Name: CATAMPONGAN, JOSHUA G.

Course and Section: BSCE-3C

Subject Course: EM 315__NUMERICAL SOLUTIONS TO CE PROBLEMS
Name of Activity: MIDTERM PROJECT

LESSON 5 – RO
Method Use: CLOSED METHOD (INCREMENTAL SEARCH METHOD)

EQ
INCREMENT
"𝒇(𝒙) = " 𝟓𝒙^𝟒 " + " 𝟐𝒙^𝟑 " − " 𝟏𝟐𝒙^𝟐 " + 20x − " 𝟓
Determine the positive real root of

Using Closed Method

Note: 𝑥_𝐿=0 and 𝑥_𝑈=1

Incremental Search Method (up to 6 decimal places n

(SOLUTIONS) :
 "𝒇(𝒙) = " 𝟓𝒙^𝟒 " + "
Note: 𝑥_𝐿=0 and

1ST ITERATION

Δx = 0.1

x f(x)
0 -5
0.1 -3.1175
0.2 -1.456
0.3 0.0145
0.4 1.336
0.5 2.5625
0.6 3.76
0.7 5.0065
0.8 6.392
0.9 8.0185
1 10

SOLVE FOR VALIDITY |εa|≤|εs|:

|"ε" _𝑎
|=(𝑐𝑢𝑟𝑟𝑒𝑛𝑡−𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠)/𝑐𝑢𝑟𝑟𝑒𝑛𝑡
×100%
|εs| = (0.50 x 102-n)%

Subtitute the values:

|εs| = 0.00005 |εa| =

Hence,
FALSE

3RD ITERATION

Δx = 0.001

x f(x)
0.29 -0.12505795
0.291 -0.11103329919
 0.292 -0.09702407552
0.293 -0.08303023199
0.294 -0.06905172152
0.295 -0.05508849687
0.296 -0.04114051072
0.297 -0.02720771559
0.298 -0.01329006392
0.299 0.00061249201
0.3 0.0145

SOLVE FOR VALIDITY |εa|≤|εs|:

|"ε" _𝑎
|=(𝑐𝑢𝑟𝑟𝑒𝑛𝑡−𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠)/𝑐𝑢𝑟𝑟𝑒𝑛𝑡
×100%
|εs| = (0.50 x 102-n)%

Subtitute the values:

|εs| = 0.00005 |εa| =

Hence,
FALSE

5TH ITERATION

Δx = 0.00001

x f(x)
0.2989 -0.00077708564
0.29891 -0.0006381211
0.29892 -0.00049915807
0.29893 -0.00036019654
0.29894 -0.00022123652
0.29895 -8.2278005E-05
0.29896 5.66790069E-05
0.29897 0.00019563451
0.29898 0.00033458852
0.29899 0.00047354101
0.299 0.00061249201

SOLVE FOR VALIDITY |εa|≤|εs|:

|"ε" _𝑎
|=(𝑐𝑢𝑟𝑟𝑒𝑛𝑡−𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠)/𝑐𝑢𝑟𝑟𝑒𝑛𝑡
×100%
|εs| = (0.50 x 102-n)%

Subtitute the values:

|εs| = 0.00005 |εa| =
 Hence,
FALSE

7TH ITERATION

Δx = 0.0000001

x f(x)
0.298955 -1.2799311E-05
0.2989551 -1.1409741E-05
0.2989552 -1.0020171E-05
0.2989553 -8.6306012E-06
0.2989554 -7.2410316E-06
0.2989555 -5.8514622E-06
0.2989556 -4.4618929E-06
0.2989557 -3.0723237E-06
0.2989558 -1.6827547E-06
0.2989559 -2.931859E-07
0.298956 1.09638279E-06

SOLVE FOR VALIDITY |εa|≤|εs|:

|"ε" _𝑎
|=(𝑐𝑢𝑟𝑟𝑒𝑛𝑡−𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠)/𝑐𝑢𝑟𝑟𝑒𝑛𝑡
×100%
|εs| = (0.50 x 102-n)%

Subtitute the values:

|εs| = 0.00005 |εa| =

Hence,
TRUE
E PROBLEMS

N 5 – ROOTS OF NONLINEAR
METHOD)

EQUATIONS
REMENTAL SEARCH METHOD
𝟐 " + 20x − " 𝟓

6 decimal places near True Value)

= " 𝟓𝒙^𝟒 " + " 𝟐𝒙^𝟑 " − " 𝟏𝟐𝒙^𝟐 " + 20x − " 𝟓
Note: 𝑥_𝐿=0 and 𝑥_𝑈=1

2ND ITERATION

Δx = 0.01

x f(x)
0.2 -1.456
0.21 -1.30095395
0.22 -1.1477912
0.23 -0.99647395
0.24 -0.8469632
0.25 -0.69921875
0.26 -0.5531992
0.27 -0.40886195
0.28 -0.2661632
0.29 -0.12505795
0.3 0.0145

SOLVE FOR VALIDITY |εa|≤|εs|:

𝑒𝑛𝑡−𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠)/𝑐𝑢𝑟𝑟𝑒𝑛𝑡 |εs| = (0.50 x 102-n)%

Subtitute the values:

33.3333333333 |εs| =

4TH ITERATION

Δx = 0.0001

x f(x)
0.298 -0.0132900639
0.2981 -0.0118991298
 0.2982 -0.0105083466
0.2983 -0.0091177143
0.2984 -0.0077272327
0.2985 -0.006336902
0.2986 -0.0049467219
0.2987 -0.0035566926
0.2988 -0.0021668138
0.2989 -0.0007770856
0.299 0.000612492

SOLVE FOR VALIDITY |εa|≤|εs|:

𝑒𝑛𝑡−𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠)/𝑐𝑢𝑟𝑟𝑒𝑛𝑡 |εs| = (0.50 x 102-n)%

Subtitute the values:

0.33444816054 |εs| =

6TH ITERATION

Δx = 0.000001

x f(x)
0.29895 -8.2278E-05
0.298951 -6.838224E-05
0.298952 -5.448648E-05
0.298953 -4.059074E-05
0.298954 -2.669502E-05
0.298955 -1.279931E-05
0.298956 1.0963828E-06
0.298957 1.4992061E-05
0.298958 2.8887725E-05
0.298959 4.2783373E-05
0.29896 5.6679007E-05

SOLVE FOR VALIDITY |εa|≤|εs|:

𝑒𝑛𝑡−𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠)/𝑐𝑢𝑟𝑟𝑒𝑛𝑡 |εs| = (0.50 x 102-n)%

Subtitute the values:

0.00334492909 |εs| =
𝑒𝑛𝑡−𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠)/𝑐𝑢𝑟𝑟𝑒𝑛𝑡

3.34497384E-05
INEAR

OD
 0.01

f(x)
-1.456
-1.30095395
-1.1477912
-0.99647395
-0.8469632
-0.69921875
-0.5531992
-0.40886195
-0.2661632
-0.12505795
0.0145

FOR VALIDITY |εa|≤|εs|:

|"ε" _𝑎
|=(𝑐𝑢𝑟𝑟𝑒𝑛𝑡−𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠)/𝑐𝑢𝑟𝑟𝑒𝑛𝑡
×100%
|εs| = (0.50 x 102-n)%

te the values:
0.00005 |εa| = 3.333333

Hence,
FALSE

0.0001

f(x)
-0.0132900639
-0.0118991298
 -0.0105083466
-0.0091177143
-0.0077272327
-0.006336902
-0.0049467219
-0.0035566926
-0.0021668138
-0.0007770856
0.000612492

FOR VALIDITY |εa|≤|εs|:

|"ε" _𝑎
|=(𝑐𝑢𝑟𝑟𝑒𝑛𝑡−𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠)/𝑐𝑢𝑟𝑟𝑒𝑛𝑡
×100%
|εs| = (0.50 x 102-n)%

te the values:
0.00005 |εa| = 0.033445

Hence,
FALSE

0.000001

f(x)
-8.2278E-05
-6.838224E-05
-5.448648E-05
-4.059074E-05
-2.669502E-05
-1.279931E-05
1.0963828E-06
1.4992061E-05
2.8887725E-05
4.2783373E-05
5.6679007E-05

FOR VALIDITY |εa|≤|εs|:

|"ε" _𝑎
|=(𝑐𝑢𝑟𝑟𝑒𝑛𝑡−𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠)/𝑐𝑢𝑟𝑟𝑒𝑛𝑡
×100%
|εs| = (0.50 x 102-n)%

te the values:
0.00005 |εa| = 0.000334
Hence,
FALSE
Name: CATAMPONGAN, JOSHUA G.
Course and Section: BSCE-3C
Subject Course: EM 315__NUMERICAL SOLUTIONS TO CE PROBLEMS
Name of Activity: MIDTERM PROJECT

LESSON 5 – RO
Method Use: CLOSED METHOD (BISECTION METHOD)

EQU
BISECT
"𝒇(𝒙) = " 𝟓𝒙^𝟒 " + " 𝟐𝒙^𝟑 " − " 𝟏𝟐𝒙^𝟐 " + 20x − " 𝟓
Determine the positive real root of

Using Closed Method

Note: 𝑥_𝐿=0 and 𝑥_𝑈=1

Bisection Method (Make an iteration up to True Value)

(SOLUTIONS) :
 "𝒇(𝒙) = " 𝟓𝒙^𝟒 " + " 𝟐
Note: 𝑥_𝐿=0 and

ITERATION XL XU f(xl) f(xu)

1 0 1 -5 10
2 0 0.5 -5 2.5625
3 0.25 0.5 -0.69921875 2.5625
4 0.25 0.375 -0.69921875 1.016845703125
5 0.25 0.3125 -0.69921875 0.186843872070313
6 0.28125 0.3125 -0.24843883514404 0.186843872070313
7 0.296875 0.3125 -0.02894848585129 0.186843872070313
8 0.296875 0.3046875 -0.02894848585129 0.079398531466722
9 0.296875 0.30078125 -0.02894848585129 0.025339176179841
10 0.298828125 0.30078125 -0.00177593754779 0.025339176179841
11 0.298828125 0.2998046875 -0.00177593754779 0.011788776338108
12 0.298828125 0.29931640625 -0.00177593754779 0.00500821144243
13 0.298828125 0.299072265625 -0.00177593754779 0.001616585307733
14 0.2989501953125 0.299072265625 -7.95639863744E-05 0.001616585307733
15 0.2989501953125 0.29901123046875 -7.95639863744E-05 0.00076853868865
16 0.2989501953125 0.298980712890625 -7.95639863744E-05 0.000344494358811
17 0.2989501953125 0.298965454101563 -7.95639863744E-05 0.000132466938222
18 0.2989501953125 0.298957824707031 -7.95639863744E-05 2.64519139357E-05
19 0.298954010009766 0.298957824707031 -2.65559267154E-05 2.64519139357E-05
20 0.298955917358398 0.298957824707031 -5.19790139819E-08 2.64519139357E-05
21 0.298955917358398 0.298956871032715 -5.19790139819E-08 1.31999743047E-05
22 0.298955917358398 0.298956394195557 -5.19790139819E-08 6.57399935644E-06
23 0.298955917358398 0.298956155776978 -5.19790139819E-08 3.26101059844E-06
24 0.298955917358398 0.298956036567688 -5.19790139819E-08 1.6045158997E-06
25 0.298955917358398 0.298955976963043 -5.19790139819E-08 7.7626846906E-07
26 0.298955917358398 0.298955947160721 -5.19790139819E-08 3.62144734645E-07
27 0.298955917358398 0.29895593225956 -5.19790139819E-08 1.55082862108E-07
28 0.298955917358398 0.298955924808979 -5.19790139819E-08 5.1551924507E-08
29 0.298955921083689 0.298955924808979 -2.13544737449E-10 5.1551924507E-08
30 0.298955921083689 0.298955922946334 -2.13544737449E-10 2.56691894407E-08
31 0.298955921083689 0.298955922015011 -2.13544737449E-10 1.27278223516E-08
32 0.298955921083689 0.29895592154935 -2.13544737449E-10 6.25713880709E-09
33 0.298955921083689 0.298955921316519 -2.13544737449E-10 3.02179703482E-09
34 0.298955921083689 0.298955921200104 -2.13544737449E-10 1.4041257046E-09
35 0.298955921083689 0.298955921141896 -2.13544737449E-10 5.95290927663E-10
36 0.298955921083689 0.298955921112793 -2.13544737449E-10 1.90873095107E-10
37 0.298955921098241 0.298955921112793 -1.13358211706E-11 1.90873095107E-10
38 0.298955921098241 0.298955921105517 -1.13358211706E-11 8.97681928791E-11
39 0.298955921098241 0.298955921101879 -1.13358211706E-11 3.92166299434E-11
40 0.298955921098241 0.29895592110006 -1.13358211706E-11 1.39399602972E-11
41 0.298955921098241 0.29895592109915 -1.13358211706E-11 1.30206956328E-12
42 0.298955921098695 0.29895592109915 -5.01731989289E-12 1.30206956328E-12
43 0.298955921098923 0.29895592109915 -1.85806925401E-12 1.30206956328E-12
44 0.298955921099036 0.29895592109915 -2.77999845366E-13 1.30206956328E-12
45 0.298955921099036 0.298955921099093 -2.77999845366E-13 5.11590769747E-13
46 0.298955921099036 0.298955921099065 -2.77999845366E-13 1.172395514E-13
47 0.298955921099051 0.298955921099065 -7.9936057773E-14 1.172395514E-13
48 0.298955921099051 0.298955921099058 -7.9936057773E-14 1.86517468137E-14
49 0.298955921099054 0.298955921099058 -3.10862446895E-14 1.86517468137E-14
50 0.298955921099056 0.298955921099058 0 1.86517468137E-14
51 0.298955921099057 0.298955921099058 0 1.86517468137E-14
52 0.298955921099057 0.298955921099058 0 1.86517468137E-14
53 0.298955921099058 0.298955921099058 0 1.86517468137E-14
54 0.298955921099058 0.298955921099058 1.7763568394E-14 1.86517468137E-14
55 0.298955921099058 0.298955921099058 1.7763568394E-14 1.86517468137E-14
56 0.298955921099058 0.298955921099058 1.86517468137E-14 1.86517468137E-14
CE PROBLEMS

5 – ROOTS OF NONLINEAR
EQUATIONS
BISECTION METHOD
𝟐 " + 20x − " 𝟓

on up to True Value)
= " 𝟓𝒙^𝟒 " + " 𝟐𝒙^𝟑 " − " 𝟏𝟐𝒙^𝟐 " + 20x − " 𝟓
Note: 𝑥_𝐿=0 and 𝑥_𝑈=1

f(xu) xr f(xl)*f(xr) f(xr) New Interval

10 0.5 -12.8125 2.5625 _____
2.5625 0.25 3.49609375 -0.69921875 0,0.5
2.5625 0.375 -0.71099758148193 1.016845703125 0.25,0.5
1.016845703125 0.3125 -0.13064473867416 0.186843872070313 0.25,0.375
0.186843872070313 0.28125 0.173713091760874 -0.24843883514404 0.25,0.3125
0.186843872070313 0.296875 0.007191928104078 -0.02894848585129 0.28125,0.3125
0.186843872070313 0.3046875 -0.00229846726478 0.079398531466722 0.296875,0.3125
0.079398531466722 0.30078125 -0.00073353078313 0.025339176179841 0.296875,0.3046875
0.025339176179841 0.298828125 5.1410702975E-05 -0.00177593754779 0.296875,0.30078125
0.025339176179841 0.2998046875 -2.09361305414E-05 0.011788776338108 0.298828125,0.30078125
0.011788776338108 0.29931640625 -8.89427074789E-06 0.00500821144243 0.298828125,0.2998046875
0.00500821144243 0.299072265625 -2.87095454721E-06 0.001616585307733 0.298828125,0.29931640625
0.001616585307733 0.2989501953125 1.41300670854E-07 -7.95639863744E-05 0.298828125,0.299072265625
0.001616585307733 0.29901123046875 -6.1148001752E-08 0.00076853868865 0.2989501953125,0.299072265625
0.00076853868865 0.298980712890625 -2.74093444705E-08 0.000344494358811 0.2989501953125,0.29901123046875
0.000344494358811 0.298965454101563 -1.05395976678E-08 0.000132466938222 0.2989501953125,0.29898071289062
0.000132466938222 0.298957824707031 -2.10461971996E-09 2.64519139357E-05 0.2989501953125,0.29896545410156
2.64519139357E-05 0.298954010009766 2.11289539135E-09 -2.65559267154E-05 0.2989501953125,0.29895782470703
2.64519139357E-05 0.298955917358398 1.38035088604E-12 -5.19790139819E-08 0.298954010009766,0.2989578247070
2.64519139357E-05 0.298956871032715 -6.86121648945E-13 1.31999743047E-05 0.298955917358398,0.2989578247070
1.31999743047E-05 0.298956394195557 -3.41710004465E-13 6.57399935644E-06 0.298955917358398,0.2989568710327
6.57399935644E-06 0.298956155776978 -1.69504115492E-13 3.26101059844E-06 0.298955917358398,0.2989563941955
3.26101059844E-06 0.298956036567688 -8.34011543847E-14 1.6045158997E-06 0.298955917358398,0.2989561557769
1.6045158997E-06 0.298955976963043 -4.0349669607E-14 7.7626846906E-07 0.298955917358398,0.2989560365676
7.7626846906E-07 0.298955947160721 -1.88239262256E-14 3.62144734645E-07 0.298955917358398,0.2989559769630
3.62144734645E-07 0.29895593225956 -8.06105425785E-15 1.55082862108E-07 0.298955917358398,0.2989559471607
1.55082862108E-07 0.298955924808979 -2.67961820474E-15 5.1551924507E-08 0.298955917358398,0.2989559322595
5.1551924507E-08 0.298955921083689 1.10998448936E-17 -2.13544737449E-10 0.298955917358398,0.2989559248089
5.1551924507E-08 0.298955922946334 -5.48152031963E-18 2.56691894407E-08 0.298955921083689,0.2989559248089
2.56691894407E-08 0.298955922015011 -2.71795948237E-18 1.27278223516E-08 0.298955921083689,0.2989559229463
1.27278223516E-08 0.29895592154935 -1.33617906374E-18 6.25713880709E-09 0.298955921083689,0.2989559220150
6.25713880709E-09 0.298955921316519 -6.45288854423E-19 3.02179703482E-09 0.298955921083689,0.2989559215493
3.02179703482E-09 0.298955921200104 -2.99843654933E-19 1.4041257046E-09 0.298955921083689,0.2989559213165
1.4041257046E-09 0.298955921141896 -1.27121244853E-19 5.95290927663E-10 0.298955921083689,0.2989559212001
5.95290927663E-10 0.298955921112793 -4.07599449807E-20 1.90873095107E-10 0.298955921083689,0.2989559211418
1.90873095107E-10 0.298955921098241 2.42070495565E-21 -1.13358211706E-11 0.298955921083689,0.2989559211127
1.90873095107E-10 0.298955921105517 -1.01759618129E-21 8.97681928791E-11 0.298955921098241,0.2989559211127
8.97681928791E-11 0.298955921101879 -4.44552703954E-22 3.92166299434E-11 0.298955921098241,0.2989559211055
3.92166299434E-11 0.29895592110006 -1.58020897055E-22 1.39399602972E-11 0.298955921098241,0.2989559211018
1.39399602972E-11 0.29895592109915 -1.47600277211E-23 1.30206956328E-12 0.298955921098241,0.2989559211000
1.30206956328E-12 0.298955921098695 5.68754410616E-23 -5.01731989289E-12 0.298955921098241,0.2989559210991
1.30206956328E-12 0.298955921098923 9.32252783052E-24 -1.85806925401E-12 0.298955921098695,0.2989559210991
1.30206956328E-12 0.298955921099036 5.16542965295E-25 -2.77999845366E-13 0.298955921098923,0.2989559210991
1.30206956328E-12 0.298955921099093 -1.4222215488E-25 5.11590769747E-13 0.298955921099036,0.2989559210991
5.11590769747E-13 0.298955921099065 -3.25925771601E-26 1.172395514E-13 0.298955921099036,0.2989559210990
1.172395514E-13 0.298955921099051 2.22222117001E-26 -7.9936057773E-14 0.298955921099036,0.2989559210990
1.172395514E-13 0.298955921099058 -1.49094711087E-27 1.86517468137E-14 0.298955921099051,0.2989559210990
1.86517468137E-14 0.298955921099054 2.48491185145E-27 -3.10862446895E-14 0.298955921099051,0.2989559210990
1.86517468137E-14 0.298955921099056 0 0 0.298955921099054,0.2989559210990
1.86517468137E-14 0.298955921099057 0 0 0.298955921099056,0.2989559210990
1.86517468137E-14 0.298955921099057 0 0 0.298955921099057,0.2989559210990
1.86517468137E-14 0.298955921099058 0 0 0.298955921099057,0.2989559210990
1.86517468137E-14 0.298955921099058 0 1.7763568394E-14 0.298955921099058,0.2989559210990
1.86517468137E-14 0.298955921099058 3.15544362088E-28 1.7763568394E-14 0.298955921099058,0.2989559210990
1.86517468137E-14 0.298955921099058 3.31321580193E-28 1.86517468137E-14 0.298955921099058,0.2989559210990
1.86517468137E-14 0.298955921099058 3.47887659202E-28 1.86517468137E-14 0.298955921099058,0.2989559210990
NEAR
 New Interval Approximate Error(%) True Error(%)
_____ _____
0,0.5 -100
0.25,0.5 33.3333333333333
0.25,0.375 -20
0.25,0.3125 -11.1111111111111
0.28125,0.3125 5.26315789473684
0.296875,0.3125 2.56410256410256
0.296875,0.3046875 -1.2987012987013
0.296875,0.30078125 -0.65359477124183
.298828125,0.30078125 0.325732899022801
298828125,0.2998046875 -0.163132137030995
98828125,0.29931640625 -0.0816326530612245
98828125,0.299072265625 -0.0408329930583912
9501953125,0.299072265625 0.0204123290467442
501953125,0.29901123046875 -0.0102072062876391
01953125,0.298980712890625 -0.00510386362476395
01953125,0.298965454101562 -0.00255199693760367
01953125,0.298957824707031 -0.00127601475073052
010009766,0.298957824707031 0.000638003304857119
917358398,0.298957824707031 0.000319000634811263
917358398,0.298956871032715 -0.00015950057180955
917358398,0.298956394195557 -7.97503495059067E-05
917358398,0.298956155776978 -3.98751906532553E-05
917358398,0.298956036567688 -1.99375993017055E-05
917358398,0.298955976963043 -9.96880064462253E-06
917358398,0.298955947160721 -4.98440057075374E-06
5917358398,0.29895593225956 -2.49220034748749E-06
917358398,0.298955924808979 -1.2461001892714E-06
921083689,0.298955924808979 6.23050090753788E-07
921083689,0.298955922946334 -3.11525046347372E-07
921083689,0.298955922015011 -1.55762523416306E-07
5921083689,0.29895592154935 -7.78812617688078E-08
921083689,0.298955921316519 -3.89406308995676E-08
921083689,0.298955921200104 -1.94703154535748E-08
921083689,0.298955921141896 -9.73515772773511E-09
921083689,0.298955921112793 -4.86757886410449E-09
921098241,0.298955921112793 2.43378943199301E-09
921098241,0.298955921105517 -1.21689471601131E-09
 921098241,0.298955921101879 -6.08447358009359E-10
5921098241,0.29895592110006 -3.04223679005605E-10
5921098241,0.29895592109915 -1.52111839503034E-10
5921098695,0.29895592109915 7.60559197514591E-11
5921098923,0.29895592109915 3.80279598757151E-11
5921099036,0.29895592109915 1.90139799378539E-11
921099036,0.298955921099093 -9.50698996892787E-12
921099036,0.298955921099065 -4.75349498446416E-12
921099051,0.298955921099065 2.37674749223202E-12
921099051,0.298955921099058 -1.18837374611603E-12
921099054,0.298955921099058 5.94186873058009E-13
921099056,0.298955921099058 0
921099057,0.298955921099058 0
921099057,0.298955921099058 0
921099058,0.298955921099058 0
921099058,0.298955921099058 0
921099058,0.298955921099058 0
921099058,0.298955921099058 0
Error(%)
Name: CATAMPONGAN, JOSHUA G.
Course and Section: BSCE-3C
Subject Course: EM 315__NUMERICAL SOLUTIONS TO CE PROBLEMS
Name of Activity: MIDTERM PROJECT

LESSON 5 – RO
Method Use: CLOSED METHOD (FALSE POSITION METHOD_REGULA-FALSI

EQU
FALSE POSITION METH
"𝒇(𝒙) = " 𝟓𝒙^𝟒 " + " 𝟐𝒙^𝟑 " − " 𝟏𝟐𝒙^𝟐 " + 20x − " 𝟓
Determine the positive real root of

Using Closed Method

Note: 𝑥_𝐿=0 and 𝑥_𝑈=1

Regula-Falsi Method (Make an iteration up to True Val

(SOLUTIONS) :
 "𝒇(𝒙) = " 𝟓𝒙^𝟒 " + " 𝟐
Note: 𝑥_𝐿=0 and

ITERATION XL XU f(xl) f(xu)

1 0 1 -5 10
2 0 0.333333333333333 -5 0.469135802469135
3 0 0.304740406320542 -5 0.080129158553831
4 0 0.29993371901522 -5 0.013579986980152
5 0 0.299121306326141 -5 0.002297935612987
6 0 0.298983897177134 -5 0.000388740944127
7 0 0.298960653527792 -5 6.57601814638E-05
8 0 0.298956721638139 -5 1.11240359626E-05
9 0 0.298956056518555 -5 1.88174686411E-06
10 0 0.298955944006672 -5 3.18317066039E-07
11 0 0.298955924974118 -5 5.38466418121E-08
12 0 0.298955921754563 -5 9.10871911231E-09
13 0 0.298955921209942 -5 1.54083590331E-09
14 0 0.298955921117814 -5 2.60648391759E-10
15 0 0.298955921102229 -5 4.40909531108E-11
16 0 0.298955921099593 -5 7.45981054706E-12
17 0 0.298955921099147 -5 1.26121335597E-12
18 0 0.298955921099072 -5 2.13162820728E-13
19 0 0.298955921099059 -5 3.64153152077E-14
CE PROBLEMS

5 – ROOTS OF NONLINEAR
HOD_REGULA-FALSI METHOD)

EQUATIONS
ON METHOD (REGULA-FALSI METHO
𝟐 " + 20x − " 𝟓

ation up to True Value)

= " 𝟓𝒙^𝟒 " + " 𝟐𝒙^𝟑 " − " 𝟏𝟐𝒙^𝟐 " + 20x − " 𝟓
Note: 𝑥_𝐿=0 and 𝑥_𝑈=1

f(xu) xr f(xl)*f(xr) f(xr) New Interval

10 0.333333333333333 -2.34567901234568 0.469135802469135 _____
0.469135802469135 0.304740406320542 -0.40064579276915 0.080129158553831 0,0.333333333333333
0.080129158553831 0.29993371901522 -0.06789993490076 0.013579986980152 0,0.304740406320542
0.013579986980152 0.299121306326141 -0.01148967806493 0.002297935612987 0,0.29993371901522
0.002297935612987 0.298983897177134 -0.00194370472063 0.000388740944127 0,0.299121306326141
0.000388740944127 0.298960653527792 -0.00032880090732 6.57601814638E-05 0,0.298983897177134
6.57601814638E-05 0.298956721638139 -5.5620179813E-05 1.11240359626E-05 0,0.298960653527792
1.11240359626E-05 0.298956056518555 -9.40873432054E-06 1.88174686411E-06 0,0.298956721638139
1.88174686411E-06 0.298955944006672 -1.5915853302E-06 3.18317066039E-07 0,0.298956056518555
3.18317066039E-07 0.298955924974118 -2.6923320906E-07 5.38466418121E-08 0,0.298955944006672
5.38466418121E-08 0.298955921754563 -4.55435955615E-08 9.10871911231E-09 0,0.298955924974118
9.10871911231E-09 0.298955921209942 -7.70417951657E-09 1.54083590331E-09 0,0.298955921754563
1.54083590331E-09 0.298955921117814 -1.30324195879E-09 2.60648391759E-10 0,0.298955921209942
2.60648391759E-10 0.298955921102229 -2.20454765554E-10 4.40909531108E-11 0,0.298955921117814
4.40909531108E-11 0.298955921099593 -3.72990527353E-11 7.45981054706E-12 0,0.298955921102229
7.45981054706E-12 0.298955921099147 -6.30606677987E-12 1.26121335597E-12 0,0.298955921099593
1.26121335597E-12 0.298955921099072 -1.06581410364E-12 2.13162820728E-13 0,0.298955921099147
2.13162820728E-13 0.298955921099059 -1.82076576039E-13 3.64153152077E-14 0,0.298955921099072
3.64153152077E-14 0.298955921099057 0 0 0,0.298955921099059
NEAR

METHOD)
 New Interval Approximate Error(%) True Error(%)
_____ _____
0,0.333333333333333 -9.3827160493827
0,0.304740406320542 -1.60258317107662
0,0.29993371901522 -0.271599739603037
0,0.299121306326141 -0.0459587122597287
0,0.298983897177134 -0.00777481888252826
0,0.298960653527792 -0.00131520362928457
0,0.298956721638139 -0.000222480719254481
0,0.298956056518555 -3.76349372838715E-05
0,0.298955944006672 -6.36634132257231E-06
0,0.298955924974118 -1.07693284447558E-06
0,0.298955921754563 -1.82174371479194E-07
0,0.298955921209942 -3.08167037306094E-08
0,0.298955921117814 -5.21296855234063E-09
0,0.298955921102229 -8.81810456296068E-10
0,0.298955921099593 -1.49196610156864E-10
0,0.298955921099147 -2.52158054253979E-11
0,0.298955921099072 -4.2707181501044E-12
0,0.298955921099059 -7.24165251539447E-13
Error(%)
Name: CATAMPONGAN, JOSHUA G.
Course and Section: BSCE-3C
Subject Course: EM 315__NUMERICAL SOLUTIONS TO CE PROBLEMS
Name of Activity: MIDTERM PROJECT

LESSON 5 – RO
Method Use: CLOSED METHOD (FIXED POINT METHOD)

EQU
FIXED P
"𝒇(𝒙) = " 𝟓𝒙^𝟒 " + " 𝟐𝒙^𝟑 " − " 𝟏𝟐𝒙^𝟐 " + 20x − " 𝟓
Determine the positive real root of

Using Open Method

Note: 〖𝑈𝑠𝑒 𝑥〗 _𝑖=0 or 𝑥_𝑖=1

Fixed Point Method (Make certain that you develop a

(SOLUTIONS) :

"𝒇(𝒙) = " 𝟓𝒙^𝟒 " + " 𝟐𝒙^𝟑 " − "

"𝒙 =
USE EQUATION:

ITERATION
1
2
3
4
5
6
7
8
9
10
11
12
13
CE PROBLEMS

5 – ROOTS OF NONLINEAR
EQUATIONS
FIXED POINT METHOD
𝟐 " + 20x − " 𝟓

that you develop a solution that converges on the root.)

" + " 𝟐𝒙^𝟑 " − " 𝟏𝟐𝒙^𝟐 " + 20x − " 𝟓

"𝒙 = " 𝟓/(𝟓𝒙^𝟑 " + " 𝟐𝒙^𝟐 " − " 𝟏𝟐"x + 20 " )

ITERATION X VALUE 0F X
1 0 0.25
2 0.25 0.290644868301544
3 0.290644868301544 0.297548720367363
4 0.297548720367363 0.298717838944154
5 0.298717838944154 0.298915645884465
6 0.298915645884465 0.298949108094931
7 0.298949108094931 0.298954768607621
8 0.298954768607621 0.298955726143118
9 0.298955726143118 0.298955888120231
10 0.298955888120231 0.298955915520345
11 0.298955915520345 0.298955920155359
12 0.298955920155359 0.29895592093942
13 0.29895592093942 0.298955921072052
NEAR
Name: CATAMPONGAN, JOSHUA G.
Course and Section: BSCE-3C
Subject Course: EM 315__NUMERICAL SOLUTIONS TO CE PROBLEMS
Name of Activity: MIDTERM PROJECT

LESSON 5 – RO
Method Use: CLOSED METHOD (NEWTON-RAPHSON METHOD)

EQU
NEWTON-R
"𝒇(𝒙) = " 𝟓𝒙^𝟒 " + " 𝟐𝒙^𝟑 " − " 𝟏𝟐𝒙^𝟐 " + 20x − " 𝟓
Determine the positive real root of

Using Open Method

Note: 〖𝑈𝑠𝑒 𝑥〗 _𝑖=𝑥_0=0

Newton-Raphson Method (Make an iteration up to Tru

(SOLUTIONS) :

"𝒇(𝒙) = " 𝟓𝒙^𝟒 " + " 𝟐𝒙^𝟑 " − "

"𝒇'(𝒙) = " 𝟐𝟎𝒙^𝟑 " + " 𝟔𝒙^𝟐 " −

DIFFERENTIATE THE EQUATION:

Note: 𝑥_𝑖=0

ITERATION XI f(x)
1 0 -5
2 0.25 -0.69921875
3 0.297606382978723 -0.01876648432148
4 0.298954933451171 -1.37240535452E-05
5 0.298955921098528 -7.3399064604E-12
6 0.298955921099056 0
CE PROBLEMS

5 – ROOTS OF NONLINEAR
ETHOD)

EQUATIONS
WTON-RAPHSON METHOD
𝟐 " + 20x − " 𝟓

n iteration up to True Value)

" + " 𝟐𝒙^𝟑 " − " 𝟏𝟐𝒙^𝟐 " + 20x − " 𝟓

^𝟑 " + " 𝟔𝒙^𝟐 " − " 𝟐𝟒"x + 20"

f'(x) xi+1 Approximate Error(%) True Error(%)

20 0.25
14.6875 0.297606382978723 15.9964253798034
13.9160414866979 0.298954933451171 0.451088214828768
13.895702187343 0.298955921098528 0.000330365544801425
13.8956873231771 0.298955921099056 1.76677753035842E-10
13.8956873231692 0.298955921099056 0
NEAR
Name: CATAMPONGAN, JOSHUA G.
Course and Section: BSCE-3C
Subject Course: EM 315__NUMERICAL SOLUTIONS TO CE PROBLEMS
Name of Activity: MIDTERM PROJECT

LESSON 5 – RO
Method Use: CLOSED METHOD (SECANT METHOD)

EQU
SECA
"𝒇(𝒙) = " 𝟓𝒙^𝟒 " + " 𝟐𝒙^𝟑 " − " 𝟏𝟐𝒙^𝟐 " + 20x − " 𝟓
Determine the positive real root of

Using Open Method

Note: 〖𝑈𝑠𝑒 𝑥〗 _(𝑖−1)=0 or 𝑥_𝑖=1

Fixed Point Method (Make an iteration up to True Valu

(SOLUTIONS) :

"𝒇(𝒙) = " 𝟓𝒙^𝟒 " + " 𝟐𝒙^𝟑 " − "

 "𝒇(𝒙) = " 𝟓𝒙^𝟒 " + " 𝟐𝒙^𝟑 " − "
Note: 𝑥_(𝑖−1)=0
𝑥_𝑖=1

ITERATION XI f(x)
-1 0
0 1 10
1 0.333333333333333 0.469135802469137
2 0.298342541436464 -0.00852616501737
3 0.29895571722489 -2.83297198944E-06
4 0.298955921099034 -3.13526982154E-13
5 0.298955921099056 0
CE PROBLEMS

5 – ROOTS OF NONLINEAR
EQUATIONS
SECANT METHOD
𝟐 " + 20x − " 𝟓

tion up to True Value)

𝑥_𝑖=1

" + " 𝟐𝒙^𝟑 " − " 𝟏𝟐𝒙^𝟐 " + 20x − " 𝟓

" + " 𝟐𝒙^𝟑 " − " 𝟏𝟐𝒙^𝟐 " + 20x − " 𝟓

f(x-1) xi+1 Approximate Error(%) True Error(%)

-5 0.333333333333333 100
13.4074074074074 0.298342541436464 -11.7283950617284
13.90492771944 0.29895571722489 0.205105891306369
13.8956903914866 0.298955921099034 6.81953860122385E-05
13.8956873231695 0.298955921099056 7.53874595192348E-12
13.8956873231692 0.298955921099056 0
NEAR
Name: CATAMPONGAN, JOSHUA G.
Course and Section: BSCE-3C
Subject Course: EM 315__NUMERICAL SOLUTIONS TO CE PROBLEMS
Name of Activity: MIDTERM PROJECT

LESSON 6 – C
Method Use: CLOSED METHOD ( POLYNOMIAL REGRESSION)

INTER
POLYNOM
(SOLUTIONS) :

Using Regression

Given the following data set:

xi yi
0 2.1
1 7.7
2 13.6
3 27.2
4 40.9
5 61.1

a. Fit a 4th-order polynomial function and calculate the new

(Note: include the tabulation of the solution process)

TABULAR PRESENTATION.

ITERATION xi yi xi2
1 0 2.1 0
2 1 7.7 1
3 2 13.6 4
4 3 27.2 9
5 4 40.9 16
n = 6 5 61.1 25
 15 152.6 55

where:
Mean, x (x̄ ) 2.5 2.5
Mean, y (ȳ) 25.4333333333333 25.4333333333333
m 4 m - order of polynomial

HENCE, THE GENERAL EQUATION IS…

n = 6
Σxi = 15
Σxi2 = 55
Σxi3 = 225
Σxi4 = 979

b. Error Analysis: Calculate the arithmetic mean, standard deviation, vari

determination, and correlation coefficient.

¯𝑦=
ARITHMETIC MEAN
(∑▒𝑦_𝑖 )/𝑛

¯𝑥= (∑▒𝑥_𝑖 )/𝑛

𝑆_𝑌=√(𝑆_𝑡/(𝑛−1)) =
STANDARD DEVIATION

√((∑▒ 〖 ( 〖𝑦 _𝑖−¯𝑦) 〗 ^2
〗 )/(𝑛−1))
VARIANCE

〖𝑆 _𝑌 〗 ^2= 𝑆_𝑡/(𝑛−1)
(∑▒ 〖 ( 〖𝑦 _𝑖−¯𝑦) 〗 ^2 〗 )/(𝑛
COEFFICIENT OF VARIATION

𝐶.𝑉.= 𝑆_𝑌/¯𝑦×100%

𝑆_(𝑦⁄𝑥)= √(𝑆_𝑟/(𝑛−
STANDARD ERROR OF THE ESTIMATE

(𝑚+1)))

COEFFICIENT OF DETERMINATION
𝑟^2= (𝑆_𝑡−𝑆_𝑟)/𝑆_
×100%
 𝑟^2= (𝑆_𝑡−𝑆_𝑟)/𝑆_
×100%

CORRELATION COEFFICIENT

𝑟=√(𝑟^2 )
CE PROBLEMS

N 6 – CURVE FITTING AND

INTERPOLATION
OLYNOMIAL REGRESSION

yi
2.1
7.7
13.6
27.2
40.9
61.1

and calculate the new values of the dependent variable.

solution process)

xi2 xi3 xi4 xi5 xi6 xi7

0 0 0 0 0 0
1 1 1 1 1 1
4 8 16 32 64 128
9 27 81 243 729 2187
16 64 256 1024 4096 16384
25 125 625 3125 15625 78125
 55 225 979 4425 20515 96825

2.5
25.4333333333333
f polynomial

Σxi = 15 Σxi2 = 55 Σxi3 =

Σxi2 = 55 Σxi3 = 225 Σxi4 =
Σxi3 = 225 Σxi4 = 979 Σxi5 =
Σxi4 = 979 Σxi5 = 4425 Σxi6 =
Σxi5 = 4425 Σxi6 = 20515 Σxi7 =

an, standard deviation, variance, coefficient of variation, standard error of the estimate, coefficient of

¯𝑦=
(∑▒𝑦_𝑖 )/𝑛 ¯𝑦
=
25.4333333333333

¯𝑥= (∑▒𝑥_𝑖 )/𝑛

¯𝑥
=
2.5

𝑆_𝑌=√(𝑆_𝑡/(𝑛−1)) =
√((∑▒ 〖 ( 〖𝑦 _𝑖−¯𝑦) 〗 ^2 𝑆_𝑌
〗 )/(𝑛−1)) =
22.4204965749349

〖𝑆 _𝑌 〗 ^2= 𝑆_𝑡/(𝑛−1) =
〖𝑆 _𝑌 502.678666666667
(∑▒ 〖 ( 〖𝑦 _𝑖−¯𝑦) 〗 ^2 〗 )/(𝑛−1)
〗 ^2
=

𝐶.𝑉.= 𝑆_𝑌/¯𝑦×100% 𝐶.𝑉. = 88.1539839119327

𝑆_(𝑦⁄𝑥)= √(𝑆_𝑟/(𝑛−
(𝑚+1))) 𝑆_(𝑦⁄𝑥 1.0547198028073
) =

𝑟^2= (𝑆_𝑡−𝑆_𝑟)/𝑆_𝑡
𝑟^2 = 0.998672192870475
×100% = 99.867%
𝑟^2= (𝑆_𝑡−𝑆_𝑟)/𝑆_𝑡
𝑟^2 =
×100%

𝑟=√(𝑟^2 ) 𝑟 = 0.99933587590483
AND

xi7 xi8 xiyi xi2yi xi3yi xi4yi

0 0 0 0 0 0
1 1 7.7 7.7 7.7 7.7
128 256 27.2 54.4 108.8 217.6
2187 6561 81.6 244.8 734.4 2203.2
16384 65536 163.6 654.4 2617.6 10470.4
78125 390625 305.5 1527.5 7637.5 38187.5
96825 462979 585.6 2488.8 11106 51086.4

225 Σxi4 = 979 Σyi = 152.6

979 Σxi5 = 4425 Σxiyi = 585.6
4425 Σxi6 = 20515 Σxi2yi = 2488.8
20515 Σxi7 = 96825 Σxi3yi = 11106
96825 Σxi8 = 462979 Σxi4yi = 51086.4
 xi4yi (yi-ȳ)2 (yi-a0-a1xi-a2x12-a3xi3-a4xi4)2
0 544.444444444444 0.013243260240916
7.7 314.471111111111 0.3310815071419
217.6 140.027777777778 1.32432602603234
2203.2 3.12111111111111 1.3243260270451
10470.4 239.217777777778 0.331081506729916
38187.5 1272.11111111111 0.0132432601963983
51086.4 2513.39333333333 3.33730158730159
MATRIX FORM:
SIMULTANEOUS POLYNOMIAL (QUARTIC EQUATION) ARE

[
6 15 55
15 55 225
55 225 979
225 979 4425
979 4425 20515

{𝑋}= 〖 [𝐴] 〗 ^(−1) ×{𝐵}

Matrix Product:

0.996031746031693 -1.81878306878237 1.0763888888881

-1.81878306878275 13.5438712522003 -12.3842592592548
1.0763888888886 -12.3842592592553 12.3593749999959
-0.254629629629542 3.68364197530745 -3.84837962962839
0.020833333333325 -0.347222222222108 0.37326388888877
L (QUARTIC EQUATION) ARE….

]{ } { =
225 979 a0
979 4425 a1
4425 20515 a2
20515 96825 a3
96825 462979 a4

1) ×{𝐵} ELIMINATIONS OF UNKNOWNS

-0.25462962962943 0.020833333333311 a0 2.21507936459651

3.68364197530722 -0.347222222222082 a1 3.99470899477006
-3.84837962962831 0.373263888888759 a2 0.651388888907917
1.23070987654281 -0.121527777777739 a3 0.284259259257439
-0.12152777777774 0.012152777777774 a4 -0.020833333332955
{ }
152.6
585.6
2488.8
11106
51086.4
Name: CATAMPONGAN, JOSHUA G.
Course and Section: BSCE-3C
Subject Course: EM 315__NUMERICAL SOLUTIONS TO CE PROBLEMS
Name of Activity: MIDTERM PROJECT

LESSON 6 – C
Method Use: MULTIPLE LINEAR REGRESSION

INTER
MULTIPLE L
(SOLUTIONS) :

Using Regression

Given the following data set:

yi x1
5 9
10 2
9 2.5
0 1
3 4
27 7

a. Fit a 4th-order multiple linear function and calculate the new values of t
(Note: include the tabulation of the solution process)

TABULAR PRESENTATION.

ITERATION yi x1 x2
1 5 9 3
2 10 2 5
3 9 2.5 6
4 0 1 4
5 3 4 3
n = 6 27 7 2
 54 25.5 23

where:
Mean, y (ȳ) 9 9
m 4 m - number of variable

HENCE, THE GENERAL EQUATION IS…

n = 6
Σx1i = 25.5
Σx2i = 23
Σx3i = 22
Σx4i = 19

b. Error Analysis: Calculate the arithmetic mean, standard deviation, vari

determination, and correlation coefficient.

¯𝑦=
ARITHMETIC MEAN
(∑▒𝑦_𝑖 )/𝑛
¯𝑥= (∑▒𝑥_𝑖 )/𝑛
𝑆_𝑌=√(𝑆_𝑡/(𝑛−1)) =
STANDARD DEVIATION

√((∑▒ 〖 ( 〖𝑦 _𝑖−¯𝑦) 〗 ^2
〗 )/(𝑛−1))
VARIANCE

〖𝑆 _𝑌 〗 ^2= 𝑆_𝑡/(𝑛−1)
(∑▒ 〖 ( 〖𝑦 _𝑖−¯𝑦) 〗 ^2 〗 )/(𝑛
COEFFICIENT OF VARIATION

𝐶.𝑉.= 𝑆_𝑌/¯𝑦×100%

𝑆_(𝑦⁄𝑥)= √(𝑆_𝑟/(𝑛−
STANDARD ERROR OF THE ESTIMATE

(𝑚+1)))

COEFFICIENT OF DETERMINATION
𝑟^2= (𝑆_𝑡−𝑆_𝑟)/𝑆_
×100%

CORRELATION COEFFICIENT

𝑟=√(𝑟^2 )
𝑟=√(𝑟^2 )
CE PROBLEMS

N 6 – CURVE FITTING AND

INTERPOLATION
LTIPLE LINEAR REGRESSION

x1 x2 x3 x4
9 3 8 1
2 5 1 6
2.5 6 2 4
1 4 3 5
4 3 6 2
7 2 2 1

calculate the new values of the dependent variable.

ocess)

x2 x3 x4 x12 x22 x32

3 8 1 81 9 64
5 1 6 4 25 1
6 2 4 6.25 36 4
4 3 5 1 16 9
3 6 2 16 9 36
2 2 1 49 4 4
 23 22 19 157.25 99 118

9
r of variable

Σx1i = 25.5 Σx2i = 23 Σx3i =

Σx1i2 = 157.25 Σx1ix2i = 82 Σx1ix3i =
Σx1ix2i = 82 Σx2i2 = 99 Σx2ix3i =
Σx1ix3i = 120 Σx2ix3i = 75 Σx3i2 =
Σx1ix4i = 51 Σx2ix4i = 85 Σx3ix4i =

an, standard deviation, variance, coefficient of variation, standard error of the estimate, coefficient of

¯𝑦=
(∑▒𝑦_𝑖 )/𝑛 ¯𝑦
=
9

¯𝑥= (∑▒𝑥_𝑖 )/𝑛

𝑆_𝑌=√(𝑆_𝑡/(𝑛−1)) =
√((∑▒ 〖 ( 〖𝑦 _𝑖−¯𝑦) 〗 ^2 𝑆_𝑌
〗 )/(𝑛−1)) =
9.57078889120432

〖𝑆 _𝑌 〗 ^2= 𝑆_𝑡/(𝑛−1) =
〖𝑆 _𝑌
(∑▒ 〖 ( 〖𝑦 _𝑖−¯𝑦) 〗 ^2 〗 )/(𝑛−1)
〗 ^2
91.6

𝐶.𝑉.= 𝑆_𝑌/¯𝑦×100% 𝐶.𝑉. = 1.06342098791159 = 106.342%

𝑆_(𝑦⁄𝑥)= √(𝑆_𝑟/(𝑛−
(𝑚+1))) 𝑆_(𝑦⁄𝑥 1.26968573538087
) =

𝑟^2= (𝑆_𝑡−𝑆_𝑟)/𝑆_𝑡
𝑟^2 = 0.989440380786269
×100% = 98.944%

𝑟=√(𝑟^2 ) 𝑟 = 0.994706178118076
𝑟=√(𝑟^2 ) 𝑟 =
AND

x32 x42 x1x2 x1x3 x1x4 x2x3

64 1 27 72 9 24
1 36 10 2 12 5
4 16 15 5 10 12
9 25 4 3 5 12
36 4 12 24 8 18
4 1 14 14 7 4
118 83 82 120 51 75

22 Σx4i = 19 Σyi = 54
120 Σx1ix4i = 51 Σx1iyi = 288.5
75 Σx2ix4i = 85 Σx2iyi = 182
118 Σx3ix4i = 51 Σx3iyi = 140
51 Σx4i2 = 83 Σx4iyi = 134
x2x3 x2x4 x3x4 x1iyi x2iyi x3iyi
24 3 8 45 15 40
5 30 6 20 50 10
12 24 8 22.5 54 18
12 20 15 0 0 0
18 6 12 12 9 18
4 2 2 189 54 54
75 85 51 288.5 182 140
x3iyi x4iyi (yi-ȳ)2 (yi-a0-a1x1i-a2x2i-a3x3i-a4x4i)2
40 5 16 0.162429526774378
10 60 1 1.10957153491312
18 36 0 0.0859131381286316
0 0 81 1.96400454145799
18 6 36 1.48409912144255
54 27 324 0.0302877371722995
140 134 458 4.83630559988898
MATRIX FORM:
SIMULTANEOUS POLYNOMIAL (QUA

[
6 25.5
25.5 157.25
23 82
22 120
19 51

{𝑋}= 〖 [𝐴] 〗 ^(−1) ×

Matrix Product:

9.32812126539141 -0.781948974163255
-0.781948974163255 0.10103393241274
-0.44760946499202 -0.000971480119353
-0.193779443172268 -0.009946105983855
-1.077413858241 0.124025628570768
 ANEOUS POLYNOMIAL (QUARTIC EQUATION) ARE….

]{ }
23 22 19 a0
82 120 51 a1
99 75 85 a2
75 118 51 a3
85 51 83 a4

= 〖 [𝐴] 〗 ^(−1) ×{𝐵} ELIMINATIONS OF UNKNOWNS

-0.447609464992021 -0.193779443172268 -1.077413858241 a0

-0.000971480119353 -0.00994610598385 0.124025628570768 a1
0.230778571924225 -0.0063146207758 -0.129397682326001 a2
-0.006314620775796 0.046461422216054 0.028388807932212 a3
-0.129397682326001 0.028388807932212 0.297548554730568 a4
} { }
=
54
288.5
182
140
134

25.1587676080773
1.97321490528067
-0.672819374089239
-4.17410311567553
-2.10339324127404
Name: CATAMPONGAN, JOSHUA G.
Course and Section: BSCE-3C
Subject Course: EM 315__NUMERICAL SOLUTIONS TO CE PROBLEMS
Name of Activity: MIDTERM PROJECT

LESSON 6 – C
Method Use: NEWTON'S DIVIDED DIFFERENCE INTERPOLATING POLYNOM

INTER
NEWTON'S DIVIDED D
(SOLUTIONS) :
POL
Using Interpolation

Given the function:

Estimate the value of the function wherein x = 0.5

i
0
1
2
3

a. Newton’s 1^𝑠𝑡, 2^𝑛𝑑 and 3^𝑟𝑑-order Divided Difference Interpolatin

Polynomial
 LINEAR INTERPOLATION (FIRST ORDER NEWTON'S INTERPO

𝑓(𝑥)=𝑥𝑒^𝑥

Value of B i xi
x0 0 0.2
x1 1 0.3

1st FINITE DIVIDED DIFFERENCE

For b0 :
b0 = f(x0) = 0.244280551632034

𝑏_1=(𝑓(𝑥_1 )
For b1 :
𝑓[𝑥_1 , 𝑥_0]
−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 )
1.60677090640767
=

𝑓_1 (𝑥)= 〖𝑓 (𝑥 〗 _0)+(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_

Thus,

(𝑥−𝑥_0 )
𝑓_1 (0.5)=0.244280552+(1.606770906)(0.5−0.2)

𝑓_1 (0.5)=0.726311824

QUADRATIC INTERPOLATION (FIRST ORDER NEWTON'S INTER

𝑓(𝑥)=𝑥𝑒^𝑥

Value of B i xi
x0 0 0.2
x1 1 0.3
 x2 2 0.45

1st FINITE DIVIDED DIFFERENCE

For b0 :
b0 = f(x0) = 0.705740483470576

𝑏_1=(𝑓(𝑥_1 )−𝑓(𝑥_0
For b1 :
))/(𝑥_1−𝑥_0 ) 𝑓[𝑥_1 , 𝑥_0]
=
(𝑓(𝑥_2 )−𝑓(𝑥_1
))/(𝑥_2−𝑥_1 ) 𝑓[𝑥_2 , 𝑥_1]
=

2nd FINITE DIVIDED DIFFERENCE

𝑏_2=((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) −

For b2 :

(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2

− 〖 𝑥〗 _0 )

𝑓_2 (𝑥)= 〖𝑓 (𝑥 〗 _0)+(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ) (𝑥−𝑥_0 )+((𝑓(𝑥_2

Thus,

(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗 _0 ) (𝑥−𝑥_0

𝑓_2 (𝑥)=0.244280552+(1.606770906)(0.5−0.2)+(1.59379214)

𝑓_2 (0.5)=0.821939352

CUBIC INTERPOLATION (FIRST ORDER NEWTON'S INTERPO

𝑓(𝑥)=𝑥𝑒^𝑥

Value of B i xi
x0 0 0.2
x1 1 0.3
x2 2 0.45
x3 3 0.6
1st FINITE DIVIDED DIFFERENCE

For b0 :
b0 = f(x0) = 0.705740483470576

𝑏_1=(𝑓(𝑥_1 )−𝑓(𝑥_0
For b1 :
𝑓[𝑥_1 , 𝑥_0]
))/(𝑥_1−𝑥_0 ) =

(𝑓(𝑥_2 )−𝑓(𝑥_1
))/(𝑥_2−𝑥_1 ) 𝑓[𝑥_2 , 𝑥_1]
=
(𝑓(𝑥_3 )−𝑓(𝑥_2
))/(𝑥_3−𝑥_2 ) 𝑓[𝑥_3 , 𝑥_2]
=

2nd FINITE DIVIDED DIFFERENCE

𝑏_2=((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) −

For b2 :
(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2
− 〖 𝑥〗 _0 )
((𝑓(𝑥_3 )−𝑓(𝑥_2 ))/(𝑥_3−𝑥_2 ) −
(𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ))/(𝑥_3
− 〖 𝑥〗 _1 )

3rd FINITE DIVIDED DIFFERENCE

𝑏_3=(𝑓[𝑥_3 , 𝑥_(2 ),𝑥_1]− 𝑓[𝑥_2 , 𝑥_(1 ),𝑥_0])/(𝑥_3 − 〖 𝑥〗 _0

For b3 :
(𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ))/(𝑥_3 − 〖 𝑥〗 _1 ) − ((𝑓(𝑥_2 )−𝑓(𝑥
))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗 _0 ))/(𝑥_3 −

Thus,

𝑓_2 (𝑥)= 〖𝑓 (𝑥 〗 _0)+(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ) (𝑥−𝑥

(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗

𝑓_2 (𝑥)=0.244280552+(1.606770906)(0.5−0.2)+(1.5

𝑓_2 (0.5)=0.8219393
CE PROBLEMS

N 6 – CURVE FITTING AND

OLATING POLYNOMIAL

INTERPOLATION
VIDED DIFFERENCE INTERPOLATIN
POLYNOMIAL

𝑓(𝑥)=𝑥𝑒^𝑥

nction wherein x = 0.50 by interpolation and its true percent relative error using the following
values of x:

i xi
0 0.2
1 0.3
2 0.45
3 0.6

ided Difference Interpolating

ER NEWTON'S INTERPOLATING POLYNOMIAL)

Estimate at 𝑓_1 (𝑥)=𝑏_0+𝑏_1 (𝑥−𝑥_0 )

𝑏_0= 〖𝑓 (𝑥 〗 _0)
x = 0.5 where,

xi f(xi) FIRST 𝑏_1=(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_

𝑓_1 (𝑥)= 〖𝑓 (𝑥 〗 _0)+(𝑓(𝑥_1 )−𝑓(
0.2 0.244280551632034 1.60677090640767
0.3 0.404957642272801 ))/(𝑥_1−𝑥_0 ) (𝑥−𝑥_0 )

WHERE
TRUE VALUE
x y = f(x)
0.5 0.824360635350064

1.60677090640767
ANSWER:
f1(0.5) 0.726311823554335

_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 )
𝑥_0 )
06770906)(0.5−0.2)

RDER NEWTON'S INTERPOLATING POLYNOMIAL)

𝑓
Estimate at
x = 0.5

𝑏_0= 〖𝑓 (𝑥 〗 _0
where,
xi
𝑏_1=(𝑓(𝑥_1 )−𝑓
f(xi) FIRST SECOND

𝑏_2=((𝑓(𝑥_2 )
0.2 0.244280551632034 1.60677090640767 1.59379213964332

〖 𝑥〗 _0 )
0.3 0.404957642272801 2.0052189413185

𝑓_2 (𝑥)= 〖𝑓 (𝑥 〗 _0)+(

(𝑓(𝑥
 𝑏_0= 〖𝑓 (𝑥 〗 _0
where,

𝑏_1=(𝑓(𝑥_1 )−𝑓
𝑏_2=((𝑓(𝑥_2 )
0.45 0.705740483470576 〖 𝑥〗 _0 )
𝑓_2 (𝑥)= 〖𝑓 (𝑥 〗 _0)+(
(𝑓(𝑥

WHERE
TRUE VALUE
x
0.5
1.60677090640767

ANSWER:
2.0052189413185
f2(0.5)

𝑓[𝑥_2 ,
𝑥_(1 ),𝑥_0] 1.59379213964332
=

𝑥_1−𝑥_0 ) (𝑥−𝑥_0 )+((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) −

_0 ))/(𝑥_2 − 〖 𝑥〗 _0 ) (𝑥−𝑥_0 )(𝑥−𝑥_1 )

06)(0.5−0.2)+(1.59379214)(0.5−0.2)(0.5−0.3)

.5)=0.821939352

ER NEWTON'S INTERPOLATING POLYNOMIAL)

Estimate at
x = 0.5

xi f(xi) FIRST SECOND THIRD

0.2 0.244280551632034 1.60677090640767 1.59379213964332 0.834850515666931
0.3 0.404957642272801 2.0052189413185 1.92773234591009
0.45 0.705740483470576 2.58353864509153
0.6 1.09327128023431
 1.60677090640767

2.0052189413185

2.58353864509153

𝑓[𝑥_2 ,
𝑥_(1 ),𝑥_0] 1.59379213964332
=
𝑓[𝑥_2 ,
𝑥_(1 ),𝑥_0] 1.92773234591009
=

𝑥_(1 ),𝑥_0])/(𝑥_3 − 〖 𝑥〗 _0 )=(((𝑓(𝑥_3 )−𝑓(𝑥_2 ))/(𝑥_3−𝑥_2 ) −

− 〖 𝑥〗 _1 ) − ((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) − (𝑓(𝑥_1 )−𝑓(𝑥_0 𝑓[ 〖𝑥 _3 ,𝑥 〗 _
))/(𝑥_2 − 〖 𝑥〗 _0 ))/(𝑥_3 − 〖 𝑥〗 _0 ) 2 , 0.834850515666931
𝑥_(1 ),𝑥_0]
=

𝑓_3 (𝑥)= 〖𝑓 (𝑥 〗 _0)+(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ) (𝑥−𝑥_0 )+

))/(𝑥_2−𝑥_1 ))/(𝑥_3 − 〖 𝑥〗 _1 ) − ((
)−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ) (𝑥−𝑥_0 )+((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) −
0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗 _0 ) (𝑥−𝑥_0 )(𝑥−𝑥_1 ) 𝑓_3 (𝑥)=0.244280552+(1.606770906)(0.5−0.2)+(1.59379214)(0.5−0.
+(1.606770906)(0.5−0.2)+(1.59379214)(0.5−0.2)(0.5−0.3)

𝑓_2 (0.5)=0.821939352
AND

POLATING

the following
𝑥)=𝑏_0+𝑏_1 (𝑥−𝑥_0 )

〗 _0)
_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 )
〖𝑓 (𝑥 〗 _0)+(𝑓(𝑥_1 )−𝑓(𝑥_0
/(𝑥_1−𝑥_0 ) (𝑥−𝑥_0 )

𝑓_2 (𝑥)=𝑏_0+𝑏_1 (𝑥−𝑥_0 )+𝑏_2 (𝑥−𝑥_0 )(𝑥−𝑥_1 )

𝑏_0= 〖𝑓 (𝑥 〗 _0)
where,

𝑏_1=(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 )

𝑏_2=((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) − (𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2
〖 𝑥〗 _0 )
−

𝑓_2 (𝑥)= 〖𝑓 (𝑥 〗 _0)+(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ) (𝑥−𝑥_0 )+((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) −

(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗 _0 ) (𝑥−𝑥_0 )(𝑥−𝑥_1 )
 𝑏_0= 〖𝑓 (𝑥 〗 _0)
where,

𝑏_1=(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 )

𝑏_2=((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) − (𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2
〖 𝑥〗 _0 )
−

𝑓_2 (𝑥)= 〖𝑓 (𝑥 〗 _0)+(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ) (𝑥−𝑥_0 )+((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) −

(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗 _0 ) (𝑥−𝑥_0 )(𝑥−𝑥_1 )

TRUE VALUE
y = f(x)
0.824360635350064

0.821939351932934

𝑓_3 (𝑥)=𝑏_0+𝑏_1 (𝑥−𝑥_0 )+𝑏_2 (𝑥−𝑥

𝑏_0= 〖𝑓 (𝑥 〗 _0)
where,

𝑏_1=(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 )

𝑏_2=((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) − (𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 −
𝑏_3=(((𝑓(𝑥_3 )−𝑓(𝑥_2 ))/(𝑥_3−𝑥_2 ) − (𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ))/(𝑥_3 − 〖 𝑥〗 _
− 〖 𝑥〗 _0 )
𝑓_3 (𝑥)= 〖𝑓 (𝑥 〗 _0)+(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ) (𝑥−𝑥_0 )+((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−
))/(𝑥_3−𝑥_2 ) − (𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ))/(𝑥_3 − 〖 𝑥〗 _1 ) − ((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) −
 𝑏_2=((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) − (𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 −
𝑏_3=(((𝑓(𝑥_3 )−𝑓(𝑥_2 ))/(𝑥_3−𝑥_2 ) − (𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ))/(𝑥_3 − 〖 𝑥〗 _
− 〖 𝑥〗 _0 )
𝑓_3 (𝑥)= 〖𝑓 (𝑥 〗 _0)+(𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ) (𝑥−𝑥_0 )+((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−
))/(𝑥_3−𝑥_2 ) − (𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ))/(𝑥_3 − 〖 𝑥〗 _1 ) − ((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) −

WHERE
TRUE VALUE
x y = f(x)
0.5 0.824360635350064

ANSWER:
f3(0.5) 0.824443903479935

𝑥_0 ))/(𝑥_1−𝑥_0 ) (𝑥−𝑥_0 )+((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) − (𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/( 𝑥_2 − 〖 𝑥〗 _0 ) (𝑥−𝑥_0 )(𝑥−𝑥_1 )+(((
𝑥_1 ))/(𝑥_3 − 〖 𝑥〗 _1 ) − ((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) − (𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗 _0 ))/(𝑥_3 − 〖 𝑥〗 _0 ) (𝑥−𝑥

0.5−0.2)+(1.59379214)(0.5−0.2)(0.5−0.3)+(0.834850516)(0.5−0.2)(0.5−0.3)(0.5−0.45)

𝑓_3 (0.5)=0.824443903
−𝑥_1 ) −
−𝑥_1 ) −

𝑏_1 (𝑥−𝑥_0 )+𝑏_2 (𝑥−𝑥_0 )(𝑥−𝑥_1 )+𝑏_3 (𝑥−𝑥_0 )(𝑥−𝑥_1 )(𝑥−𝑥_2 )

𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗 _0 )

/(𝑥_2−𝑥_1 ))/(𝑥_3 − 〖 𝑥〗 _1 ) − ((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) − (𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗 _0 ))/(𝑥_3

)+((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) − (𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗 _0 ) (𝑥−𝑥_0 )(𝑥−𝑥_1 )+(((𝑓(𝑥_3 )−𝑓(𝑥_2
𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) − (𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗 _0 ))/(𝑥_3 − 〖 𝑥〗 _0 ) (𝑥−𝑥_0 )(𝑥−𝑥_1 )(𝑥−𝑥_2 )
𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗 _0 )
/(𝑥_2−𝑥_1 ))/(𝑥_3 − 〖 𝑥〗 _1 ) − ((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) − (𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗 _0 ))/(𝑥_3

)+((𝑓(𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) − (𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗 _0 ) (𝑥−𝑥_0 )(𝑥−𝑥_1 )+(((𝑓(𝑥_3 )−𝑓(𝑥_2
𝑥_2 )−𝑓(𝑥_1 ))/(𝑥_2−𝑥_1 ) − (𝑓(𝑥_1 )−𝑓(𝑥_0 ))/(𝑥_1−𝑥_0 ))/(𝑥_2 − 〖 𝑥〗 _0 ))/(𝑥_3 − 〖 𝑥〗 _0 ) (𝑥−𝑥_0 )(𝑥−𝑥_1 )(𝑥−𝑥_2 )

𝑥〗 _0 ) (𝑥−𝑥_0 )(𝑥−𝑥_1 )+(((𝑓(𝑥_3 )−𝑓(𝑥_2 ))/(𝑥_3−𝑥_2 ) − (𝑓(𝑥_2 )−𝑓(𝑥_1

𝑥〗 _0 ))/(𝑥_3 − 〖 𝑥〗 _0 ) (𝑥−𝑥_0 )(𝑥−𝑥_1 )(𝑥−𝑥_2 )
𝑥_2 − 〖 𝑥〗 _0 ))/(𝑥_3

)+(((𝑓(𝑥_3 )−𝑓(𝑥_2
0 )(𝑥−𝑥_1 )(𝑥−𝑥_2 )
𝑥_2 − 〖 𝑥〗 _0 ))/(𝑥_3

)+(((𝑓(𝑥_3 )−𝑓(𝑥_2
0 )(𝑥−𝑥_1 )(𝑥−𝑥_2 )
Name: CATAMPONGAN, JOSHUA G.
Course and Section: BSCE-3C
Subject Course: EM 315__NUMERICAL SOLUTIONS TO CE PROBLEMS
Name of Activity: MIDTERM PROJECT

LESSON 6 – C
Method Use: LAGRANGE INTERPOLATING POLYNOMIAL

INTER
LAGRANGE INTER
(SOLUTIONS) :

Using Interpolation

Given the function:

Estimate the value of the function wherein x = 0.5

i
0
1
2
3

b. 1^𝑠𝑡, 2^𝑛𝑑 and 3^𝑟𝑑-order Lagrange Interpolating Polynomial

GENERAL EQUATION:
𝑓_𝑛 (𝑥)=∑_(𝑖=

𝐿_𝑖
Where,

(𝑥)=∏24_█(𝑗
−𝑥_𝑗 )
 (𝑥)=∏24_█(𝑗
−𝑥_𝑗 )

1st LAGRANGE INTERPOLATING POLYNOMIAL

𝑓(𝑥)=𝑥𝑒^𝑥

i
x0 0
x1 1
x2 2
x3 3

FORMULA:

𝑓_1 (𝑥)=(𝑥−𝑥_1)/(𝑥_0−𝑥_1 ) 𝑓(𝑥_

(𝑥−𝑥_0)/(𝑥_1−𝑥_0 ) 𝑓(𝑥_1 )

Substituting the given values…

𝑓_1 (𝑥)=(𝑥−𝑥_1)/(𝑥_0−𝑥_1 ) 𝑓(𝑥

𝑓(𝑥_1 )

𝑓_1 (0.5)=(0.5−0.3)/(0.2−0.3) (0.2442

(0.404957642)

𝑓_1 (0.5)=0.726311824

2nd LAGRANGE INTERPOLATING POLYNOMIAL

𝑓(𝑥)=𝑥𝑒^𝑥

i
x0 0
x1 1
x2 2
x3 3

𝑓_2 (𝑥)=((𝑥−𝑥_1)(𝑥−𝑥_2))/( 〖 (𝑥 〗 _0−

FORMULA:

((𝑥−𝑥_0)(𝑥−𝑥_2))/( 〖 (𝑥 〗 _1−𝑥_0) 〖 (𝑥 〗 _1
〖 (𝑥 〗 _2−𝑥_1)) 𝑓(𝑥_2 )
Substituting the given values…

𝑓_2 (𝑥)=((𝑥−𝑥_1)(𝑥−𝑥_2))/( 〖 (𝑥 〗 _0−𝑥_1) 〖 (

((𝑥−𝑥_0)(𝑥−𝑥_1))/( 〖 (𝑥 〗 _2−𝑥_0) 〖 (𝑥 〗 _2−𝑥

𝑓_2 (0.5)=((0.5−0.3)(0.5−0.45))/((0.2−0.3)(0.2
(0.5−0.2)(0.5−0.3)/(0.45−0.2)(0.45−0.3) (0.7

𝑓_2 (0.5)=0.821939352
 𝑓_2 (𝑥)=((𝑥−𝑥_1)(𝑥−𝑥_2))/( 〖 (𝑥 〗 _0−𝑥_1) 〖 (
((𝑥−𝑥_0)(𝑥−𝑥_1))/( 〖 (𝑥 〗 _2−𝑥_0) 〖 (𝑥 〗 _2−𝑥

𝑓_2 (0.5)=((0.5−0.3)(0.5−0.45))/((0.2−0.3)(0.2
(0.5−0.2)(0.5−0.3)/(0.45−0.2)(0.45−0.3) (0.7

𝑓_2 (0.5)=0.821939352

3rd LAGRANGE INTERPOLATING POLYNOMIAL

𝑓(𝑥)=𝑥𝑒^𝑥

i
x0 0
x1 1
x2 2
x3 3

FORMULA:

𝑓_3 (𝑥)=((𝑥−𝑥_1)(𝑥−𝑥_2)(𝑥−𝑥_3))/( 〖 (𝑥
〖 (𝑥 〗 _1−𝑥_2) 〖 (𝑥 〗 _1−𝑥_3)) 𝑓(𝑥_1 ) + ((𝑥
( 〖 (𝑥 〗 _3−𝑥_0) 〖 (𝑥 〗 _3−𝑥_1) 〖 (𝑥 〗 _3−𝑥_2))
Substituting the given values…

𝑓_3 (𝑥)=((𝑥−𝑥_1)(𝑥−𝑥_2)(𝑥−𝑥_3))/( 〖 (𝑥 〗 _0−𝑥_

((𝑥−𝑥_0)(𝑥−𝑥_1)(𝑥−𝑥_3))/( 〖 (𝑥 〗 _2−𝑥_0) 〖 (𝑥 〗 _2−𝑥

𝑓_3 (0.5)=((0.5−0.3)(0.5−0.45)(0.5−0.6))/((0.2−0.3)(

𝑓_3 (0.5)=0.824443903
CE PROBLEMS

N 6 – CURVE FITTING AND

INTERPOLATION
GE INTERPOLATING POLYNOMIAL

𝑓(𝑥)=𝑥𝑒^𝑥

nction wherein x = 0.50 by interpolation and its true percent relative error using the following
values of x:

i xi
0 0.2
1 0.3
2 0.45
3 0.6

rpolating Polynomial

𝑓_𝑛 (𝑥)=∑_(𝑖=0)^𝑛▒ 〖𝐿 _𝑖 (𝑥)𝑓(𝑥_𝑖) 〗

𝐿_𝑖
Where,

(𝑥)=∏24_█(𝑗=0@𝑗≠𝑖)^𝑛▒(𝑥−𝑥_𝑗)/(𝑥_𝑖
−𝑥_𝑗 )
 (𝑥)=∏24_█(𝑗=0@𝑗≠𝑖)^𝑛▒(𝑥−𝑥_𝑗)/(𝑥_𝑖
−𝑥_𝑗 )

POLYNOMIAL

Estimate at
x = 0.5

i xi f(xi) WHERE
0 0.2 0.244280551632034 TRUE VALUE
1 0.3 0.404957642272801 x y = f(x)
2 0.45 0.705740483470576 0.5 0.824360635350064
3 0.6 1.09327128023431

ANSWER:

−𝑥_1)/(𝑥_0−𝑥_1 ) 𝑓(𝑥_0 )+
f1(0.5) 0.726311823554335
1−𝑥_0 ) 𝑓(𝑥_1 )

𝑥−𝑥_1)/(𝑥_0−𝑥_1 ) 𝑓(𝑥_0 )+ (𝑥−𝑥_0)/(𝑥_1−𝑥_0 )

0.5−0.3)/(0.2−0.3) (0.244280552)+ (0.5−0.2)/(0.3−0.2)

642)

726311824

POLYNOMIAL

Estimate at
x = 0.5

i xi f(xi)
0 0.2 0.244280551632034
1 0.3 0.404957642272801
2 0.45 0.705740483470576
3 0.6 1.09327128023431

𝑥_1)(𝑥−𝑥_2))/( 〖 (𝑥 〗 _0−𝑥_1) 〖 (𝑥 〗 _0−𝑥_2)) 𝑓(𝑥_0 ) +

2))/( 〖 (𝑥 〗 _1−𝑥_0) 〖 (𝑥 〗 _1−𝑥_2)) 𝑓(𝑥_1 ) + ((𝑥−𝑥_0)(𝑥−𝑥_1))/( 〖 (𝑥 〗 _2−𝑥_0)
𝑓(𝑥_2 )

(𝑥−𝑥_2))/( 〖 (𝑥 〗 _0−𝑥_1) 〖 (𝑥 〗 _0−𝑥_2)) 𝑓(𝑥_0 ) + ((𝑥−𝑥_0)(𝑥−𝑥_2))/( 〖 (𝑥 〗 _1−𝑥_0) 〖 (𝑥 〗 _1−𝑥_2)) 𝑓(𝑥_1 ) +

)/( 〖 (𝑥 〗 _2−𝑥_0) 〖 (𝑥 〗 _2−𝑥_1)) 𝑓(𝑥_2 )

0.3)(0.5−0.45))/((0.2−0.3)(0.2−0.45)) (0.244280552) + (0.5−0.2)(0.5−0.45)/(0.3−0.2)(0.3−0.45) (0.404957642) +

)/(0.45−0.2)(0.45−0.3) (0.705740483)

39352
 (𝑥−𝑥_2))/( 〖 (𝑥 〗 _0−𝑥_1) 〖 (𝑥 〗 _0−𝑥_2)) 𝑓(𝑥_0 ) + ((𝑥−𝑥_0)(𝑥−𝑥_2))/( 〖 (𝑥 〗 _1−𝑥_0) 〖 (𝑥 〗 _1−𝑥_2)) 𝑓(𝑥_1 ) +
)/( 〖 (𝑥 〗 _2−𝑥_0) 〖 (𝑥 〗 _2−𝑥_1)) 𝑓(𝑥_2 )

0.3)(0.5−0.45))/((0.2−0.3)(0.2−0.45)) (0.244280552) + (0.5−0.2)(0.5−0.45)/(0.3−0.2)(0.3−0.45) (0.404957642) +

)/(0.45−0.2)(0.45−0.3) (0.705740483)

39352

POLYNOMIAL

Estimate at
x = 0.5 WHERE
TRUE VALUE
i xi f(xi) x
0 0.2 0.244280551632034 0.5
1 0.3 0.404957642272801
2 0.45 0.705740483470576
3 0.6 1.09327128023431 ANSWER:
f3(0.5)

1)(𝑥−𝑥_2)(𝑥−𝑥_3))/( 〖 (𝑥 〗 _0−𝑥_1) 〖 (𝑥 〗 _0−𝑥_2) 〖 (𝑥 〗 _0−𝑥_3)) 𝑓(𝑥_0 ) + ((𝑥−𝑥_0)(𝑥−𝑥_2)(𝑥−𝑥_3))/( 〖 (𝑥 〗 _1−𝑥_0)

〗 _1−𝑥_3)) 𝑓(𝑥_1 ) + ((𝑥−𝑥_0)(𝑥−𝑥_1)(𝑥−𝑥_3))/( 〖 (𝑥 〗 _2−𝑥_0) 〖 (𝑥 〗 _2−𝑥_1) 〖 (𝑥 〗 _2−𝑥_3)) 𝑓(𝑥_2 )+((𝑥−𝑥_0)(𝑥−𝑥_1)(𝑥
𝑥 〗 _3−𝑥_1) 〖 (𝑥 〗 _3−𝑥_2)) 𝑓(𝑥_3 )

_2)(𝑥−𝑥_3))/( 〖 (𝑥 〗 _0−𝑥_1) 〖 (𝑥 〗 _0−𝑥_2) 〖 (𝑥 〗 _0−𝑥_3)) 𝑓(𝑥_0 ) + ((𝑥−𝑥_0)(𝑥−𝑥_2)(𝑥−𝑥_3))/( 〖 (𝑥 〗 _1−𝑥_0) 〖 (𝑥 〗 _1

( 〖 (𝑥 〗 _2−𝑥_0) 〖 (𝑥 〗 _2−𝑥_1) 〖 (𝑥 〗 _2−𝑥_3)) 𝑓(𝑥_2 )+((𝑥−𝑥_0)(𝑥−𝑥_1)(𝑥−𝑥_2))/( 〖 (𝑥 〗 _3−𝑥_0) 〖 (𝑥 〗 _3−𝑥_1) 〖 (𝑥 〗 _3−𝑥_

.5−0.45)(0.5−0.6))/((0.2−0.3)(0.2−0.45)(0.2−0.6)) (0.244280552)+ (0.5−0.2)(0.5−0.45)(0.5−0.6)/(0.3−0.2)(0.3−0.45)(0.3−0.6) (0.404957

(0.705740483)+(0.5−0.2)(0.5−0.3)(0.5−0.45)/(0.6−0.2)(0.6−0.3)(0.6−0.45) (1.09327128
AND

OMIAL

the following
VALUE
y = f(x)
0.824360635350064

0.726311823554335

WHERE
TRUE VALUE
x y = f(x)
0.5 0.824360635350064

ANSWER:
f2(0.5) 0.821939351932934

)) 𝑓(𝑥_1 ) +

404957642) +
)) 𝑓(𝑥_1 ) +

404957642) +

TRUE VALUE
x y = f(x)
0.5 0.824360635350064

ANSWER:
f3(0.5) 0.824443903479935

2)(𝑥−𝑥_3))/( 〖 (𝑥 〗 _1−𝑥_0)
𝑓(𝑥_2 )+((𝑥−𝑥_0)(𝑥−𝑥_1)(𝑥−𝑥_2))/

3))/( 〖 (𝑥 〗 _1−𝑥_0) 〖 (𝑥 〗 _1−𝑥_2) 〖 (𝑥 〗 _1−𝑥_3)) 𝑓(𝑥_1 ) +

〖 (𝑥 〗 _3−𝑥_1) 〖 (𝑥 〗 _3−𝑥_2)) 𝑓(𝑥_3 )

.3−0.45)(0.3−0.6) (0.404957642)+ (0.5−0.2)(0.5−0.3)(0.5−0.6)/(0.45−0.2)(0.45−0.3)(0.45−0.6)

3)(0.6−0.45) (1.09327128)
Name: CATAMPONGAN, JOSHUA G.
Course and Section: BSCE-3C
Subject Course: EM 315__NUMERICAL SOLUTIONS TO CE PROBLEMS
Name of Activity: MIDTERM PROJECT

LESSON 6 – C
Method Use: LINEAR SPLINE INTERPOLATION

INTER
LINEAR SPLI
(SOLUTIONS) :

Using Interpolation

Given the function:

Estimate the value of the function wherein x = 0.5

i
0
1
2
3

b. Linear Spline Interpolation

LINEAR SPLINE INTERPOLATION

𝑓(𝑥)=𝑥𝑒^𝑥

i
0
1
2
3

INTERVAL
0.45≤𝑥 ≥0.6

0.45≤ 0.5 ≥0.6

Therefore,

𝑓_1 (𝑥)=𝑓(𝑥_0 )+𝑚_𝑖

𝑓_1 (𝑥)=𝑓(𝑥_0 )+(𝑓(𝑥_(𝑖+

𝑓_1 (0.5)=0.705740483

𝑓_1 (0.5)=
CE PROBLEMS

N 6 – CURVE FITTING AND

INTERPOLATION
EAR SPLINE INTERPOLATION

𝑓(𝑥)=𝑥𝑒^𝑥

nction wherein x = 0.50 by interpolation and its true percent relative error using the following
values of x:

i xi
0 0.2
1 0.3
2 0.45
3 0.6
INTERPOLATION

𝑥)=𝑥𝑒^𝑥
Estimate at
x = 0.5

i xi f(xi)
0 0.2 0.244280551632034
1 0.3 0.404957642272801
2 0.45 0.705740483470576
3 0.6 1.09327128023431

SLOPE, m
0.45≤𝑥 ≥0.6
𝑚_𝑖=(𝑓(𝑥_(𝑖+1) )
0.45≤ 0.5 ≥0.6 −𝑓(𝑥_𝑖 ))/(𝑥_(𝑖+1)−𝑥_𝑖 )

m = 2.58353864509153

𝑓_1 (𝑥)=𝑓(𝑥_0 )+𝑚_𝑖 (𝑥−𝑥_0)

𝑓_1 (𝑥)=𝑓(𝑥_0 )+(𝑓(𝑥_(𝑖+1) )−𝑓(𝑥_𝑖 ))/(𝑥_(𝑖+1)−𝑥_𝑖 ) 𝑥−𝑥_0)

𝑓_1 (0.5)=0.705740483+(2.583538645)(0.5−0.45)

𝑓_1 (0.5)=

f1(0.5) = 0.834917415725152
AND

the following
𝑓(𝑥_(𝑖+1) )
))/(𝑥_(𝑖+1)−𝑥_𝑖 )

2.58353864509153
Name: CATAMPONGAN, JOSHUA G.
Course and Section: BSCE-3C
Subject Course: EM 315__NUMERICAL SOLUTIONS TO CE PROBLEMS
Name of Activity: MIDTERM PROJECT

LESSON 6 – C
Method Use: QUADRATIC SPLINE INTERPOLATION

INTER
QUADRATIC SP
(SOLUTIONS) :

Using Interpolation

Given the function:

Estimate the value of the function wherein x = 0.5

i
0
1
2
3

b. Quadratic Spline Interpolation

QUADRATIC SPLINE INTERPOLATION

𝑓(𝑥)=𝑥𝑒^𝑥

i
0
1
2
3

INTERVAL
0.2≤𝑥 ≥0.3

0.3≤𝑥 ≥0.45

0.45≤𝑥 ≥0.6

1ST STEP.

2ND STEP.
 0.2≤𝑥 ≥0.3

0.3≤𝑥 ≥0.45

0.45≤𝑥 ≥0.6

3RD STEP.

4TH STEP.

5TH STEP.

GENERAL EQUATIONS:

𝑎_1=0
0.04 𝑎_1+0.2 𝑏_1+ 𝑐_1=0.244280552

0.09 𝑎_1+0.3 𝑏_1+ 𝑐_1=0.404957642

0.09 𝑎_2+0.3 𝑏_2+ 𝑐_2=0.404957642

𝑎_1=0
0.04 𝑎_1+0.2 𝑏_1+ 𝑐_1=0.244280552

0.09 𝑎_1+0.3 𝑏_1+ 𝑐_1=0.404957642

0.09 𝑎_2+0.3 𝑏_2+ 𝑐_2=0.404957642

0.2025 𝑎_2+0.45 𝑏_2+ 𝑐_2=0.707540483

0.2025 𝑎_3+0.45 𝑏_3+ 𝑐_3=0.707540483

0.36 𝑎_3+0.6 𝑏_3+ 𝑐_3=1.09327128

0.6𝑎_1+𝑏_1−0.6 𝑎_2−𝑏_2=0

0.9𝑎_2+𝑏_2−0.9𝑎_3−𝑏_3=0
CE PROBLEMS

N 6 – CURVE FITTING AND

INTERPOLATION
RATIC SPLINE INTERPOLATION

𝑓(𝑥)=𝑥𝑒^𝑥

nction wherein x = 0.50 by interpolation and its true percent relative error using the following
values of x:

i xi
0 0.2
1 0.3
2 0.45
3 0.6
NE INTERPOLATION

𝑥)=𝑥𝑒^𝑥
Estimate at
x = 0.5

i xi f(xi)
0 0.2 0.244280551632034
1 0.3 0.404957642272801
2 0.45 0.705740483470576
3 0.6 1.09327128023431

0.2≤𝑥 ≥0.3

0.3≤𝑥 ≥0.45

0.45≤𝑥 ≥0.6

𝑓(𝑥)=𝑎𝑥^2+𝑏𝑥+𝑐
1ST STEP.

𝑛+1=4
𝑛=3 (3 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠)

3𝑛=3(3)
3𝑛=9 (9 𝑢𝑛𝑘𝑛𝑜𝑤𝑛𝑠)

2ND STEP.
2𝑛−2=2(3)−2
2𝑛−2=4 (4 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠) CONDITIONS

𝑎_(𝑖−1) 〖𝑥 _(𝑖−1) 〗 ^2+𝑏_(𝑖−1)

𝑥_(𝑖−1)+𝑐_(𝑖−1)= 〖𝑓 (𝑥 〗 _(𝑖−1))

𝑎_𝑖 〖𝑥 _(𝑖−1) 〗 ^2+𝑏_𝑖 𝑥_(𝑖−1)+𝑐_𝑖= 〖𝑓 (𝑥 〗

_(𝑖−1))
 𝑎_(𝑖−1) 〖𝑥 _(𝑖−1) 〗 ^2+𝑏_(𝑖−1)
𝑥_(𝑖−1)+𝑐_(𝑖−1)= 〖𝑓 (𝑥 〗 _(𝑖−1))

𝑎_𝑖 〖𝑥 _(𝑖−1) 〗 ^2+𝑏_𝑖 𝑥_(𝑖−1)+𝑐_𝑖= 〖𝑓 (𝑥 〗

_(𝑖−1))
0.2≤𝑥 ≥0.3
0.04 a1 + 0.2 b1 + c1 = 0.244280551632034
0.09 a1 + 0.3 b1 + c2 = 0.404957642272801

0.3≤𝑥 ≥0.45
0.09 a2 + 0.3 b2 + c2 = 0.404957642272801
0.2025 a2 + 0.45 b2 + c2 = 0.705740483470576

0.45≤𝑥 ≥0.6
0.2025 a3 + 0.45 b3 + c3 = 0.705740483470576
0.36 a3 + 0.6 b3 + c3 = 1.09327128023431

3RD STEP.
CONDITIONS

𝑎_1 〖𝑥 _0 〗 ^2+𝑏_1 𝑥_0+𝑐_0= 〖𝑓 (𝑥 〗

_0)

𝑎_𝑛 〖𝑥 _𝑛 〗 ^2+𝑏_𝑛 𝑥_𝑛+𝑐_𝑛= 〖𝑓 (𝑥 〗 _𝑛)

0.04 a1 + 0.2 b1 + c1 = 0.244280551632034

0.36 a3 + 0.6 b3 + c3 = 1.09327128023431

4TH STEP.
𝑛−1=3−1
𝑛−1=2 (2 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠) CONDITION (FIRST DERIVATIVE OF INTERIOR KNOTS)
𝑓^′ (𝑥)=2𝑎𝑥+𝑏=2𝑎_(𝑖−1)
𝑥_(𝑖−1)+𝑏_(𝑖−1)

1ST DERIVATIVE OF QUADRATIC SPLINE

2𝑎_1 𝑥_1+𝑏_1=2𝑎_2 𝑥_2+𝑏_2

at x = 0.3
0.6 a1 + b1 = 0.6 a2 + b2
at x = 0.45
0.9 a2 + b2 = 0.9 a3 + b3

5TH STEP.

𝑎_1=0
MATRIX FORM:

b1 c1

44280552

957642

57642
44280552

[
957642
0.2 1
57642 0.3 1
0 0
.707540483
0 0
07540483 0 0
27128
0 0
1 0
0 0

Matrix Product:

[
-10 10
3 -2
66.66667 -66.66667
-50 50
9 -9
-66.66667 66.66667
70 -70
-18 18

0.2≤𝑥 ≥0.3 𝑓_2 (𝑥)=𝑎_1 𝑥^2+𝑏_1

0.3≤𝑥 ≥0.45 𝑓_2 (𝑥)=𝑎_2 𝑥^2+𝑏_2

0.45≤𝑥 ≥0.6 𝑓_2 (𝑥)=𝑎_3 𝑥^2+𝑏_3

Therefore,
AND

ON

the following
) 〗 ^2+𝑏_(𝑖−1)
(𝑥 〗 _(𝑖−1))

_𝑖 𝑥_(𝑖−1)+𝑐_𝑖= 〖𝑓 (𝑥 〗
) 〗 ^2+𝑏_(𝑖−1)
(𝑥 〗 _(𝑖−1))

_𝑖 𝑥_(𝑖−1)+𝑐_𝑖= 〖𝑓 (𝑥 〗

0.244280551632034
0.404957642272801

0.404957642272801
0.705740483470576

0.705740483470576
1.09327128023431

0.244280551632034
1.09327128023431

ATIVE OF INTERIOR KNOTS)

=2𝑎_(𝑖−1)

a2 b2 c2 a3 b3 c3

[𝑨
]
 ]{
[𝑨
0 0 ] 0 0 0 0
0 0 0 0 0 0
0.09 0.3 1 0 0 0
0.2025 0.45 1 0 0 0
0 0 0 0.2025 0.45 1
0 0 0 0.36 0.6 1
-0.6 -1 0 0 0 0
0.9 1 0 -0.9 -1 0

{𝑋}= 〖 [𝐴] 〗 ^(−1)

×{𝐵}
〖 [𝐴]

]{
0 0 〗 ^(−
0 0 0 0
0 0 0 0 0 0
1)
-44.44444 44.44444 0 0 6.66666666667 0
26.66667 -26.66667 0 0 -5 0
-3 4 0 0 0.9 0
88.88889 -88.88889 -44.44444 44.444444 -6.6666666667 6.6666666667
-93.33333 93.33333 40 -40 7 -7
24 -24 -8 9 -1.8 1.8

𝑓_2 (𝑥)=𝑎_1 𝑥^2+𝑏_1 𝑥+𝑐_1

= 0 x2 + 1.6067709064 x + -0.0770736296

𝑓_2 (𝑥)=𝑎_2 𝑥^2+𝑏_2 𝑥+𝑐_2

= 2.6563202 x2 + 0.0129787668 x + 0.1619951913

𝑓_2 (𝑥)=𝑎_3 𝑥^2+𝑏_3 𝑥+𝑐_3

= 1.1991445 x2 + 1.3244369631 x + -0.1330829029

Therefore,
𝑓_2 (𝑥)=1.199144459
𝑥^2+1.324436963𝑥−0.133082903
f2(5) = 0.82892169343
{𝑿 {𝑩
} }
 =
{} { }
{𝑿 {𝑩
}b
1
}
0.244280551632034
c1 0.404957642272801
a2 0.404957642272801
b2 0.705740483470576
c2 0.705740483470576
a3 1.09327128023431
b3 0
c3 0

=
{} { }
{𝑿 {𝑩
}b
1
}
1.60677090640767
c1 -0.0770736296495
a2 2.65632023273887
b2 0.012978766764348
c2 0.161995191296998
a3 1.19914445908132
b3 1.32443696305614
c3 -0.133082902868656
}
}
Name: CATAMPONGAN, JOSHUA G.
Course and Section: BSCE-3C
Subject Course: EM 315__NUMERICAL SOLUTIONS TO CE PROBLEMS
Name of Activity: MIDTERM PROJECT

LESSON 6 – CURV
Method Use: CUBIC SPLINE INTERPOLATION

CUBIC
(SOLUTIONS) :

Using Interpolation

Given the function:

Estimate the value of the function wherein x =

i
0
1
2
3

b. Cubic Spline Interpolation

CUBIC SPLINE INTERPOLATION

𝑓(𝑥)=𝑥𝑒^𝑥

i
0
1
2
3

INTERVAL
0.2≤𝑥 ≥0.3

0.3≤𝑥 ≥0.45

0.45≤𝑥 ≥0.6

1ST STEP.

2ND STEP.

0.2≤𝑥 ≥0.3
 0.3≤𝑥 ≥0.45

0.45≤𝑥 ≥0.6

3RD STEP.

4TH STEP.

5TH STEP.
 6TH STEP.

GENERAL EQUATIONS:

0.008 𝑎_1+0.04 𝑏_1+0.2 𝑐_1+

𝑑_1=0.244280552

0.027 𝑎_1+0.09 𝑏_1+0.3 𝑐_1+

𝑑_1=0.404957642

0.027 𝑎_2+0.09 𝑏_2+0.3 𝑐_2+

𝑑_2=0.404957642

0.091125 𝑎_2+0.2025 𝑏_2+0.45 𝑐_2+

𝑑_2=0.707540483

0.091125 𝑎_3+0.2025 𝑏_3+0.45 𝑐_3+

𝑑_3=0.707540483

0.216 𝑎_3+0.36 𝑏_3+0.6 𝑐_3+ 𝑑_3=1.09327128

0.27𝑎_1+0.6𝑏_1+𝑐_1−0.27𝑎_2−0.6 𝑏_2−𝑐_2=0

0.6075𝑎_2+0.9𝑏_2+𝑐_2−0.6 〖 075𝑎 〗 _3−0.9𝑏_3

−𝑐_3=0

1.8𝑎_1+2𝑏_1−1.8𝑎_2−2𝑏_2=0

2.7𝑎_2+2𝑏_2−2.7𝑎_3− 〖 2𝑏 〗 _3=0

1.2𝑎_1+2𝑏_1=0

3.6𝑎_3+2𝑏_3=0
CE PROBLEMS

CURVE FITTING AND INTERP

CUBIC SPLINE INTERPOLATION

𝑓(𝑥)=𝑥𝑒^𝑥

he function wherein x = 0.50 by interpolation and its true percent relative error using the following valu

i xi
0 0.2
1 0.3
2 0.45
3 0.6
NTERPOLATION

𝑥)=𝑥𝑒^𝑥
Estimate at
x = 0.5

i xi f(xi)
0 0.2 0.244280551632034
1 0.3 0.404957642272801
2 0.45 0.705740483470576
3 0.6 1.09327128023431

0.2≤𝑥 ≥0.3

0.3≤𝑥 ≥0.45

0.45≤𝑥 ≥0.6

𝑓(𝑥)=𝑎𝑥^3+𝑏𝑥^2+𝑐𝑥+𝑑
1ST STEP.

𝑛+1=4
𝑛=3 (3 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠)

4𝑛=4(3)
4𝑛=12 (12 𝑢𝑛𝑘𝑛𝑜𝑤𝑛𝑠)

2ND STEP.
2𝑛−2=2(3)−2
2𝑛−2=4 (4 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠) CONDITIONS
𝑎_(𝑖−1) 〖𝑥 _(𝑖−1) 〗 ^3+𝑏_(𝑖−1)
〖𝑥 _(𝑖−1) 〗 ^2+𝑐_(𝑖−1)
𝑥+𝑑_(𝑖−1)= 〖𝑓 (𝑥 〗 _(𝑖−1))

𝑎_𝑖 〖𝑥 _(𝑖−1) 〗 ^3+𝑏_𝑖 〖𝑥 _(𝑖−1) 〗 ^2+𝑐_

𝑥+𝑑_𝑖= 〖𝑓0.2
(𝑥c〗1 _(𝑖−1))
0.2≤𝑥 ≥0.3
0.008 a1 + 0.04 b1 + +
0.027 a1 + 0.09 b1 + 0.3 c1 +
 0.3≤𝑥 ≥0.45
0.027 a2 + 0.09 b2 + 0.3 c2 +
0.091125 a2 + 0.2025 b2 + 0.45 c2 +

0.45≤𝑥 ≥0.6
0.091125 a3 + 0.2025 b3 + 0.45 c3 +
0.216 a3 + 0.36 b3 + 0.6 c3 +

3RD STEP.
CONDITIONS

𝑎_1 〖𝑥 _0 〗 ^3+𝑏_1 〖𝑥 _0 〗 ^2+𝑐_1

𝑥_0+𝑑_0= 〖𝑓 (𝑥 〗 _0)

𝑎_𝑛 〖𝑥 _𝑛 〗 ^3+𝑏_𝑛 〖𝑥 _𝑛 〗 ^2+𝑐_𝑛

𝑥_𝑛+𝑑_𝑛= 〖𝑓 (𝑥 〗 _𝑛)
0.008 a1 + 0.04 b1 + 0.2 c1 +
0.216 a3 + 0.36 b3 + 0.6 c3 +

4TH STEP.
𝑛−1=3−1
𝑛−1=2 (2 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠) CONDITION (FIRST DERIVATIVE OF

𝑓^′ (𝑥)=3𝑎𝑥^2+2𝑏𝑥+𝑐

1ST DERIVATIVE OF CUBIC SPLINE

3𝑎_1 𝑥^2+ 〖 2𝑏 〗 _1 𝑥+𝑐_1=3𝑎_2

𝑥^2+ 〖 2𝑏 〗 _2 𝑥+𝑐_2
at x = 0.3
0.27 a1 + 0.6 b1 + c1 = 0.27 a2 +
at x = 0.45
0.6075 a2 + 0.9 b2 + c2 = 0.6075 a3 +

5TH STEP.
𝑛−1=3−1
𝑛−1=2 (2 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠) CONDITION (SECOND DERIVATIVE OF

𝑓^′′ (𝑥)=6𝑎𝑥+2𝑏

2ND DERIVATIVE OF CUBIC SPLINE

6𝑎_1 𝑥+2𝑏_1=0
at x = 0.3
1.8 a1 + 2 b1 = 1.8 a2 + 2
at x = 0.45
 2.7 a2 + 2 b2 = 2.7 a3 + 2

6TH STEP.
𝑛−1=3−1
𝑛−1=2 (2 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠) CONDITION (SECOND DERIVATIVE
𝑓^′′ (𝑥)=6𝑎𝑥+2𝑏

2ND DERIVATIVE OF CUBIC SPLINE

6𝑎_1 𝑥+2𝑏_1=0

at x = 0.2
1.2 a1 + 2 b1 = 0
at x = 0.6
3.6 a3 + 2 b3 = 0

MATRIX FORM:

2 𝑐_1+

[
a1 b1

3 𝑐_1+

3 𝑐_2+
0.008 0.04
0.027 0.09

_2+0.45 𝑐_2+
0 0
0 0

_3+0.45 𝑐_3+
0 0
0 0
0.27 0.6
6 𝑐_3+ 𝑑_3=1.09327128 0 0

.27𝑎_2−0.6 𝑏_2−𝑐_2=0
1.8 2
0 0
−0.6 〖 075𝑎 〗 _3−0.9𝑏_3 1.2 2
0 0

𝑏_2=0

2𝑏 〗 _3=0

Matrix Product:
[
216.216216216 -216.21621622
-129.72972973 129.72972973
13.7837837838 -13.783783784
1.7027027027 -0.7027027027
-180.18018018 180.18018018
227.027027027 -227.02702703
-93.243243243 93.2432432432
12.4054054054 -12.405405405
36.036036036 -36.036036036
-64.864864865 64.8648648649
38.1081081081 -38.108108108
-7.2972972973 7.2972972973

0.2≤𝑥 ≥0.3

0.3≤𝑥 ≥0.45

0.45≤𝑥 ≥0.6
NTERPOLATION
ION

the following values of x:

−1) 〗 ^3+𝑏_(𝑖−1)
(𝑖−1)
𝑥 〗 _(𝑖−1))

+𝑏_𝑖 〖𝑥 _(𝑖−1) 〗 ^2+𝑐_𝑖

𝑖−1)) d1 = 0.244280551632034
d2 = 0.404957642272801
 d2 = 0.404957642272801
d2 = 0.705740483470576

d3 = 0.705740483470576
d3 = 1.09327128023431

d1 = 0.244280551632034
d3 = 1.09327128023431

N (FIRST DERIVATIVE OF INTERIOR KNOTS)

+2𝑏𝑥+𝑐

0.6 b2 + c3

0.9 b3 + c3

SECOND DERIVATIVE OF INTERIOR KNOTS)

b2
 b3

N (SECOND DERIVATIVE OF END KNOTS)

𝑏

ORM:

c1 d1 a2 b2 c2 d2

[𝑨
0.2 1 0 0 ]0 0
0.3 1 0 0 0 0
0 0 0.027 0.09 0.3 1
0 0 0.091125 0.2025 0.45 1
0 0 0 0 0 0
0 0 0 0 0 0
1 0 -0.27 -0.6 -1 0
0 0 0.6075 0.9 1 0
0 0 -1.8 -2 0 0
0 0 2.7 2 0 0
0 0 0 0 0 0
0 0 0 0 0 0

{𝑋}= 〖 [𝐴] 〗 ^(−1) ×{𝐵}

 {𝑋}= 〖 [𝐴] 〗 ^(−1) ×{𝐵}
〖 [𝐴]
〗 ^(−1
-180.18018018 180.18018018 36.036036036 -36.036036036 21.62162162162 -5.405405405406
108.108108108 -108.10810811 -21.6216216216 21.6216216216 -12.972972973 3.2432432432434
) 2.378378378378 -0.594594594595
-19.81981982 19.8198198198 3.96396396396 -3.963963964
1.08108108108 -1.0810810811 -0.21621621622 0.21621621622 -0.12972972973 0.0324324324324
224.224224224 -224.22422422 -104.104104104 104.104104104 -18.018018018 15.615615615616
-255.85585586 255.855855856 104.504504505 -104.5045045 22.7027027027 -15.67567567568
89.3693693694 -89.369369369 -33.8738738739 33.8738738739 -9.32432432432 5.0810810810812
-8.8378378378 9.83783783784 3.56756756757 -3.5675675676 1.240540540541 -0.535135135135
-104.1041041 104.104104104 80.0800800801 -80.08008008 3.603603603604 -12.01201201201
187.387387387 -187.38738739 -144.144144144 144.144144144 -6.48648648649 21.621621621622
-110.09009009 110.09009009 78.018018018 -78.018018018 3.810810810811 -12.7027027027
21.0810810811 -21.081081081 -12.2162162162 13.2162162162 -0.72972972973 2.4324324324325

0.2≤𝑥 ≥0.3 𝑓_3 (𝑥)=𝑎_1 𝑥^3+𝑏_1

𝑥^2+𝑐_1 𝑥+𝑑_1 = 5.4890401938934

0.3≤𝑥 ≥0.45 𝑓_3 (𝑥)=𝑎_2 𝑥^3+𝑏_2 𝑥^2+𝑐_2

𝑥+𝑑_2 = 1.8515743247892

0.45≤𝑥 ≥0.6 𝑓_3 (𝑥)=𝑎_3 𝑥^3+𝑏_3

𝑥^2+𝑐_3 𝑥+𝑑_3 = -5.510934454051

Therefore,
𝑓_2 (𝑥)=0.540972099 𝑥^2+1.294501247
𝑥^2+0.773953063𝑥+0.04602902
f2(5) =
 ]{
a3 b3 c3 d3

0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0.091125 0.2025 0.45 1
0.216 0.36 0.6 1
0 0 0 0
-0.6075 -0.9 -1 0
0 0 0 0
-2.7 -2 0 0
0 0 0 0
3.6 2 0 0
 ]{
0.945945945946 -0.27027027027 -2.027027027027 0.135135135135
-0.567567567568 0.162162162162 1.7162162162162 -0.081081081081
0.104054054054 -0.02972972973 -0.472972972973 0.014864864865
-0.005675675676 0.001621621622 0.0421621621622 -0.000810810811
0.600600600601 0.780780780781 0.3003003003003 -0.39039039039
-0.756756756757 -0.783783783784 -0.3783783783784 0.391891891892
0.310810810811 0.254054054054 0.1554054054054 -0.127027027027
-0.041351351351 -0.026756756757 -0.0206756756757 0.013378378378
-0.12012012012 0.510510510511 -0.0600600600601 1.411411411411
0.216216216216 -0.918918918919 0.1081081081081 -2.040540540541
-0.127027027027 0.539864864865 -0.0635135135135 0.967567567568
0.024324324324 -0.103378378378 0.0121621621622 -0.150810810811

x3 + -3.293424116336 x2 + 2.210565327736 x + -0.11000787081

x3 + -0.019704834142 x2 + 1.228449543078 x + -0.01179629235

x3 + 9.919682017293 x2 + -3.244274540068 x + 0.65911232012

099 𝑥^2+1.294501247
𝑥+0.04602902
0.828028747658
{ }{ }
{𝑿 {𝑩
}a }

=
1 0.244280551632034
b1 0.404957642272801
c1 0.404957642272801
d1 0.705740483470576
a2 0.705740483470576
b2 1.09327128023431
c2 0
d2 0
a3 0
b3 0
c3 0
d3 0
{ }{ }
{𝑿 {𝑩

=
} a
1
}
5.48904019389336
b1 -3.293424116336
c1 2.21056532773595
d1 -0.110007870812862
a2 1.85157432478926
b2 -0.019704834142331
c2 1.22844954307784
d2 -0.011796292347051
a3 -5.51093445405149
b3 9.91968201729269
c3 -3.24427454006791
d3 0.659112320124814
}
}
Name: CATAMPONGAN, JOSHUA G.
Course and Section: BSCE-3C
Subject Course: EM 315__NUMERICAL SOLUTIONS TO CE PROBLEMS
Name of Activity: MIDTERM PROJECT

LESSON 7 – NUME
Method Use: FINITE-DIVIDED DIFFERENCE APPROXIMATION OF DERIVATIV

FINITE-DIVIDED
(SOLUTIONS) :

Numerical Differentiation

Given the function:

Find the 1st, 2nd, 3rd, and 4th der

i
0
1
2
3

a. Using Finite-Divided Difference Approximation of Derivatives.

i. Forward (Accuracy = 1)
ii. Backward (Accuracy = 1)
iii. Centered (Accuracy = 2)
 FORWARD - FINITE DIVIDED DIFFERENCE

1ST DERIVATIVE OF THE FUNCTION

FORMULA:

𝑓(𝑥)=𝑥 sin⁡𝑥

i
xi-5 -5
xi-4 -4
xi-3 -3
xi-2 -2
xi-1 -1
xi 0
xi+1 1
xi+2 2
xi+3 3
xi+4 4
xi+5 5

THEREFORE,
 BACKWARD - FINITE DIVIDED DIFFERENCE

1ST DERIVATIVE OF THE FUNCTION

FORMULA:

𝑓(𝑥)=𝑥 sin⁡𝑥

i
xi-5 -5
xi-4 -4
xi-3 -3
xi-2 -2
xi-1 -1
xi 0
xi+1 1
xi+2 2
xi+3 3
xi+4 4
xi+5 5

THEREFORE,

CENTERED - FINITE DIVIDED DIFFERENCE

1ST DERIVATIVE OF THE FUNCTION

FORMULA:

𝑓(𝑥)=𝑥 sin⁡𝑥
 𝑓(𝑥)=𝑥 sin⁡𝑥

i
xi-5 -5
xi-4 -4
xi-3 -3
xi-2 -2
xi-1 -1
xi 0
xi+1 1
xi+2 2
xi+3 3
xi+4 4
xi+5 5

THEREFORE,
CE PROBLEMS

– NUMERICAL DIFFERENTIAT
TION OF DERIVATIVES

INTEGRATION
DIVIDED DIFFERENCE APPROXIMATI
DERIVATIVES

𝑓(𝑥)=𝑥 sin⁡𝑥

1st, 2nd, 3rd, and 4th derivatives of the function wherein x = π/4 and interval h = π/12 and its true
percent relative error.

i xi
0 0.2
1 0.3
2 0.45
3 0.6

on of Derivatives.
DIVIDED DIFFERENCE

FORMULA:

𝑥)=𝑥 sin⁡𝑥
Estimate at
x = 0.785398163397448 h = 0.261799387799149
x = 0.785398163397448
i xi f(xi)
-5 -0.5235987755983 0.261799387799149
1ST DERIVATIVE OF THE FUNCTION
-4 -0.26179938779915 0.0677586675586004
-3 0 0 𝑓(𝑥)=𝑥 sin⁡𝑥

(𝑑(𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
-2 0.26179938779915 0.0677586675586004 Product Rule of Differentiation:
-1 0.523598775598299 0.261799387799149
0 0.785398163397448 0.555360367269796 𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥
1 1.0471975511966 0.906899682117109
2 1.30899693899575 1.26439394990933
3 1.5707963267949 1.5707963267949 TRUE VALUE 1.26246714845634
4 1.83259571459405 1.77015152987306
5 2.0943951023932 1.81379936423422

THEREFORE,
𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥
PERCENT RELATIVE ERROR:
f'(π/4) = 1.34278127157811 εT = -6.36168024015287
 DIVIDED DIFFERENCE

FORMULA:

𝑥)=𝑥 sin⁡𝑥
Estimate at
x = 0.785398163397448 h = 0.261799387799149
x = 0.785398163397448
i xi f(xi)
-5 -0.5235987755983 0.261799387799149
1ST DERIVATIVE OF THE FUNCTION

𝑓(𝑥)=𝑥 sin⁡𝑥
-4 -0.26179938779915 0.0677586675586004
-3 0 0

(𝑑(𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
-2 0.26179938779915 0.0677586675586004 Product Rule of Differentiation:
-1 0.523598775598299 0.261799387799149
0 0.785398163397448 0.555360367269796 𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥
1 1.0471975511966 0.906899682117109
2 1.30899693899575 1.26439394990933
3 1.5707963267949 1.5707963267949 TRUE VALUE 1.26246714845634
4 1.83259571459405 1.77015152987306
5 2.0943951023932 1.81379936423422

THEREFORE,
𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥
PERCENT RELATIVE ERROR:
f'(π/4) = 1.12132034355964 εT = 11.1802358635062

DIVIDED DIFFERENCE

FORMULA:

𝑥)=𝑥 sin⁡𝑥
Estimate at
𝑥)=𝑥 sin⁡𝑥
x = 0.785398163397448 h = 0.261799387799149
x = 0.785398163397448
i xi f(xi)
-5 -0.5235987755983 0.261799387799149
1ST DERIVATIVE OF THE FUNCTION

𝑓(𝑥)=𝑥 sin⁡𝑥
-4 -0.26179938779915 0.0677586675586004
-3 0 0

(𝑑(𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
-2 0.26179938779915 0.0677586675586004 Product Rule of Differentiation:
-1 0.523598775598299 0.261799387799149
0 0.785398163397448 0.555360367269796 𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥
1 1.0471975511966 0.906899682117109
2 1.30899693899575 1.26439394990933
3 1.5707963267949 1.5707963267949 TRUE VALUE 1.26246714845634
4 1.83259571459405 1.77015152987306
5 2.0943951023932 1.81379936423422

THEREFORE,
𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥
PERCENT RELATIVE ERROR:
f'(π/4) = 1.23205080756888 εT = 2.40927781167669
NTIATION AND

XIMATION OF

and its true

 2ND DERIVATIVE OF THE FUNCTION

𝑓(𝑥)=𝑥 sin⁡𝑥
0.261799387799149
0.785398163397448

xi-5
THE FUNCTION
xi-4
xi-3
erentiation: xi-2
))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
xi-1
os⁡𝑥 xi
xi+1
xi+2
1.26246714845634 xi+3
xi+4
xi+5

THEREFORE,

-6.36168024015287
 2ND DERIVATIVE OF THE FUNCTION

𝑓(𝑥)=𝑥 sin⁡𝑥
0.261799387799149
0.785398163397448

xi-5
THE FUNCTION
xi-4
xi-3
erentiation: xi-2
))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
xi-1
os⁡𝑥 xi
xi+1
xi+2
1.26246714845634 xi+3
xi+4
xi+5

THEREFORE,

11.1802358635062

2ND DERIVATIVE OF THE FUNCTION

𝑓(𝑥)=𝑥 sin⁡𝑥
 𝑓(𝑥)=𝑥 sin⁡𝑥
0.261799387799149
0.785398163397448

xi-5
THE FUNCTION
xi-4
xi-3
erentiation: xi-2
))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
xi-1
os⁡𝑥 xi
xi+1
xi+2
1.26246714845634 xi+3
xi+4
xi+5

THEREFORE,

2.40927781167669
 FORMULA:

(𝑥)=𝑥 sin⁡𝑥
Estimate at
x = 0.785398163397448

i xi f(xi)
-5 -0.523598775598299 0.261799387799149
-4 -0.261799387799149 0.0677586675586004
-3 0 0
-2 0.26179938779915 0.0677586675586004
-1 0.523598775598299 0.261799387799149
0 0.785398163397448 0.555360367269796
1 1.0471975511966 0.906899682117109
2 1.30899693899575 1.26439394990933
3 1.5707963267949 1.5707963267949
4 1.83259571459405 1.77015152987306
5 2.0943951023932 1.81379936423422

THEREFORE,
𝑓^′′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥
⁡

f''(π/4) = 0.086884254851373
 FORMULA:

(𝑥)=𝑥 sin⁡𝑥
Estimate at
x = 0.785398163397448

i xi f(xi)
-5 -0.523598775598299 0.261799387799149
-4 -0.261799387799149 0.0677586675586004
-3 0 0
-2 0.26179938779915 0.0677586675586004
-1 0.523598775598299 0.261799387799149
0 0.785398163397448 0.555360367269796
1 1.0471975511966 0.906899682117109
2 1.30899693899575 1.26439394990933
3 1.5707963267949 1.5707963267949
4 1.83259571459405 1.77015152987306
5 2.0943951023932 1.81379936423422

THEREFORE,
𝑓^′′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥
⁡

f''(π/4) = 1.45202550646835

FORMULA:

(𝑥)=𝑥 sin⁡𝑥
Estimate at
(𝑥)=𝑥 sin⁡𝑥
x = 0.785398163397448

i xi f(xi)
-5 -0.523598775598299 0.261799387799149
-4 -0.261799387799149 0.0677586675586004
-3 0 0
-2 0.26179938779915 0.0677586675586004
-1 0.523598775598299 0.261799387799149
0 0.785398163397448 0.555360367269796
1 1.0471975511966 0.906899682117109
2 1.30899693899575 1.26439394990933
3 1.5707963267949 1.5707963267949
4 1.83259571459405 1.77015152987306
5 2.0943951023932 1.81379936423422

THEREFORE,
𝑓^′′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥
⁡

f''(π/4) = 0.845918433500587
0.785398163397448 h = 0.261799387799149
x = 0.785398163397448

2ND DERIVATIVE OF THE FUNCTION

𝑓(𝑥)=𝑥 sin⁡𝑥
𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥

(𝑑(𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
Product Rule of Differentiation:

𝑓^′′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥

⁡

TRUE VALUE 0.858853195103299

PERCENT RELATIVE ERROR:

εT = 89.8836896285956
0.785398163397448 h = 0.261799387799149
x = 0.785398163397448

2ND DERIVATIVE OF THE FUNCTION

𝑓(𝑥)=𝑥 sin⁡𝑥
𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥

(𝑑(𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
Product Rule of Differentiation:

𝑓^′′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥

⁡

TRUE VALUE 0.858853195103299

PERCENT RELATIVE ERROR:

εT = -69.0656231759962
0.785398163397448 h = 0.261799387799149
x = 0.785398163397448

2ND DERIVATIVE OF THE FUNCTION

𝑓(𝑥)=𝑥 sin⁡𝑥
𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥

(𝑑(𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
Product Rule of Differentiation:

𝑓^′′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥

⁡

TRUE VALUE 0.858853195103299

PERCENT RELATIVE ERROR:

εT = 1.50605035603975
3RD DERIVATIVE OF THE FUNCTION

FORMULA:

𝑓(𝑥)=𝑥 sin⁡𝑥
Estimate at
x = 0.785398163397448

i xi f(xi)
xi-5 -5 -0.5235987755983 0.261799387799149
xi-4 -4 -0.26179938779915 0.0677586675586
xi-3 -3 0 0
xi-2 -2 0.26179938779915 0.0677586675586
xi-1 -1 0.523598775598299 0.261799387799149
xi 0 0.785398163397448 0.555360367269796
xi+1 1 1.0471975511966 0.906899682117109
xi+2 2 1.30899693899575 1.26439394990933
xi+3 3 1.5707963267949 1.5707963267949
xi+4 4 1.83259571459405 1.77015152987306
xi+5 5 2.0943951023932 1.81379936423422

THEREFORE,
𝑓^′′′ (𝑥)= 〖− 3 sin⁡𝑥−𝑥 〗⁡cos⁡𝑥

f'''(π/4) = -3.17925777389845
3RD DERIVATIVE OF THE FUNCTION

FORMULA:

𝑓(𝑥)=𝑥 sin⁡𝑥
Estimate at
x = 0.785398163397448

i xi f(xi)
xi-5 -5 -0.5235987755983 0.261799387799149
xi-4 -4 -0.26179938779915 0.0677586675586
xi-3 -3 0 0
xi-2 -2 0.26179938779915 0.0677586675586
xi-1 -1 0.523598775598299 0.261799387799149
xi 0 0.785398163397448 0.555360367269796
xi+1 1 1.0471975511966 0.906899682117109
xi+2 2 1.30899693899575 1.26439394990933
xi+3 3 1.5707963267949 1.5707963267949
xi+4 4 1.83259571459405 1.77015152987306
xi+5 5 2.0943951023932 1.81379936423422

THEREFORE,
𝑓^′′′ (𝑥)= 〖− 3 sin⁡𝑥−𝑥 〗⁡cos⁡𝑥

f'''(π/4) = -1.49145218439533

3RD DERIVATIVE OF THE FUNCTION

FORMULA:

𝑓(𝑥)=𝑥 sin⁡𝑥
Estimate at
 𝑓(𝑥)=𝑥 sin⁡𝑥
x = 0.785398163397448

i xi f(xi)
xi-5 -5 -0.5235987755983 0.261799387799149
xi-4 -4 -0.26179938779915 0.0677586675586
xi-3 -3 0 0
xi-2 -2 0.26179938779915 0.0677586675586
xi-1 -1 0.523598775598299 0.261799387799149
xi 0 0.785398163397448 0.555360367269796
xi+1 1 1.0471975511966 0.906899682117109
xi+2 2 1.30899693899575 1.26439394990933
xi+3 3 1.5707963267949 1.5707963267949
xi+4 4 1.83259571459405 1.77015152987306
xi+5 5 2.0943951023932 1.81379936423422

THEREFORE,
𝑓^′′′ (𝑥)= 〖− 3 sin⁡𝑥−𝑥 〗⁡cos⁡𝑥

f'''(π/4) = -3.17925777389845
 4TH DERIVATIVE OF THE FUNCTION

0.785398163397448 h = 0.261799387799149
x = 0.785398163397448

xi-5
3RD DERIVATIVE OF THE FUNCTION

𝑓(𝑥)=𝑥 sin⁡𝑥
xi-4

𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥

xi-3
𝑓^′′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥
⁡ xi-2
xi-1
(𝑑(𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
Product Rule of Differentiation:
xi

𝑓^′′′ (𝑥)= 〖− 3 sin⁡𝑥−𝑥 〗⁡cos⁡𝑥

xi+1
xi+2
xi+3
xi+4
xi+5
TRUE VALUE -2.67668071082944

PERCENT RELATIVE ERROR:

εT = -18.7761304901127
 4TH DERIVATIVE OF THE FUNCTION

0.785398163397448 h = 0.261799387799149
x = 0.785398163397448

xi-5
3RD DERIVATIVE OF THE FUNCTION

𝑓(𝑥)=𝑥 sin⁡𝑥
xi-4

𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥

xi-3
𝑓^′′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥
⁡ xi-2
xi-1
(𝑑(𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
Product Rule of Differentiation:
xi

𝑓^′′′ (𝑥)= 〖− 3 sin⁡𝑥−𝑥 〗⁡cos⁡𝑥

xi+1
xi+2
xi+3
xi+4
xi+5
TRUE VALUE -2.67668071082944

PERCENT RELATIVE ERROR:

εT = 44.2797873365641

4TH DERIVATIVE OF THE FUNCTION

0.785398163397448 h = 0.261799387799149
x = 0.785398163397448

xi-5
3RD DERIVATIVE OF THE FUNCTION

𝑓(𝑥)=𝑥 sin⁡𝑥
xi-4

𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥

xi-3
𝑓^′′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥
⁡ xi-2
xi-1
(𝑑(𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
Product Rule of Differentiation:
xi

𝑓^′′′ (𝑥)= 〖− 3 sin⁡𝑥−𝑥 〗⁡cos⁡𝑥

xi+1
xi+2
xi+3
xi+4
xi+5
TRUE VALUE -2.67668071082944

PERCENT RELATIVE ERROR:

εT = -18.7761304901127
OF THE FUNCTION

FORMULA:

𝑓(𝑥)=𝑥 sin⁡𝑥
Estimate at
x = 0.785398163397448 h =
x =
i xi f(xi)
xi-5 -5 -0.5235987755983 0.261799387799149
4TH DERIVATIVE OF THE FUNCTION
-4 -0.26179938779915 0.0677586675586
𝑓(𝑥)=𝑥 sin⁡𝑥
xi-4

𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥

xi-3 -3 0 0

𝑓^′′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥

xi-2 -2 0.26179938779915 0.0677586675586
𝑓^′′′ (𝑥)= 〖− 3 sin⁡𝑥−𝑥 〗⁡cos
⁡
xi-1 -1 0.523598775598299 0.261799387799149
0 0.785398163397448 0.555360367269796
(𝑑(𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+
xi Product Rule of Differentiati
xi+1 1 1.0471975511966 0.906899682117109

𝑓^((4)) (𝑥)= 〖− 4 cos⁡𝑥+

xi+2 2 1.30899693899575 1.26439394990933
xi+3 3 1.5707963267949 1.5707963267949
xi+4 4 1.83259571459405 1.77015152987306
xi+5 5 2.0943951023932 1.81379936423422
TRUE VALUE
THEREFORE,
𝑓^((4)) (𝑥)= 〖− 4 cos⁡𝑥+𝑥 〗⁡sin⁡𝑥
PERCENT RELATIVE ERROR:
f(4)(π/4) = 0.232366482795009 εT =
OF THE FUNCTION

FORMULA:

𝑓(𝑥)=𝑥 sin⁡𝑥
Estimate at
x = 0.785398163397448 h =
x =
i xi f(xi)
xi-5 -5 -0.5235987755983 0.261799387799149
4TH DERIVATIVE OF THE FUNCTION

𝑓(𝑥)=𝑥 sin⁡𝑥
xi-4 -4 -0.26179938779915 0.0677586675586

𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥

xi-3 -3 0 0

𝑓^′′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥

xi-2 -2 0.26179938779915 0.0677586675586
𝑓^′′′ (𝑥)= 〖− 3 sin⁡𝑥−𝑥 〗⁡cos
⁡
xi-1 -1 0.523598775598299 0.261799387799149
0 0.785398163397448 0.555360367269796
(𝑑(𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+
xi Product Rule of Differentiati
xi+1 1 1.0471975511966 0.906899682117109

𝑓^((4)) (𝑥)= 〖− 4 cos⁡𝑥+

xi+2 2 1.30899693899575 1.26439394990933
xi+3 3 1.5707963267949 1.5707963267949
xi+4 4 1.83259571459405 1.77015152987306
xi+5 5 2.0943951023932 1.81379936423422
TRUE VALUE
THEREFORE,
𝑓^((4)) (𝑥)= 〖− 4 cos⁡𝑥+𝑥 〗⁡sin⁡𝑥
PERCENT RELATIVE ERROR:
f(4)(π/4) = -3.73096318405578 εT =

OF THE FUNCTION

FORMULA:

𝑓(𝑥)=𝑥 sin⁡𝑥
Estimate at
 𝑓(𝑥)=𝑥 sin⁡𝑥
x = 0.785398163397448 h =
x =
i xi f(xi)
xi-5 -5 -0.5235987755983 0.261799387799149
4TH DERIVATIVE OF THE FUNCTION

𝑓(𝑥)=𝑥 sin⁡𝑥
xi-4 -4 -0.26179938779915 0.0677586675586

𝑓^′ (𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥

xi-3 -3 0 0

𝑓^′′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥

xi-2 -2 0.26179938779915 0.0677586675586
𝑓^′′′ (𝑥)= 〖− 3 sin⁡𝑥−𝑥 〗⁡cos
⁡
xi-1 -1 0.523598775598299 0.261799387799149
0 0.785398163397448 0.555360367269796
(𝑑(𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+
xi Product Rule of Differentiati
xi+1 1 1.0471975511966 0.906899682117109

𝑓^((4)) (𝑥)= 〖− 4 cos⁡𝑥+

xi+2 2 1.30899693899575 1.26439394990933
xi+3 3 1.5707963267949 1.5707963267949
xi+4 4 1.83259571459405 1.77015152987306
xi+5 5 2.0943951023932 1.81379936423422
TRUE VALUE
THEREFORE,
𝑓^((4)) (𝑥)= 〖− 4 cos⁡𝑥+𝑥 〗⁡sin⁡𝑥
PERCENT RELATIVE ERROR:
f(4)(π/4) = -2.23124477164447 εT =
 0.261799387799149
0.785398163397448

IVATIVE OF THE FUNCTION

𝑥)=𝑥 sin⁡𝑥
(𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥
′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥
′′ (𝑥)= 〖− 3 sin⁡𝑥−𝑥 〗⁡cos⁡𝑥
⁡

𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
duct Rule of Differentiation:

((4)) (𝑥)= 〖− 4 cos⁡𝑥+𝑥 〗⁡sin⁡𝑥

-2.27306675747639

ATIVE ERROR:
110.22259825985
 0.261799387799149
0.785398163397448

IVATIVE OF THE FUNCTION

𝑥)=𝑥 sin⁡𝑥
(𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥
′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥
′′ (𝑥)= 〖− 3 sin⁡𝑥−𝑥 〗⁡cos⁡𝑥
⁡

𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
duct Rule of Differentiation:

((4)) (𝑥)= 〖− 4 cos⁡𝑥+𝑥 〗⁡sin⁡𝑥

-2.27306675747639

ATIVE ERROR:
-64.1378622860146
 0.261799387799149
0.785398163397448

IVATIVE OF THE FUNCTION

𝑥)=𝑥 sin⁡𝑥
(𝑥)=sin⁡𝑥+𝑥 cos⁡𝑥
′ (𝑥)= 〖 2 cos⁡𝑥−xsin 〗𝑥
′′ (𝑥)= 〖− 3 sin⁡𝑥−𝑥 〗⁡cos⁡𝑥
⁡

𝑢𝑣))/𝑑𝑥=(𝑢(𝑑𝑣))/𝑑𝑥+(𝑣(𝑑𝑢))/𝑑𝑥
duct Rule of Differentiation:

((4)) (𝑥)= 〖− 4 cos⁡𝑥+𝑥 〗⁡sin⁡𝑥

-2.27306675747639

ATIVE ERROR:
1.83989254580262