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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

hi Stefy,

Thanks for the reply. The idea that √ and 'square root' don't mean quite the same thing was new to me so I've been thinking about it.

So how does the principle value concept work then?

Bob

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

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**zetafunc.****Guest**

From complex analysis I think you write the number in polar form, then use De Moivre's theorem to find the principal value, with the angle between -π and π (and r is non-negative).

In this case, we get 1 + i√3 as the principal value.

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,092

Thanks zetafunc.

So, in practice, does it really matter?

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: 14,809

Does it matter in which sense?

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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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 230

This is a solved example from the book, but has no negative and positive sign in it finall answer though I expect;

Example 10.26

Make u the subject of the relation

V = 1/4(25-U^2)^1/2.

And

Make V the subject.

E = V + 1/2MV^2.

What do you make of this?

Why do suppose the book did not bring both signs at all, in each case?

Thanks.

*Last edited by EbenezerSon (2013-08-06 01:04:17)*

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**zetafunc.****Guest**

bob bundy wrote:

Thanks zetafunc.

So, in practice, does it really matter?

Bob

If one is not careful, it can lead to incorrect solutions to integrals like the one in #48 (if the method of evaluation is to use a contour with the residue theorem).

**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 230

EbenezerSon wrote:

Make u the subject of the relation

V = 1/4(25-U^2)^1/2.

And

Make V the subject.

E = V + 1/2MV^2.

The above is a problem in a book, it is solved in it, but the letter which should be the subject in each case carries a square, and to my suprise the right hand side eqution has no positive and negative sign in front of the root signs and it really baffles me why the signs are absent.

Do anyone have an idea why the signs where omitted by the book?

Thanks.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,418

Hi;

I am getting

**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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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 230

Yes that's fine.

But I suppose when you solve it down the RHS must take 'the square root sign'

and a positive and a negative sign as well. Strangely only the root sign is present.

What do you think is like that in the book?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,418

In that case it was not necessary to take a square root at any time.

**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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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 230

So you mean there will be no square at the RHS at all? The book has it there.

My problem with it is, why there is no positive and negative sign in front of the root sign at the RHS in the book.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,418

The above is a problem in a book, it is solved in it, but the letter which should be the subject in each case carries a square

You said the subject is squared.

I took your quote to mean you wanted to solve for U^2. If you want to solve for U then we will take a root.

**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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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 230

But if I am solving for U, and take root of both sides, I think there will come a root sign to the the R.H.S, in front of the root sign there must be positive and negative sign present, is that not it?

Thank you Bobbym.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,418

Yes, if you solve for U, then there will be a ± in front.

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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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 230

Thanks, Bobbym

I cant tell why the book did not bring that.

Thanks once more.

Amen.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,418

It just means one is positive and the other is negative.

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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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 230

It seems you did not get what I mean.

I mean the book solved for U but there was no positive and negative sign in front of 'the root sign' to the R.H.S.

what do you think it did not have it?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,418

Because sometimes it is left out.

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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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 230

If for inştance, a student does that would not a teacher mark him down?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,418

Yes he/she might mark them down.

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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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 230

Thanks.

In the case of pythagoras theorem, I rarely see those sign in the final solution.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,418

When dealing with the Pythagorean theorem you are usually working on a triangle. Triangles can not have negative lengths.

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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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 230

Okay, Thanks bobbym.

Then I suppose, in change of subject, speed, velocity, time etc. Can't be negative in practise and therefore, if one is asked to make any of the above subject of a relation, should be in the postive form instead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,418

Speed is a scalar, so it can not be negative. Velocity is a vector, it can be negative. According to the physicists, time can take a negative value.

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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**EbenezerSon****Member**- Registered: 2013-07-04
- Posts: 230

On what basis can time take a negative value?

I haven't heard that before, did he prove why?

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