A level Further Pure Maths help (Polynomials)

[usman94]

1. The problem statement, all variables and given/known data

Find the values of Σ(a^2), Σ(1/a), Σ(a^2)(B^2) and ΣaB(a + B) for: x^4 - x^3 + 2x + 3 = 0

2. Relevant equations

Σa = 1, ΣaB = 0, ΣaBC = -2, aBCD = 3

3. The attempt at a solution

I found the Σ(a^2) and Σ(1/a) successfully correct bt could neither find Σ(a^2)(B^2) nor (Σa)(ΣaB):
'ΣaB(a + B) = 6' is given as the answer but my own answer comes 4. Please somebody post the complete solution for ΣaB(a + B) explaining each step, given that i found ΣaB(a + B) = (Σa)(ΣaB) - 2(ΣaBc).
It seems i should have got ΣaB(a + B) = (Σa)(ΣaB) - 3(ΣaBc) in order to get the correct answer '6' instead of the erroneous '4'.
If perhaps this is true, then prove that ΣaB(a + B) = (Σa)(ΣaB) - 3(ΣaBc)

I found Σ(a^2)(B^2) = (ΣaB)^2 - 2Σ(a.B^2.C) - 4aBCD. Now 4m here i can't proceed forward to find Σ(a.B^2.C). Perhaps, i hav done it wrong or there exists an alternative easier way.
NB: a,B,C,D represent the roots alpha, beta, gamma and the 4th root (partial derivative sign) respectively.

[Mark44]

1. The problem statement, all variables and given/known data

Find the values of Σ(a^2), Σ(1/a), Σ(a^2)(B^2) and ΣaB(a + B) for: x^4 - x^3 + 2x + 3 = 0
I don't understand the question. What do the summations have to do with the polynomial equation? That is, how are a and B connected with x^4 - x^3 + 2x + 3 = 0?

2. Relevant equations

Σa = 1, ΣaB = 0, ΣaBC = -2, aBCD = 3

3. The attempt at a solution

I found the Σ(a^2) and Σ(1/a) successfully correct bt could neither find Σ(a^2)(B^2) nor (Σa)(ΣaB):
'ΣaB(a + B) = 6' is given as the answer but my own answer comes 4. Please somebody post the complete solution for ΣaB(a + B) explaining each step, given that i found ΣaB(a + B) = (Σa)(ΣaB) - 2(ΣaBc).
It seems i should have got ΣaB(a + B) = (Σa)(ΣaB) - 3(ΣaBc) in order to get the correct answer '6' instead of the erroneous '4'.
If perhaps this is true, then prove that ΣaB(a + B) = (Σa)(ΣaB) - 3(ΣaBc)

I found Σ(a^2)(B^2) = (ΣaB)^2 - 2Σ(a.B^2.C) - 4aBCD. Now 4m here i can't proceed forward to find Σ(a.B^2.C). Perhaps, i hav done it wrong or there exists an alternative easier way.
NB: a,B,C,D represent the roots alpha, beta, gamma and the 4th root (partial derivative sign) respectively.

[usman94]

ohhh m terribly sorry missing to mention dat. a, B, C and D are the 4 roots of the equation: x^4 - x^3 + 2x + 3 = 0

[Mark44]

OK, then what do Σ(a^2) and the other summations mean in the context of this problem?

I'm still clueless as to what this problem is asking for.

[usman94]

Exercise 9d Q2(d) of this book:
http://books.google.com.pk/books?id=gB5QyMiCOGsC&pg=PA294&dq=zFurther+Pure+Maths&hl=en&ei=wy-VTNazLovJcfbivKQF&sa=X&oi=book_result&ct=result&resnum=2&ved=0CC4Q6AEwAQ

Here is an attachment to the question:
http://www.physicsforums.com/showthread.php?t=430362

[usman94]

these summations are of the roots e.g. Σa means a + B + C + D. Likewise ΣaB represents aB + BC + CD + DA

[Mark44]

The first look opens p. 294 of this book. In a quick search I didn't find exercise 9d Q2(d). The attachment in your other thread for this problem doesn't give me enough context to know what the summations are adding.

[statdad]

Are you working on functions of the roots of a polynomial (theory of equations?)

[epenguin]

It may be helpful to write the thing out in full and then you will see there are three not two products giving you \alpha\beta\gamma

I think you are agreeing with me (?) that \Sigma\alpha\beta(\alpha + \beta) is just 2\Sigma\alpha^2\beta

You can get that out of \Sigma\alpha\beta\Sigma\alpha.

\Sigma\alpha\beta\Sigma\alpha

= 0 because

\Sigma\alpha\beta = 0 .

On the other hand \Sigma\alpha\beta\Sigma\alpha also equals in full

(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta)(\alpha + \beta + \gamma + \delta)

and there are two products \alpha^2\beta and three products \alpha\beta\gamma

So you've got essentially your answer

0 = 2\alpha^2\beta - 3\alpha\beta\gamma


At first you find them. Later you think you need to look in the first bracket only at the ones without \delta. There are three of them. And they each have their partner in the second bracket to make \alpha\beta\gamma. Later you realise you don't need to look at each of the three because of the symmetry. And you realise there not just are but must be three - and that the \Sigma notation is quite nifty - however you can always fall back on the full formulae if out of practice or something not working.

(I tried to bring them out by colouring the three pairs but it is already rubbish enough to try and write ordinary tex because of defects I mentioned earlier, nearly impossible to edit :grumpy:, however when you write on paper you can use colour underlining if it helps.)

[epenguin]

Edit: That last line should be of course

0 = 2Σα2β - 3Σαβγ