NM MEI A-level Maths Coursework

Report for the Numerical Methods module of OCR AS/A Level GCE Further Mathematics (MEI)

February 2015

FIND THE VOLUME OF GOLD

REQUIRED FOR A SOLID GOLD

PRISMATIC BUTTERFLY

SCULPTURE

By

Maurice Yap

Peter Symonds College

Find the volume of gold required for a solid gold prismatic butterfly sculpture

Course unit | Maurice Yap 6946 ii

TABLE OF CONTENTS 1. Problem specification ..........................................................................1

2. Strategy ...............................................................................................1

2.1. Trapezium rule......................................................................................................................................... 2

2.2. MId-point rule ......................................................................................................................................... 2

3. Working ............................................................................................. 4

4. Error analysis ..................................................................................... 7

5. Conclusion ......................................................................................... 9

6. Reference List .................................................................................. 10

7. Bibliography ................................................. Error! Bookmark not defined.

8. Appendices ........................................................................................ 11

TABLE OF FIGURES Figure 1: Graphic representation of x6+y6=x2 .................................................................................................... 1

Figure 2: Trapezium rule with four strips ......................................................................................................... 3

Figure 3: Mid-point rule with four strips........................................................................................................... 3

Figure 4: A graph of y=(18 x4+2)/(9 (x4-1) (x2-x6)^(5/6)), the second derivative of f(x) ................................ 10


GCE Further Mathematics (MEI) – Numerical Methods | Maurice Yap 6946 1

1.PROBLEM SPECIFICATION The North Korean dictator has decided to build a butterfly sanctuary in the capital, Pyongyang. He wants a

solid gold sculpture of a butterfly to be built at its entrance. Because gold is expensive, the exact quantity

required must be determined.

Architects have decided that the sculpture will take the shape of a, 18-cm deep butterfly prism, where its

“base” is modelled by the following implicit equation (known as the butterfly curve), with each unit

representing a metre (Barile & Weisstein 2002):

𝑦6 + 𝑥6 = 𝑥2

This is shown graphically in figure 1. To accommodate for bolts to attach the gold prism to its supporting

plinth, 5cm either end of the prism must be made from a different metal. For the equation to be true, all x

values are in the range [-1, 1]. This means that the x value must be limited to a new domain of [-0.95, 0.95].

The volume of this prism can be calculated by multiplying the depth by the cross sectional area. By

rearranging the implicit equation to have y as the subject, the cross sectional area is an integral with limits

of -0.95 and 0.95. Therefore (in units, metres cubed):

𝑉 = 0.18 ∫ ±√𝑥2 − 𝑥660.95

−0.95

𝑑𝑥

FIGURE 1: GRAPHIC REPRESENTATION OF X6+Y6=X2

2.STRATEGY Because I am not able to solve the integral using analytical methods (e.g. by parts, substitution), it is

necessary to estimate it using numerical integration methods.



The equation which I must integrate is (rearrangement of original implicit equation):

𝑦 = ±√𝑥2 − 𝑥66

Figure 1 appears to show that the two axes are lines of symmetry. For the y-axis, this can be proven

algebraically:

𝑊ℎ𝑒𝑛 (−𝑥)𝑖𝑠 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒𝑑 𝑖𝑛 𝑝𝑙𝑎𝑐𝑒 𝑜𝑓 𝑥:

𝑦 = ± √(−𝑥)2 − (−𝑥)66= ±√𝑥2 − 𝑥66

⇒ 𝑦 ℎ𝑎𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑓𝑜𝑟 𝑥 𝑤ℎ𝑒𝑛 𝑥 𝑖𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 − 1∎

Furthermore, for the x-axis:

𝑅𝑒𝑎𝑟𝑟𝑎𝑛𝑔𝑖𝑛𝑔 𝑖𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑦:

𝑥2 − 𝑥6 = ±√𝑦6

𝑊ℎ𝑒𝑛 (−𝑦)𝑖𝑠 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒𝑑 𝑖𝑛 𝑝𝑙𝑎𝑐𝑒 𝑜𝑓 𝑦:

𝑥2 − 𝑥6 = ±√(−𝑦)6± √(𝑦)6

⇒ (𝑥2 − 𝑥6) ℎ𝑎𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑓𝑜𝑟 𝑦 𝑤ℎ𝑒𝑛 𝑦 𝑖𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑒𝑑 𝑏𝑦 − 1∎

This means that it is possible to integrate just the portion of the curve that is in the positive quadrant, then

multiply it by four in order to calculate the total cross sectional area.

An equivalent formula for the volume of gold required, as deduced from the above, is:

𝑉 = 4 × 0.18 ∫ √𝑥2 − 𝑥660.95

0

𝑑𝑥 = 0.72 ∫ √𝑥2 − 𝑥660.95

0

𝑑𝑥

The three methods available to me are the trapezium rule, the mid-point rule and Simpson’s rule. For the

sake of simplicity in the explanations, please note that the coefficient of 0.72 is disregarded for the

remainder of section 2.

2.1.TRAPEZIUM RULE

Applying the trapezium rule involves taking the area under the curve and splitting it into several trapezia

with equal width. The sum of the areas gives an approximation to the integral’s solution. Figure 2 shows

this with four equal strips of width 0.2375 (0.95 ÷ 4). It can be seen from the shaded region that this

method gives an under-estimate for the integral. Moreover, it can be deduced that as the number of strips

increases (and so, the thinner each strip is), the extent of the under-estimation given by the total area of the

trapeziums decreases, and in turn, the magnitude of the error in the approximation decreases.

2.2.MID-POINT RULE

The mid-point rule splits the area under the curve into equal-width rectangles, with the centre of the top of

each of these rectangles being a point on the curve. Again, the total area of each of these rectangles

approximates the solution of the integral. Figure 3 shows this rule applied to my integral. This provides an

over-estimate as the curve is “well-behaved” and is concave (Lissaman & West 2004, p.76).



FIGURE 2: TRAPEZIUM RULE WITH FOUR STRIPS

FIGURE 3: MID-POINT RULE WITH FOUR STRIPS

2.3.SIMPSON’S RULE

Under the notion that the trapezium rule will always be an underestimate of the integral as the number of

strips increases and the mid-point rule continues to produce over-estimates, upper and lower bounds can

be deduced for the integral. As the number of strips increases, the error (extent to which the area is under-

or over-estimated) will become less.

Simpson’s rule capitalises on this and is an average of these two integration approximations, with the mid-

point rule being weighted twice as heavily as the trapezium rule. It is therefore a more accurate

approximation for the area under the curve than the other two approximations. Because its convergence

towards the true value for the area under the curve is derived from two other, also converging

approximations, Simpson’s rule converges much quicker and requires less strips, and consequently, fewer

calculations, to produce a more accurate value.



3.WORKING I will first approximate the integral below first using the three numerical methods identified in section 2,

before multiplying it by 0.72 to approximate the volume of gold required for my problem.

∫ √𝑥2 − 𝑥660.95

0

𝑑𝑥

Hence, the following definition is given:

𝑓(𝑥) = √𝑥2 − 𝑥66

The formula for the trapezium rule is the following:

∫ 𝑓(𝑥)𝑑𝑥𝑏

𝑎

≈ 𝑇𝑛 =ℎ

2[𝑓(𝑎) + 𝑓(𝑏) + 2(𝑓(𝑎 + ℎ) + 𝑓(𝑎 + 2ℎ) + ⋯ + 𝑓(𝑎 + ℎ(𝑛 − 2)) + 𝑓(𝑎 + ℎ(𝑛 − 1)))]

where n is the number of strips and h = (b - a) ÷ n, the width of each strip.

Hence for my integral: (where h = ((0.95 - 0) ÷ n) = (0.95 ÷ n))

𝑇𝑛 =0.95

2𝑛[𝑓(0) + 𝑓(0.95) + 2 (𝑓(

0.95

𝑛) + 𝑓(

2 × 0.95

𝑛) + ⋯ + 𝑓(

0.95(𝑛 − 2)

𝑛) + 𝑓(

0.95(𝑛 − 1)

𝑛))]

The formula for the mid-point rule is the following:

∫ 𝑓(𝑥)𝑑𝑥𝑏

𝑎

≈ 𝑀𝑛 = ℎ(𝑓(𝑎 +ℎ

2) + 𝑓(𝑎 +

3ℎ

2) + ⋯ + 𝑓(𝑎 +

ℎ(2𝑛 − 3)

2) + 𝑓(𝑎 +

ℎ(2𝑛 − 1)

2))

where n is the number of strips and h = (b - a) ÷ n, the width of each strip.

Hence for my integral: (where h = ((0.95-0) ÷ n) = (0.95 ÷ n))

𝑀𝑛 =0.95

𝑛(𝑓(

0.95

2𝑛) + 𝑓(

3 × 0.95

2𝑛) + ⋯ + 𝑓(

0.95(2𝑛 − 3)

2𝑛) + 𝑓(

0.95(2𝑛 − 1)

2𝑛))

The trapezium rule can easily be calculated using Tn and Mn using this formula:

𝑆𝑛 =2𝑀𝑛 + 𝑇𝑛

3

where n is the number of strips.

I have elected to write a computer program to produce the trapezium rule, mid-point rule and Simpson’s

rule approximations to the integral. This is because it is much more efficient at outputting approximations

with large numbers of strips and is much quicker to implement than a spreadsheet.

It is written in Python and has three input variables: ‘a’ and ‘b’, the lower and upper limits of the integral

respectively, and ‘p’, the total number of n values to be calculated:



The trapezium rule is calculated using a function named ‘trap’. Within this function, ‘h’ is the width of each

strip, ‘v’ is the variable for the multiplier of (h ÷ 2) and ‘u’ is the variable that generates and represents each

of the values of x to be used for the different values of f(x) that are added together and multiplied by two.

Finally, ‘t’ is the output value of the function, the trapezium rule approximation.

The initial value of v is calculated by adding together f(a) and f(b) on line 4. Lines 6 to 8 are a loop that

generates values of u and f(u). The value of f(u) is multiplied by two and added to the ‘v’ variable on line 8

on each iteration of the loop. On line 10, the function is completed by multiplying the final value of the ‘v’

variable by half of ‘h’ to create the value of ‘t’, which is outputted on line 11. Please note that a double

asterisk (**) is the exponent operator in Python (equivalent to ^ in most spreadsheet programs).

The ‘midOrd’ function calculates the mid-ordinate rule approximation. The width of each strip is again

represented by the ‘h’ variable; ‘v’ is the variable which accumulates the values of f(x) and ‘u’ produces the

values of x for which f(x) must be calculated. The calculated approximation is the variable, ‘m’.

Lines 6 to 8 are a loop which generates all values of ‘u’ (line 7) and accumulates f(u) onto the ‘v’ variable.

Line 10 calculates ‘m’ by finding the product of the final value of ‘v’ and ‘h’. Line 11 outputs ‘m’.

The code which outputs and displays all the desired trapezium rule, mid-point rule and Simpson’s rule

approximations is shown below.

Lines 3 to 5 display the approximations for the trapezium rule. Lines 4 and 5 are a loop that runs the ‘trap’

function using the lower and upper limits of the integral with a number of strips, starting from one, that is

1 #Input variables

2 p=int(eval(input('Number of n values in approximations')))

3 a=float(eval(input('Lower bound')))

4 b=float(eval(input('Upper bound')))

1 #Trapezium rule

2 def trap(a,b,n):

3 h=(b-a)/n

4 v=(((a**2)-(a**6))**(1/6.0))+(((b**2)-(b**6))**(1/6.0))

5 u=0

6 for i in range(1,n):

7 u=(a+(i*h))

8 v+=2*((u**2)-(u**6))**(1/6.0)

9

10 t=(h/2.0)*v

11 return t

1 #Mid ordinate rule

2 def midOrd(a,b,n):

3 h=(b-a)/n

4 v=0.0

5 u=0

6 for i in range(1,n+1):

7 u=(a+(((2.0*i)-1.0)*h)/2.0)

8 v+=(((u**2)-(u**6))**(1/6.0))

9

10 m=v*h

11 return m



doubled each time (the number of times it is doubled depends on the value of ‘p’). Line 5 creates the label

for each approximation so that each can be identified by type of approximation and number of strips.

Lines 7 to 9 do the same thing, using the ‘midOrd’ function to obtain the mid-ordinate rule approximations.

Lines 11 to 13 calculate the Simpson’s rule approximations by inserting the Tn and Mn values into the

formula for it. They are obtained using the ‘trap’ and ‘midOrd’ functions respectively.

It should be noted that for each time the number of strips is doubled, the width of each strip is halved.

Running the program with the user inputs of a = 0, b = 0.95 and p = 20 gives the following approximations

(the raw output of the program can be found in the appendix):

n (number of strips)

h (width of each strip)

Tn (trapezium rule approximation)

Mn (mid-point rule approximation)

Sn (Simpson's rule approximation)

1 0.95 0.3526330592938719 0.7348064575728066 0.6074153248131617 2 [A3] 0.475 [B3] 0.5437197584333392 0.6976945615912183 0.6463696272052587 4 0.2375 0.6207071600122788 0.6803066221069612 0.6604401347420671 8 0.11875 0.65050689105962 0.6731615980211872 0.6656100290339982 16 0.059375 0.6618342445404037 0.6704370430236609 0.6675694435292417 32 0.0296875 0.6661356437820323 0.6694314248300444 0.6683328311473736 64 0.01484375 0.6677835343060384 0.6690596489693071 0.6686342774148842 128 0.007421875 0.6684215916376727 0.6689198846426211 0.668753786974305 256 0.0037109375 0.6686707381401477 0.6688664403164426 0.6688012062576777 512 0.00185546875 0.6687685892282945 0.6688457415256341 0.6688200240931876 1024 0.000927734375 0.6688071653769639 0.6688376551967419 0.6688274919234826 2048 0.0004638671875 0.6688224102868545 0.6688344781543543 0.6688304555318544 4096 0.00023193359375 0.6688284442206045 0.66883322535043 0.6688316316404882 8192 0.000115966796875 0.6688308347855151 0.668832730176492 0.6688320983794998 16384 5.79833984375 × 10-5 0.6688317824810049 0.6688325341669943 0.6688322836049978 32768 2.899169921875 × 10-5 0.6688321583240022 0.6688324565056877 0.6688323571117926 65536 1.4495849609375 × 10-5 0.6688323074148468 0.6688324257170426 0.6688323862829773 131072 7.2479248046875 × 10-6 0.6688323665659334 0.6688324135063952 0.6688323978595746 262144 3.62396240234375 × 10-6 0.6688323900361666 0.6688324086625402 0.668832402453749 524288 1.811981201171875 × 10-6 0.6688323993493503 0.668832406740728 0.6688324042769355

The second column’s cells’ values were generated by dividing 0.95 by the cell to its left. Spreadsheet

software (Microsoft Excel) was used for this because of the speed and convenience offered by its “drag-and-

fill” function. For example, the formula in cell B3 was:

=0.95/A3

1 #Output

2

3 print('Trapezium rule approximations:')

4 for x in range(p):

5 print('T',(2**x),'=',trap(a,b,(2**x)))

6

7 print('Midpoint rule approximations:')


9 print('M',(2**x),'=',midOrd(a,b,(2**x)))

10

11 print('Simpsons rule approximations:')


13 print('S',(2**x),'=',((2*(midOrd(a,b,(2**x)))+(trap(a,b,(2**x))))/3))



Under the notion that trapezium rule and mid-point rule approximations are under- and over-estimations

respectively, the actual solution to the integral must lie in the interval (0.6688323993493503,

0.668832406740728) and therefore, is 0.6688324 to seven significant figures since both bounds round to

this value.

4.ERROR ANALYSIS I attempted to calculate the order of convergence for each of the approximations. I did this using the ratio of

differences (shown for Simpson’s rule, but the same for trapezium and mid-point rules; column D):

𝑅𝑎𝑡𝑖𝑜 𝑜𝑓 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠 (𝑅𝑜𝐷𝑠) =𝑆4𝑛 − 𝑆2𝑛

𝑆2𝑛 − 𝑆𝑛

This was compared to the textbook’s ratio of differences. My values converged towards roughly 0.397 for all

approximating methods, directly contradicting the expected values of 0.25 for the trapezium and mid-point

rules and 0.0625 for Simpson’s rule. This suggests that the order of convergence for my integration

approximations are not second-order for trapezium and mid-point and fourth-order for Simpson’s. Using

natural logarithms, I calculated the orders of convergence (column E):

𝑅𝑜𝐷𝑠 = (ℎ𝑛+1

ℎ𝑛)

𝑂𝑟𝑑𝑒𝑟 𝑜𝑓 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒(𝑂𝑜𝐶)

𝐵𝑒𝑐𝑎𝑢𝑠𝑒 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑖𝑠 ℎ𝑎𝑙𝑣𝑒𝑑 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑛𝑒𝑥𝑡 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒:

𝑅𝑜𝐷𝑠 =1

2

𝑂𝑜𝐶

⟹ ln(𝑅𝑜𝐷𝑠) = ln1

2

𝑂𝑜𝐶= 𝑂𝑜𝐶 × ln

1

2

⟹ 𝑂𝑜𝐶 =ln(𝑅𝑜𝐷𝑠)

ln1

2

A spreadsheet was used for this. The formulas are shown below:

A B C D E

1 n h

Approximations (Sn,

Tn, Mn etc.)

Ratio of

differences

Order of

convergence

2 1 0.95 xxxxxxxxxxx

3 2 0.475 xxxxxxxxxxx

4 4 0.2375 xxxxxxxxxxx =(C4-C3)/(C3-C2) =LN(D4)/LN(1/2)



︙ ︙ ︙ ︙ ︙ ︙



The following values of each value’s order of convergence were produced (the full spreadsheets of values are

in the appendix):

n (number of strips)

Tn order of convergence Mn order of convergence

Sn order of convergence

1

2

4 1.311533077 1.093794728 1.46910831

8 1.369323069 1.2830762 1.444467682

16 1.39548848 1.390918261 1.399712164

32 1.396932881 1.437937957 1.359934926

64 1.384185642 1.435577652 1.340514921

128 1.368862436 1.411436924 1.334774844

256 1.356691739 1.386887476 1.333580257

512 1.348334522 1.368490305 1.333373166

1024 1.342878778 1.355989702 1.333339646

2048 1.339381493 1.347800545 1.333334329

4096 1.337156861 1.342523702 1.33333349

8192 1.335747277 1.339153347 1.333333363

16384 1.334856104 1.337011819 1.333333343

32768 1.334293444 1.335655674 1.333333189

65536 1.333938465 1.334797612 1.333333625

131072 1.333715075 1.334258723 1.333332764

262144 1.333572518 1.333912072 1.333333534

524288 1.333485776 1.33368843 1.333344002

For all three methods, the orders of convergence appear to converge towards 1.3333…, or 4

3. It is therefore

reasonable to conclude that the orders of convergence for my integral are 4

3. The similarity in the orders of

convergence suggests that the speeds of convergence between my three methods are roughly the same.

Because I will use my Simpson’s rule approximations to generate an improved approximation to my

integral, it can be deduced that the absolute error of each approximation is given by ε = kh4/3, where k is the

constant of proportionality.

If I is the theoretical exact solution to the integral, I = Sn – ε = Sn – kh4/3.

Combining together the general approximations of Sn and S2n and inserting the general expression for each

strip width, the respective values of I can be combined and eliminated to find the value of k:

𝐼 = 𝑆𝑛 − 𝑘 × (0.95

𝑛)

4

3= 𝑆2𝑛 − 𝑘 × (

0.95

2𝑛)

4

3

𝑘((0.95

2𝑛)

4

3 − (0.95

𝑛)

4

3) = 𝑆2𝑛– 𝑆𝑛

𝑘 =𝑆2𝑛–𝑆𝑛

(0.95

2𝑛)

43−(

0.95

𝑛)

43

With k, I and ε can be calculated for each approximation. A spreadsheet was used for this:



A B C D E

1 n Sn k I ε of Sn

2 1 0.6074153

24813161

=(B3-B2)/(((0.95/A3))^(4/3)-

((0.95/A2))^(4/3))

=B2-C2*(0.95/A2)^(4/3) =C2*(0.95/A2)^(4/3)

3 2 0.6463696

27205258

=(B4-B3)/(((0.95/A4))^(4/3)-

((0.95/A3))^(4/3))

=B3-C3*(0.95/A3)^(4/3) =C3*(0.95/A3)^(4/3)

4 4 0.6604401

34742067

=(B5-B4)/(((0.95/A5))^(4/3)-

((0.95/A4))^(4/3))

=B4-C4*(0.95/A4)^(4/3) =C4*(0.95/A4)^(4/3)

5 8 0.6656100

29033998

=(B6-B5)/(((0.95/A6))^(4/3)-

((0.95/A5))^(4/3))

=B5-C5*(0.95/A5)^(4/3) =C5*(0.95/A5)^(4/3)

6 16 0.6675694

43529241

=(B7-B6)/(((0.95/A7))^(4/3)-

((0.95/A6))^(4/3))

=B6-C6*(0.95/A6)^(4/3) =C6*(0.95/A6)^(4/3)

︙ ︙ ︙ ︙ ︙ ︙

n Sn k I ε of Sn

1 0.6074153248131617 -0.069156363451686 0.672000120266008 -0.064584795452847 2 0.6463696272052587 -0.062944792688284 0.669698009319140 -0.023328382113882 4 0.6604401347420671 -0.058278053144106 0.669011628642698 -0.008571493900631 8 0.6656100290339982 -0.055657412984714 0.668858665939591 -0.003248636905593 16 0.6675694435292417 -0.054640559611278 0.668835112021312 -0.001265668492071 32 0.6683328311473736 -0.054369240394011 0.668832617931674 -0.000499786784301 64 0.6686342774148842 -0.054314942921677 0.668832419852969 -0.000198142438085 128 0.668753786974305 -0.054305647474960 0.668832406395756 -0.000078619421451 256 0.6688012062576777 -0.054304148118075 0.668832405534333 -0.000031199276656 512 0.6688200240931876 -0.054303910507442 0.668832405480157 -0.000012381386970 1024 0.6688274919234826 -0.054303873044472 0.668832405476768 -0.000004913553286 2048 0.6688304555318544 -0.054303867137927 0.668832405476556 -0.000001949944702 4096 0.6688316316404882 -0.054303866011072 0.668832405476540 -0.000000773836052 8192 0.6688320983794998 -0.054303865655806 0.668832405476538 -0.000000307097039 16384 0.6688322836049978 -0.054303871100461 0.668832405476550 -0.000000121871553 32768 0.6688323571117926 -0.054303860120849 0.668832405476540 -0.000000048364748 65536 0.6688323862829773 -0.054303881539811 0.668832405476548 -0.000000019193571 131072 0.6688323978595746 -0.054303873971120 0.668832405476547 -0.000000007616972 262144 0.668832402453749 -0.054303472399519 0.668832405476524 -0.000000003022775 524288 0.668832404276935

From these calculations, the column labelled I gives the extrapolated value of the integral as

0.6688324054765 to 13 significant figures because from n = 4096 onwards, all but one of the values for I

round to this figure.

5.CONCLUSION The solution to ∫ √𝑥2 − 𝑥660.95

0𝑑𝑥, using 524288 strips is 0.6688324054765 to 13 significant figures.

This means that the integral’s solution lies in the interval [0.66883240547645, 0.66883240547655). In the

context of my problem involving a solid gold prism, this means that the volume of gold required, in metres

cubed, lies in the interval [0.481559331943044, 0.481559331943116). This was obtained by multiplying the

integral’s bounds by 0.72.

The volume of gold required for the sculpture is therefore 0.481559331943 m3 to 12 significant figures. It is

probably true that this level of precision is excessive in the context of measuring an amount of gold;

equipment would likely not be able to measure to this high degree of precision for the volume.



My chosen graphical function has near vertical portions (i.e. large gradients) towards the limits of x = -1 and

x = 1. If these had been the limits of my integral, the accuracy of my approximations may have been severely

compromised because f(-1) and f(0) both have values of zero. This would cause trapezium rule estimates to

be under-estimated to a much larger extent because of the zero value that is included in all calculations for

any number of strips; this would cause the approximations to have a much slower rate on convergence to

the true value because increasing the number of strips would lessen the effect that the zero value would

have on the approximation. Consequently, this would also cause the Simpson’s rule estimate (partly

calculated using the trapezium rule estimate) to have a slower rate of convergence. Fortunately, because of

the real-world nature of my problem and hypothetical engineering constraints, this did not turn out to be a

problem for me.

My method for numerical integration is valid for finding the area beneath my chosen curve; the graph of y =

f’’(x) in figure 4 proves that the graph is always convex in my range of x values. The value of the second

derivative is less than zero for the entirety of my domain. Therefore, it is correct to say that the trapezium

and mid-point rule approximations are under- and over-estimates respectively, and so the true value of the

integral lies somewhere in between the estimates given by each of these two methods.

FIGURE 4: A GRAPH OF Y=18𝑥4+2

9(𝑥4−1)(𝑥2−𝑥6)56

, THE SECOND DERIVATIVE OF F(X)

6.REFERENCE LIST BARILE, M. & WEISSTEIN, E. W., 2002 Butterfly Curve. [Online]. [Accessed 30 January 2015]. Available

from: http://mathworld.wolfram.com/ButterflyCurve.html

LISSAMAN, R. AND WEST, E. (2004) MEI Numerical Methods (MEI Structured Mathematics (A+AS

Level)). 3rd ed. United Kingdom: Hodder & Stoughton.



7.APPENDICES 7.1.RAW OUTPUT OF THE COMPUTER PROGRAM

Trapezium rule approximations:

T 1 = 0.3526330592938719

T 2 = 0.5437197584333392

T 4 = 0.6207071600122788

T 8 = 0.65050689105962

T 16 = 0.6618342445404037

T 32 = 0.6661356437820323

T 64 = 0.6677835343060384

T 128 = 0.6684215916376727

T 256 = 0.6686707381401477

T 512 = 0.6687685892282945

T 1024 = 0.6688071653769639

T 2048 = 0.6688224102868545

T 4096 = 0.6688284442206045

T 8192 = 0.6688308347855151

T 16384 = 0.6688317824810049

T 32768 = 0.6688321583240022

T 65536 = 0.6688323074148468

T 131072 = 0.6688323665659334

T 262144 = 0.6688323900361666

T 524288 = 0.6688323993493503

Midpoint rule approximations:

M 1 = 0.7348064575728066

M 2 = 0.6976945615912183

M 4 = 0.6803066221069612

M 8 = 0.6731615980211872

M 16 = 0.6704370430236609

M 32 = 0.6694314248300444

M 64 = 0.6690596489693071

M 128 = 0.6689198846426211

M 256 = 0.6688664403164426

M 512 = 0.6688457415256341

M 1024 = 0.6688376551967419

M 2048 = 0.6688344781543543

M 4096 = 0.66883322535043

M 8192 = 0.668832730176492

M 16384 = 0.6688325341669943

M 32768 = 0.6688324565056877

M 65536 = 0.6688324257170426

M 131072 = 0.6688324135063952

M 262144 = 0.6688324086625402

M 524288 = 0.668832406740728

Simpsons rule approximations:

S 1 = 0.6074153248131617

S 2 = 0.6463696272052587

S 4 = 0.6604401347420671

S 8 = 0.6656100290339982

S 16 = 0.6675694435292417

S 32 = 0.6683328311473736

S 64 = 0.6686342774148842

S 128 = 0.668753786974305

S 256 = 0.6688012062576777

S 512 = 0.6688200240931876

S 1024 = 0.6688274919234826

S 2048 = 0.6688304555318544

S 4096 = 0.6688316316404882

S 8192 = 0.6688320983794998

S 16384 = 0.6688322836049978

S 32768 = 0.6688323571117926

S 65536 = 0.6688323862829773

S 131072 = 0.6688323978595746

S 262144 = 0.668832402453749

S 524288 = 0.6688324042769355



7.2.SPREADSHEET TABLES FOR RATES OF

CONVERGENCE

7.2.1.TRAPEZIUM RULE A B C D E

1 n (number of

strips)

h (width of

each strip)

Tn (trapezium rule

approximation)

Ratio of

differences

Order of

convergence

2 1 0.95 0.352633

3 2 0.475 0.54372

4 4 0.2375 0.620707 0.402893 1.311533

5 8 0.11875 0.650507 0.387073 1.369323

6 16 0.059375 0.661834 0.380116 1.395488

7 32 0.0296875 0.666136 0.379736 1.396933

8 64 0.01484375 0.667784 0.383106 1.384186

9 128 0.007421875 0.668422 0.387196 1.368862

10 256 0.003710938 0.668671 0.390477 1.356692

11 512 0.001855469 0.668769 0.392745 1.348335

12 1024 0.000927734 0.668807 0.394233 1.342879

13 2048 0.000463867 0.668822 0.39519 1.339381

14 4096 0.000231934 0.668828 0.3958 1.337157

15 8192 0.000115967 0.668831 0.396187 1.335747

16 16384 5.79834 × 10-5 0.668832 0.396432 1.334856

17 32768 2.89917 × 10-5 0.668832 0.396586 1.334293

18 65536 1.44958 × 10-5 0.668832 0.396684 1.333938

19 131072 7.24792 × 10-6 0.668832 0.396745 1.333715

20 262144 3.62396 × 10-6 0.668832 0.396784 1.333573

21 524288 1.81198 × 10-6 0.668832 0.396808 1.333486

7.2.2.MID-POINT RULE A B C D E

1 n (number of

strips)

h (width of

each strip)

Mn (mid-point rule

approximation)

Ratio of

differences

Order of

convergence

2 1 0.95 0.734806458

3 2 0.475 0.697694562

4 4 0.2375 0.680306622 0.468527383 1.093794728

5 8 0.11875 0.673161598 0.410918389 1.2830762

6 16 0.059375 0.670437043 0.381322017 1.390918261

7 32 0.0296875 0.669431425 0.369094474 1.437937957

8 64 0.01484375 0.669059649 0.369698821 1.435577652

9 128 0.007421875 0.668919885 0.375937067 1.411436924

10 256 0.003710938 0.66886644 0.382388893 1.386887476

11 512 0.001855469 0.668845742 0.387296319 1.368490305

12 1024 0.000927734 0.668837655 0.390666729 1.355989702

13 2048 0.000463867 0.668834478 0.392890572 1.347800545

14 4096 0.000231934 0.668833225 0.394330252 1.342523702

15 8192 0.000115967 0.66883273 0.395252544 1.339153347

16 16384 5.79834 × 10-5 0.668832534 0.39583969 1.337011819

17 32768 2.89917 × 10-5 0.668832457 0.396211958 1.335655674

18 65536 1.44958 × 10-5 0.668832426 0.39644768 1.334797612

19 131072 7.24792 × 10-6 0.668832414 0.396595792 1.334258723

20 262144 3.62396 × 10-6 0.668832409 0.396691098 1.333912072

21 524288 1.81198 × 10-6 0.668832407 0.396752597 1.33368843



7.2.3.SIMPSON’S RULE A B C D E

1 n (number of

strips)

h (width of

each strip)

Sn (Simpson's rule

approximation)

Ratio of

differences

Order of

convergence

2 1 0.95 0.607415325

3 2 0.475 0.646369627

4 4 0.2375 0.660440135 0.361205481 1.46910831

5 8 0.11875 0.665610029 0.367427705 1.444467682

6 16 0.059375 0.667569444 0.37900475 1.399712164

7 32 0.0296875 0.668332831 0.389599863 1.359934926

8 64 0.01484375 0.668634277 0.394879692 1.340514921

9 128 0.007421875 0.668753787 0.396453937 1.334774844

10 256 0.003710938 0.668801206 0.396782346 1.333580257

11 512 0.001855469 0.668820024 0.396839306 1.333373166

12 1024 0.000927734 0.668827492 0.396848527 1.333339646

13 2048 0.000463867 0.668830456 0.396849989 1.333334329

14 4096 0.000231934 0.668831632 0.39685022 1.33333349

15 8192 0.000115967 0.668832098 0.396850255 1.333333363

16 16384 5.79834 × 10-5 0.668832284 0.39685026 1.333333343

17 32768 2.89917 × 10-5 0.668832357 0.396850303 1.333333189

18 65536 1.44958 × 10-5 0.668832386 0.396850183 1.333333625

19 131072 7.24792 × 10-6 0.668832398 0.39685042 1.333332764

20 262144 3.62396 × 10-6 0.668832402 0.396850208 1.333333534

21 524288 1.81198 × 10-6 0.668832404 0.396847328 1.333344002

Documents

NM MEI A-level Maths Coursework