520 Reputation

7 Badges

7 years, 224 days

MaplePrimes Activity

These are replies submitted by Zeineb


thank you


Please, how can I select from your code only the solution that verify some assymption ( positive parameter ) as added in the code 

Thank you for your answer, 
You say you beleave that this can be done more simply, can you propose another idea 



The recursive equation is proposed in my code 
B_^p is defined using B_i^{p-1} 
So, for p=1 B_^0 is known so we can determine B_i^1

B := proc (i, p, t) if p = 0 then piecewise(t < t[i-2] or t[i-1] <= t, 0, 1) else (t-t[i-2])*B_recursive(i, p-1, t)/(p*h)+(t[i+p-1]-t

@Carl Love 

Thanks for looking to my question. 

First :  the sequence alpha is defined using a general formula 
alpha[i]=(-1)^{i+1}  15/(5i-4), for i from 1 to 5 

I would like to get a simular formula like alpha[i] so that 
beta[3] =alpha[1]
beta[4] =alpha[4]
beta[5]= alpha[5]

can I have general form of beta[i]=.......... formula that depend only on i 


Thank you, 
In the system of equations A , B and C must be replaced by A(t), B(t) and C(t) 
because they unknown functions 

I run the same code I get 
Error, (in ODEtools/ODESolStruc/info) incorrect ODESolStruc: expected 1 old dependent variable in the 2nd set of transformation equations; received: 0

I have 8 unknows functions, \Zeta_i(t,x,y,z), i=0,1,2,3 and A(t), B(t), C(t), \psi(t,x,y,z)

Thank you for your help 


The difference equation is the first equation in the maple code : system defined at three different times (n+1), n and n-1 



thank you


I think everything is well now, in my code. 



In the definition of beta_1 the loop from 1 to q is correct, beta_1 is a coefficient used  in A 

and everything run correctly 


@Carl Love 

so only one i a field, so can not get a homomorphism between them


This means that in geneal case, for a given x,y,z and w , we can always find n, a,b and c solution of the system 

Since we have many choice of x , y, z and we so for each choice we have a solution 

there are infintely many solution of the system 

@Axel Vogt 

But Maple return a zero solution 
Instead of a nonzero solution 

for a fixed x=1, y=1, z=6 and w=-1 we have 

n=4, a=-1, b-0, c=5 

is a solution for this fixed vector defined by x,y,z and w 


Big thanks 


The function f is defined only in the unit ball centered at zero and outside is zero. 
So, we have singularity at zero, and it is integrable in the vicinity of zero since 2< 3 


Using polar coordinate you will integrate over the range [0,1]  the function r^(-2) r^(3-1) =1  and so integrable 

In general case, f(x)=|x|^(-alpha) times indicator of unit ball, is integrable over the unit ball if alpha< dimension of space  ( in our example alpha equal 2 < 3 ), so the function is integrable 

1 2 3 4 5 6 7 Last Page 1 of 11