I'm trying to understand the various ways to do pattern matching. The problem is that sometimes the patterns seem to matched against a normal form that's different from the one displayed. For example can somebody explain to me why the last pattern match fails below?

> patmatch((1/2)*Zeta+(1/3)*Zeta^5, conditional(Zeta^n::posint*y::anything+z::anything, n >= 3), 's'); s;
true
[n = 5, y = 1/3, z = (1/2)*Zeta]
> patmatch(Zeta^2+(1/3)*Zeta^5, conditional...

Thank you for Assisting me to solve the PDE, Now please would you tell me what F5 means (in other words, explain to me the solution) and also could you how to test if the solution satisfies the initial and boundary conditions.

Please help me plot the graphs of the following equations:-

eq1 := diff(E[X(t)], t) = lambda-delta*E[X(t)]-beta*E[X(t)*V(t)];

 eq2 := diff(E[Y(t)], t) = -delta*E[Y(t)]+beta*exp(-rho*tau)*E[X(t)*V(t)];
                                                    
eq3 := diff(E[V(t)], t) = gamma+N*kappa*E[Y(t)]-mu*E[V(t)]-beta*E[X(t)*V(t)]

NB:- E[X(t)]= the expected value of variable X at time t. 

Please help me know how to input this...

> restart;
> with(LinearAlgebra);
> tolerance = 10^(-3); x1km1 := 0; x2km1 := 0; x3km1 := 0; for k to 2 do k := k; x1k := 1/4*(5-x2km1+x3km1); x2k := 1/3*(-4+x1km1-x3km1); x3k := 1/5*(1-2*x1km1-2*x2km1); evalf(x1k); evalf(x2k); evalf(x3k) end do;
                               1
                               5
                               -
                               4
                               -4
                               --

q1.mw

This equations should have numerical solution, but they can’t be calculated. The result is too long, and include ‘RootOf’

It just like this:

> t := 15000; S := solve({n = b+2*c+f+2*g, (a+b+c)/(a+b+c+p+q+f+g) = 5*(1/10), q^2-3.47*p*10^27*t^(3/2)*e^(-10.97*10^4/t) = 0, n*b-(2.407*a*10^21*8)*t^(3/2)*exp(-18.27*10^4/t) = 0, n*c-(1/2)*(2.407*b*10^21*5)*t^(3/2)*exp(-32.035*10^4/t...

Regards,

I need help with solving an equation. I have the following equation:

24*b[0]+24*b[1]*xi+24*b[2]*xi^2+(2*(-12+24*xi))*(b[1]+2*b[2]*xi)+(2*(-12*xi+12*xi^2))*b[2]+P[cl]*L^2*(-12*xi+12*xi^2)

to this equation I applied the comand: series(%,xi) and as a result obtained:

24*b[0]-24*b[1]+(72*b[1]-72*b[2]-12*P[cl]*L^2)*xi+(144*b[2]+12*P[cl]*L^2)*xi^2

My question is: the way to solve this equation in the manner I need is setting the coefficients...

Hi, experts

I have an array which is empty, say

f:= []:

is there away to check if it's empty? I want to do something like in Matlab:

if isempty(f)

......

end

Many thanks

Kyle

Hi together,

i wanted to assume the sequence term A(n) of the first Perrin-Pseude-Prime n = 271441.

It´s not that difficult, but i do not understand the Maple-output, and i´ve never found an explenation for it.

the input was: > MatrixVectorMultiply(M^90480, v);  --> its the Matrix formula for the recurrence.

the output gives me a vector (3x1) with the following entry in the second line (witch is the line i want to know):

Hello colleagues! :)

I have some problems after create two objects, with array row in object definition...

The result - two same object...

Use the Gram-Schmidt process to transform the basis
 
           {Vector(3, {(1) = 0, (2) = 1, (3) = 1}), 

             Vector(3, {(1) = 1, (2) = 1, (3) = 1}), 

             Vector(3, {(1) = 1, (2) = 2, (3) = 3})}

for the Euclidean space R3 into an orthonormal basis for R3

                               
Show 3D result.

I am having a great deal of difficulty with this exercise.

Clues are

>irem(floor(evalf(K*10^n,N)),10) can be used to find the nth digits of the decimal expansion of K

A given process.

The following procedure P(n) counts the number of prime numbers ≤ n.
> P:=proc(n) local c,i;
c:=0; # initialize the counter c=0
for i from 1 to n do
if type(i,prime) then c:=c+1;
end if; # increase c by 1 if i is prime

I have a multivariate polynomial equation, in that somehow I know the coeffcients, using this information, I want to extract the variables. This will be the opposite of coeffs function.

for e.g. I have 3*x3 + 5*x4

Given 3 and 5, I want to extract x3 and x4.

Thanks in advance.

Satya

I believe this is a result of my ignorance rather then the flaws of the system. Can anyone teach me how to do this question?

Given 

>sum(1/n, n = 1 .. N)

I want to find the value of N such that sum(1/n, n = 1 .. N) is greater then 20.12.

>solve(sum(1/n, n = 1 .. N) > 20.12, N)

Warning, solutions may have been lost

Help please! Thanks.

Hi, experts,

I have an equations:

How to solve a, and b? It's easy to know that a = 2, b =5 though.

Many thanks! 

In the name of God

Any one knows how I can summerize this:

v1:=Vector(3):v2:=Vector(3):v3:=Vector(3):v4:=vector(3):v5:=Vector(3):

these 5 are independent.

Maybe some help page of maple helps me.

Hossayni