Solving Technique |[pic] |Let [pic], then |By integrating factor |

| |Separate x and y |[pic] |[pic] |

| |from both sides completely. |Yield [pic] |and the solution is |

| | | |[pic] |

|Further technique：Inspection (e.g. you should know xdy/dx + y is d(xy)/dx immediately) and Substitution |

e.g. 1 (Separable)

ANS:

e.g. 2 (Homogeneous)

i. ANS:

ii. ANS:

e.g. 3 (Linear First Order with Integrating factor)

ANS:

B. Second Order

Homogeneous

Solve [pic].

|Step 1：Solve the auxiliary equation |

|[pic] |

|Hence [pic]= [pic], [pic] |

|Step 2：Write down the solution according to the following three cases. |

|Distinct real roots [pic],[pic] |Equal roots |Complex roots |

| |[pic] = [pic] = [pic] |[pic], [pic]= [pic] |

|Solution： |Solution： |Solution： |

|[pic] |[pic] [pic] |[pic]or |

| | |[pic] |

Non-Homogeneous

Solve [pic].

|Step 1：Solve the homogeneous DE. Obtain the solution yc. |

|Solve [pic]by the method mentioned above and obtain the solution yc. |

|Step 2：Guess a particular solution yp by the method of undetermined coefficients. |

|By observing f(x), we can guess out yp easily in the following ways: |

| |f(x) |yp |

|Exponential |[pic] |[pic] |

|Polynomials |[pic] |[pic] |

|Sine and Cosine |[pic]or [pic] |[pic] |

|Step 3：Modification for yp ? |

|If there is an expression in the guessing yp occurs in yc, just multiply the guessing yp by x. Continue this |

|"multiplication process" until no term in yp occurs in yc. |

|Step 4：Substitute yp into the original DE to determine the coefficients in the yp. |

|Step 5：The general solution to the DE is y = yc + yp. |

More examples in



e.g. 4

ANS:

e.g. 5

ANS:

e.g. 6 (01-3) substitution

e.g. 7 (00-2) drug injection

e.g. 8 (05-7) drug injection

e.g. 9 (01-7) flowing

e.g. 10 (02-9) flowing (current)

Exercises (* means the question is a school past paper)

DE1. ANS:

DE2. ANS:

DE3. ANS:

*DE4. (a) Solve [pic].

b) Evaluate [pic], where y is a function of x.

c) Using the results in (a) and (b), solve

[pic].

DE5. (96-7) flowing (element)

DE6. (04-9) flowing

DE7. (00-9) modeling

*DE8. flowing (molecules)

The figure shows that there is a small hole between two closed regions: A and B. The number of gas molecules moves from A to B per second and that from B to A per second are proportional to the numbers of molecules in A and B respectively; the proportionality constants are k1 and k2 respectively (k1 , k2 > 0). Let the numbers of molecules in A and B at time t be x and y respectively. At t = 0, x = n and y = 0.

(a) (i) Write down [pic] in terms of k1 , k2 x and y.

(ii) Show that [pic].

(iii) Hence express x in terms of k1 , k2 , n and t.

(iv) Find x in terms of k1 , k2 and n as t tends to infinity.

(b) After some time, the holes between B and C are opened. Let z be the number of molecules in C at time t. The number of molecules moves from B to C per second and that moves form C to B per second are proportional to y and z respectively; the proportionality constants are k3, where k3 is a positive constant.

(i) Write down [pic] in terms of k3 , y, and z.

(ii) Without solving differential equation, find x in terms of k1 , k2 and n as t tends to infinity.

(c) After some time, the hole between C and outside is opened. The number of

molecules moves from C to outside per second is [pic].

(i) Express [pic] in terms of k1 , k2 , k3 , x , y and z.

(ii) Hence, using (a)(i), show that [pic].

*DE9. Using the transformation [pic], show that the equation

[pic] can be transformed into the equation

[pic]. Hence find y in terms of x.

*DE10. Figure 1a shows a tank initially contains [pic] cm3 of sugar solution with concentration [pic] kg/cm3. There is a small hole at the bottom and the rate of solution flowing out is proportional to [pic], where V cm3 is the volume of solution in the tank and the proportionality constant is k. When the solution starts to flow out of the tank, sugar powder drops from above at a constant rate [pic]kg/s. Let x be the number of kilograms of sugar inside the tank t seconds after the solution starts to flow out.

(a) (i) Show that [pic].

(ii) Solve the above differential equation.

(iii) Find t when [pic]. (8 marks)

(b) Suppose when [pic], some sugar solution with concentration [pic] kg/cm3 flows into the tank (see Figure 1b) and there is no sugar powder dropping from above. The rate of solution flowing in is proportional to [pic] and the proportionality constant is also k.

i) Express [pic] in terms of k and V.

Express [pic] in terms of k, x, [pic], V0 and [pic].

Hence, express x in terms of [pic], V0 and [pic].

ii) Without solving any differential equations, find V as [pic]. (7 marks)

Probabilities I

Simple counting

For n distinct objects,

number of permutations of r objects = nPr ;

number of combinations of r objects = nCr.

e.g. 11

n persons a1, a2, …, an, lining up in a queue, how many ways of permutation such that a1, a2 and a3 stand next to each other? ANS: (n ( 2)!(3! (Regard 3 guys as a whole, like n ( 2 guys lining up, ways: (n ( 2)!. Then the 3 guys can permute as well, ways: 3!)

e.g. 12

Refer to the figure below. A man starts moving from A to B along grid lines. In every grid points, he may choose to go either north or east. How many different routes are there? ANS: 10C4 (Out of 10 steps, select 4 steps going north and the remaining going east.)

Number of ways of n persons sitting in a round table = [pic] = (n ( 1)!

Number of ways of lining up objects A, A, A, B, B, B, B, C, C is [pic].

e.g. 13

n persons a1, a2, …, an, sitting on a round table randomly. Find the probability that a1, a2 and a3 sit next to each other.

(method 1) Regard the 3 guys as a whole, there are n ( 2 guys sitting on the round table. Ways: 3!(n ( 2 ( 1)! The 3! means the 3 guys permute themselves. Hence the required probability = 3!(n ( 3)!/(n ( 1)! = 6/(n ( 1)(n ( 2).

(method 2)

a1 sits first, out of the remaining n ( 1 seats, a2 can choose to (1) sit either right next to a1 with the probability = 2/(n ( 1) or (2) leave one seat between a1 and a2 and the probability = 2/(n ( 1). For case (1), a3 has two positions for choosing, hence the probability that a3 sits next to them is 2/(n ( 2). For case (2), a3 can only choose the seat at the middle, hence the probability = 1/(n ( 2). Hence the required probability = 2/(n ( 1) ( 2/(n ( 2) + 2/(n ( 1) ( 1/(n ( 2) = 6/(n ( 1)(n ( 2).

Further examples

Let n, k be positive integers with n ( k.

Number of positive integral solutions to x1 + x1 + ... + xk = n is nCk ( 1.

Number of non-negative integral solutions to x1 + x1 + ... + xk = n is n + k ( 1Cn.

P(A∪B) = P(A) + P(B) ( P(A∩B)

P(A∪B∪C) = P(A) + P(B) + P(C) ( P(A∩B) ( P(B∩C) ( P(C∩A) + P(A∩B∩C)

Exclusive events: P(A∩B) = 0 => P(A∪B) = P(A) + P(B)

Complimentary events: P(A') = 1 ( P(A)

Conditional probability

- By formula P(A | B) = [pic] => P(A∩B) = P(B)P(A | B)

- By tree diagram (Bayes’ theorem)

P(A | E) = [pic]

P(B | E) = [pic]

P(A | F) = [pic]

P(B | F) = [pic]

- By reduced sample spaces

e.g. P(roll a fair die, obtain 6 before 1) = [pic]

Independent events: P(A | B) = P(A) or P(B | A) = P(B) => P(A∩B) = P(A)P(B)

e.g. 14 (05-3) Mobile phone

Random variable (= variate) X

| |Discrete |Continuous |

|Distribution |Probability function (p.f.) |Probability density function (p.d.f.) |

|f(x) |- table form or |P(a < X < b) = [pic] |

| |- P(X = x1) = f(x1) |f(x) ( 0 and [pic] = 1 |

| |0 ( f(x) ( 1 and[pic] = 1 | |

|Expectation |E(X) = [pic] |E(X) = [pic] |

|( | | |

|Variance |Var(X) = E[(X ( ()2] |Var(X) = [pic] |

|( 2 |= [pic] | |

|Hence, Var(X) = E(X 2) ( ( 2 |

|Cumulative |F(x) = P(X ( x) = [pic] |F(x) = P(X ( x) = [pic] |

|distribution | | |

|function (c.d.f.) | | |

|F(x) | | |

|Find f(x) from F(x).|For a ( x < b, F(x) = p. |f(x) = [pic]F(x) |

| |For b ( x < c, F(x) = q, | |

| |then f(b) = p ( q. | |

Note 1: Distribution = p.f. (or p.d.f.)

Distribution function = c.d.f.

THUS, Distribution ( Distribution function

Note 2: Expectation with condition

e.g. For discrete X, E(X | the value of X ( x1) = [pic].

Properties of Expectation and Variance

1. E(a) = a

2. E(aX + bY) = aE(X) + bE(Y)

3. If X and Y are independent, E(XY) = E(X)E(Y).

(Note : E(XY) ( E(X)E(Y) in general)

4. Var(a) = 0

5. Var(aX) = a 2Var(X) (Further: s.d. of aX = |a|(s.d. of X)

6. Var(X ( Y) = Var(X) + Var(Y) for X and Y are independent

e.g. 15 (05-6) distribution function

e.g. 16 (02-11) machine game

e.g. 17 (00-11) 2 black, 13 white balls

e.g. 18 (04-10) drawing balls in grand sale

e.g. 19 (05-10) token in fun fair

Recurrence

Type A

e.g. In a series of games, P(win | last game win) = p, P(win | last game lose) = q

Let Pn = P(win at the nth game), then

Pn = pPn ( 1 + q(1 ( Pn ( 1) = (p ( q)Pn ( 1 + q

By sum of G.P., we can find the general formula for Pn as

Pn = (p ( q)n ( 1P1 + q[pic]

Putting P1 = 1 or 0 indicating the first game being win or lose respectively.

Type B

e.g. A fair die is rolled.

Let Pn be the probability that no two consecutive sixes occur in the first n rolling.

Express Pn in terms of Pn ( 1 and Pn ( 2 (n > 2).

Pn = [pic]Pn ( 1 + [pic]Pn ( 2

e.g. 20 (02-2) number of ‘six’ in n thrown

Exercises

PROB-1.1 (06-6) dart game

PROB-1.2 (99-6) probability functions

*PROB-1.3

A continuous random variable X has the continuous cumulative distribution function: F(x) = [pic]

(a) Determine the values of k , c and the probability density function of X in

terms of a .

(b) Evaluate the median of X in terms of a .

PROB-1.4 (03-12) TV game show

PROB-1.5 (06-10) shooting game, pdf

PROB-1.6 (98-4) knock-out tournament

*PROB-1.7

In the figure, there is a ball at 0 initially. In every step, it can either move to the

left or to the right by 1 unit. The probabilities for it to move to the left and that to

the right are the same.

(a) Find the probability that it goes back to 0 after 4 moves.

(b) Find the probability that it is at the right of 0 after n moves.

*PROB-1.8 [In this question, give your answers in exact values.]

In a Super Football Tournament, group A consists of 4 teams:[pic], a, b and c; group B

consists of another 4 teams: [pic], d, e and f, where [pic] and [pic] are seeded teams and the rest are non-seeded teams. Each match is independent with the following probabilities.

If both [pic] and [pic] are both seeded or both non-seeded,

probability that [pic] wins = probability that [pic] loses [pic].

If [pic] is seeded and [pic] is non-seeded,

probability that [pic] wins [pic], probability that [pic] loses [pic].

In Round 1, each team has to play every other team in the same group once. In each

match, 3 points, 0 point and 1 point will be awarded to the team which wins, loses and

has a draw respectively. The 2 teams with highest points will go to Round 2. If 2, 3 or 4 teams get the same point, each team has to draw a card from the bag containing

cards numbered 1, 2, 3 and 4. The team(s) gets the largest number will go to Round 2.

(a) In group A, 4 matches have been played. The results are as follows:

[pic] defeated a, [pic] defeated b, a defeated c and a had a draw with b.

Find the probability that c can go to Round 2.

(b) In group B, no match has been played.

(i) Find the probability that [pic] has 9 points.

ii) Find the probability that [pic] has 6 points.

iii) Find the probability that [pic] has 6 points and [pic] cannot go to Round 2.

iv) If [pic] wins 1 match and loses 2, find the probability that it can go to Round 2.

c) In Round 2, there is a play-off game. Suppose [pic], b and e can go to Round 2 and they are arranged as shown in Figure 2. If there is a draw in any match, a fair coin will be flipped to decide which team wins.

(i) Find the probability that [pic] is the champion.

ii) If [pic] is the champion, find the probability that [pic] plays with [pic].

PROB-1.9 (01-11) traveling time of train

PROB-1.10 (92-5) recurrence, drawing white and black balls

*PROB-1.11

An elevator starts carrying five persons at the ground floor and then goes up. It can stop at any floor of the building (from the first floor to the third floor). Events that people leaving the elevator are assumed to be independent. Let X be the number of ‘stop’ of the elevator during a “going-up” journey.

(a) Find P(X = 1) and P(X = 2).

(b) Find E(X).

PROB-1.12 (88-5) minimizing expected loss, pdf

The manager of a supermarket knows, from past experience, that the demand for a certain product for the coming week is x kilograms, where x lies between 0 and A , with the following probability density function :

Suppose the manager stocks S kilograms of the product to meet the demand. If S > x, the supermarket will suffer a loss of $C1 for every kilogram overstocked. If S  0 on [a , b] => f concave upwards on [a , b] => IT > I (over-estimate)

f’’(x) < 0 on [a , b] => f concave downwards on [a , b] => IT < I (over-estimate)

Simpson’s rule

(1) 2 subintervals

[pic]

(2) 2n (n > 1) subintervals

IS = [pic]

where 2n = number of sub-intervals

h = (equal) width of sub-interval = [pic]

f0 = f(a) , f2n = f(b) , fi = f(a + ih) for i = 1, 2, …, 2n ( 1

P.S. Don’t forget : the number of sub-intervals is even.

Error in Simpson’s rule

E = I ( IS = ([pic]f (4)(() where a < ( < b

P.S. To use the formula to find the number of subintervals needed to meet a required accuracy, try to replace h by [pic] in the formula.

e.g. 30 (05-9) Bobby’s method in trapezoidal rule

e.g. 31 (03-8) Interpolation and integration

e.g. 32 (06-7) Trapezoidal rule and infinite series

e.g. 33 (02-7) George’s method in Simpson’s rule

Exercise

NA-I.1 (01-2) Trapezoidal rule and error

NA-I.2 (00-8) Trapezoidal rule and odd function

NA-I.3 (99-8) Simpson’s rule and cubic polynomial

NA-I.4 (04-7) Trapezoidal rule, error and solving D.E.

NA-I.5 (96-9) Trapezoidal rule, concavity

NA-I.6 (95-7) Trapezoidal rule, substitution

*NA-I.7

Let [pic].

(a) (i) Show that [pic].

(ii) Hence find the bound for [pic].

(Correct your answer to 2 decimal places.) (3 marks)

(b) (i) Use Trapezoidal Rule with 3 subintervals to estimate I.

(Correct your answer to 3 decimal places.)

(ii) Find the error bound in (b)(i). (Correct your answer to 4 decimal places.)

(c) Taylor's Trapezoidal Rule: For each subinterval [pic], use Taylor Polynomial

of degree 1 about [pic] to estimate I. (See the figure below.)

(i) Use Taylor's Trapezoidal Rule with 3 subintervals to estimate I .

(Correct your answer to 3 decimal places.)

(ii) Find the error bound in (c)(i). (Correct your answer to 4 decimal places.)

(d) With an aid of a graph, after considering [pic] for x in [1, 1.6], using

(b)(i) and (c)(i), write down the upper bound and the lower bound for I.

(Correct your answer to 2 decimal places.)

*NA-I.8

Let I = [pic]and In be the trapezoidal estimate of I with n subintervals.

(a) Evaluate I8 . (Give 8 significant figures)

Let En = I ( In .

(b) (i) By finding I exactly, evaluate E8 . (Give 7 decimal places)

(ii) Give that I16 = 1.3774327. Find a natural number k such that [pic] ( k .

(iii) By using the value k in (b)(ii), determine a better estimate than I16 for I .

(Give 8 significant figures)

(c) Let J = [pic].

Is it true that I2003 < J ? Explain without direct evaluation on J .

(D) Solving equations

To prove f(x) = 0 has exactly one root in [a,b], we need to show

(1) f(a)f(b) < 0 (2) f(x) is continuous on [a,b] (3) f(x) is monotonic on [a,b]

(D1) Bisection method

Algorithm

Having compute and f(xr)

then, (i) is the root of f(x) ’ 0 if

(ii)

Error

Suppose [a0 , b0] is the starting interval, after n steps, the interval width will be

bn ( an = [pic](bn ( 1 ( an ( 1) = [pic](bn ( 2 ( an ( 2) = … = [pic](b0 ( a0). Let s be a

root of f(x) = 0, then after n steps, xn = [pic], | xn ( s | ( [pic](bn ( 1 ( an ( 1)

= [pic](b0 ( a0). Hence, we can estimate the error.

e.g. 34 Estimate the number of iterations required to obtain the root to having an

accuracy to 4 decimal places. Starting interval is [0 , 1].

Solution [pic](1 ( 0) < 0.5 ( 10(4 => n > 14.29 => 15 iterations required.

Remark : drawback --- irregular error.

(D2) Regular Falsi method

By similar triangle, [pic] => xr + 1 = [pic]

Algorithm

Having [ar , br] compute xr + 1 = [pic].

then, (i) xr + 1 is the root of f(x) ’ 0 if f(xr + 1) ’ 0,

(ii) [ar + 1 , br + 1] = [pic]

(D3) Secant

Algorithm

xr + 2 = xr + 1 ( f(xr + 1)[pic] ; r ( 0

e.g. 35 (04-3) method of false position

(D4) Newton-Raphson's Method

By considering the slope of tangent at x = xr, we have

f’(xr) = [pic] => xr + 1 = xr ( [pic]

Remark : We cannot locate the true root s if s is complex, f’’(s) = 0 or the ‘guessing’ value is far away from s.

e.g. 36 (03-11a) Newton’s method and probability

Error estimation

Suppose ( is the unique root of f(x) = 0 in the interval [a , b].

Let m = [pic] and M = [pic].

Suppose {xn} be the sequence generated by Newton’s method.

Then | xn ( ( | ( [pic]| xn ( 1 ( ( |2 .

[pic]( [pic]

And hence, if |M/2m| < 1, xn converges to (; also the error can be estimated.

(D5) Fixed-Point Iteration Method

To solve f(x) = 0, try to convert it into x = g(x); where g(x) is called an iteration function. i.e. xn = g(xn ( 1) ; n ( N.

Note that g(x) may not be unique for the same f(x).

Theorem 1 (Existence of solution)

If g(x) is continuous on [a , b] such that g(x) ( [a , b] ( x ( [a , b], then x = g(x)

has at least one solution ( ( [a , b].

Proof of theorem 1

Let f(x) = x ( g(x) on [a , b].

∵ g(x) is cont. on [a , b] ( f(x) is cont. on [a , b]

∵ a ( g(x) ( b ( x ( [a , b] ( f(a) = a ( g(a) ( 0 and f(b) = b ( g(b) ( 0

By intermediate value theorem, ( ( ( [a , b] s.t. f(() = 0 => ( = g(().

Theorem 2 (Unique solution and convergence)

If g(x) and g’(x) are continuous on [a , b] such that g(x) ( [a , b] ( x ( [a , b], and ( a positive constant K s.t. [pic] = K < 1. then

(1) x = g(x) has a unique solution ( ( [a , b];

(2) {xn} converges to ( for any initial guess x0 ( [a , b].

Proof of theorem 2

(1) By theorem 1, x = g(x) has at least one root ( ( [a , b]. Suppose there are two roots ( and ( ( [a , b]; i.e. ( = g(() and ( = g((); thus,

( ( ( = g(() ( g(() = g’(c)(( ( () where c lies between ( and ( [mean value theorem]

Now, |( ( (| = |g’(c)||( ( (| => |g’(c)| = 1 if ( ( ( (contradicting to the fact that [pic] < 1) hence ( = (.

(2) Claim that xn ( [a , b] for n = 0,1,2,…

x0 ( [a , b] (given)

Suppose xk ( [a , b], consider xk + 1 = g(xk)

and since g(x) ( [a , b] for any x ( [a , b]

hence g(xk) ( [a , b] by the fact that xk ( [a , b],

i.e. xk + 1 ( [a , b], by M.I., xn ( [a , b]

Now,

|xn ( (| = |g(xn ( 1) ( g(()|

= |g’(cn)||xn ( 1 ( (| where cn lies between ( and xn ( 1 [mean value theorem]

( K|xn ( 1 ( (| (∵ xn ( 1 , ( ( [a , b] => cn ( [a , b] => |g’(cn)| ( K)

( K2|xn ( 2 ( (|

…

( Kn|x0 ( (| (inductively)

Since 0 < K < 1, Kn ( 0 as n ( ( ; hence {xn} converges to (.

Theorem 3 (Error estimation)

Let K = [pic] < 1. In the n-th iteration,

(1) |xn ( (| ( [pic]|x1 ( x0| ( n ( N;

(2) |xn ( (| ( [pic]|xn ( xn ( 1| ( n ( N;

Proof of theorem 3

(1) |x1 ( (| = |g(x0) ( g(()|

= |g’(c)||x0 ( (| where c lies between ( and x0 [mean value theorem]

( K|x0 ( (|

Hence |x0 ( (| = |(x0 ( x1) + (x1 ( ()| ( |x0 ( x1| + |x1 ( (| ( |x0 ( x1| + K|x0 ( (|

=> |x0 ( (| ( [pic]|x0 ( x1|

By the proof of theorem 2 (2), we have

|xn ( (| ( Kn|x0 ( (| ( [pic]|x1 ( x0|

(2) |xn ( (| = |g(xn ( 1) ( g(()|

= |g’(cn)||xn ( 1 ( (| where cn lies between ( and xn ( 1 [mean value theorem]

( K|xn ( 1 ( (|

= K|(xn ( 1 ( xn) + (xn ( ()|

( K|xn ( 1 ( xn| + K| xn ( (|

Thus, |xn ( (| ( [pic]|xn ( xn ( 1|.

e.g. 37 (01-9) fixed-point iteration, choosing suitable (

e.g. 38 (05-8) fixed-point iteration, choosing suitable iteration function

e.g. 39 (03-9) fixed-point iteration, coordinate geometry

Exercise

NA-SE.1 (00-7) fixed-point iteration, proof

NA-SE.2 (02-8) fixed-point iteration, proof

NA-SE.3 (04-8) fixed-point iteration, proof

*NA-SE.4

Let [pic] and let equation (*) be f(x) = 0.

(a) Show that (*) has a root in [1.5, 2]

(b) If Newton’s Method is employed to estimate the root of (*) and it is considered as a Fixed-Point Iteration, by considering the derivative of the iteration function, show that Newton’s Method is not suitable.

(c) Use Secant Method with x1 = 1.5 and x2 = 2 to approximate the root of (*) until

[pic]. (Correct your answer to 4 decimal places.)

NA-SE.6 (06-8) fixed-point iteration, proof

*NA-SE.7

Let [pic]

(a) (i) Show that [pic] has a root in the interval [pic].

ii) For [pic], find the minimum value of [pic].

(iii) For [pic], find the maximum value of [pic].

(b) Figure 3 shows the curve of [pic].

Given that [pic], [pic] [pic],

[pic] and for any [pic], [pic].

AB cuts the x-axis at [pic].

i) Prove that [pic],

where [pic] and [pic].

(You may assume Mean Value Theorem is applicable.)

ii) Hence, show that [pic].

iii) Using the above result, explain when method of false position is employed to estimate the root of the equation [pic], b is always one of the end points of the sequence of intervals containing the root, i.e., the sequence of intervals containing the root are [pic].

iv) From (b)(i), show that [pic],

where [pic] and [pic]for any [pic].

(You may assume Mean Value Theorem is applicable.) (7 marks)

c) (i) Using the result (b)(iii), apply method of false position to estimate the root of the equation [pic] with initial interval [pic] up to 2 more intervals, correct your answer to 4 decimal places.

ii) Using the above results, estimate the error bound, correct your answer to 6 decimal places.

NA-SE.9 (85-4)

(a) Show that the equation

x3 + x + 1 ’ 0

has only one root α in the interval -1  ................
................

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