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Dear All

 

Suppose that you have a pdf for the number of request receiving to a system with mean m. Suppose that it is not a common distribution.

Is it correct that there is a time between request with a specific distribution and the mean value of distribution is 1/m.

Is there any math rule that can prove it?

 

Best Regards,

Saeid

I've got the problem of warnings polluting my text outputs.

If I use interface(warnlevel = 0) the warnings are suppressed in the terminal but not in the files exported by the writeto command.

A sample:

> restart;with(CodeGeneration); with(plots); with(ListTools); with(FileTools);
> Matlab(u(1) = 3);
Warning, the function names {u} are not recognized in the target language
cg = u(1) == 3;
> interface(warnlevel = 0); Matlab(u(1) = 3);

I found following bug:

p := proc (k) sum((-1)^n*n/(factorial(k-n)*factorial(n)), n = 1 .. k) end proc; 
sum(p(k), k = 1 .. 1); 
sum(p(1), k = 1 .. 1); 
p(1)

Results must be same, but output is 0 (bug!), -1 (correct), -1 (correct)

 

Dear All,

 

Hi. I am interested to find the pdf of this moment function: phi(t) = exp(a * b / p (e^(a*t) -1))

This is a moment of random variable Y = aX where X is a Poisson random variable with rate (a*b/p)

I want to know is it possible to find a pdf from its moment in maple?

If so how?

 

Best Regards,

Saeid

Sorry to bother you again,fellow primers.A few days ago I asked a similar kind of q.,but here is some change, & I'm stuck again.

 

int(p^(D-2)/(exp(beta*p)-1), p = 0 .. infinity) assuming beta::positive, D::posint #D>0 & is an integer,this two are the only restriction on D,nothing else.

 I'm not getting any result (correct/incorrect).

The  result is  GAMMA(D-2)*Zeta(D-2)/beta^(D-2).  #Zeta is the Riemann zeta function.

can someone give help or a hint on how to overcome the faulty result in the pdsolve problem with radiation bc described in the following worksheet? I tried also with different method, but no success. the result from pdsolve makes physical no sense, because under pure radiation bc (h=0) the body should cool down and not heat up!

maple_heat_rad_pdsol.mw

thanks :-)

btw, the problem was reported befor, see:

How to compute the highest total degree of a polynomial in which a variable, say x, appears?

For example, f := x^2*y+3*x*y^2+x^3*y^3-x^5+y^4*z^5;
then the highest total degree in x is 6.

The degree command does not seem to do what I would like to do, since degree(f,x) will give 5 instead.

Thank you in advance.

Dyson_Mclean_V3.mw

Dyson_Mclean_V5.mw

Hi All,

I have a question that I see has been seldom discussed except certain posts by Mr. Lopez. This concerns the estimation of optimum parameters of the solution of a system of differential equations to fit experimental data. I referred to the Maple document provided by Mr. Lopez on the Maplesoft website.

Good morning.

I want to do a program in a worksheet that it can recall other workeet for plotting with display.

I don´t know how can recall other worksheets.

Please help me.

P.D.: I send a worksheet 01-06-2011-robot-man.mw

Hi,

I am new to Maple so I don't know if this is expected behavior or simply a setting or command I need to set.

 

When I calculate (12x2y2)/4xy, I get (3x2y2)/xy instead of the simplified 3xy. Is this normal?

 

Thanks,

 

Perry

Hi,

I am tring to solve two coupled pde's with piecewise/Heaviside conditions. I am puzzled with the maple error...Rhop, Rhos are constants n given some values..

> pde_sys:=[diff(c(m,t),t)=(2/m)*diff(c(m,t),m)+diff(c(m,t),m,m)-(convert(piecewise(d(m,t)<Rhop,(Rhop-d(m,t))*d(m,t)*c(m,t)),Heaviside)+(Rhos-d(m,t))*c(m,t)*convert(piecewise(d(m,t)>Rhop,Rhop,d(m,t)),Heaviside)), diff(d(m,t),t)=convert(piecewise(d(m,t)<Rhop,(Rhop-d(m,t))*d(m,t)*c(m,t)),Heaviside...

Like when I type

1/2^2;

it gives

1/4

but if I type

0.5^2;

it gives

0.25

I want it to consider 0.5 as 1/2, because sometimes I make computations on large equations and I want to have access to the symbolic output, not the numeric output.

Hello all,

I am trying to solve a pde system with piecewise conditions numerically but unsucessful. Here's what i am doing:

> restart;

> sys:=[diff(c(x,t),t)=v(x,t)*diff(c(x,t),x$2),diff(v(x,t),t)=piecewise(c(x,t)>v(x,t),c(x,t)*diff(v(x,t),x$2),0)];

> bc:={c(0,t)=2,c(1,t)=1,c(x,0)=0,v(0,t)=1,v(1,t)=2,v(x,0)=0};

> sol:=pdsolve(sys,bc,type=numeric,time=t);

I am getting the following error.......


Error, (in pdsolve/numeric/process_PDEs...

Hello,

I am trying to solve a transient heat transfer problem.

The problem is that I have an insulated pipe that is immersed in cold water.  At time 0, the fluid filling the pipe is at a constant temperature.  The insulation at one end of the pipe is different to the rest of the insulation.  I need to find out the time that it would take any fluid to reach a certain temperature, called the cool down time.

A graphic representation of the...

Is there an easy way to find the limit of the following equations as Pec_i approches infinity.

 Equation 1

(Q*(R*S+1-S)/(1-R*R3*k*(1-R1)/C/Pec_i*S)*(H_vap+1/(1-exp(Pec_i)))*exp(C*Pec_i/(R*S+1-S)*(1-R*R3*k*(1-R1)/C/Pec_i*S)*S/k)-H_vap)*(1-exp(Pec_i*(1-S)/(R*S+1-S)*(1-R*R3*k*(1-R1)/C/Pec_i*S))) = 1;

 Equation 2

 Theta0/C = (exp(Pec_i/(R*S+1-S)*(1-R*R3*k*(1-R1)/C/Pec_i*S)*(1-S))-1)*(H_liq+1/(1-exp(-C/k*Pec_i/(R*S+1-S)*(1-R*R3*k*(1-R1)/C/Pec_i*S)*S)));

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