Zeineb

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4 years, 275 days

MaplePrimes Activity


These are replies submitted by Zeineb

@acer 

I appreciate your answer thank you

@acer 

thank you. Now, i get the solution with the modified code. 

 

@vv 

Thank you for your codes 

when I run your example two, I obtain the following error 
Error, missing operator or `;`

Example_two.mw

 

@dharr 

tank you for your answer, I tried but it does not work

solve(f>0, assuming  1<=x, x<=4, 0<y,y<1)
 

@Kitonum 

Thanks in advance for  your remarks

I want to verify the convolution product theorem using an example

convolution.mw

many thanks

@vv 

Thanks for your remarks.

I study the function f(y), I proved that this function is always positive by hand and it has a minimum at y_star= 1/4*log(5/15*4))

so according to your idea, 1/f(y) will be always positive and bounded above by 1/f(y_star).

thus

x-0=int(1/f(y), y=y0 .. y(x)) <= y(x)-y(0)/f(y_star)

so, for any y(0) the solution will always diverge to +infty
maybe this can be the answer why the solution diverges. Hope this idea conclude my question

many thanks

@vv 

Thanks .. very nice computation

@vv 

Thanks for your answer.
All quantities are obtained by hand. How can I get the coefficient of  1/z in the quantity H defined in maple code 

By_hand.mw

thanks

@tomleslie 

I appreciate all answer proposed for this question. thanks

@acer 

in this case : y=0 and 3<=x<=4, we get g =0<=0  ( this case is true)

I  want to show that max{g(x,y)} <=0 less or equal zero.

@acer 

Thank you

this can be proved by hand, its enough to use that 1-y/1+y always less than one. after that we obtain a function which depends only on x and simple derivative give us the critical point. 

 

@Ramakrishnan 

Thank you for your time to see my question.

The solution proposed by maple satify all conditions maybe its the unique solution

@tomleslie 

My vector is [u[1, 0], u[2, 0], u[3, 0], u[1, 1], u[2, 1], u[3, 1], u[1, 2], u[2, 2], u[3, 2], u[1, 3], u[2, 3], u[3, 3]]
the quanties


alpha=5;

beta=10;

u[0, 0]=  u[0, 1]= u[0, 2]= u[0, 3]=100

so all these parameters are fixed.

 

@tomleslie 

the quantities f[1, 0], f[1, 1], f[1, 2], f[1, 3], f[2, 0], f[2, 1], f[2, 2], f[2, 3], f[3, 0], f[3, 1], f[3, 2], f[3, 3]   are constant , we can drop them from the sytem if you want, my goal is obtain the matrix whose entries are coefficient of u[i,j]

 

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