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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,257

hi

The screen shot below shows what Wolfram does to

Is there a way to tell the software that r is a function of h ?

You may recognise this from another thread.

Thanks,

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

It depends on the type of dependence between r and h...

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,257

Just r = kh (k a constant)

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,610

Hi;

Why not just substitute that

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

You mean (kh)^2, bobbym?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,257

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

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,610

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.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,257

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,610

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)

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,257

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,610

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.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,257

hi Igor

Welcome to the forum.

If anyone will know it'll be bobbym. He's not on-line at the moment.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

Hi Bob

If I were you, I would not reply to automated messages.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,257

hi Stefy,

What made you think it was automated?

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

Just a feeling.

Here lies the reader who will never open this book. He is forever dead.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,257

You may be right, but, as 'he' isn't advertising anything, I don't see the point.

Bob

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,859

That is true, but replying to "him" doesn't have a point either.

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,610

Hi;

anonimnystefy is correct. If he gets replied to he usually deletes his post. This is how he amuses himself.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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