The same plot as above, refined to log-log, and with gridlines at key locations.

T:= 10/7:  K:= k= 0..3:
plot(
     L[1..7^4-1],
     axis[1]= [mode= log, tickmarks= ['7^k=``(7)^k' $ K], gridlines= [7^k $ K]],
     axis[2]= [
          mode= log,
          tickmarks= ['T^k = (evalf(T^k)=``(T)^k)' $ K],
          gridlines= [T^k $ K]
     ]
);

The same plot as above, refined to log-log, and with gridlines at key locations.

T:= 10/7:  K:= k= 0..3:
plot(
     L[1..7^4-1],
     axis[1]= [mode= log, tickmarks= ['7^k=``(7)^k' $ K], gridlines= [7^k $ K]],
     axis[2]= [
          mode= log,
          tickmarks= ['T^k = (evalf(T^k)=``(T)^k)' $ K],
          gridlines= [T^k $ K]
     ]
);

@abbeykabir wrote:

the only diffrence I notice was in my error.

Does that mean that you still have an error?

you make a copy of the LinearAlgebra package and named it LA, having all the functions and sub-packages of the original LinearaAlgebra.

Not quite. There is no copy made. It is just a substitution of the abbreviation LA for the name LinearAlgebra.

So instead of using the whole LinearAlgebra, I can just us LA.

Yes, instead of typing the whole name LinearAlgebra, you can just type LA. I do this because I don't like using the with command for several reasons:

1. with makes it hard to see which packages commands come from and hard to look up help about them.
2. with does not work inside procedures.
3. with might overload operators (for example, change the meaning of +) without you noticing it. For example, with(LinearAlgebra) assigns a meaning to &x.

I have some questions for you I my previous thread. Don't think it will be good enough to ask you here.

Please make a separate thread, a new Question. It is too difficult to notice when things are added to an old thread.

@abbeykabir wrote:

the only diffrence I notice was in my error.

Does that mean that you still have an error?

you make a copy of the LinearAlgebra package and named it LA, having all the functions and sub-packages of the original LinearaAlgebra.

Not quite. There is no copy made. It is just a substitution of the abbreviation LA for the name LinearAlgebra.

So instead of using the whole LinearAlgebra, I can just us LA.

Yes, instead of typing the whole name LinearAlgebra, you can just type LA. I do this because I don't like using the with command for several reasons:

1. with makes it hard to see which packages commands come from and hard to look up help about them.
2. with does not work inside procedures.
3. with might overload operators (for example, change the meaning of +) without you noticing it. For example, with(LinearAlgebra) assigns a meaning to &x.

I have some questions for you I my previous thread. Don't think it will be good enough to ask you here.

Please make a separate thread, a new Question. It is too difficult to notice when things are added to an old thread.

It's still a bit kludgy, but it's less kludgy than mine. I made your Reply into an Answer so that I could vote up. If you mind, I'll put it back.

The Dirac solution can be evaluated with two-argument eval:

eval(h, f= 3);

I tried to do something similar with Physics:-KroneckerDelta(i,1), but I couldn't make it work.

It's still a bit kludgy, but it's less kludgy than mine. I made your Reply into an Answer so that I could vote up. If you mind, I'll put it back.

The Dirac solution can be evaluated with two-argument eval:

eval(h, f= 3);

I tried to do something similar with Physics:-KroneckerDelta(i,1), but I couldn't make it work.

It may be possible to represent the diagram within Maple's geometry package and have Maple say that the answer cannot be determined.

It may be possible to represent the diagram within Maple's geometry package and have Maple say that the answer cannot be determined.

@Kitonum The timing of your Tuples is impressive, no doubt. But it is specialized to the Asker's specific problem---a Cartesian product of identical lists. The other procedure, CartProdSeq, will do the product of any list of lists. So it is the equivalent of the Mathematica command mentioned.

What does w = (u,v) mean?

I've read your worksheet, and I understand the problem. But could you make a smaller example of it not working? I am measuring by the number of substitutions. So, could you make an example with, say, three substitutions, where the iterative application of simplify with side relations does not yield the same results as the all-at-once application?

Note to other readers: I am not saying that the iterative application should give the same results as the all-at-once! However, in an earlier very similar problem that Casper had, I was able to order the side relations in such a way that the results were the same, with the iterative way being thousands of times faster.

What exactly do you mean by Components? Do you mean embedded components? I can offer a great deal of advice about your worksheet if that's what's wanted.

(I am tempted to use my editorial discretion to make this a Question, but this time I'll leave it to someone else to make the final decision.)

Suppose we take the circle as fixed, b = c, and a < < b. As a --> 0, b and c approach a diameter of the circumcircle. So, b --> 2*R and c --> 2*R. Thus R*p = R*(a+b+c)/2 --> 2*R^2 and 2*a*R + b*c --> 4*R^2. So, R*p/(2*a*R+b*c) --> 1/2.

OTOH, suppose the circle is fixed, b = c, and a is at its maximal value, 2*R. Then it's a right triangle and b = c = sqrt(2)*R. Then R*p/(2*a*R+b*c) = (1+sqrt(2))/6 = 0.402, which is just slightly larger than 2/5.

While neither of these arguments is a proof of either side of the inequality, they do make it seem plausible. Somewhere we need to use the acuteness condition. Is acuteness equivalent to  a^2 < b^2 + c^2 with a being the longest side? Is acuteness equivalent to the center of the circle being inside the triangle?

How about trying to use Maple in the c# program? If you get an error, then trap it.

@Alejandro Jakubi FWIW, lprint shows that the 2D if in prefix form is translated to 1D `if` as an automatic simplification.