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- bobbym
- 2013-03-03 06:07:35
Hi;
anonimnystefy is correct. If he gets replied to he usually deletes his post. This is how he amuses himself.
- anonimnystefy
- 2013-03-03 01:27:32
That is true, but replying to "him" doesn't have a point either.
- bob bundy
- 2013-03-03 01:22:11
You may be right, but, as 'he' isn't advertising anything, I don't see the point.
Bob
- anonimnystefy
- 2013-03-03 01:15:31
Just a feeling.
- bob bundy
- 2013-03-03 01:13:46
hi Stefy,
What made you think it was automated?
Bob
- anonimnystefy
- 2013-03-03 01:09:15
Hi Bob
If I were you, I would not reply to automated messages.
- bob bundy
- 2013-03-03 00:14:24
hi Igor
Welcome to the forum.
If anyone will know it'll be bobbym. He's not on-line at the moment.
Bob
- bobbym
- 2013-01-15 07:06:21
Hi;
Here is another example, looks like you can get Alpha to output anything you like, plug this in.
Integrate[ pi r^2 ] with r =f(h)
I have for you I hereby grant you lifetime permission to use Bob's constant whenever you wish, free of charge and without the requirement to acknowledge whose constant it is.
Well, at least I have that going for me.
At least you are on this planet!
Wish I could get off but they won't take me. There is one major difference between me and that fellow, if you hit me on the head three times hard with a hammer you can change my mind.
- bob bundy
- 2013-01-15 06:55:18
hi bobbym
bobbym wrote:
Sorry, but if there is a bigger bungler than the poster you are talking about it is me.
WHAT! You must be joking! At least you are on this planet! And in honour (honor?) of the respect I have for you I hereby grant you lifetime permission to use Bob's constant whenever you wish, free of charge and without the requirement to acknowledge whose constant it is.
Can't say fairer than that, can I ?
Got to go and do the washing up now, see you later. 
Bob
- bobbym
- 2013-01-15 06:50:22
Hi bob;
Sorry, but if there is a bigger bungler than the poster you are talking about it is me.
I have been playing with alpha and entered this
Integrate pi r^2 , h assuming r = f(h)
- bob bundy
- 2013-01-15 06:44:02
hi bobbym
bobbym wrote:
Nope, k is a constant, so I am thinking k = k^2.
Not in my world, please. That k is important to me. I'm thinking of naming it "Bob's constant". So you'll have to pay royalties just to use it!
Bob
- bobbym
- 2013-01-15 06:34:36
Hi;
anonimnystefy wrote:
You mean (kh)^2, bobbym?
Nope, k is a constant, so I am thinking k = k^2.
bob bundy wrote:
Is there any way to get 'help' on what Wolfram can do? I see no 'help' link.
Not that I know of.
- bob bundy
- 2013-01-15 06:18:37
hi bobbym and Stefy
bobbym wrote:
Why not just substitute .......
Well it's not exactly a tough integral is it? So you could also say "Why not just do it yourself?" And I have, of course!
A certain poster has read that post and used Wolfram to show I have done it incorrectly. And I have; if there's no relationship between r and h. So I just wondered if there was a way to get Wolfram to do it with the relationship. Substituting would probably be inadmissible.
Alternative question: Is there any way to get 'help' on what Wolfram can do? I see no 'help' link.
Bob
- anonimnystefy
- 2013-01-15 06:05:25
You mean (kh)^2, bobbym?
- bobbym
- 2013-01-15 05:21:38
Hi;
Why not just substitute that