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

Hello guys long time no see,got a new exercise:

On R[X] f=X^4-aX^3-aX+1

Proove that if |a|<1 then |x1|=|x2|=|x3|=|x4|=1;

Good luck.

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

Ok guys a second problem

f(1)=8

f(2)=-5

f(3)=0

f(4)=35

f=?

The thing is i know how to solve the problem but i cannot really solve the system.

f(3)=0 -> x1=3 solution ->f = (X-3)(aX^2+bX+c)==>

==>f=aX^3+(b-3a)X^2+(c-3b)X-3c

so f(3)=0 ==>-3c=0 =>c=0

f(1)=8 ==> a+b=-4

f(2)=-5 ==>-4a-2b=-5==>4a+2b=5

f(4)=35 ==> 16a+4b=35

Can anyone solve the system or at least tell me where i'm wrong ? I'm getting very mad here.

*Last edited by Yusuke00 (2014-02-03 09:56:58)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi;

f(1)=9

f(2)=-5

f(3)=0

f(4)=35

**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

bobbym wrote:

Hi;

f(1)=9

f(2)=-5

f(3)=0

f(4)=35

Could you describe your solution step by step please? For f(1)=8 not 9 sorry.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Hi;

After substitution into the general form you get.

Can you solve that or do you need a little more?

**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

Ok i got those ecuations either.W/E i'm stupid.Thank you.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,238

Are you okay with getting the solutions? if you need help just post back.

**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

Yes i had the same ecuations,but i've been tired and just couldn't focus yesterday.Got them now thank you.

By the way,any clues on 1st problem?

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

hi Yusuke00

I didn't post earlier because I do not understand what you mean by X1 X2 etc.

Bob

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

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

x1 are the solutions to the ecuation like...

f(x1)=0,f(x2)=0,f(x3)=0,f(x4)=0;

Sorry that's how i know to note them.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

hi Yusuke00

That's what I thought you meant and that's where my difficulty began.

If you go to this page of the MIF site

http://www.mathsisfun.com/data/function … x)&func2=2

you will find a very useful function grapher. It even allows you to put in an 'a' and adjust its value using the slider.

This shows no roots at all.

Bob

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

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

For a>=1 there are real solutions.For a<1 there are not.That function grapher is for R not for C.

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

But it's very helpful,i like it.Thank you very much ^_^.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,426

On R[X] f=X^4-aX^3-aX+1

So what does R[X] mean ?

Bob

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

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

It means that f(x) =y,y is part of R.

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