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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

Hey guys.I have a little question or just a personal wonder about them.

Question is: for any ?

What about ?

Test:

*Last edited by Yusuke00 (2014-02-09 04:57:39)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,215

Hi;

Please adjust the latex.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

Done yeah cool now i found out how it works. hehe

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,215

Hi;

How can K be an element of R?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

It's N* sorry math. I don't really know how to write N* or R*+.How you do that?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,215

You can just put N or R. You can use

http://www.codecogs.com/latex/eqneditor.php

for all ypur latexing.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

Not any opinions/ideas yet?

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,507

Yusuke00 wrote:

Hey guys.I have a little question or just a personal wonder about them.

Question is: Is for any ?

What about ?

Fixed the post a bit.

Why would you think they cannot be positive?

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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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

You got it wrong.

In my opinion the first one is always positive for any x real but i don't really know how to prove it.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,215

Hi;

That is the gf for that sum. It is obviously positive when x >=0. You might now try to prove that the numerator and denominator have the same sign for x<0.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

That would be quite hard.

I know the problem is to prove that x does not have roots on (-1,0) because it's easy to see on the other cases.Ideas?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 85,215

Hi;

Is it really that hard?

Take the numerator when x<0. It is obvious that x^(2k+1) is always negative and therefore x^(2k)-1 is always negative.

Now the denominator is obviously negative for x<0, so we have (-) / (-) which is always positive.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Yusuke00****Member**- Registered: 2013-11-19
- Posts: 43

Indeed,good point you are right.

http://www.mathsisfun.com/data/function-grapher.php?func1=sqrt%28x%29&func2=2

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,507

Or, you could just say that, for any integer n, x^(2n)>0.

Also, I was confused by the question. It seems, by what you wanted, that it should have been "Is f>0 for every x in R?".

*Last edited by anonimnystefy (2014-02-10 15:22:08)*

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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