dharr

Dr. David Harrington

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20 years, 107 days
University of Victoria
Professor or university staff
Victoria, British Columbia, Canada

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I am a retired professor of chemistry at the University of Victoria, BC, Canada. My research areas are electrochemistry and surface science. I have been a user of Maple since about 1990.

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These are replies submitted by dharr

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);
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);
That's interesting, and very mysterious. Do you get it OK with my example? I am using Maple 10 under Windows. The only other thing I can think of would be that using a list might force the order, and I was just lucky skipping it. display([points, curve]); display([curve,points]);
That's interesting, and very mysterious. Do you get it OK with my example? I am using Maple 10 under Windows. The only other thing I can think of would be that using a list might force the order, and I was just lucky skipping it. display([points, curve]); display([curve,points]);
Thanks, this works. I look forward to Maple 11 then.
Thanks, this works. I look forward to Maple 11 then.
Yes, sorry about that.
Yes, sorry about that.
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