Unanswered Questions

This page lists MaplePrimes questions that have not yet received an answer

Hi:

how i can plot the function in terms of different values of n in a figure:

f(x)=1+2*(1+x)^n

n=1,2,3,4

x-axis->x

y-axis->f(x)

 

Good Afternoon Everyone,

Would anybody be kind to help me with the interpolation of two variables in a table to obtain a value for f18.1 (Stiffness) and f18.2 (Damping) given SR_ratio and Lamda_ratio.

 

Regards

Moses


 

NULL

restart

with(CurveFitting)

with(ExcelTools):

``

Slenderness Ratio (Variable 1):

SR__ratio := 84.01

``

Ratio of modulus of elasticity of Pile vs shear modulus of soil (Variable 2):

`λ__ratio` := 229.58

``

Slenderness ratio of pile from stiffness of pile table:

SR__stiff := [10.8696, 21.7391, 32.6087, 43.4783, 46.7391, 54.3478, 65.2174, 76.0870, 86.9565, 100.0]

``

Ratio of Young's Modulus of Pile to Shear modulus of soil:

EG__ratio := [250.0, 500.0, 1000.0, 2500.0, 5000.0, 10000.0]

``

Vertical stiffness coefficient for floating pile as per Novak (1983) f18.1.250:

`f__18.1.250` := [0.332e-1, 0.509e-1, 0.571e-1, 0.582e-1, 0.582e-1, 0.582e-1, 0.582e-1, 0.582e-1, 0.582e-1, 0.582e-1]

``

Vertical stiffness coefficient for floating pile as per Novak (1983) f18.1.500:

`f__18.1.500` := [0.187e-1, 0.301e-1, 0.364e-1, 0.405e-1, 0.416e-1, 0.416e-1, 0.416e-1, 0.416e-1, 0.416e-1, 0.416e-1]

``

Vertical stiffness coefficient for floating pile as per Novak (1983) f18.1.1000:

`f__18.1.1000` := [0.104e-1, 0.166e-1, 0.218e-1, 0.26e-1, 0.27e-1, 0.281e-1, 0.291e-1, 0.301e-1, 0.301e-1, 0.301e-1]

``

Vertical stiffness coefficient for floating pile as per Novak (1983) f18.1.2500:

`f__18.1.2500` := [0.52e-2, 0.83e-2, 0.104e-1, 0.125e-1, 0.135e-1, 0.145e-1, 0.166e-1, 0.177e-1, 0.187e-1, 0.197e-1]

``

Vertical stiffness coefficient for floating pile as per Novak (1983) f18.1.10000:

`f__18.1.10000` := [0.21e-2, 0.31e-2, 0.42e-2, 0.42e-2, 0.52e-2, 0.52e-2, 0.62e-2, 0.62e-2, 0.73e-2, 0.83e-2]

``

Slenderness ratio of pile from damping of pile table:

SR__damp := [10.8696, 16.3043, 21.7391, 27.1739, 32.6087, 38.0435, 43.4783, 48.913, 54.3478, 59.7826, 65.2174, 70.6522, 76.0870, 81.5217, 86.9565, 92.3913, 100.0]

``

Vertical damping coefficient for floating pile as per Novak (1983) f18.2.250:

`f__18.2.250` := [.1032, .1137, .1126, .1095, .1053, .1021, 0.989e-1, 0.979e-1, 0.979e-1, 0.979e-1, 0.979e-1, 0.979e-1, 0.979e-1, 0.979e-1, 0.979e-1, 0.989e-1, 0.989e-1]

``

Vertical damping coefficient for floating pile as per Novak (1983) f18.2.500:

`f__18.2.500` := [0.558e-1, 0.695e-1, 0.811e-1, 0.832e-1, 0.811e-1, 0.789e-1, 0.758e-1, 0.737e-1, 0.726e-1, 0.716e-1, 0.705e-1, 0.695e-1, 0.695e-1, 0.695e-1, 0.695e-1, 0.705e-1, 0.705e-1]

``

 Vertical damping coefficient for floating pile as per Novak (1983) f18.2.1000:

`f__18.2.1000` := [0.295e-1, 0.421e-1, 0.495e-1, 0.537e-1, 0.568e-1, 0.589e-1, 0.579e-1, 0.568e-1, 0.558e-1, 0.537e-1, 0.526e-1, 0.516e-1, 0.516e-1, 0.505e-1, 0.505e-1, 0.505e-1, 0.495e-1]

``

 Vertical damping coefficient for floating pile as per Novak (1983) f18.2.2500:

`f__18.2.2500` := [0.126e-1, 0.179e-1, 0.232e-1, 0.263e-1, 0.305e-1, 0.326e-1, 0.347e-1, 0.368e-1, 0.379e-1, 0.379e-1, 0.379e-1, 0.379e-1, 0.379e-1, 0.368e-1, 0.358e-1, 0.358e-1, 0.337e-1]

``

   Vertical damping coefficient for floating pile as per Novak (1983) f18.2.10000:

`f__18.2.10000` := [0.32e-2, 0.53e-2, 0.74e-2, 0.84e-2, 0.105e-1, 0.116e-1, 0.137e-1, 0.147e-1, 0.147e-1, 0.168e-1, 0.168e-1, 0.179e-1, 0.189e-1, 0.189e-1, 0.200e-1, 0.211e-1, 0.211e-1]

``

 

NULL

``

``

`f__18.1` := ArrayInterpolation(SR__ratio, `λ__ratio`, SR__stiff, EG__ratio, `f__18.1.250`)

`f__18.2__ _` := ArrayInterpolation(SR__ratio, `λ__ratio`, SR__damp, EG__ratio, `f__18.2.250`)

NULL

NULL


Download Dynamic_Table.mwDynamic_Table.mw

Good Afternoon Everyone,

Would anybody be kind to help me with the interpolation of two variables in a table to obtain a value for f18.1 (Stiffness) and f18.2 (Damping) given SR_ratio and Lamda_ratio.

 

Regards

Moses

Hi:

how i can plot the function f(x) in term of 1/x in maple?for example i have the function f(x)=x^2 and x=0..1,now i plot f(x) in term of 1/x.

It is possible to add groups of questions to an assignment in Maple T.A. But how do you see the grades divided into these groups? As an example, suppose we have two groups of questions say 5 questions in the group "algebra" and 5 questions in the group "geometry". The class grades show the grades for all 10 questions all together, but I would like to see the grades for two groups individually.

 

 

When Matrix or Vector output that has variables with assumptions on them (and therefore with tildes on them) is copied and pasted to the expression argument of a map( ) function, the map code cannot find the assumptions. Of course, when it is pasted in appears in the input with the tildes.  It then seems to not recognize the tildes as indicating assumptions on the base name.

If, on the other hand, the output is directly saved to a name, and that name is used in the input to the map( ) function, then the assumptions are recognized and the output is correct. It seems that in this case, the tilde is an output phenomenon that is not recognized on input. Apparently, the save output is not saved with the tilde.

Here is a simple example:

> assume(a>0, b<0);
   v:=<a,b>;

   see_assump:=proc(x)
     print(getassumptions(x));
     x;
   end proc:

map(see_assump, v);

map(see_assump, <a~, b~>;
map(see_assump, <copied from output of v assignment statement above>);

The first map statement shows the assumptions, the second two just give empty brackets { }.

Attempting to use more complicated procs that require the assumptions in the map( ) function leads to incorrect answers or failure to evaluate in the last two cases.

I'm trying to limit the solutions of an equation to only positive values. I already found "with(RealDomain)" to ignore complex solutions.

 

Now I try something like

 

 

but that does not work, I still get both +sqrt(2) and -sqrt(2).

Hi,

 

  I tried to obtain power of a series. I have the following input

 

***

assume(c<-10);
eq1:=2*x^a*y^b*(x+y)^c+3*x^(a+1)*y^(b+1)*(x+y)^(c+1);


f1:=map(t -> `if`(match(t = d*x^e*y^f*(x+y)^g, x, 's'), subs(s,e), NULL),
convert(eq1, list));


f2:=map(t -> `if`(match(t = d*x^e*y^f*(x+y)^g, y, 's'), subs(s,f), NULL),
convert(eq1, list));


f3:=map(t -> `if`(match(t = d*x^e*y^f*(x+y)^g, x+y, 's'), subs(s,g), NULL),
convert(eq1, list));

print(f1);

 ****

 

The output is

****

a b c (a + 1) (b + 1) (c + 1)
2 x y (x + y) + 3 x y (x + y)
[]
[]
Error, (in unknown) invalid input: match expects its 2nd argument, vv, to be of type {name, set(name)}, but received x+y
[]

***

 

  How to get the correct powers of x, y, and (x+y)? Since the power of x+y is negative, it cannot be absorbed into x and y.

 

Thank you!

How do you create a "rating" question type in Maple T.A.? I.e. a question of the following kind:

"rate this and that" and ask the students to select an answer from the list {1, 2, 3, 4, 5}. To this kind of question there is no correct answer. The grade should be the chosen number. 

Edit: I just found that this kind of question is called a "Likert-scale question".

 

As I am trying to solve this integration:

restart; with(linalg); with(stats); with(plots); with(Statistics); with(LinearAlgebra); with(Optimization);
lambda0 := proc (t) options operator, arrow; gamma0+gamma1*t+gamma2*t^2 end proc;
lambda := lambda0(t)*exp(beta*s);
t1 := 145; t3 := 250; t2 := (t1+t3)*(1/2);
s := 1/(273.16+50); s1 := 1/(273.16+t1); s3 := 1/(273.16+t3); s2 := 1/(273.16+t2); gamma0 := 0.1e-3; gamma1 := .5; gamma2 := 0; beta := -3800;
c := 300; n := 200;
Theta := solve(1-exp(-(gamma0*tau1+(1/2)*gamma1*tau1^2+(1/3)*gamma2*tau1^3)*exp(beta*s1)) = 1-exp(-(gamma0*a+(1/2)*gamma1*a^2+(1/3)*gamma2*a^3)*exp(beta*s2)), a);

a := Theta[1];

Delta := solve(1-exp(-(gamma0*(a+tau2-tau1)+(1/2)*gamma1*(a+tau2-tau1)^2+(1/3)*gamma2*(a+tau2-tau1)^3)*exp(beta*s2)) = 1-exp(-(gamma0*b+(1/2)*gamma1*b^2+(1/3)*gamma2*b^3)*exp(beta*s3)), b);

b := Delta[1];

A1 := `assuming`([unapply(int(exp(beta*s1)*exp(-(gamma0*t+(1/2)*gamma1*t^2+(1/3)*gamma2*t^3)*exp(beta*s1))/(gamma0+gamma1*t+gamma2*t^`2`), t = N .. M), N, M)], [N > 0, M > 0]);
A2 := unapply(int(exp(beta*s2)*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2+(1/3)*gamma2*(a+t-tau1)^3)*exp(beta*s2))/(gamma0+gamma1*(a+t-tau1)+gamma2*(a+t-tau1)^2), t = N .. M), N, M);
A3 := unapply(int(exp(beta*s3)*exp(-(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2+(1/3)*gamma2*(b+t-tau2)^3)*exp(beta*s3))/(gamma0+gamma1*(b+t-tau2)+gamma2*(b+t-tau2)^2), t = N .. M), N, M);
B1 := `assuming`([unapply(int(t^2*exp(beta*s1)*exp(-(gamma0*t+(1/2)*gamma1*t^2+(1/3)*gamma2*t^3)*exp(beta*s1))/(gamma2*t^2+gamma1*t+gamma0), t = N .. M), N, M)], [N > 0, M > 0]);
B2 := unapply(int((a+t-tau1)^2*exp(beta*s2)*exp(-(gamma0*(a+t-tau1)+(1/2)*gamma1*(a+t-tau1)^2+(1/3)*gamma2*(a+t-tau1)^3)*exp(beta*s2))/(gamma0+gamma1*(a+t-tau1)+gamma2*(a+t-tau1)^2), t = N .. M), N, M);
B3 := unapply(int((b+t-tau2)^2*exp(beta*s3)*exp(-(gamma0*(b+t-tau2)+(1/2)*gamma1*(b+t-tau2)^2+(1/3)*gamma2*(b+t-tau2)^3)*exp(beta*s3))/(gamma0+gamma1*(b+t-tau2)+gamma2*(b+t-tau2)^2), t = N .. M), N, M);

F0 := A1(0, tau1)+A2(tau1, tau2)+A3(tau2, c);
F1 := B1(0, tau1)+B2(tau1, tau2)+B3(tau2, c);

NLPSolve(1/(n^3*(F0*F1-F1)), tau1 = 115 .. 201, tau2 = 237 .. 273);

I need to have tau1 tau2 as varibles to get there optimal values ..

But this error keeps coming :


Error, (in Optimization:-NLPSolve) integration range or variable must be specified in the second argument, got HFloat(1.0) = HFloat(158.0) .. HFloat(255.0)

Please Help ..

I am a problem with solve differential equation, please help me: THANKS 

g := (y^2-1)^2; I4 := int(g^4, y = -1 .. 1); I5 := 2*(int(g^3*(diff(g, y, y)), y = -1 .. 1)); I6 := int(g^3*(diff(g, y, y, y, y)), y = -1 .. 1); with(Student[Calculus1]); I10 := ApproximateInt(6/(1-f(x)*g)^2, y = -1 .. 1, method = simpson);

dsys3 := {I4*f(x)^2*(diff(f(x), x, x, x, x))+I5*f(x)^2*(diff(f(x), x, x))+I6*f(x)^3 = I10, f(-1) = 0, f(1) = 0, ((D@@1)(f))(-1) = 0, ((D@@1)(f))(1) = 0};

dsol5 := dsolve(dsys3, numeric, output = array([0.]));

              Error, (in dsolve/numeric/bvp) system is singular at left endpoint, use midpoint method instead

****************FORMAT TWO ********************************************************

g := (y^2-1)^2; I4 := int(g^4, y = -1 .. 1); I5 := 2*(int(g^3*(diff(g, y, y)), y = -1 .. 1)); I6 := int(g^3*(diff(g, y, y, y, y)), y = -1 .. 1); with(Student[Calculus1]); I10 := ApproximateInt(6/(1-f(x)*g)^2, y = -1 .. 1, method = simpson);
dsys3 := {I4*f(x)^2*(diff(f(x), x, x, x, x))+I5*f(x)^2*(diff(f(x), x, x))+I6*f(x)^3 = I10, f(-1) = 0, f(1) = 0, ((D@@1)(f))(-1) = 0, ((D@@1)(f))(1) = 0};

dsol5 := dsolve(dsys3, method = bvp[midrich], output = array([0.]));
%;
                                   Error, (in dsolve) too many levels of recursion

I DONT KNOW ABOUT THIS ERROR

PLEASE HELP ME

THANKS A LOT

 

what conditions are necessary that a group G has stabilisers?

Hi,

 

Say I run a procedure, it could take about 3 minutes to run, or maybe hours, or longer.

Is it possible to trigger a sound, like "ding", after it's finished?

I may like to read something while the program is running, but I want to see the results immediately.

It would be nice to have such function.

 

Also, for procedures that runs for a "long" time, could I get 10 (or 15) minutes sound notice?

 

Thanks,

 

casper

I'd like to try the new Maple IDE

http://www.maplesoft.com/products/toolboxes/IDE/index.aspx

Maple IDE - Integrated Development Environment

Is there a way to obtain a trial version even for a week just to try it out? I do not want to spend all this money then find out it does not work well.

any one here uses this and can recommend it or comment on how well it works? For example, can one run Maple code directly from it and have the code execute in Maple, or must one load the file manually from the editor into Maple to run it each time they make any changes to the code?

Hi all

I have a mathematical problem and I asked it in various sites but the answers till yet are not correct.

Assume that we have:

T[m]:=t->t^m:
b[n,m]:=unapply(piecewise(t>=(n-1)*tj/N and t<n*tj/N, T[m](N*t-(n-1)*tj), 0), t):

where n,N,tj are known constants. furthermore assume that we want to comute the following integral:

for following approximations:

I have written the following code but it seems to be incorrect:

V1:=Vector([seq(seq(b[n,m](t),m=0..1),n=1..3)]);
V:=evalf(V1.Transpose(V1));

the original program is :

taaylor.mws

I will be so grateful if any one can help me to solve it by maple

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

Hi all

 

I am trying to maximize a function f(x,y,z,w) in terms of x. (Only x is treated as a variable, and the others are treated as parameters).

However, all I know is that y,z,w they are parameters and they are non-negative. I have already tried with the "optmization help page" from maplesoft's website, and it looks like it will search the range of x,y,z,w, and it will return numerical values at which this function is maximized. 

 

 

However, what I want is instead a close-form solution of x=g(y,z,w) that will maximize the function.   In other words, I would like to keep the parameter in symbolic forms. 

 

Can Maple do that?

First 257 258 259 260 261 262 263 Last Page 259 of 363