hi everyone , i need your point of view in my question,any help would be appreciated in advance .

we have a discrete function named g(t) and a continous function f(t) in in convolution integral just like this :
int(f(t-x)*g(x),x=0..t) ; 
we have just g(x) in some special points int the interval (0..t) , thus i need to convert this integral to a series.
how should i do this ? can anyone help or any idea ? i need at first a mathematical solution or idea about how to do this and then, how to do this in a software ?
tnx again.

I am trying to define some functions by recursion relations but I get this error:

Error, (in type/realcons) too many levels of recursion

Please see the attached file for details:

File.mw

restart:

Eq2 := diff(T(y), y, y)+(diff(sigma(y), y))*(diff(T(y), y))+(diff(T(y), y))^2+(exp(-y)

+exp(y))^2 = 0;

Eq3 := diff(sigma(y), y, y)+(diff(T(y), y, y))= 0;

bcs2:=T(h1)=0,T(h2)=1,sigma(h1)=0,sigma(h2)=1;

dsolve({Eq2,Eq3,bcs2});

This give me nothing. Please help me out. 

Dear all:

hello everybody;

I need your help to solve the system f(x,y)=0, and g(x,y)=0, such that there some parameter in the system, also all the parameter are positive and also our unkowns  x and y are also positive.

I try to write this code. I feel that under some condition we can have four solution or three or two. I need your help. Many thinks.

Systemsolve.mw

follow Computing non-commutative Groebner bases and Groebner bases for modules

in maple 12

Error, (in Groebner:-Basis) the first argument must be a list or set of polynomials or a PolynomialIdeal

then i find in maple 15 help file is changed from module M := [seq(Vector(subsop(i+1 = 1, [F[i], 0, 0, 0])), i = 1 .. 3)]

to array M := [seq( s^3*F[i] + s^(3-i), i=1..3)];

though it can run, but when apply other example can not run

such as

restart;
with(Groebner):
F := [2*x^2+3*y+z^2, x^2*z^2+z+2*x, x^4*y^7+3*x];
M := [seq( s^3*F[i] + s^(3-i), i=1..3)];
with(Ore_algebra);
A := poly_algebra(x,y,z,s);
T := MonomialOrder(A, lexdeg([s], [x,y,z]), {s});
G := Groebner[Basis](M, T);
Error, (in Groebner:-Basis) the first argument must be a list or set of polynomials or a PolynomialIdeal

G1 := select(proc(a) evalb(degree(a,s)=3) end proc, G);
[seq(Vector([seq(coeff(j,s,3-i), i=0..3)]), j=G1)];
C := Matrix([seq([seq(coeff(j,s,3-i), i=1..3)], j=G1)]);
GB := map(expand, convert(C.Vector(F), list));
Groebner[Basis](F, tdeg(x,y,z));

follow Computing non-commutative Groebner bases and Groebner bases for modules

using LeadingMonomials(f, lexdeg([s],[x,y]));

f:=[y^2+2*x^2*y, y^2];
f1:=[x+y^2,x];
f2:=[x,y];

g:=[f1,f2];

LeadingMonomial(g[1], lexdeg([s],[x,y]));
LeadingMonomial(g[2], lexdeg([s],[x,y]));
LeadingMonomial(g[1], lexdeg([s],[y,x]));
LeadingMonomial(g[2], lexdeg([s],[y,x]));

no result is (0,y^2)

where is wrong?

I`ve written this program:



It determines the power of the Ithfactor, only one problem. When I try to compute the factor of 512 I get 2_2 (subscript) which is extremly stupid. I want to know how I can change my program so that it can give the power of every ith factor. Thanks!

If I want to make a display of data, say 2 columns of data, with a heading on top of each column, how would one go about this in Maple? This will be for display purposes only (say for final result of a homework result, etc..)  

In Mathematica, I use the Grid command. I am not sure how to do this in Maple. I could not find Grid like command in Maple.

I can generate the data in 2 sequences  in Maple, but need to insert a heading, and also need to insert frame around eveything.

Here is an example:

f:=x->x^2;
g:=x->x^3;
data=[seq([i,f(i),g(i)],i=0..5)];
    #data = [[0, 0, 0], [1, 1, 1], [2, 4, 8], [3, 9, 27], [4, 16, 64], [5, 25, 125]]

The above data has 3 columns, 6 rows. I need to insert first row also called {"index","f","g"} and then display all in nice table form. This is how I would do the above in Mathematica to give idea of what I looking for

f[x_] := x^2;
g[x_] := x^3;
data = Table[{i, f[i], g[i]}, {i, 0, 5}]
    (* {{0, 0, 0}, {1, 1, 1}, {2, 4, 8}, {3, 9, 27}, {4, 16, 64}, {5, 25, 125}} *)

And this is below the part I do not know how to emulate in Maple:

data = Insert[data, {"i", "f(i)", "g(i)"}, 1]
    #   {{"i", "f(i)", "g(i)"}, {0, 0, 0}, {1, 1, 1}, {2, 4, 8}, {3, 9, 27}, {4, 16, 64}, {5, 25, 125}}
    # now use Grid to format with frame around everything
Grid[data, Frame -> All]

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;