I have this matrix

and I would like to evaluate the variables like this

I tried this

but it didnt work.

Any suggestions??

Thank you so much

I want to compute some matrix multiplications and i need this expression to be 1 always, i.e,

x^2+y^2+z^2+w^2=1

for every calculation I do.

I have tried x^2+y^2+z^2+w^2:=1 and assign(x^2+y^2+z^2+w^2,1) but it doesn't work.

What I should type to make it work?

Thank you

Hi,

In line with my previous questions

http://www.mapleprimes.com/questions/201944-Convergence-Problem-In-My-Algorithm-DSOLVE

I write a code with different variables and odes. However I tried to get the solution, I cannot get the results. I think it should be some problems with my Guess procedure. The code is attached. I would be most grateful if you help me on this problem.

code.mw

Thank you.

Amir

The engineering design process involves numerous steps that allow the engineer to reach his/her final design objectives to the best of his/her ability. This process is akin to creating a fine sculpture or a great painting where different approaches are explored and tested, then either adopted or abandoned in favor of better or more developed and fine-tuned ones. Consider the x-ray of an oil painting. X-rays of the works of master artists reveal the thought and creative processes of their minds as they complete the work. I am sure that some colleagues may disagree with the comparison of our modern engineering designs to art masterpieces, but let me ask you to explore the innovations and their brilliant forms, and maybe you will agree with me even a little bit.

Design Process

Successful design engineers must have the very best craft, knowledge and experience to generate work that is truly worthy of being incorporated in products that sell in the tens, or even hundreds, of millions. This is presently achieved by having cross-functional teams of engineers work on a design, allowing cross checking and several rounds of reviews, followed by multiple prototypes and exhaustive preproduction testing until the team reaches a collective conclusion that “we have a design.” This is then followed by the final design review and release of the product. This necessary and vital approach is clearly a time consuming and costly process. Over the years I have asked myself several times, “Did I explore every single detail of the design fully”? “Am I sure that this is the very best I can do?” And more importantly, “Does every component have the most fine-tuned value to render the best performance possible?” And invariably I am left with a bit of doubt. That brings me to a tool that has helped me in this regard.

A Great New Tool

I have used Maple for over 25 years to dig deeply into my designs and understand the interplay between a given set of parameters and the performance of the particular circuit I am working on. This has always given me a complete view of the problem at hand and solidly pointed me in the direction of the best possible solutions.

In recent years, a new feature called “Explore” has been added to Maple. This amazing feature allows the engineer/researcher to peer very deeply into any formula and explore the interaction of EVERY variable in the formula. 

Take for example the losses in the control MOSFET in a synchronous buck converter. In order to minimize these losses and maximize the power conversion efficiency, the most suitable MOSFET must be selected. With thousands of these devices being available in the market, a dozen of them are considered very close to the best at any given time. The real question then is, which one is really the very best amongst all of them? 

There are two possible approaches - one, build an application prototype, test a random sample of each and choose the one that gives you the best efficiency.  Or, use an accurate mathematical model to calculate the losses of each and chose the best. The first approach lacks the variability of each parameter due to the six sigma statistical distribution where it is next to impossible to get a device laying on the outer limits of the distribution. That leaves the mathematical model approach. If you take this route, you can have built-in tolerances in the equations to accommodate all the variabilities and use a simplified equation for the control MOSFET losses (clearly you can use a very detailed model should you chose to) to explore these losses. Luckily you can explore the losses using the Explore function in Maple.

The figure below shows a three dimensional plot, plus five other variables in the formula that the user can change using sliders that cover the range of values of interest including Minima and Maxima, while observing in real time the effects of the change on the power loss.

This means that by changing the values of any set of variables, you can observe their effect on the function. To put it simply, this single feature helps you replace dozens of plots with just one, saving you precious time and cost in fine-tuning your design. In my opinion, this is equivalent to an eight-dimensional/axes plot.

I used this amazing feature in the last few weeks and I was delighted at how simple it is to use and how much it simplifies the study of my approach and my components selection, in record times!

with(DynamicSystems)

T:= a vector

Iwant to make a single plot of:

DiscretePlot(T,x1,stile=stair); DisdretPlot(T,x2,stile=stair)

The usal way: DiscreePlot(T,[x1,x2]... ain't work

Thansk for helping

I am trying to solve a PDE which is converted to ODE when we assign one of the  variables some value. The boundary conditions given to the PDE are numerical values given for fixed numerical values to the two independent variables. I am trying to solve the PDE with the staandard syntax:

pds := pdsolve(pde,[ibc],numeric,time=z,range=0..beta);

The error message I get is:  

Error, (in pdsolve/numeric/process_PDEs) PDEs can only contain dependent variables with direct dependence on the independent variables of the problem, got {theta(z, 0)}

The pde and boundary conditions are as follows:

PDE:   pde := diff(theta(z, 0), z, z)+theta(z, 0)

Where zero is the fixed value for an independent variable

Boundary Condition:  ibc:={theta(0,0)=beta,D[1](theta)(0,0)=0};

When I try to solve it as an ODE the error is:

Error, (in dsolve) not an ODE system, please try pdsolve

Hello

i was stuck with some simple problem. I tried using if condition in program that involves a truth condition. If maple cannot determine the condition is true or false i want to assign o to a particular varable. I  show a similar program.

restart;

h[1] := -1;

for i from 1 by 1 while(i<150 and h[i]<0) do

p1[i] := .989347189582843*x^2-0.139423979061219e-1*x-1.82559474469870*10^8*x^15+1.30761381361453*10^8*x^16-6.88520063191821*10^7*x^17+2.51079317463498*10^7*x^18-5.66094206949155*10^6*x^19+5.94129446612678*10^5*x^20-6812.74182685426*x^5+59230.0931084044*x^6-3.83520584559500*10^5*x^7+1.90126822307036*10^6*x^8-7.34991883857609*10^6*x^9+2.24203561757434*10^7*x^10-5.43284775909785*10^7*x^11+1.04806113793011*10^8*x^12-1.60600324339222*10^8*x^13+1.94090536353833*10^8*x^14+559.557918804679*x^4-30.6576714427729*x^3-3.93727537464007*10^(-15)-i^2-i; fsolve(p1[i]-0.312e-1, x, -1 .. 1);

if fsolve(p1[i]-0.312e-1, x, -.2 .. 0) < 0 then h[i+1] := fsolve(p1[i]-0.312e-1, x, -.2 .. 0) elif FAIL then h[i+1] := 0 end if;

print(i, h[i])

end do;

i got an error message

Error, cannot determine if this expression is true or false: () < 0

how can i get a value of 0 for h[i] if the error comes.

Thanks.

Aditya

Hello,

I would like to change the numbering of my equation.

For the moment, the numbering of my equations is : 4.2.4 for Section 4, Subsection 2 and 4th equations.

I would like to keep the number for the section but to remove the numbers for the subsections.

Consequently, I would like to have this kind of result only : 4.4 for Section 4, Equation 4.

Do you have some ideas to do this changement ?

Thanks a lot for your help.

i have to compute lie algebra condition by using maple 15, but currently i have code in maple 5.,

i already try to run in maple 15 but it say 'error,unable to match delimeters'
i try to find error, but i cant find it..
the coding are here...

check[lie]:=proc(A,n)

local  i,j,k,l,m;  

for i from 1 by 1 to n do

for j from 1 by 1 to n do  

for k  from 1 by 1 to n do    

if A[i,i,k]<>0 then  

RETURN ('Input is NOT a Lie algebra (',i,i,k,')=',A[i,i,k], 'is not zero');  

elif A[i,j,k]+A[j,i,k]<>0 then  

RETURN ('Input is NOT a Lie algebra,(',i,j,k,')+(',j,i,k,')=',A[i,j,k]+A[j,i,k],'is not zero');  

else  

for 1 from 1 by 1 to n do

if  

simplify(sum(A[i,j,m]*A[m,l,k]+A[j,l,m]*A[m,i,k]+A[l,i,m]*A[m,j,k],   m=1..n))<>0  

then  

RETURN('Input is NOT a Lie algebra---the Jac(',i,j,l,') is not zero');    

fi;  

od;  

fi;  

od;  

od;  

od;  

print('Yes,input IS a Lie algebra');  

end:

can anyone help me here? Thank You..

Here, i attached the result in printscreen

Hi,

I ues the optimization tool in Maple2015.

In my experience, Maple optimizatiom tool just shows the optimum values, not a history of the optimizatiom process.

I want to know how I can obtain the history of the optimization process, such as values of the objective, constraints, and tolerance at every iteration.

Is there anyone who knows how to obtain the history data of the optimization process?

Hi everyone,

I am trying to solve the equation of heat tranfer, time dependent, with particular Initial and boundary conditions but I am stuck by technical problems both in getting an analytical solution and a numerical one.

The equation

the equation.

I defined a and b numerically. domain is : and I defined surf_power numerically.

The initial condition is : , T0 defined numerically

The boundary condition is : , because it has a shperical symetry.

To me, it looks like a well posed problem. Does it look fine ?

Problem in analytical solution :

It doesn't accept the boundary condition so I only input the initial condition and it actually gives me back an expression that can be evaluated but it never does : I can't reduce it more than an expression of fourier which I can't eval. The solution :
The solution calculated in (0,0). I was hoping T0...

Are you familiar with these problems ? What would be the perfect syntax you would use to solve this ?

The numerical solution problems :

Sometimes it tells me that my boundary condition is equivalent  to 0 = 0, and I don't see why. Some other times it tells me I only gave 1 boundary/initial condition even if I wrote both. Here is what I wrote for example :

(because it kept asking me to add these two options : 'time' and 'range')

Are you familiar with these problems ? What would be the perfect syntax you would use to solve this ? I must at least have syntax problems because even if I keep reading the Help, it's been a long time since I used Maple.

Thank very much for any indication you could give me !

Simon

I want to refer to the first element in a list such as x:=[1,2,3] as x[0] not x[1].

How to do it?

 my answer are 143.2030067,143.2030339,143.2030610

but, it is wrong. can anyone tell me where is the wrong part?

please!!

I am attempting to solve a system of second order ODEs. I place conditions on the solutions and use the solve command to figure the correct constants for the general solutions of the ODEs; however, the conditions do not appear to hold after I substitute the constants back into the general solutions. Any help would be greatly appreciated. Here's the code and an explanation:

First some constants

> A := 1; B := 9/10;
> j := 1-1/B;

 This is our homogeneous odes. I will give the general solutions of the inhomogeneous system momentarily 

> eqnv1 := diff(v1(x), `$`(x, 2)) = (1-1/(j+1))*v1(x)+v2(x)/(j+1);
> eqnv2 := diff(v2(x), `$`(x, 2)) = -v1(x)/(A*(j+1))+(B/A+1/(A*(j+1)))*v2(x);

Next we get the general solution of this sytem of odes.

> soln := dsolve([eqnv1, eqnv2])

Next we have our solutions of the inhomogeneous problem1. Basically solution v1neg, v2neg on [0,xi] and v1pos, v2pos on [xi,1]. We will assume v1,v2 are C^1 across xi; however, the location of xi is not known at this time so they must remain split.

> v1neg := op([1, 2], soln)-1;
> v2neg := op([2, 2], soln)-1/B;
> v1pos := op([1, 2], soln)+1;
> v2pos := op([2, 2], soln)+1/B;

There's probably a better way to do this, but I relabeled the constants:

> v1negc := subs([_C1 = a[1], _C2 = a[2], _C3 = a[3], _C4 = a[4]], v1neg);
> v2negc := subs([_C1 = a[1], _C2 = a[2], _C3 = a[3], _C4 = a[4]], v2neg);
>
> v1posc := subs([_C1 = a[5], _C2 = a[6], _C3 = a[7], _C4 = a[8]], v1pos);
> v2posc := subs([_C1 = a[5], _C2 = a[6], _C3 = a[7], _C4 = a[8]], v2pos);

Next we have eight conditions the solutions must satisfy. Namely v1, v2 are C^1 across xi and v1',v2' are 0 at {0,1}.

> syscon1 := subs(x = xi, v1negc) = subs(x = xi, v1posc);
> syscon2 := subs(x = xi, v2negc) = subs(x = xi, v2posc);
> syscon3 := subs(x = xi, diff(v1negc, x)) = subs(x = xi, diff(v1posc, x));
> syscon4 := subs(x = xi, diff(v2negc, x)) = subs(x = xi, diff(v2posc, x));
> syscon5 := subs(x = 0, diff(v1negc, x)) = 0;
> syscon6 := subs(x = 0, diff(v2negc, x)) = 0;
> syscon7 := subs(x = 1, diff(v1posc, x)) = 0;
> syscon8 := subs(x = 1, diff(v2posc, x)) = 0;

We solve to get the constants for the solutions.

> constants := simplify(evalf(solve({syscon1, syscon2, syscon3, syscon4, syscon5, syscon6, syscon7, syscon8}, {a[1], a[2], a[3], a[4], a[5], a[6], a[7], a[8]})));
>

We substitute the values for the constants.

> a[1] := op([1, 2], constants); a[2] := op([2, 2], constants); a[3] := op([3, 2], constants); a[4] := op([4, 2], constants); a[5] := op([5, 2], constants); a[6] := op([6, 2], constants); a[7] := op([7, 2], constants); a[8] := op([8, 2], constants);

Lastly we try to verify that the conditions from earlier hold:

> evalf(subs(xi = .2, subs(x = xi, v1negc-v1posc)));
-1.7597825261536669519
> evalf(subs(xi = .2, subs(x = xi, v2negc-v2posc)));
-1.8936659961101033997
> evalf(subs([x = 0, xi = .2], diff(v1negc, x)));
-0.38633519704430619686

They should hold for any xi, but they don't appear to. All of these should be 0. For a large xi, the numbers get very large so I was thinking perhaps roundoff error, but even when I do an exact solution and then evalf just at the end, I still have large error so I'm not sure what the problem is. Sorry for the long question. Thanks so much for the help.

Hello!

Vectors are defined by a lenth and an angle.

The arrow command wants me to give it a starting point and an endpoint in coordinates, for example:

with(LinearAlgebra);

with(plots);

gu := arrow([5, 10], [15, 20], .2, .4, .1, color = green)

I'm dealing with vectors from electrical circuits, so the information I have would be the lenth of the vector, and an angle given in degrees (which I could convert to radians if needed for the plot).

The goal is to plot all the vectors in the circuit on the same display so that it can be seen how they are in relation to each other, and to visualise the solution. This means I would need to find out the endpoint of the arrows to start the next arrow at that location with its angle in reference to the same axis. Does it make sence? I'm not sure about the wording of this.

Here's a picture of a calculated vector lenth and angle: http://imgur.com/hJtNeGg

So, the -45 degrees would need to be in reference to the y-axis for example, then the next vector can be placed at lets say perpendicular in reference to the vector sol_1, starting at its endpoint.

The only other way I have to draw this, excluding doing it by hand, is using the object tools from microsoft word, and to be honset I dislike using word for math stuff. I would much rather learn the syntax for Maple to do this.

I hope someone knows what I mean, Please ask if somethig is unclear,

Thanks!