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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,922

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

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

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

Just r = kh (k a constant)

Bob

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,132

Hi;

Why not just substitute that

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,922

You mean (kh)^2, bobbym?

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

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,132

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.**

**If it ain't broke, fix it until it is.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

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: 95,132

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.**

**If it ain't broke, fix it until it is.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

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: 95,132

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.

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

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: 15,922

Hi Bob

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

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

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

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: 15,922

Just a feeling.

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

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,926

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: 15,922

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

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,132

Hi;

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

**In mathematics, you don't understand things. You just get used to them.**

**If it ain't broke, fix it until it is.**

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