ANANDMUNAGALA

Mr. ANAND MUNAGALA

175 Reputation

9 Badges

11 years, 6 days
MRIET
Associate Professor in Mathematics
I have more than a decade of teaching experience in Mathematics for various Undergraduate programs in reputed educational institutions in Hyderabad, Andhra Pradesh, INDIA. Currently working as Assistant Professor in Mathematics and as Managed Network expert for Chegg E-Learning services private limited, Kolkatta, INDIA, for two years for solution authoring and content creation in Mathematics.

MaplePrimes Activity


These are replies submitted by ANANDMUNAGALA

@Carl Love 

 

Thank you sir

@Carl Love 

 

Dear sir, thank you very much for your swift response in this regard.

Can you provide the constarints clearly in algebgraic format to solve manually using Simplex method. As it need to be solved by showing the constraints.

 

With thanks & regards

 

Mr M ANAND

@Rouben Rostamian  

 

Dear sir, thank you very much for your kind and swift response to my query.

 

With best regards.

 

Mr M ANAND

@rlopez 

 

Thank you very much sir. I tried it many times manually and now it is very clear.

Dear sir,

 

It is exactly what I am looking for.Thanks for your kind and swift response.

Can I expect some more examples with squares and other cross-sections ?. Kindly support in this regard,

 

 

With best regards,

 

Mr M ANAND

@Kitonum

Good evening sir, the maple commands you suggested is very excellent and thank you very much for your kind & swift response.

 

With thanks & regards.

M.Anand

@vv 

 

Thank you very much sir.

@Kitonum 

 

That is very exact sir. Thank you very much for your kind support to my query.

 

With thanks & regards.

 

Mr.M.Anand

@Kitonum 

 

Thank you very much for your kind & swift response sir. If we need to plot piecewise line segments in 3D, what is the appropriate procedure? could you please help in this?.

 

With thanks & regards.

 

Mr.M.Anand

Associate Professor in Mathematics

@Carl Love 

 

Thank you very much for your swift response to my query.

 

With thanks & regards.

 

M.ANAND

@Carl Love 

 

Good morning sir.

 

I submitted this procedure to get the necessary changes so that the new procedure can be applied to find iterated inverse interpolation.

 

If it is not needed then I extremely apologize for sending this code.

 

With thanks & regards.

 

M.ANAND

@Carl Love 

 

I here by submitting you the Maple procedure for iterated  interpolation by Necille's method for your kind notice.

 

> restart;
> # NEVILLE'S ITERATED INTERPOLATION ALGORITHM
> #
> # To evaluate the interpolating polynomial P on the
> # (n+1) distinct numbers x(0), ..., x(n) at the number x
> # for the function f:
> #
> # INPUT:   numbers x(0),..., x(n) as XX(0),...,XX(N);
> #          number x; values of f as the first column of Q
> #          or may be computed if function f is supplied.
> #
> # OUTPUT:  the table Q with P(x) = Q(N+1,N+1).
> alg031 := proc() local TRUE, FALSE, OK, FLAG, N, I, XX, Q, A, NAME, INP, F, X, D, J, OUP;
> printf(`This is Neville's Method.\n`);
> OK := FALSE;
> while OK = FALSE do
> printf(`Choice of input method:\n`);
> printf(`1. Input entry by entry from keyboard\n`);
> printf(`2. Input data from a text file\n`);
> printf(`3. Generate data using a function F\n`);
> printf(`Choose 1, 2, or 3 please\n`);
> FLAG := scanf(`%d`)[1];
> if FLAG = 1 or FLAG = 2 or FLAG = 3 then
> OK := TRUE;
> fi;
> od;
> if FLAG = 1 then
> OK := FALSE;
> while OK <> TRUE do
> printf(`Input n\n`);
> N := scanf(`%d`)[1];
> if N > 0 then
> OK := TRUE;
> for I from 0 to N do
> printf(`Input X(%d) and F(X(%d)) `, I, I);
> printf(`separated by a space\n`);
> XX[I] := scanf(`%f`)[1];
> Q[I,0] := scanf(`%f`)[1];
> od;
> else
> printf(`Number must be a positive integer\n`);
> fi;
> od;
> fi;
> if FLAG = 2 then
> printf(`Has a text file been created with the data in two columns ?\n`);
> printf(`Enter Y or N\n`);
> A := scanf(`\n%c`)[1];
> if A = "Y" or A = "y" then
> printf(`Input the file name in the form - `);
> printf(`drive:\\name.ext\n`);
> printf(`For example:   A:\\DATA.DTA\n`);
> NAME := scanf(`%s`)[1];
> INP := fopen(NAME,READ,TEXT);
> OK := FALSE;
> while OK = FALSE do
> printf(`Input N\n`);
> N := scanf(`%d`)[1];
> if N > 0 then
> for I from 0 to N do
> XX[I] := fscanf(INP, `%f`)[1];
> Q[I,0] := fscanf(INP, `%f`)[1];
> od;
> fclose(INP);
> OK := TRUE;
> else
> printf(`Number must be a positive integer\n`);
> fi;
> od;
> else
> printf(`Please create the input file in two column `);
> printf(`form with the X values and\n`);
> printf(`F(X) values in the corresponding columns.\n`);
> printf(`The program will end so the input file can `);
> printf(`be created.\n`);
> OK := FALSE;
> fi;
> fi;
> if FLAG = 3 then
> printf(`Input the function F(x) in terms of x\n`);
> printf(`For example: cos(x)\n`);
> F := scanf(`%a`)[1];
> F := unapply(F,x);
> OK := FALSE;
> while OK = FALSE do
> printf(`Input n\n`);
> N := scanf(`%d`)[1];
> if N > 0 then
> for I from 0 to N do
> printf(`Input X(%d)\n`, I);
> XX[I] := scanf(`%f`)[1];
> Q[I,0] := F(XX[I]);
> od;
> OK := TRUE;
> else
> printf(`Number must be a positive integer\n`);
> fi;
> od;
> fi;
> if OK = TRUE then
> printf(`Input the point at which the polynomial is to be evaluated\n`);
> X := scanf(`%f`)[1];
> fi;
> if OK = TRUE then
> # Step 1
> D[0] := evalf(X-XX[0]);
> for I from 1 to N do
> D[I] := evalf(X-XX[I]);
> for J from 1 to I do
> Q[I,J] := evalf((D[I]*Q[I-1,J-1]-D[I-J]*Q[I,J-1])/(D[I]-D[I-J]));
> od;
> od;
> # Step 2
> printf(`Select output destination\n`);
> printf(`1. Screen\n`);
> printf(`2. Text file\n`);
> printf(`Enter 1 or 2\n`);
> FLAG := scanf(`%d`)[1];
> if FLAG = 2 then
> printf(`Input the file name in the form - drive:\\name.ext\n`);
> printf(`For example:   A:\\OUTPUT.DTA\n`);
> NAME := scanf(`%s`)[1];
> OUP := fopen(NAME,WRITE,TEXT);
> else
> OUP := default;
> fi;
> fprintf(OUP, `NEVILLE'S METHOD\n`);
> fprintf(OUP, `Table for P evaluated at X = %12.8f , follows: \n`, X);
> fprintf(OUP, `Entries are XX(I), Q(I,0), ..., Q(I,I) `);
> fprintf(OUP, `for each I = 0, ..., N where N = %3d\n\n`, N);
> for I from 0 to N do
> fprintf(OUP, `%11.8f `, XX[I]);
> for J from 0 to I do
> fprintf(OUP, `%11.8f `, Q[I,J]);
> od;
> fprintf(OUP, `\n`);
> od;
> if OUP <> default then
> fclose(OUP);
> printf(`Output file %s created successfully`,NAME);
> fi;
> fi;
> RETURN(0);
> end;
> alg031();

 

 

With thanks & regards.

 

M.ANAND

 

@Carl Love 

 

Please find the enclosed Maple file.Powerfit.mw

with(Statistics):

X := Vector([0.17e-1, 0.87e-1, .174, 1.11, 1.74, 4.09, 5.45, 5.96, 0.25e-1, .111, .211, .999, 3.02, 4.28, 4.58, 4.68, 0.2e-1, 0.85e-1, .171, 1.29, 3.04, 4.29, 5.30, 0.20e-1, .119, .210, 1.32, 3.34, 5.48, 0.25e-1, .233, .783, 1.35, 1.69, 2.75, 4.83, 5.53], datatype = float)

X := Vector(37, {(1) = 0.17e-1, (2) = 0.87e-1, (3) = .174, (4) = 1.11, (5) = 1.74, (6) = 4.09, (7) = 5.45, (8) = 5.96, (9) = 0.25e-1, (10) = .111, (11) = .211, (12) = .999, (13) = 3.02, (14) = 4.28, (15) = 4.58, (16) = 4.68, (17) = 0.2e-1, (18) = 0.85e-1, (19) = .171, (20) = 1.29, (21) = 3.04, (22) = 4.29, (23) = 5.3, (24) = 0.2e-1, (25) = .119, (26) = .21, (27) = 1.32, (28) = 3.34, (29) = 5.48, (30) = 0.25e-1, (31) = .233, (32) = .783, (33) = 1.35, (34) = 1.69, (35) = 2.75, (36) = 4.83, (37) = 5.53}, datatype = float[8])

(1)

Y := Vector([.154, .296, .363, .531, 2.23, 3.58, 3.52, 2.40, .23, .357, .366, .771, 2.01, 3.28, 2.96, 5.10, .181, .260, .334, .87, 3.59, 3.40, 3.88, .180, .299, .428, 1.15, 2.83, 4.15, .234, .537, 1.47, 2.48, 1.44, 1.84, 4.66, 6.94], datatype = float)

Y := Vector(37, {(1) = .154, (2) = .296, (3) = .363, (4) = .531, (5) = 2.23, (6) = 3.58, (7) = 3.52, (8) = 2.4, (9) = .23, (10) = .357, (11) = .366, (12) = .771, (13) = 2.01, (14) = 3.28, (15) = 2.96, (16) = 5.1, (17) = .181, (18) = .26, (19) = .334, (20) = .87, (21) = 3.59, (22) = 3.4, (23) = 3.88, (24) = .18, (25) = .299, (26) = .428, (27) = 1.15, (28) = 2.83, (29) = 4.15, (30) = .234, (31) = .537, (32) = 1.47, (33) = 2.48, (34) = 1.44, (35) = 1.84, (36) = 4.66, (37) = 6.94}, datatype = float[8])

(2)

R := PowerFit(X, Y, t)

HFloat(1.3029717779462269)*t^HFloat(0.5756426027724945)

(3)

R__1 := evalf(subs(t = 0.17e-1, R))

HFloat(0.12482627780112172)

(4)

R__2 := evalf(subs(t = 0.87e-1, R))

HFloat(0.31950459726743996)

(5)

R__3 := evalf(subs(t = .174, R))

HFloat(0.47617084589354475)

(6)

R__4 := evalf(subs(t = 1.110, R))

HFloat(1.383645536941253)

(7)

R__5 := evalf(subs(t = 1.740, R))

HFloat(1.7922781261482401)

(8)

R__6 := evalf(subs(t = 4.090, R))

HFloat(2.931358980304867)

(9)

R__7 := evalf(subs(t = 5.450, R))

HFloat(3.458089333871749)

(10)

R__8 := evalf(subs(t = 5.960, R))

HFloat(3.640824825820044)

(11)

R__9 := evalf(subs(t = 0.25e-1, R))

HFloat(0.15585510278516226)

(12)

R__10 := evalf(subs(t = .111, R))

HFloat(0.36760570590574204)

(13)

R__11 := evalf(subs(t = .211, R))

HFloat(0.5320631765027012)

(14)

R__12 := evalf(subs(t = .211, R))

HFloat(0.5320631765027012)

(15)

R__13 := evalf(subs(t = 3.020, R))

HFloat(2.4617698876348713)

(16)

R__14 := evalf(subs(t = 4.280, R))

HFloat(3.00899138140134)

(17)

R__15 := evalf(subs(t = 4.580, R))

HFloat(3.128652530858454)

(18)

R__16 := evalf(subs(t = 4.680, R))

HFloat(3.167795054518516)

(19)

R__17 := evalf(subs(t = 0.20e-1, R))

HFloat(0.13706781627398337)

(20)

R__18 := evalf(subs(t = 0.85e-1, R))

HFloat(0.3152556894134705)

(21)

R__19 := evalf(subs(t = .171, R))

HFloat(0.4714274792895653)

(22)

R__20 := evalf(subs(t = 1.290, R))

HFloat(1.5086731380097755)

(23)

R__21 := evalf(subs(t = 3.040, R))

HFloat(2.4711415075963346)

(24)

R__22 := evalf(subs(t = 4.290, R))

HFloat(3.013036348458928)

(25)

R__23 := evalf(subs(t = 5.300, R))

HFloat(3.402977329111567)

(26)

R__24 := evalf(subs(t = 0.20e-1, R))

HFloat(0.13706781627398337)

(27)

R__25 := evalf(subs(t = .1190, R))

HFloat(0.38263126697923644)

(28)

R__26 := evalf(subs(t = .2100, R))

HFloat(0.5306101579436261)

(29)

R__27 := evalf(subs(t = 1.3200, R))

HFloat(1.5287712284729584)

(30)

R__28 := evalf(subs(t = 3.3400, R))

HFloat(2.608709912722286)

(31)

R__29 := evalf(subs(t = 5.4800, R))

HFloat(3.4690341300545136)

(32)

R__30 := evalf(subs(t = 0.250e-1, R))

HFloat(0.15585510278516226)

(33)

R__31 := evalf(subs(t = .2330, R))

HFloat(0.5633238319331533)

(34)

R__32 := evalf(subs(t = .7830, R))

HFloat(1.1318262269649904)

(35)

R__33 := evalf(subs(t = 1.3500, R))

HFloat(1.5486763948651998)

(36)

R__34 := evalf(subs(t = 1.6900, R))

HFloat(1.762447970110909)

(37)

R__35 := evalf(subs(t = 2.7500, R))

HFloat(2.3325643000863123)

(38)

R__36 := evalf(subs(t = 4.8300, R))

HFloat(3.225849602641803)

(39)

R__37 := evalf(subs(t = 5.5300, R))

HFloat(3.487219117025346)

(40)

Y__1 := .154

.154

(41)

Y__2 := .296

.296

(42)

Y__3 := .363

.363

(43)

Y__4 := .531

.531

(44)

Y__5 := 2.230

2.230

(45)

Y__6 := 3.580

3.580

(46)

Y__7 := 3.520

3.520

(47)

Y__8 := 2.400

2.400

(48)

Y__9 := .230

.230

(49)

Y__10 := .357

.357

(50)

Y__11 := .366

.366

(51)

Y__12 := .771

.771

(52)

Y__13 := 2.010

2.010

(53)

Y__14 := 3.280

3.280

(54)

Y__15 := 2.960

2.960

(55)

Y__16 := 5.100

5.100

(56)

Y__17 := .181

.181

(57)

Y__18 := .260

.260

(58)

Y__19 := .334

.334

(59)

Y__20 := .870

.870

(60)

Y__21 := 3.590

3.590

(61)

Y__22 := 3.400

3.400

(62)

Y__23 := 3.880

3.880

(63)

Y__24 := .180

.180

(64)

Y__25 := .299

.299

(65)

Y__26 := .428

.428

(66)

Y__27 := 1.150

1.150

(67)

Y__28 := 2.830

2.830

(68)

Y__29 := 4.150

4.150

(69)

Y__30 := .234

.234

(70)

Y__31 := .537

.537

(71)

Y__32 := 1.470

1.470

(72)

Y__33 := 2.480

2.480

(73)

Y__34 := 1.440

1.440

(74)

Y__35 := 1.840

1.840

(75)

Y__36 := 4.660

4.660

(76)

Y__37 := 6.940

6.940

(77)

E := sum((Y__n-R__n)^2, n = 1 .. 37)

37*(Y__n-R__n)^2

(78)

``


Download Powerfit.mw

@Kitonum 

 

 

Good evening sir.

 

Thank you very much for your kind suggestion.

 

With thanks & regards.

 

M.ANAND

@Carl Love 

 

I apologize for this, I resubmitting the file please find the same.

Problem.docx

1 2 3 4 5 6 7 Last Page 2 of 10