Dear people in Mapleprimes,

alpha-gamma*q[k]-eta*(int(q[i], i = 0 .. M__n))-eta*q[k] = p[k];

Then,

algsubs(-eta*(int(q[i], i = 0 .. M__n))-eta*q[k] = -eta*(int(q[i], i = 0 .. M__n)), alpha-gamma*q[k]-eta*(int(q[i], i = 0 .. M__n))-eta*q[k] = p[k]);

This works fine.

But,

applyrule(-eta*(int(q[i], i = 0 .. M__n))-eta*q[k] = -eta*(int(q[i], i = 0 .. M__n)), alpha-gamma*q[k]-eta*(int(q[i], i = 0 .. M__n))-eta*q[k] = p[k]);

This doesn't bring an right replacement.

What difference is there between applyrule and algsubs?

Best wishes

taro

Dear all,

I woul line integrate e gradient moltiply to a vector. I will try to explain better.

I have define this function:

and I have a vector a

v=(0,1)

I would like to applu a dot product between the gradient of phi an the vector and integrate the results. 

I have already try in many way but without succeed. Someone could please help me?

Thanks 

For some reason it always bugs me that Maple pulls a negative sign out in the solution.

eq:=c=a+b*exp(-k)

solve(eq,k)

The solution is correct, but one would expect it to be written like ..

.. after all that is a more conventional solution, isn't it? 

If I type the last equation into Maple it evaluates to

..again slightly unconventional but still correct.  Only if back quotes around the `c-a` are used can we achieve what we would write out on paper.  It would be nice if Maple would output the answer in a conventional manner ... at least in my opinion. 

Can we manipulate Maple to evaluate that solution into the conventional answer? 

Afterall, Maple is advertising it's "typsetting appears like it would appear in a textbook" and uses standard Math notation.  All I am trying to say is that if Maple advertises as a math standard could we also present solutions in the same manner? 

But not to get sidetracked from the question I posed, is there a way we can manipulate that solution into a "handwritten" convention?

Hi all

I'm having a hard time, making Maple plot a pretty huge expression in my project.

I have solved a differential equation with initial conditions with method=laplace. The differential equation contains a fourier serie equation, so the more accurate i want the equation to be, the larger the differential equation will be.

Maple solves the equation just fine, and i can plot the solution with 2-4 fourier parts, but when i go higher as i need, the graph ends up empty?

with 20 parts i get the following equation: 

0.*sin(52.88*t)+0.*cos(74.03*t)-0.*sin(74.03*t)-0.*cos(52.88*t)+0.*cos(200.95*t)-0.*sin(200.95*t)+0.*cos(158.65*t)-5.55*10^(-8)*sin(105.76*t)-0.*sin(116.34*t)+0.*cos(31.73*t)-.45*sin(10.58*t)+1.02*cos(10.58*t)+0.*sin(95.19*t)+0.*cos(116.34*t)+0.*sin(179.80*t)-0.*cos(179.80*t)+0.*sin(137.49*t)-0.*sin(31.73*t)-0.*cos(95.19*t)+5.53*10^(-993)*(-1.13*10^992*cos(10.61*t)+8.14*10^991*sin(10.61*t))*exp(-0.7e-1*t)+4.23*10^(-7)*cos(211.53*t)-6.69*10^(-7)*cos(63.46*t)-6.11*10^(-7)*cos(105.76*t)+5.79*10^(-7)*cos(126.92*t)+6.67*10^(-8)*sin(42.31*t)-5.88*10^(-8)*sin(148.07*t)+5.88*10^(-8)*sin(211.53*t)+7.09*10^(-7)*cos(42.31*t)+5.45*10^(-8)*sin(84.61*t)+6.40*10^(-7)*cos(84.61*t)+5.72*10^(-8)*sin(126.92*t)-9.01*10^(-7)*cos(21.15*t)+5.97*10^(-8)*sin(169.22*t)+5.06*10^(-7)*cos(169.22*t)-5.98*10^(-8)*sin(190.38*t)-4.65*10^(-7)*cos(190.38*t)-5.44*10^(-7)*cos(148.07*t)-1.33*10^(-7)*sin(21.15*t)-5.61*10^(-8)*sin(63.46*t)-0.*cos(137.49*t)-0.*sin(158.65*t)

if i plot that expression, the graph ends up empty?

I did also try to solve the equation numerical to plot it with odeplot, but when i try to solve it without the laplace method i get this error message:
"Error, (in dsolve) found the following equations not depending on the unknows of the input system:"

The differential equation is:

ode:=diff(Theta(t), t, t)+2*Zeta*omega[balanceue]*(diff(Theta(t), t))+omega[balanceue]^2*Theta(t) = M[p]/m[balanceue]

and the initial conditions:

ICS := Theta(0) = (1/8)*Pi, (D(Theta))(0) = 0;

when i do:

dsolve({ICS, ode}, Theta(t), method = laplace) it solves just fine.

but when i try with:

dsolve({ICS, ode}, Theta(t))

or

dsolve({ICS, ode}, Theta(t),numeric)

I get the message: 

Error, (in dsolve) found the following equations not depending on the unknowns of the input system: {Theta(0) = (1/8)*Pi, (D(Theta))(0) = 0}

It doesnt seem logical at all, is it a bug? Or can anybody help me with this problem?

Regards

Nicolai

Determine the exact solution to the initial value problem

y'(x)=   (y(x)(20-y(x)))/80 , y(0)=1

 Compute a polynomial approximation to y(x). Plot this polynomial approximation together with y(x) on the same axes for x∈[0,20]. Choose different colours and linestyles for each curve.

Investigate whether or not it is possible to choose Order to be large enough to ensure that the plots of the polynomial approximation and y(x) are indistinguishable over the [0,20] interval? If this is possible, determine the minimum value of Order required. If you think that it is not possible, explain why not.

I tried

des := diff(y(x), x) = (1/80)*(y(x))(20-y(x))

and

ics := y(0) = 1

then i type

soln := dsolve({des, ics}, {y(x)})

came up with

y(x) = RootOf(x-(Int(80/_a(20-_a), _a = _b .. _Z))+80*(Int(1/_a(20-_a), _a = _b .. 1)))

then i tried 

Y := convert(rhs(soln), polynom)

it gives me the same thing

i put

PY := plot(y, x = 0 .. 20)

then it's error...

what should I do next?

A function f is defined on R by

f(x):= (1+a|x|)^1/x      , x<0

         B                      ,  x=0

         ln(1+(a^2)|x|)/x , x>0

where α and β are constants. Investigate whether it is possible to choose α and β so
as to ensure that f is real-valued and continuous at x = 0. Compute any such values
for α and β correct to 10 significant figures. Make use of the piecewise command in
plotting a graph of any resulting continuous function(s) f over the range −20 ≤ x ≤ 20.

I used the help in Maple and manage to get 

f = piecewise(x < 0, (1+alpha*abs(x))^(1/x), x = 0, beta, x > 0, ln(1+alpha^2*abs(x))/x)

Not sure about how to compute a and B...

What does it mean by  f is real-valued and continuous at x = 0?

I have a three paramter ode problem that involves three tanks with given initial concentrations.  Overtime the concentration equalizes but one of the steps is to determine all bifurcation values.  Not sure how to continue with this number of variables.

 This is our given system with initial conditions

sys_ode := diff(x(t),t) = (-r*x(t))/100+0+(r*z(t))/50, 
> diff(y(t),t) = (r*x(t))/100+(-r*y(t))/25+0,
> diff(z(t),t) = 0+(r*y(t))/25+(-r*z(t))/50;
> x0:=0; y0 := 200; z0:=0;

I have denotation like A[0], A[1], A[2], A[3]... But one package doesn't allow to use indexed variables.

I'd like to change denotation. For example, to A0, A1, A2, A3, but I don't know how to do it automatically...

Hello,

I would like to copy/ paste my maplesim model in microsoft visio.

The idea is to create a vectorial illustration for presentation.

I manage to do this with simulink but not with Maplesim.

Apparently, when I go in visio, there is no possibility to copy like a metafile.

Is there a possibility to copy the object of a Maplesim model in a software like visio or illustrator so as to be able to create vectorial illustration for presentation ?

Thank you for your help.

> restart:
> m:=2; k:=1.0931; a:=k-m; b:=k+m-1;
m := 2
k := 1.0931
a := -0.9069
b := 2.0931
> z:=(k*m)/10^(0.1*10);
z := 0.2186200000
> simplify(((10^(0.1*yo))^((b-a+2*p-1)/2)*z^((b-a+2*p+1)/2)*GAMMA((1-(b-a+2*p))/2))/(p!*GAMMA(p-a+1)*GAMMA(1+((1-(b-a+2*p))/2))));
1 / 
---------------------- \0.3183098861 sin(3.141592654 p + 3.141592654) exp(
GAMMA(p + 1.906900000) 
p 
-3.040840432 - 1.520420216 p + 0.2302585095 yo) (exp(0.2302585095 yo)) GAMMA(
\
-1. - 1. p)/
> [seq(limit(.3183098861*sin(3.141592654*p+3.141592654)*exp(-3.040840432-1.520420216*p+.2302585095*yo)*(exp(.2302585095*yo))^p*GAMMA(-1.-1.*p)/GAMMA(p+1.906900000),p=k),k=0..10)]
Warning, inserted missing semicolon at end of statement, ...=k),k=0..10)];
[ / 1 / 
[limit|---------------------- \0.3183098861 sin(3.141592654 p + 3.141592654) 
[ \GAMMA(p + 1.906900000)

p 
exp(-3.040840432 - 1.520420216 p + 0.2302585095 yo) (exp(0.2302585095 yo))

\ \ / 1 / 
GAMMA(-1. - 1. p)/, p = 0|, limit|---------------------- \0.3183098861 sin(
/ \GAMMA(p + 1.906900000)

3.141592654 p + 3.141592654) exp(-3.040840432 - 1.520420216 p

p \ \ 
+ 0.2302585095 yo) (exp(0.2302585095 yo)) GAMMA(-1. - 1. p)/, p = 1|, 
/

/ 1 / 
limit|---------------------- \0.3183098861 sin(3.141592654 p + 3.141592654) 
\GAMMA(p + 1.906900000)

p 
exp(-3.040840432 - 1.520420216 p + 0.2302585095 yo) (exp(0.2302585095 yo))

\ \ / 1 / 
GAMMA(-1. - 1. p)/, p = 2|, limit|---------------------- \0.3183098861 sin(
/ \GAMMA(p + 1.906900000)

3.141592654 p + 3.141592654) exp(-3.040840432 - 1.520420216 p

p \ \ 
+ 0.2302585095 yo) (exp(0.2302585095 yo)) GAMMA(-1. - 1. p)/, p = 3|, 
/

/ 1 / 
limit|---------------------- \0.3183098861 sin(3.141592654 p + 3.141592654) 
\GAMMA(p + 1.906900000)

p 
exp(-3.040840432 - 1.520420216 p + 0.2302585095 yo) (exp(0.2302585095 yo))

\ \ / 1 / 
GAMMA(-1. - 1. p)/, p = 4|, limit|---------------------- \0.3183098861 sin(
/ \GAMMA(p + 1.906900000)

3.141592654 p + 3.141592654) exp(-3.040840432 - 1.520420216 p

p \ \ 
+ 0.2302585095 yo) (exp(0.2302585095 yo)) GAMMA(-1. - 1. p)/, p = 5|, 
/

/ 1 / 
limit|---------------------- \0.3183098861 sin(3.141592654 p + 3.141592654) 
\GAMMA(p + 1.906900000)

p 
exp(-3.040840432 - 1.520420216 p + 0.2302585095 yo) (exp(0.2302585095 yo))

\ \ / 1 / 
GAMMA(-1. - 1. p)/, p = 6|, limit|---------------------- \0.3183098861 sin(
/ \GAMMA(p + 1.906900000)

3.141592654 p + 3.141592654) exp(-3.040840432 - 1.520420216 p

p \ \ 
+ 0.2302585095 yo) (exp(0.2302585095 yo)) GAMMA(-1. - 1. p)/, p = 7|, 
/

/ 1 / 
limit|---------------------- \0.3183098861 sin(3.141592654 p + 3.141592654) 
\GAMMA(p + 1.906900000)

p 
exp(-3.040840432 - 1.520420216 p + 0.2302585095 yo) (exp(0.2302585095 yo))

\ \ / 1 / 
GAMMA(-1. - 1. p)/, p = 8|, limit|---------------------- \0.3183098861 sin(
/ \GAMMA(p + 1.906900000)

3.141592654 p + 3.141592654) exp(-3.040840432 - 1.520420216 p

p \ \ 
+ 0.2302585095 yo) (exp(0.2302585095 yo)) GAMMA(-1. - 1. p)/, p = 9|, 
/

/ 1 / 
limit|---------------------- \0.3183098861 sin(3.141592654 p + 3.141592654) 
\GAMMA(p + 1.906900000)

p 
exp(-3.040840432 - 1.520420216 p + 0.2302585095 yo) (exp(0.2302585095 yo))

\ \]
GAMMA(-1. - 1. p)/, p = 10|]
/]

why the solution is in limit approaches to form??? need to have a closed form expression. any help..????

I am trying to perform the following manipulation (This is a minimum working example).

a < b  < c;
(1)*2;

Error, invalid terms in product: a < b and b < c

 Can anyone tell if it is possible to manipulate inequalities exactly as it is the case with equations?

I have two Reissner Nordstrom black holes that are near extreme. How do I show they move? 

When i copy expression and past it in word, i can change the size of the picture whitout loosing the detials.

How can i export the expression to a file, such that when i will open it in word i could change the size without loosing details? much thnks :)

Dear Maple users,

My problem is as follows:

I have a factor base [2,3,5,7,11,33,34,35,36,37,38,39,40]

The numbers from 2 till 11 are primes, the rest is not. 

Then I have to factor (H+c1)(H+c2) in numbers of the factor base , where c1 and c2 go from 1 to some pre-defined limit. H=32 in my case.
And then I have to put the powers of the numbers of the factor base in a matrix. For example: (H+1)(H+1)=33² but also (H+1)(H+1)=3²*11².

That will become in matrix form [0 , 2, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0 ] but also (!) [0 , 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0 ].

This is not what I want! I want no double representations....

What I want is that (H+c1)(H+c2) should be represented in primes in the matrix if possible and else just represented as the other numbers.

hope you guys can help me!

Hi,

I have a system of equations containing curls, divergence and gradients of variables. 

How can I extract the coefficients of the equations (i.e. coefficients of d/dt rho, d/dx p) and form a matrix?

thanks.