@Alejandro Jakubi 

But the second part of my example shows that assuming is not following this design principle (also from ?assuming,details ):

The output is the same as that received by successively doing the following.
  1. Calling assume (to enter assumptions on names).
  2. Entering (and so evaluating under the assumptions) the expression depending on these names.
  3. Removing the assumptions.

The example shows that it is possible to make the assumptions effectively without scanning the procedure.

@Alejandro Jakubi 

But the second part of my example shows that assuming is not following this design principle (also from ?assuming,details ):

The output is the same as that received by successively doing the following.
  1. Calling assume (to enter assumptions on names).
  2. Entering (and so evaluating under the assumptions) the expression depending on these names.
  3. Removing the assumptions.

The example shows that it is possible to make the assumptions effectively without scanning the procedure.

@abbeykabir 

Use option linestyle to the plot command:

plot(
     [M[.., [1,2]], M[.., [1,3]]],
     labels= [time, `error`], legend= [error1, error2],
     linestyle= [dash, dashdot]
);

The possible values for linestyle are solid, dot, dash, dashdot, longdash, spacedash, and spacedot. See ?plot,options .

@abbeykabir 

Use option linestyle to the plot command:

plot(
     [M[.., [1,2]], M[.., [1,3]]],
     labels= [time, `error`], legend= [error1, error2],
     linestyle= [dash, dashdot]
);

The possible values for linestyle are solid, dot, dash, dashdot, longdash, spacedash, and spacedot. See ?plot,options .

@brian bovril 

All I can recommend is practice, asking questions, and constant re-reading of the help pages. Writing programs about primes is a good place to start. My first program, 35 years ago, was a program that generated primes by trial division.

@brian bovril 

All I can recommend is practice, asking questions, and constant re-reading of the help pages. Writing programs about primes is a good place to start. My first program, 35 years ago, was a program that generated primes by trial division.

If you increase Digits over the hardware-float threshold of 15, then the rounding errors disappear.

@abbeykabir

Good question. In M[.., [1,2]], the .. means "take all rows"; and the overall expression means "take all rows and columns 1 & 2 of M". If it was M[[1,2], ..], it would mean "take rows 1 & 2 and all columns". See ?rtable_indexing .

What do you mean by "modify the shape"? The shape is determined by the data. Or do you mean the style, like dashed lines and dotted lines?

@abbeykabir

Good question. In M[.., [1,2]], the .. means "take all rows"; and the overall expression means "take all rows and columns 1 & 2 of M". If it was M[[1,2], ..], it would mean "take rows 1 & 2 and all columns". See ?rtable_indexing .

What do you mean by "modify the shape"? The shape is determined by the data. Or do you mean the style, like dashed lines and dotted lines?

Indeed, you can include the inequalities in the solve command:

solve({x+y+z=6, x^2+y^2+z^2=14, x^3+y^3+z^3=36, x >= y, y >= z});

Indeed, you can include the inequalities in the solve command:

solve({x+y+z=6, x^2+y^2+z^2=14, x^3+y^3+z^3=36, x >= y, y >= z});

@N00bstyle The interface setting displayprecision is completely separate from the computational engine. Indeed, your variables will be maintained internally at 15 digits. The setting only affects how they are displayed.

In my opinion, 15 is the best default value for Digits, rather than Maple's default of 10.

@N00bstyle The interface setting displayprecision is completely separate from the computational engine. Indeed, your variables will be maintained internally at 15 digits. The setting only affects how they are displayed.

In my opinion, 15 is the best default value for Digits, rather than Maple's default of 10.

@Kitonum 

The step length AD must be a multiple of primorial(n) where n is the progression length, except when the first prime in the progression is n itself, in which case AD must be a multiple of primorial(n-1).

Some other examples of the exceptional case are

2, 3  (AD = 1 = primorial(2-1))
3, 5, 7  (AD = 2 = primorial(3-1))
5, 11, 17, 23, 29  (AD = 6 = primorial(5-1))

@Kitonum 

The step length AD must be a multiple of primorial(n) where n is the progression length, except when the first prime in the progression is n itself, in which case AD must be a multiple of primorial(n-1).

Some other examples of the exceptional case are

2, 3  (AD = 1 = primorial(2-1))
3, 5, 7  (AD = 2 = primorial(3-1))
5, 11, 17, 23, 29  (AD = 6 = primorial(5-1))