A float can be converted to a rational, so the following does something similar (note the decimal point to convert the fraction to a real) convert(77/45.,rational,3); # gives 12/7 convert(77/45.,rational,2); # gives 5/3

A float can be converted to a rational, so the following does something similar (note the decimal point to convert the fraction to a real) convert(77/45.,rational,3); # gives 12/7 convert(77/45.,rational,2); # gives 5/3

try nops (number of operands). I would have expected it to be under ?list, but I see it isn't (in v. 10), which does seems like a serious omission.

And yet, a/bc is usually interpreted in typeset notation as a/(b*c). In the International Union of Pure and Appled Chemistry rules: "In evaluating combinations of many factors, multiplication takes precedence over division in the sense that a/bc should be interpretaed as a/(bc) rather than (a/b)c; however in complex expressions it is desirable to use brackets to eliminate any ambiguity" Usually the IUPAP (physics) and IUPAC (chemistry) rules are in agreement, and when I write papers with my mathematics colleagure, we also use this convention, so I think it is fairly universal. (I am thinking as it it written on the page, i.e., output, here; I am not suggesting input a/b*c be interpreted this way, since (a/b)*c is the standard programming interpretation of input.)

You have solved x4=0 and x5=0, but x4=x2-x1 and x5=x3-x1, so no matter what x1 is, we must have x2=x3. So we really only have one equation x2=x3. If you solve this for C, you can get C as a function of L, but to get both L and C you need some more information.

Of course, I agree it is not a proof in the usual sense of the word; I was using it loosely. George used the word "verify", which would have been better. If the Logic package tests all the possibilities in a truth table is that more verified that if it hard codes that knowledge? I take the point of view with Maple that if it is hard coded in, then it is a piece of knowledge that has been proved elsewhere. Now perhaps there is a typo in some code that means that Maple is wrong, but that is a bit like an unfound error in a published proof. I accept that Maple is right as often as mathematicians, i.e., almost all of the time. Cheers, David.

Of course, I agree it is not a proof in the usual sense of the word; I was using it loosely. George used the word "verify", which would have been better. If the Logic package tests all the possibilities in a truth table is that more verified that if it hard codes that knowledge? I take the point of view with Maple that if it is hard coded in, then it is a piece of knowledge that has been proved elsewhere. Now perhaps there is a typo in some code that means that Maple is wrong, but that is a bit like an unfound error in a published proof. I accept that Maple is right as often as mathematicians, i.e., almost all of the time. Cheers, David.

This works without the logic package: evalb((not(a and b))=((not a) or (not b))); evalb((not(a or b))=((not a) and (not b))); again both returning true. These boolean operators can return fail, so are different from the others, but that isn't a problem here since you get true back. The = is "equivalent" in the context of evalb. I don't think you need implies here. Cheers, David.

This works without the logic package: evalb((not(a and b))=((not a) or (not b))); evalb((not(a or b))=((not a) and (not b))); again both returning true. These boolean operators can return fail, so are different from the others, but that isn't a problem here since you get true back. The = is "equivalent" in the context of evalb. I don't think you need implies here. Cheers, David.

I didn't look in the programming guide, but to show de Morgan's laws I would just do: with(Logic): Equivalent(¬(a &and b), (¬ a) &or (¬ b)); Equivalent(¬(a &or b), (¬ a) &and (¬ b)); Since both are true this is a proof. Cheers, David.

I didn't look in the programming guide, but to show de Morgan's laws I would just do: with(Logic): Equivalent(¬(a &and b), (¬ a) &or (¬ b)); Equivalent(¬(a &or b), (¬ a) &and (¬ b)); Since both are true this is a proof. Cheers, David.

note that x=10 (log(x)=1) is plotted one unit of length to the left of x=100 (log(x)=2). So moving left one unit each time, we plot x values of 100 10 1 0.1 0.01 1e-3 1e-4 1e-5 1e-6 Since log(0)= -infinity, very far left we get very small positive values, but we never get to negative x values.

note that x=10 (log(x)=1) is plotted one unit of length to the left of x=100 (log(x)=2). So moving left one unit each time, we plot x values of 100 10 1 0.1 0.01 1e-3 1e-4 1e-5 1e-6 Since log(0)= -infinity, very far left we get very small positive values, but we never get to negative x values.

This perfectly represents the Maple learning curve, and is very much the way I go about things, i.e., you eventually get it, and then contract it down. In simple cases like this, where I know where I am headed and just want to check it, I simplify the difference of my starting expression and my guess and check for zero; in your case that would be: simplify((tan(a)+tan(b))/(1-tan(a)*tan(b))-tan(a+b)); But it doesn't work! A quick fiddle and it does simplify(expand(tan(a)+tan(b))/(1-tan(a)*tan(b))-tan(a+b))); Cheers, David.

Glad it worked for you. Just a tip: diff(y(x), x, x, x, x); is simpler as diff(y(x), x, x, x, x); which can be further abbreviated as diff(y(x), x$4);