two matlab assignments

profilestrength
finite.zip

finite/1st Order ODES - annotated.pdf

Sensitivity: Internal

1st Order Ordinary Differential Equations (ODES)

Sensitivity: Internal

Disclaimer

We are merely looking to get an idea of the topic, we are not studying this rigorously! For a proper introduction there are plenty of online resources or textbooks.

You will also see these throughout the course in more detail.

We just want an idea so that we can compare a numerical method that we are going to learn.

Sensitivity: Internal

What is a 1st Order Ordinary Differential Equation (ODE) An equation involving the independent variable and the first derivative

𝑑𝑦

𝑑𝑥 + 2𝑦 = 𝑥 + 5

𝑑𝑦

𝑑𝑥 = 𝑥2

𝑑𝑦

𝑑𝑥 = 2𝑦

Sensitivity: Internal

Solutions to 1st Order Differential Equations

Find a function 𝑦(𝑥) which solves the following equations

𝑑𝑦

𝑑𝑥 = 𝑥2 (1)

Solution 𝑦(𝑥) = 𝑥3

3 + 𝐶 (2)

The derivative of the solution is 𝑑𝑦

𝑑𝑥 = 𝑥2

Why is it a solution, because it solves equation (1)

𝐿𝐻𝑆 = 𝑑𝑦

𝑑𝑥 = 𝑥2 = 𝑅𝐻𝑆

Sensitivity: Internal

Solutions to 1st Order Differential Equations

Find a function 𝑦 𝑥 which solves the following equations

𝑑𝑦

𝑑𝑥 = 2𝑦 (1)

Solution 𝑦 = 𝐴𝑒2𝑥 (2)

The derivative of the solution is 𝑑𝑦

𝑑𝑥 = 2𝐴𝑒𝑥

Why does a solution, because it solves equation (1)

𝐿𝐻𝑆 = 𝑑𝑦

𝑑𝑥 = 2𝐴𝑒𝑥 = 2𝑦 = 𝑅𝐻𝑆

Sensitivity: Internal

Worked Solution 1

𝑑𝑦

𝑑𝑥 = 𝑥2 (1)

Solve Complimentary 𝑑𝑦

𝑑𝑥 = 0

𝑦𝑐 = 𝐶1

Solve Particular

න𝑑𝑦 = න𝑥2𝑑𝑥

𝑦𝑝 = 𝑥3

3 + 𝐶2

Thus 𝑦 = 𝑦𝑐 + 𝑦𝑝

𝑦 = 𝑥3

3 + 𝐶1 + 𝐶2 +

𝑥3

3 + 𝐴

Check 𝑦 solves equations (1)

𝐿𝐻𝑆 = 𝑑𝑦

𝑑𝑥 = 𝑥2 = 𝑅𝐻𝑆

Sensitivity: Internal

Worked Solution 2

𝑑𝑦

𝑑𝑥 = 2𝑦 1

Sometime with extra information about the solution, known as an initial condition.

𝑦 0 = 2 (2)

Separate Variables

න 𝑑𝑦

𝑦 = න2𝑑𝑥

ln 𝑦 = 2𝑥 + 𝐶 𝑦 = 𝑒2𝑥+𝐶 = 𝐴𝑒2𝑥

To find 𝐴 use the initial condition 𝑦 0 = 2

𝑦 0 = 𝐴𝑒0 = 𝐴 = 2 ∴ 𝑦 = 2𝑒2𝑥

Check, we must check that our solution 𝑦 satisfies both equations (1) and (2)

𝐿𝐻𝑆 = 𝑑𝑦

𝑑𝑥 = 4𝑒2𝑥 = 2 2𝑒2𝑥 = 2𝑦 = 𝑅𝐻𝑆

And, 𝐿𝐻𝑆 = 𝑦 0 = 2𝑒2⋅0 = 2 = 𝑅𝐻𝑆

Sensitivity: Internal

Worked Solution 3

𝑑𝑦

𝑑𝑥 + 2𝑦 = 𝑥 + 5

Solve Complimentary

𝑑𝑦

𝑑𝑥 + 2𝑦 = 0

න 𝑑𝑦

𝑦 = න−2𝑑𝑥

ln 𝑦 = −2𝑥 + 𝐶 𝑦𝑐 = 𝐴𝑒

−2𝑥

Solve 𝑑𝑦

𝑑𝑥 + 2𝑦 = 𝑥 + 5

Assume 𝑦𝑝 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐

2𝑎𝑥 + 𝑏 + 2 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 𝑥 + 5 2𝑎𝑥2 + 2𝑎 + 2𝑏 𝑥 + (𝑏 + 2𝑐) = 𝑥 + 5

Equate coefficients

2𝑎 + 2𝑏 = 1 = 1

2

𝑏 + 2𝑐 = 5,⇒ 𝑐 = 9

4

Thus 𝑦 = 𝑦𝑐 + 𝑦𝑝

𝑦 = 𝐴𝑒−2𝑥 + 𝑥

2 + 9

4

Check!

𝐿𝐻𝑆 = −2𝐴𝑒−2𝑥 + 1

2 + 2 𝐴𝑒−2𝑥 +

𝑥

2 + 9

4 = 𝑥 + 5 = 𝑅𝐻𝑆

Sensitivity: Internal

Springs

Imagine the movement of a spring, it goes up and down. If you plot this as time passes you get the following.

https://www.youtube.com/watch?v=P-Umre5Np_0

Sensitivity: Internal

Newtons Second Law

We are not studying physics, but this is important!

You can model a spring (undamped) using newtons second law

𝐹 = 𝑚𝑎

𝐹 = 𝑚 𝑑2𝑦

𝑑𝑡2

and hook’s force law 𝐹 = 𝑘𝑦

Equating we have,

−𝑘𝑦 = 𝑚 𝑑2𝑦

𝑑𝑡2 (1)

𝑑2𝑦

𝑑𝑡2 + 𝑘

𝑚 𝑦 = 0

Let 𝜔 = 𝑘

𝑚 𝑦 𝑡 = 𝑅𝑐𝑜𝑠(𝜔𝑡 + 𝛿)

Why? 𝑑𝑦

𝑑𝑡 = −𝜔𝑅𝑠𝑖𝑛 𝜔𝑡 + 𝛿

𝑑2𝑦

𝑑𝑡2 = −𝜔2𝑅𝑐𝑜𝑠 𝜔𝑡 + 𝛿

Sub into (1)

𝐿𝐻𝑆 = −𝜔2𝑅𝑐𝑜𝑠 𝜔𝑡 + 𝛿 + 𝑘

𝑚 𝑅𝑐𝑜𝑠 𝜔𝑡 + 𝛿

= −𝜔2𝑅𝑐𝑜𝑠 𝜔𝑡 + 𝛿 + 𝜔2𝑅𝑐𝑜𝑠 𝜔𝑡 + 𝛿 = 𝑅𝐻𝑆

Sensitivity: Internal

The solution describes the springs movement in time!

𝑦 𝑡 = 𝑅𝑐𝑜𝑠(𝜔𝑡 + 𝛿)

𝑅 – amplitude

𝜔 - frequency

𝛿 - displacement

Interactive demo

https://www.desmos.com/calculator/dytbibashl

http://hyperphysics.phy-astr.gsu.edu/hbase/shm2.html

finite/1st Order ODES.pdf

Sensitivity: Internal

1st Order Ordinary Differential Equations (ODES)

Sensitivity: Internal

Disclaimer

We are merely looking to get an idea of the topic, we are not studying this rigorously! For a proper introduction there are plenty of online resources or textbooks.

You will also see these throughout the course in more detail.

We just want an idea so that we can compare a numerical method that we are going to learn.

Sensitivity: Internal

What is a 1st Order Ordinary Differential Equation (ODE) An equation involving the independent variable and the first derivative

𝑑𝑦

𝑑𝑥 = 𝑓 𝑥

e.g. 𝑑𝑦

𝑑𝑥 = 𝑥2

𝑑𝑦

𝑑𝑥 = 𝑓(𝑥,𝑦)

𝑑𝑦

𝑑𝑥 + 2𝑦 = 𝑥 + 5

𝑑𝑦

𝑑𝑥 = 2𝑦

Note that here 𝑓 𝑥,𝑦 = 0𝑥 + 2𝑦

Sensitivity: Internal

Solutions to 1st Order Differential Equations

Find a function 𝑦(𝑥) which solves the following equation

𝑑𝑦

𝑑𝑥 = 𝑥2 (1)

Solution 𝑦(𝑥) = 𝑥3

3 + 𝐶 (2)

The derivative of the solution is 𝑑𝑦

𝑑𝑥 = 𝑥2

Why is it a solution, because it solves equation (1)

𝐿𝐻𝑆 = 𝑑𝑦

𝑑𝑥 = 𝑥2 = 𝑅𝐻𝑆

Sensitivity: Internal

Solutions to 1st Order Differential Equations

Find a function 𝑦 𝑥 which solves the following equation

𝑑𝑦

𝑑𝑥 = 2𝑦 (1)

Solution 𝑦 = 𝐴𝑒2𝑥 (2)

The derivative of the solution is 𝑑𝑦

𝑑𝑥 = 2𝐴𝑒𝑥

Why does a solution, because it solves equation (1)

𝐿𝐻𝑆 = 𝑑𝑦

𝑑𝑥 = 2𝐴𝑒𝑥 = 2𝑦 = 𝑅𝐻𝑆

Sensitivity: Internal

Worked Solution 1

𝑑𝑦

𝑑𝑥 = 𝑥2 (1)

Solve Complimentary 𝑑𝑦

𝑑𝑥 = 0

𝑦𝑐 = 𝐶1

Solve Particular

න𝑑𝑦 = න𝑥2𝑑𝑥

𝑦𝑝 = 𝑥3

3 + 𝐶2

Thus 𝑦 = 𝑦𝑐 + 𝑦𝑝

𝑦 = 𝑥3

3 + 𝐶1 + 𝐶2 +

𝑥3

3 + 𝐴

Check 𝑦 solves equations (1)

𝐿𝐻𝑆 = 𝑑𝑦

𝑑𝑥 = 𝑥2 = 𝑅𝐻𝑆

Sensitivity: Internal

Worked Solution 2

𝑑𝑦

𝑑𝑥 = 2𝑦 1

Sometime with extra information about the solution, known as an initial condition.

𝑦 0 = 2 (2)

Separate Variables

න 𝑑𝑦

𝑦 = න2𝑑𝑥

ln 𝑦 = 2𝑥 + 𝐶 𝑦 = 𝑒2𝑥+𝐶 = 𝐴𝑒2𝑥

To find 𝐴 use the initial condition 𝑦 0 = 2

𝑦 0 = 𝐴𝑒0 = 𝐴 = 2 ∴ 𝑦 = 2𝑒2𝑥

Check, we must check that our solution 𝑦 satisfies both equations (1) and (2)

𝐿𝐻𝑆 = 𝑑𝑦

𝑑𝑥 = 4𝑒2𝑥 = 2 2𝑒2𝑥 = 2𝑦 = 𝑅𝐻𝑆

And, 𝐿𝐻𝑆 = 𝑦 0 = 2𝑒2⋅0 = 2 = 𝑅𝐻𝑆

Sensitivity: Internal

Worked Solution 3

𝑑𝑦

𝑑𝑥 + 2𝑦 = 𝑥 + 5

Solve Complimentary

𝑑𝑦

𝑑𝑥 + 2𝑦 = 0

න 𝑑𝑦

𝑦 = න−2𝑑𝑥

ln 𝑦 = −2𝑥 + 𝐶 𝑦𝑐 = 𝐴𝑒

−2𝑥

Solve 𝑑𝑦

𝑑𝑥 + 2𝑦 = 𝑥 + 5

Assume 𝑦𝑝 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐

2𝑎𝑥 + 𝑏 + 2 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 𝑥 + 5 2𝑎𝑥2 + 2𝑎 + 2𝑏 𝑥 + (𝑏 + 2𝑐) = 𝑥 + 5

Equate coefficients

2𝑎 + 2𝑏 = 1 = 1

2

𝑏 + 2𝑐 = 5,⇒ 𝑐 = 9

4

Thus 𝑦 = 𝑦𝑐 + 𝑦𝑝

𝑦 = 𝐴𝑒−2𝑥 + 𝑥

2 + 9

4

Check!

𝐿𝐻𝑆 = −2𝐴𝑒−2𝑥 + 1

2 + 2 𝐴𝑒−2𝑥 +

𝑥

2 + 9

4 = 𝑥 + 5 = 𝑅𝐻𝑆

Sensitivity: Internal

Springs

Imagine the movement of a spring, it goes up and down. If you plot this as time passes you get the following.

https://www.youtube.com/watch?v=P-Umre5Np_0

Sensitivity: Internal

Newtons Second Law

We are not studying physics, but this is important!

You can model a spring (undamped) using newtons second law

𝐹 = 𝑚𝑎

𝐹 = 𝑚 𝑑2𝑦

𝑑𝑡2

and hook’s force law 𝐹 = 𝑘𝑦

Equating we have,

−𝑘𝑦 = 𝑚 𝑑2𝑦

𝑑𝑡2 (1)

𝑑2𝑦

𝑑𝑡2 + 𝑘

𝑚 𝑦 = 0

Let 𝜔 = 𝑘

𝑚 𝑦 𝑡 = 𝑅𝑐𝑜𝑠(𝜔𝑡 + 𝛿)

Why? 𝑑𝑦

𝑑𝑡 = −𝜔𝑅𝑠𝑖𝑛 𝜔𝑡 + 𝛿

𝑑2𝑦

𝑑𝑡2 = −𝜔2𝑅𝑐𝑜𝑠 𝜔𝑡 + 𝛿

Sub into (1)

𝐿𝐻𝑆 = −𝜔2𝑅𝑐𝑜𝑠 𝜔𝑡 + 𝛿 + 𝑘

𝑚 𝑅𝑐𝑜𝑠 𝜔𝑡 + 𝛿

= −𝜔2𝑅𝑐𝑜𝑠 𝜔𝑡 + 𝛿 + 𝜔2𝑅𝑐𝑜𝑠 𝜔𝑡 + 𝛿 = 𝑅𝐻𝑆

Sensitivity: Internal

The solution describes the springs movement in time!

𝑦 𝑡 = 𝑅𝑐𝑜𝑠(𝜔𝑡 + 𝛿)

𝑅 – amplitude

𝜔 - frequency

𝛿 - displacement

Interactive demo

https://www.desmos.com/calculator/dytbibashl

http://hyperphysics.phy-astr.gsu.edu/hbase/shm2.html

finite/excel_finite_differences - example 1 (2).xlsx

Sheet1

k x_k y_k y Error h 0.1
1 0 2 1 1
2 0.1 2.4 1.2214027582 1.1785972418
3 0.2 2.88 1.4918246976 1.3881753024
4 0.3 3.456 1.8221188004 1.6338811996
5 0.4 4.1472 2.2255409285 1.9216590715
6 0.5 4.97664 2.7182818285 2.2583581715
7 0.6 5.971968 3.3201169227 2.6518510773
8 0.7 7.1663616 4.0551999668 3.1111616332
9 0.8 8.59963392 4.9530324244 3.6466014956
10 0.9 10.319560704 6.0496474644 4.2699132396
11 1 12.3834728448 7.3890560989 4.9944167459
12 1.1 14.8601674138 9.0250134994 5.8351539143
13 1.2 17.8322008965 11.0231763806 6.8090245159
14 1.3 21.3986410758 13.463738035 7.9349030408
15 1.4 25.678369291 16.4446467711 9.2337225199
16 1.5 30.8140431492 20.0855369232 10.728506226
17 1.6 36.976851779 24.5325301971 12.4443215819
18 1.7 44.3722221348 29.9641000474 14.4081220874
19 1.8 53.2466665618 36.5982344437 16.6484321181
20 1.9 63.8959998741 44.7011844933 19.1948153808
21 2 76.6751998489 54.5981500331 22.0770498158
22 2.1 92.0102398187 66.6863310409 25.3239087778
23 2.2 110.4122877825 81.450868665 28.9614191175
24 2.3 132.494745339 99.4843156419 33.0104296971
25 2.4 158.9936944068 121.5104175187 37.483276888
26 2.5 190.7924332881 148.4131591026 42.3792741856
27 2.6 228.9509199458 181.2722418752 47.6786780706
28 2.7 274.7411039349 221.4064162042 53.3346877307
29 2.8 329.6893247219 270.4264074262 59.2629172957
30 2.9 395.6271896663 330.2995599096 65.3276297566
31 3 474.7526275995 403.4287934927 71.3238341068
32 3.1 569.7031531194 492.7490410933 76.9541120262
33 3.2 683.6437837433 601.8450378721 81.7987458713
34 3.3 820.372540492 735.095189242 85.27735125
35 3.4 984.4470485904 897.8472916504 86.59975694
36 3.5 1181.3364583085 1096.6331584285 84.70329988
37 3.6 1417.6037499702 1339.4307643944 78.1729855758
38 3.7 1701.1244999642 1635.9844299959 65.1400699683
39 3.8 2041.3493999571 1998.1958951041 43.1535048529
40 3.9 2449.6192799485 2440.6019776245 9.017302324
41 4 2939.5431359382 2980.9579870417 -41.4148511036
42 4.1 3527.4517631258 3640.9503073324 -113.4985442066
43 4.2 4232.942115751 4447.0667476999 -214.1246319489
44 4.3 5079.5305389012 5431.659591363 -352.1290524618
45 4.4 6095.4366466814 6634.2440062779 -538.8073595965
46 4.5 7314.5239760177 8103.0839275754 -788.5599515577
47 4.6 8777.4287712212 9897.1290587439 -1119.7002875227
48 4.7 10532.9145254654 12088.380730217 -1555.4662047515
49 4.8 12639.4974305585 14764.7815655772 -2125.2841350187
50 4.9 15167.3969166702 18033.7449278285 -2866.3480111582
51 5 18200.8763000043 22026.4657948066 -3825.5894948024
52 5.1 21841.0515600051 26903.1860742974 -5062.1345142923

y_k 2 2.4 2.88 3.456 4.1471999999999998 4.9766399999999997 5.9719679999999995 7.1663615999999992 8.5996339199999987 10.319560703999999 12.383472844799998 14.860167413759998 17.832200896511996 21.398641075814393 25.678369290977269 30.814043149172722 36.976851779007262 44.372222134808716 53.246666561770461 63.895999874124549 76.675199848949461 92.01023981873935 110.41228778248721 132.49474533898464 158.99369440678157 190.79243328813789 228.95091994576546 274.74110393491856 329.68932472190227 395.62718966628273 474.75262759953927 569.70315311944705 683.64378374333648 820.37254049200374 984.44704859040439 1181.3364583084851 1417.6037499701822 1701.1244999642186 2041.3493999570621 2449.6192799484743 2939.5431359381691 3527.4517631258027 4232.9421157509632 5079.5305389011555 6095.4366466813863 7314.5239 760176637 8777.4287712211953 10532.914525465434 12639.49743055852 15167.396916670223 18200.876300004267 21841.051560005119 y 1 1.2214027581601699 1.4918246976412703 1.8221188003905091 2.2255409284924679 2.7182818284590451 3.3201169227365472 4.0551999668446745 4.953032424395114 6.0496474644129448 7.3890560989306486 9.0250134994341185 11.023176380641601 13.463738035001692 16.444646771097055 20.085536923187675 24.532530197109363 29.964100047397036 36.598234443678024 44.701184493300872 54.598150033144286 66.686331040925211 81.450868664968212 99.484315641933946 121.51041751873508 148.41315910257686 181.27224187515154 221.40641620418756 270.42640742615328 330.29955990964947 403.42879349273619 492.74904109325763 601.84503787208382 735.09518924197528 897.84729165042074 1096.6331584284626 1339.4307643944228 1635.9844299959329 1998.1958951041261 2440.6019776245093 2980.9579870417388 3640.9503073323649 4447.0667476998651 5431.6595913629881 6634.2440062778896 8103.0839275753842 9897.1290587439089 12088.380730216968 14764.781565577241 18033.744927828458 22026.465794806638 26903.186074297446

Error 1 1.1785972418398301 1.3881753023587295 1.6338811996094909 1.9216590715075319 2.2583581715409546 2.6518510772634523 3.1111616331553247 3.6466014956048847 4.269913239587054 4.9944167458693496 5.835153914325879 6.8090245158703944 7.934903040812701 9.2337225198802138 10.728506225985047 12.444321581897899 14.40812208741168 16.648432118092437 19.194815380823677 22.077049815805175 25.32390877781414 28.961419117519 33.010429697050697 37.48327688804649 42.37927418556103 47.678678070613927 53.334687730730991 59.262917295748991 65.327629756633257 71.323834106803076 76.954112026189421 81.798745871252663 85.27735125002846 86.599756939983649 84.703299880022541 78.172985575759412 65.140069968285616 43.153504852936067 9.017302323964941 -41.414851103569617 -113.4985442065622 -214.12463194890188 -352.12905246183254 -538.80735959650337 -788.55995155772052 -1119.7002875227136 -1555.466204751534 -2125.2841350187209 -2866.3480111582357 -3825.5894948023706 -5062.1345142923274

y_k 2 2.4 2.88 3.456 4.1471999999999998 4.9766399999999997 5.9719679999999995 7.1663615999999992 8.5996339199999987 10.319560703999999 12.383472844799998 14.860167413759998

finite/excel_finite_differences - example 2.xlsx

Sheet1

k x_k y_k Analytical Error h 0.1
1 0 2 2 0
2 0.1 2.4 2.4481562059 -0.0481562059
3 0.2 2.88 3.0066055697 -0.1266055697
4 0.3 3.456 3.6997673009 -0.2437673009
5 0.4 4.1472 4.5574670891 -0.4102670891
6 0.5 4.97664 5.616134114 -0.639494114
7 0.6 5.971968 6.9202630762 -0.9482950762
8 0.7 7.1663616 8.5241999254 -1.3578383254
9 0.8 8.59963392 10.4943229549 -1.8946890349
10 0.9 10.319560704 12.9117067949 -2.5921460909
11 1 12.3834728448 15.8753762226 -3.4919033778
12 1.1 14.8601674138 19.5062803737 -4.64611296
13 1.2 17.8322008965 23.9521468564 -6.1199459599
14 1.3 21.3986410758 29.3934105788 -7.9947695029
15 1.4 25.678369291 36.050455235 -10.372085944
16 1.5 30.8140431492 44.1924580772 -13.378414928
17 1.6 36.976851779 54.1481929435 -17.1713411645
18 1.7 44.3722221348 66.3192251066 -21.9470029718
19 1.8 53.2466665618 81.1960274983 -27.9493609365
20 1.9 63.8959998741 99.3776651099 -35.4816652358
21 2 76.6751998489 121.5958375746 -44.9206377256
22 2.1 92.0102398187 148.7442448421 -56.7340050233
23 2.2 110.4122877825 181.9144544962 -71.5021667137
24 2.3 132.494745339 222.4397101944 -89.9449648554
25 2.4 158.9936944068 271.9484394172 -112.9547450104
26 2.5 190.7924332881 332.4296079808 -141.6371746927
27 2.6 228.9509199458 406.3125442191 -177.3616242733
28 2.7 274.7411039349 496.5644364594 -221.8233325245
29 2.8 329.6893247219 606.8094167088 -277.1200919869
30 2.9 395.6271896663 741.4740097967 -345.8468201304
31 3 474.7526275995 905.9647853587 -431.2121577591
32 3.1 569.7031531194 1106.8853424598 -537.1821893404
33 3.2 683.6437837433 1352.3013352122 -668.6575514689
34 3.3 820.372540492 1652.0641757944 -831.6916353024
35 3.4 984.4470485904 2018.2064062135 -1033.759357623
36 3.5 1181.3364583085 2465.424606464 -1284.0881481556
37 3.6 1417.6037499702 3011.6692198874 -1594.0654699173
38 3.7 1701.1244999642 3678.8649674909 -1977.7404675266
39 3.8 2041.3493999571 4493.7907639843 -2452.4413640272
40 3.9 2449.6192799485 5489.1544496551 -3039.5351697067
41 4 2939.5431359382 6704.9054708439 -3765.3623349057
42 4.1 3527.4517631258 8189.8381914978 -4662.386428372
43 4.2 4232.942115751 10003.5501823247 -5770.6080665737
44 4.3 5079.5305389012 12218.8340805667 -7139.3035416656
45 4.4 6095.4366466814 14924.5990141253 -8829.1623674439
46 4.5 7314.5239760177 18229.4388370446 -10914.914861027
47 4.6 8777.4287712212 22265.9903821738 -13488.5616109526
48 4.7 10532.9145254654 27196.2566429882 -16663.3421175227
49 4.8 12639.4974305585 33218.1085225488 -20578.6110919903
50 4.9 15167.3969166702 40573.226087614 -25405.8291709438
51 5 18200.8763000043 49556.7980383149 -31355.9217383107
52 5.1 21841.0515600051 60529.3686671692 -38688.3171071641
y_k 2 2.4 2.88 3.456 4.1471999999999998 4.9766399999999997 5.9719679999999995 7.1663615999999992 8.5996339199999987 10.319560703999999 12.383472844799998 14.860167413759998 17.832200896511996 21.398641075814393 25.678369290977269 30.814043149172722 36.976851779007262 44.372222134808716 53.246666561770461 63.895999874124549 76.675199848949461 92.01023981873935 110.41228778248721 132.49474533898464 158.99369440678157 190.79243328813789 228.95091994576546 274.74110393491856 329.68932472190227 395.62718966628273 474.75262759953927 569.70315311944705 683.64378374333648 820.37254049200374 984.44704859040439 1181.3364583084851 1417.6037499701822 1701.1244999642186 2041.3493999570621 2449.6192799484743 2939.5431359381691 3527.4517631258027 4232.9421157509632 5079.5305389011555 6095.4366466813863 7314.5239760176637 8777.4287712211953 10532.914525465434 12639.49743055852 15167.396916670223 18200.876300004267 21841.051560005119 Analytical 2 2.4481562058603825 3.0066055696928582 3.6997673008786456 4.5574670891080524 5.6161341140328513 6.920263076157231 8.5241999254005183 10.494322954889006 12.911706794929126 15.87537622259396 19.506280373726767 23.952146856443601 29.393410578753809 36.050455234968368 44.192458077172269 54.14819294349607 66.31922510664333 81.196027498275555 99.377665109926951 121.59583757457465 148.74424484208171 181.91445449617848 222.43971019435136 271.94843941715396 332.42960798079793 406.31254421909097 496.56443645942198 606.80941670884488 741.47400979671124 905.9647853586564 1106.88534 24598296 1352.3013352121886 1652.0641757944443 2018.2064062134466 2465.4246064640411 3011.6692198874512 3678.8649674908493 4493.7907639842842 5489.154449655146 6704.9054708439126 8189.8381914978208 10003.550182324696 12218.834080566723 14924.599014125251 18229.438837044614 22265.990382173797 27196.256642988181 33218.108522548791 40573.226087614035 49556.798038314933 60529.368667169249

Error 0 -4.8156205860382606E-2 -0.1266055696928583 -0.24376730087864562 -0.41026708910805265 -0.63949411403285161 -0.94829507615723152 -1.3578383254005191 -1.8946890348890069 -2.592146090929127 -3.4919033777939621 -4.6461129599667697 -6.1199459599316057 -7.9947695029394161 -10.372085943991099 -13.378414927999547 -17.171341164488808 -21.947002971834614 -27.949360936505094 -35.481665235802403 -44.920637725625184 -56.734005023342363 -71.502166713691267 -89.94496485536672 -112.95474501037239 -141.63717469266004 -177.3616242733255 -221.82333252450343 -277.12009198694261 -345.84682013042851 -431.21215775911713 -537.18218934038259 -668.65755146885215 -831.69163530244055 -1033.7593576230422 -1284.0881481555559 -1594.065469917269 -1977.7404675266307 -2452.4413640272223 -3039.5351697066717 -3765.3623349057434 -4662.3864283720177 -5770.6080665737327 -7139.3035416655675 -8829.1623674438652 -10914.914861026951 -13488.561610952602 -16663.342117522749 -20578.611091990271 -25405.829170943813 -31355.921738310666 -38688.317107164126

y_k 2 2.4 2.88 3.456 4.1471999999999998 4.9766399999999997 5.9719679999999995 7.1663615999999992 8.5996339199999987 10.319560703999999 12.383472844799998 14.860167413759998

finite/Finite Differences - annotated.pdf

Sensitivity: Internal

Finite Differences

Sensitivity: Internal

Rationale

There are differential equations we just can’t solve.

𝑑𝑦

𝑑𝑥 + sin 𝑥 𝑦 + 3𝑦 + 𝑥𝑦 = 𝑥 + 5

Sensitivity: Internal

Starter Problem

We will find a numerical answer to the following 𝑑𝑦

𝑑𝑥 = 2𝑦 1

𝑦 0 = 2 (2)

Sensitivity: Internal

Approximations via Calculus

We can make use of some basic calculus definitions

𝑑𝑦

𝑑𝑥 = lim

ℎ→0

𝑓 𝑥 + ℎ − 𝑓(𝑥)

Sensitivity: Internal

Difference Quotient

We can make use of some basic calculus definitions

𝑑𝑦

𝑑𝑥 = lim

ℎ→0

𝑓 𝑥+ℎ −𝑓(𝑥)

Therefore we can approximate 𝑑𝑦

𝑑𝑥 ≈

𝑓 𝑥+ℎ −𝑓(𝑥)

𝑥 𝑥 + ℎ

𝑓(𝑥)

𝑓(𝑥 + ℎ)

Sensitivity: Internal

Notation

Let ℎ be a fixed constant

𝑑𝑦

𝑑𝑥 ≈

𝑓 𝑥𝑘+1 −𝑓(𝑥𝑘)

OR we use 𝑦𝑘+1 and 𝑦𝑘

𝑑𝑦

𝑑𝑥 ≈

𝑦𝑘+1− 𝑦𝑘

𝑥1 𝑥2

𝑦𝑘 = 𝑓(𝑥𝑘)

𝑦𝑘+1 = 𝑓 𝑥𝑘+1

𝑥𝑛𝑥𝑘 𝑥𝑘+1

Sensitivity: Internal

Forward Difference Scheme

𝑑𝑦

𝑑𝑥 ≈ 𝑦𝑘+1 − 𝑦𝑘

We can use this to set up a scheme for 𝑑𝑦

𝑑𝑥 = 2𝑦

𝑦𝑘+1 − 𝑦𝑘 ℎ

= 2𝑦𝑘

Rearrange so that 𝑦𝑘+1 is the subject 𝑦𝑘+1 = 2ℎ𝑦𝑘 + 𝑦𝑘

𝑦𝑘+1 = 1 + 2ℎ 𝑦𝑘

Sensitivity: Internal

How do we start???

𝑦𝑘+1 = 1 + 2ℎ 𝑦𝑘

Let 𝑘 = 0 𝑦1 = 1 + 2ℎ 𝑦0 (3)

To get 𝑦1 we need 𝑦0 This comes from the initial condition 𝑦 0 = 2

𝑦0 = 2

From (3) 𝑦1 = 1 + 2ℎ ⋅ 2

e.g. if ℎ = 0.1 𝑦1 = 1 + 2 ⋅ 0.1 ⋅ 2 = 2.4

Sensitivity: Internal

Results

𝑦𝑘+1 = 1 + 2ℎ 𝑦𝑘 𝑦0 = 2

Sensitivity: Internal

What is a 1st Order Differential Equation

𝑑𝑦

𝑑𝑥 ≈ 𝑦𝑘+1 − 𝑦𝑘

We can use this to set up a scheme for 𝑑𝑦

𝑑𝑥 = 2𝑦 + 𝑥

𝑦 0 = 2

𝑦𝑘+1 − 𝑦𝑘 ℎ

= 2𝑦𝑘 + 𝑥𝑘

Rearrange so that 𝑦𝑘+1 is the subject 𝑦𝑘+1 − 𝑦𝑘 = ℎ(2𝑦𝑘 + 𝑥𝑘)

𝑦𝑘+1 = 1 + 2ℎ 𝑦𝑘 + ℎ𝑥𝑘

finite/Finite Differences.pdf

Sensitivity: Internal

Finite Differences

Sensitivity: Internal

Rationale

There are differential equations we just can’t solve.

𝑑𝑦

𝑑𝑥 + sin 𝑥 𝑦 + 3𝑦 + 𝑥𝑦 = 𝑥 + 5

Sensitivity: Internal

Starter Problem

We will find a numerical answer to the following 𝑑𝑦

𝑑𝑥 = 2𝑦 1

𝑦 0 = 2 (2)

Sensitivity: Internal

Approximations via Calculus

We can make use of some basic calculus definitions

𝑑𝑦

𝑑𝑥 = lim

ℎ→0

𝑓 𝑥 + ℎ − 𝑓(𝑥)

Sensitivity: Internal

Difference Quotient

We can make use of some basic calculus definitions

𝑑𝑦

𝑑𝑥 = lim

ℎ→0

𝑓 𝑥+ℎ −𝑓(𝑥)

Therefore we can approximate 𝑑𝑦

𝑑𝑥 ≈

𝑓 𝑥+ℎ −𝑓(𝑥)

𝑥 𝑥 + ℎ

𝑓(𝑥)

𝑓(𝑥 + ℎ)

Sensitivity: Internal

Notation

Let ℎ be a fixed constant

𝑑𝑦

𝑑𝑥 ≈

𝑓 𝑥𝑘+1 −𝑓(𝑥𝑘)

OR we use 𝑦𝑘+1 and 𝑦𝑘

𝑑𝑦

𝑑𝑥 ≈

𝑦𝑘+1− 𝑦𝑘

𝑥1 𝑥2

𝑦𝑘 = 𝑓(𝑥𝑘)

𝑦𝑘+1 = 𝑓 𝑥𝑘+1

𝑥𝑛𝑥𝑘 𝑥𝑘+1

Sensitivity: Internal

Forward Difference Scheme

𝑑𝑦

𝑑𝑥 ≈ 𝑦𝑘+1 − 𝑦𝑘

We can use this to set up a scheme for 𝑑𝑦

𝑑𝑥 = 2𝑦

𝑦𝑘+1 − 𝑦𝑘 ℎ

= 2𝑦𝑘

Rearrange so that 𝑦𝑘+1 is the subject 𝑦𝑘+1 = 2ℎ𝑦𝑘 + 𝑦𝑘

𝑦𝑘+1 = 1 + 2ℎ 𝑦𝑘

Sensitivity: Internal

How do we start???

𝑦𝑘+1 = 1 + 2ℎ 𝑦𝑘

Let 𝑘 = 0 𝑦1 = 1 + 2ℎ 𝑦0 (3)

To get 𝑦1 we need 𝑦0 This comes from the initial condition 𝑦 0 = 2

𝑦0 = 2

From (3) 𝑦1 = 1 + 2ℎ ⋅ 2

e.g. if ℎ = 0.1 𝑦1 = 1 + 2 ⋅ 0.1 ⋅ 2 = 2.4

Sensitivity: Internal

Results

𝑦𝑘+1 = 1 + 2ℎ 𝑦𝑘 𝑦0 = 2

Sensitivity: Internal

What is a 1st Order Differential Equation

𝑑𝑦

𝑑𝑥 ≈ 𝑦𝑘+1 − 𝑦𝑘

We can use this to set up a scheme for 𝑑𝑦

𝑑𝑥 = 2𝑦 + 𝑥

𝑦 0 = 2

𝑦𝑘+1 − 𝑦𝑘 ℎ

= 2𝑦𝑘 + 𝑥𝑘

Rearrange so that 𝑦𝑘+1 is the subject 𝑦𝑘+1 − 𝑦𝑘 = ℎ(2𝑦𝑘 + 𝑥𝑘)

𝑦𝑘+1 = 1 + 2ℎ 𝑦𝑘 + ℎ𝑥𝑘

finite/Forward Differences Notes with Example.pdf

Solve the following differential equation with initial condition using the forward differences scheme.

𝑑𝑦

𝑑𝑥 = 2𝑦 + 𝑥2, 𝑦(0) = 5

Equally discretised the space that you wish to approximate over. Here we are looking to

approximate the red curve over the space [0,2]

Here we have “chunked” the space into equal parts by a step size of ℎ = 0.2

𝑥0 = 0

𝑥1 = 0.2

𝑥𝑛−1 = 1.9

𝑥𝑛 = 2

We can approximate the first derivative by,

𝑑𝑦

𝑑𝑥 ≈

𝑦𝑘+1 − 𝑦𝑘 ℎ

Where,

𝑥 = 𝑥𝑘

𝑦 = 𝑦𝑘

and use this to create an iterative method to solve our differential equation.

e.g. for the following

𝑑𝑦

𝑑𝑥 = 2𝑦 + 𝑥2, 𝑦(0) = 5

𝑦𝑘+1 − 𝑦𝑘 ℎ

= 2𝑦𝑘 + 𝑥𝑘 2

𝑦𝑘+1 = 𝑦𝑘 + 2ℎ𝑦𝑘 + ℎ𝑥𝑘 2

𝑦𝑘+1 = (1 + 2ℎ)𝑦𝑘 + ℎ𝑥𝑘 2

Then for a step size of ℎ = 0.2 we have

and,

𝑦𝑘+1 = (1 + 2 ⋅ 0.2)𝑦𝑘 + 0.2𝑥𝑘 2

𝑦𝑘+1 = 1.4𝑦𝑘 + 0.2𝑥𝑘 2

e.g.

𝑘 = 0

We require 𝑥0 and 𝑦0, which we can get from our initial condition 𝑦(0) = 5, giving,

𝑥0 = 0, 𝑦0 = 5

Thus

𝑦1 = 1.4𝑦0 + 0.2𝑥0 2 = 1.4 ⋅ 5 = 7

𝑘 = 1,

𝑥1 = 𝑥0 + ℎ = 0 + 0.2 = 0.2, 𝑦1 = 7

𝑦2 = 1.4𝑦1 + 0.2𝑥1 2 = 1.4 ⋅ 7 + 0.2 ⋅ 0.22 = 9.808

etc….

𝑘 𝑥𝑘 𝑦𝑘 Analytical Error

0 0 5 5 0

1 0.2 7 7.46208 -0.46208

2 0.4 9.808 11.15409 -1.34609

3 0.6 13.7632 16.70061 -2.93741

4 0.8 19.34048 25.03342 -5.69294

5 1 27.20467 37.54254 -10.3379

finite/Task Questions.pdf

University of Derby School of Electronics, Computing and Mathematics In-course Assignment Specification

Module Code and Title: 4MA502 – Mathematical Software Assignment No. and Title: 1 – Coursework Part 2 Assessment Tutor: Sam O’Neill Weighting Towards Module Grade: 40 % Date Set: 03/04/2020 Submission Date: 11/05/2020 – 23:59

Penalty for Late Submission

Recognising that deadlines are an integral part of professional workplace practice; the University expects students to meet all agreed deadlines for submission of assessments. However, the University acknowledges that there may be circumstances which prevent students from meeting deadlines. There are now 3 distinct processes in place to deal with differing student circumstances. https://www.derby.ac.uk/about/academic-regulations/ Assessed Extended Deadline (AED) Students with disabilities or long term health issues are entitled to a Support Plan. Exceptional Extenuating Circumstances (EEC) The EEC policy applies to situations where serious, unforeseen circumstances prevent the student from completing the assignment on time or to the normal standard. Late Submission Requests for late submission (LSR) will be made to the relevant Subject Manager in the School (or Head of Joint Honours for joint honours students) who can authorise an extension of up to a maximum of one week.

Level of Collaboration:

This is an individual assignment. No collaboration with other students or anyone else is allowed.

Learning Outcome(s) covered in this assignment: Demonstrate a knowledge of the underlying concepts of optimisation and metaheuristics and apply appropriate techniques to a range of problems; Derive solutions to various problems through the application and implementation of modern optimisation techniques.

Submissions in Turnitin and Blackboard You must submit your work using your student number to identify yourself, not your name. You must not use your name in the text of the work at any point. When you submit your work in Turnitin you must submit your student number within the assignment document and in the Submission title field in Turnitin.

Marking Criteria: Credit will be awarded for:

70–100%: Excellent

Outstanding; high to very high standard; a high level of critical analysis and evaluation; commendable originality; high quality presentation; exceptional clarity of ideas; excellent coherence and logic.

60-69%: Very good

A very good standard; a very good critical analysis and evaluation; significant originality; well researched; a very good standard of presentation; pleasing clarity of ideas; thoughtful and effective presentation; very good sense of coherence and logic;

50-59%: Good

A good standard; a fairly good level of critical analysis and evaluation; some evidence of original thinking; quite well researched; a good standard of presentation; ideas generally clear and coherent, some misunderstandings; some deficiencies in presentation.

40-49%: Satisfactory

A sound standard of work; a fair level of critical analysis and evaluation; little evidence of original thinking or originality; adequately researched; a sound standard of presentation; ideas fairly clear and coherent, some significant misunderstandings and errors; some weakness in style or presentation but satisfactory overall.

35-39%: Unsatisfactory (marginal fail)

Overall marginally unsatisfactory; some sound aspects but some of the following weaknesses are evident: inadequate critical analysis and evaluation; little evidence of originality; not well researched; standard of presentation unacceptable; ideas unclear and incoherent; some significant errors and misunderstandings.

1-34%: Very poor (fail)

Well below the pass standard; a poor critical analysis and evaluation; no evidence of originality; poorly researched; standard of presentation totally unacceptable; ideas confused and incoherent, some serious misunderstandings and errors. A clear fail, well short of the pass standard...The work demonstrates nothing of merit.

IMPORTANT GUIDELINES FOR SUBMISSION

All 4 questions must be answered using either Excel or MATLAB. You are encouraged to use other methods/means to verify your Excel/MATLAB solutions.

You are required to submit a report containing a write up of your answers/results and submit any Excel or MATLAB files that were used to in answering the questions.

This is an individual assignment. No collaboration with other students or anyone else is allowed. Please be aware that this is obvious when MATLAB/Excel files and report answers are extremely similar. You MUST not collude with other students, students expected of collusion may be penalised subject to the academic regulations

Question 1 Derive and implement a Forward Differences Scheme to numerically solve the following and provide a plot of your solution between 𝑥 = 0 and 𝑥 = 5.

Use a step size ℎ = 0.1, additional marks can be gained through analysis (e.g. errors, changes in step size etc…)

a) = 4𝑦, 𝑦(0) = 6

(15 marks)

b) = 𝑥𝑐𝑜𝑠(𝑥) + 2, 𝑦(0) = 8

(15 marks)

Question 2 Use Newton’s Method to solve the following to a tolerance of 0.001 using the given starting points 𝑥 . Provide a table of results and a plot of the iterations. Investigate whether the choice of 𝑥 matters?

i.e. successive terms should differ by less than or equal to 0.001. i.e. |𝑥 − 𝑥 | ≤ 0.001

a) 𝑥 + 3𝑥 + 4𝑥 + 3 = 0, 𝑥 = 1 (10 marks)

b) 3𝑥 + 3 = 𝑒 , 𝑥 = 3 (10 marks)

Question 3 Use numerical integration to evaluate the following, provide a table to summarise your results. How could you improve the evaluations?

a) ∫ sin , use 10 intervals, 𝑁 = 10

(10 marks)

b) ∫ √1 + 𝑥 , use 8 intervals, 𝑁 = 8 (10 marks)

Question 4 Answer the following problems:

a) The product of the first five natural numbers is,

1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 = 5! = 120

The square of the sum of the first five natural numbers is,

(1 + 2 + 3 + 4 + 5) = 15 = 225

Hence the difference between the product of the first ten natural numbers and the square of the sum is 120 − 225 = −105.

Find the difference between the product of the first fifty natural numbers and the square of the sum of the first fifty natural numbers.

(6 marks)

b) By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. What is the 2021st prime number? Hint, a useful algorithm to look to implement is the Sieve of Eratosthenes

(12 marks)

c) You have the chance to play the following game. Each round cost £4 and you win £𝑛 per round, where 𝑛 is your score for that round. You can play as many rounds as you like. Each round is as follows - You roll a dice. If the dice lands on 3 then you can roll again, if it lands on anything else then you stop. Your score for the round is the total of all your rolls in that round. e.g. First round Roll 5, STOP Round Score 5 Second round Roll 3, Roll 3, Roll 1, STOP Round Score 3+3+1 = 7 Third round Roll 1, STOP Round Score 1 Total score after 3 rounds of 5+7+1 = 13 Therefore, you made a profit of £1 in total, as you played 3 rounds costing £12, but won back £13. a) Given that you have plenty of money to play with, would you keep playing the game? Use a simulation

to decide. b) If the game is changed so that you can continue to roll on a 1 instead of a 3. Would this affect the

choice of whether you would play? (12 marks)

Total 100 marks