You are not logged in.
Pages: 1
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.
Offline
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)
Offline
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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
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.
Offline
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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
Ok i got those ecuations either.W/E i'm stupid.Thank you.
Offline
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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
Offline
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?
Offline
hi Yusuke00
I didn't post earlier because I do not understand what you mean by X1 X2 etc.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Offline
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.
Offline
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
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Offline
For a>=1 there are real solutions.For a<1 there are not.That function grapher is for R not for C.
Offline
But it's very helpful,i like it.Thank you very much ^_^.
Offline
On R[X] f=X^4-aX^3-aX+1
So what does R[X] mean ?
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
Offline
It means that f(x) =y,y is part of R.
Offline
Pages: 1