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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

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.  smile

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

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