Carl Love

Carl Love

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12 years, 297 days
Himself
Wayland, Massachusetts, United States
My name was formerly Carl Devore.

MaplePrimes Activity


These are replies submitted by Carl Love

@mutaz  the domain is 0≤x1≤2pi.

I was afraid that you would say that that's the domain.... The solution we have so far is only good for 0 ≤ x1 ≤ 0.066, as Adri explained. To find solutions on the rest of the domain we need more initial conditions. Indeed, the initial condition we have for the current solution is just one that Adri made up, V(0)=0. We can do it for an arbitrary intial condition like this

dsolve({DE, V(x0)=v0}, V(x1));

and you'll get a solution only very slightly more complicated than what we already have. You will not be able to numerically evaluate this until you supply numeric values for x0 and v0. You may be able get some more information from the original PDE problem.

We also need to address the issue of the significance of x2 in your original problem. Whatever solution we get for V(x1), we need to add to it an arbitrary function of x2 but not x1: V(x1,x2) = V1(x1) + F(x2), where the V1 represents the solution from dsolve. Note that the F(x2) disappears when you take the derivative with respect to x1.

@mutaz  the domain is 0≤x1≤2pi.

I was afraid that you would say that that's the domain.... The solution we have so far is only good for 0 ≤ x1 ≤ 0.066, as Adri explained. To find solutions on the rest of the domain we need more initial conditions. Indeed, the initial condition we have for the current solution is just one that Adri made up, V(0)=0. We can do it for an arbitrary intial condition like this

dsolve({DE, V(x0)=v0}, V(x1));

and you'll get a solution only very slightly more complicated than what we already have. You will not be able to numerically evaluate this until you supply numeric values for x0 and v0. You may be able get some more information from the original PDE problem.

We also need to address the issue of the significance of x2 in your original problem. Whatever solution we get for V(x1), we need to add to it an arbitrary function of x2 but not x1: V(x1,x2) = V1(x1) + F(x2), where the V1 represents the solution from dsolve. Note that the F(x2) disappears when you take the derivative with respect to x1.

@mutaz The integral can only be evaluated numerically, for specific numeric values of x1. What is the domain for x1?

@mutaz The integral can only be evaluated numerically, for specific numeric values of x1. What is the domain for x1?

@adel-00 I don't understand what you mean. The plot command that I gave below does it all without any loops.

@eddyliu You need to get one more vote up for yourself before you can vote on others' answers.

Edit Add: Now you've got that extra vote. So now you can vote by clicking on that "thumb up" symbol that you see in the upper right corner of Answers, Questions, and Posts. There is no way to "close" a question.

acer wrote:

BTW, the values of the grayscale conversion are just one way amongst several possible for getting the "intensity". Another way is to grab the 3rd ("V"=intensity) layer of the HSV conversion.

The Y coordinate of the YUV color system is "the" classic black-and-white representation of the image. See ?ImageTools,RGBtoYUV and the Wikipedia article on YUV.

acer wrote:

BTW, the values of the grayscale conversion are just one way amongst several possible for getting the "intensity". Another way is to grab the 3rd ("V"=intensity) layer of the HSV conversion.

The Y coordinate of the YUV color system is "the" classic black-and-white representation of the image. See ?ImageTools,RGBtoYUV and the Wikipedia article on YUV.

@HDN46 This requires only a slight modification of my earlier code.

restart;
t:= Array(-1..20, [0,1]):  # [0,1] are initial values
Soln:= table():
abc:= combinat:-cartprod([[$-20..20] $ 3]):  #Iterator
while not abc[finished] do
    assign(('a','b','c')= abc[nextvalue]()[]);
    for n from 0 to 19 do        
        t[n+1]:= iquo((a*n*(n+1)+b)*t[n] + c*n^2*t[n-1], (n+1)^2, 'r');
        if r <> 0 or t[n+1] = 0 and t[n] = 0 then  break  end if;
    end do;
    if n = 20 then  
        Soln[a,b,c]:= convert(t,list);
        print(a,b,c)
    end if
end do:
[indices](Soln);
eval(Soln);

There are 34 solutions. Some are interesting.

@HDN46 This requires only a slight modification of my earlier code.

restart;
t:= Array(-1..20, [0,1]):  # [0,1] are initial values
Soln:= table():
abc:= combinat:-cartprod([[$-20..20] $ 3]):  #Iterator
while not abc[finished] do
    assign(('a','b','c')= abc[nextvalue]()[]);
    for n from 0 to 19 do        
        t[n+1]:= iquo((a*n*(n+1)+b)*t[n] + c*n^2*t[n-1], (n+1)^2, 'r');
        if r <> 0 or t[n+1] = 0 and t[n] = 0 then  break  end if;
    end do;
    if n = 20 then  
        Soln[a,b,c]:= convert(t,list);
        print(a,b,c)
    end if
end do:
[indices](Soln);
eval(Soln);

There are 34 solutions. Some are interesting.

Your post does not have the file attached.

Why does it seem impossible here to suppress or bypass evalhf? Doesn't library code usually try evalhf and then fall back to evalf if that fails? Also, setting Digits to a value significantly larger than evalhf(Digits) does not help, nor does setting UseHardwareFloats to false.

Why does it seem impossible here to suppress or bypass evalhf? Doesn't library code usually try evalhf and then fall back to evalf if that fails? Also, setting Digits to a value significantly larger than evalhf(Digits) does not help, nor does setting UseHardwareFloats to false.

@emma hassan Here's an update to your worksheet using the `.` operator and other modern Maple Matrix concepts. You can see that the values generated for the U vectors are the same as what you had. The option in the plot heights= histogram didn't make sense to me because I think that you want a surface, so I took it out.

y_not.mw

@emma hassan Here's an update to your worksheet using the `.` operator and other modern Maple Matrix concepts. You can see that the values generated for the U vectors are the same as what you had. The option in the plot heights= histogram didn't make sense to me because I think that you want a surface, so I took it out.

y_not.mw

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