maple2015

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@Carl Love 

The above ODE is the motion equation of the one degree of freedom system for lumped mass connected to the viscous medium. The parameter C denotes damping coefficient. I want to know more about behavior of system when the damping  coefficient is assumed to be complex. 

For example in page 61 of following reference this concept is investigated:
https://nehrpsearch.nist.gov/static/files/NSF/PB92197235.pdf

 

@Carl Love 

I check the problem by employing DTM.

By assigning  0.1i to C instead of 0.1, the complex value for u(t) will be obtained.

It seem that for complex value of C, there is no way to get trigonometric solution or atleast a real amount for u(t) when initial conditions are real (real initial displacement U[0] and real initial velocity U[1])!

restart;

M := 1; C :=0.1; K := 5; m := 40;

U[0] := -1; U[1] := 0.1;

for k from 0 to m do

U[k+2] := solve(M*factorial(k+2)*U[k+2]/factorial(k)+C*factorial(k+1)*U[k+1]/factorial(k)+K*U[k], U[k+2]) end do;

u := add(U[k]*x^k, k = 0 .. m);

plot(u, x = 0 .. 7)

@Rouben Rostamian  

Thanks alot for your hints

@Markiyan Hirnyk 

Thank you

@Carl Love 

I use17 concrete mixing plans with 4 speciment for each plan (68 samples).

This is common to have a limmited number of plans for this types of researchs.

I think, as the Markiyan Hirnyk said, for statistical analyses you should have a sufficintly great data (say 30 plans in the case of current work).

I think for great amounts of data using statistical analysis (or other analyses like the artificial neural network (ANN) or adaptive neuro-fuzzy inference system (ANFIS)) is more reasonable. For example, in the case of sesmic performance of the RC buildings, 256 four to seven story buildings are investigated via ANN. For small available data, from my point of view, using something like the interpolation is more better way to predict Z at some ratios (i.e. (X,Y)) due to lack of enough data.

 

@Markiyan Hirnyk

I just want to calculate an approximated function of two variables (nu and mu) with a known values (Z) at the certain points (X,Y).

I find out that for this purpose the command Statistics:-Fit  can be helpful, but I have no more information about the magic number, 30. It is a good hint.

I find the following function by using interpolation on duplicate values in X.

M = (-53097.22220*nu^5+4.711493057*10^5*nu^4-1.407843057*10^6*nu^3+

1.276592369*10^6*nu^2+9.779029872*10^5*nu-1.638753190*10^6)*mu^2+

(29447.04860*nu^5-2.446397570*10^5*nu^4+6.161949664*10^5*nu^3-

1.068835818*10^5*nu^2-1.506857065*10^6*nu+1.483546194*10^6)*mu-3765.520832*nu^5+27451.77085*nu^4-40941.93771*nu^3-

1.245605199*10^5*nu^2+4.144527100*10^5*nu-3.218281111*10^5

This function takes the values Z at certain points (nu,mu)=(X, Y)

For example, Z(X=2.2,Y=0.2)= 33.7  and M(nu=2.2,mu=0.2) = 33.7005

 

@Carl Love 

It seems your new proposed model has tolerable errors.

From an engineering point of view, this is more accurate than previous model !

I will exam same models for similar problems.

 

Thank you for taking your time.

 

 

@mmcdara 

Thank you again for your hints,  specially for conducted tests (LinearFit2.mw)

I will check your mentioned issues.

 

@mmcdara 

Thank you very much for your kindness and taking your time

If it is possible, please work on data.

I have not enough knowledge about kriging model.

 

@Carl Love

I am working on revision of a manuscript according to the reviewers comments.

X vector is the ratio of water to cement, Y vector is the ratio of bentonite to cement and Z vector is hydraulic conductivity of cracked plastic concrete (nanometer/sec).

I have some experimental data and it is required to find a two variable function with some known values at certain points.

For example, it is known that the value of the function at (x,y)=(2.2,0.2) must be equal to 33.7.

Your model (i.e. -2057.7133+988.24655*x+6195.9003*y-2973.2493*x*y) takes the value 47.379478 at (x,y)=(2.2,0.2).

The function which is calculated in attached file (i.e. Model.mw) gives the value 33.7005 for (nu,mu)=(2.2,0.2).

I use curvefitting for duplicate values (i.e. 0.2, 0.3 and 0.35 in Y)

@mmcdara 

Thanks a lot for your great comments.

Please download attached file and check the interpolated function.

Model.mw


 

restart; Y := `<,>`(.2, .2, .2, .2, .2, .3, .3, .3, .3, .3, .3, .35, .35, .35, .35, .35, .35); X := `<,>`(2, 2.2, 2.4, 2.6, 2.8, 2, 2.2, 2.4, 2.6, 2.8, 3, 2, 2.2, 2.4, 2.6, 2.8, 3); Z := `<,>`(15, 33.7, 62.8, 188, 394, 5.47, 5.82, 6.21, 8.3, 11.5, 24.1, .372, .485, .675, 1.11, 1.27, 1.35); ans := Statistics:-Fit(add(add(a[k, n-k]*x^k*y^(n-k), k = 0 .. n), n = 0 .. 2), `<|>`(X, Y), Z, [x, y], summarize = embed)

-HFloat(814.3815773809906)+HFloat(1163.6625396826184)*y+HFloat(10379.76111111104)*y^2+HFloat(489.5947767857354)*x-HFloat(3239.053571428586)*x*y+HFloat(117.2734374999967)*x^2

(1)

err = (eval(ans, [x = 2., y = .35])-Z[12])/Z[12]

err = HFloat(120.95390104966586)

(2)

IN1 := CurveFitting:-PolynomialInterpolation([seq([X[k], Z[k]], k = 1 .. 5)], nu);

-2630.208328*nu^4+25983.33329*nu^3-94873.54151*nu^2+152197.4164*nu-90668.99986

(3)

INP := `$`(eval(IN1, nu = 2+.2*k), k = 0 .. 4):

Student:-Statistics:-ChiSquareGoodnessOfFitTest([seq(Z[k], k = 1 .. 5)], [INP], output = both)

Chi-Square Test for Goodness-of-Fit
-----------------------------------
Null Hypothesis:
Observed sample does not differ from expected sample
Alt. Hypothesis:
Observed sample differs from expected sample
 
Categories:              5
Distribution:            ChiSquare(4)
Computed Statistic:      .7231672199e-9
Computed p-value:        1.
Critical Values:         9.48772903678116
 
Result: [Accepted]
This statistical test does not provide enough evidence to conclude that the null hypothesis is false.
Histogram Type:  default
Data Range:      15. .. 394.
Bin Width:       12.6333333333333
Number of Bins:  30
Frequency Scale: relative
Histogram Type:  default
Data Range:      14.9999 .. 394.

Bin Width:       12.6333366666667
Number of Bins:  30
Frequency Scale: relative

 

Student:-Statistics:-ChiSquareGoodnessOfFitTest([seq(Z[k], k = 1 .. 5)], [INP], output = plot)

 

IN2 := CurveFitting:-PolynomialInterpolation([seq([X[k], Z[k]], k = 6 .. 11)], nu);

289.8437500*nu^5-3536.718750*nu^4+17210.67708*nu^3-41732.28124*nu^2+50406.85916*nu-24252.04000

(4)

INP := `$`(eval(IN2, nu = 2+.2*k), k = 0 .. 5):

Student:-Statistics:-ChiSquareGoodnessOfFitTest([seq(Z[k], k = 6 .. 11)], [INP], output = both)

Chi-Square Test for Goodness-of-Fit

-----------------------------------
Null Hypothesis:
Observed sample does not differ from expected sample
Alt. Hypothesis:
Observed sample differs from expected sample
 
Categories:              6
Distribution:            ChiSquare(5)
Computed Statistic:      .4230272580e-8
Computed p-value:        1.
Critical Values:         11.0704974062099
 
Result: [Accepted]
This statistical test does not provide enough evidence to conclude that the null hypothesis is false.
Histogram Type:  default
Data Range:      5.47 .. 24.1
Bin Width:       .621
Number of Bins:  30
Frequency Scale: relative
Histogram Type:  default
Data Range:      5.4699 .. 24.09985
Bin Width:       .620998333333333
Number of Bins:  30
Frequency Scale: relative

 

Student:-Statistics:-ChiSquareGoodnessOfFitTest([seq(Z[k], k = 6 .. 11)], [INP], output = plot)

 

IN3 := CurveFitting:-PolynomialInterpolation([seq([X[k], Z[k]], k = 12 .. 17)], nu);

36.53645836*nu^5-456.3541670*nu^4+2265.526044*nu^3-5587.208338*nu^2+6845.853006*nu-3334.209004

(5)

INP := `$`(eval(IN3, nu = 2+.2*k), k = 0 .. 5):

Student:-Statistics:-ChiSquareGoodnessOfFitTest([seq(Z[k], k = 12 .. 17)], [INP], output = both)

Chi-Square Test for Goodness-of-Fit

-----------------------------------
Null Hypothesis:
Observed sample does not differ from expected sample
Alt. Hypothesis:
Observed sample differs from expected sample
 
Categories:              6
Distribution:            ChiSquare(5)
Computed Statistic:      .1321775288e-8
Computed p-value:        1.
Critical Values:         11.0704974062099
 
Result: [Accepted]
This statistical test does not provide enough evidence to conclude that the null hypothesis is false.
Histogram Type:  default
Data Range:      .372 .. 1.35
Bin Width:       .0326
Number of Bins:  30
Frequency Scale: relative
Histogram Type:  default
Data Range:      .37201 .. 1.350015
Bin Width:       .0326001666666667
Number of Bins:  30
Frequency Scale: relative

 

Student:-Statistics:-ChiSquareGoodnessOfFitTest([seq(Z[k], k = 12 .. 17)], [INP], output = plot)

``

 

ans := CurveFitting:-PolynomialInterpolation([[.2, IN1], [.3, IN2], [.35, IN3]], mu)

(-53097.22220*nu^5+471149.3057*nu^4-1407843.057*nu^3+1276592.369*nu^2+977902.9872*nu-1638753.190)*mu^2+(29447.04860*nu^5-244639.7570*nu^4+616194.9664*nu^3-106883.5818*nu^2-1506857.065*nu+1483546.194)*mu-3765.520832*nu^5+27451.77085*nu^4-40941.93771*nu^3-124560.5199*nu^2+414452.7100*nu-321828.1111

(6)

err = (eval(ans, [nu = 2., mu = .35])-Z[12])/Z[12]

err = 0.2688172043e-3

(7)

``


 

Download Model.mw

@Carl Love 

Thank you very much for your valuable hints

@Mac Dude 

Thank you for your hints

In the case of more than two variables, is it possible to use  "ChiSquareGoodnessOfFitTest" ?

@tomleslie 

Thank you

@Carl Love 

I hope you have a good sleep.

I attach a PDF to explain the proposed algorithm which is used in the "ADAMS.mw" worksheet. As I mentioned before, the convergence is decreased by moving from vicinity of the initial point.

I will be grateful, if I can know more about digging into the Maple code (method= classical[adambash]) ).

ADAMS.pdf

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