Further Mathematics: Post your doubts here!

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XPF has Mathematics threads in almost every CIE section. This one was highly needed; the complicated nature of the subject makes any external help all the more welcome! So post all your Further Maths problems here from now on. :) Anyone who happens to (be awesome enough to :p ) have taken the subject, please let it be known and attempt a solution at the questions asked. It may not receive the same traffic an average Maths thread gets, all efforts are definitely 'thank-worthy'. :wink:
Any notes, whether scanned or in document form, would be very much appreciated. :Bravo:
 
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Re: The Further Maths Thread

Thank you, Nibz! :D
Here's the first problem:
If the roots of the equation x^4 - px^3 + qx^2 - pq x + 1 = 0 are α, ß, Γ and delta(D) show that:
( α + ß + Γ) ( α + ß + D) ( α + Γ + D) ( ß + Γ + D) = 1.
 
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Re: The Further Maths Thread

I guess this is a question from the book, coz I haven't been able to solve it. And probably, it is beyond our comprehension (quite common for the questions in book), atleast it is beyond mine. If you do get the solution, please post it here.
 
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Re: The Further Maths Thread

am...do u even have notes for Further mechanics??
would be very thankful if your can upload it!
 
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No, I uploaded everything I had on FM.
 
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Re: The Further Maths Thread

U CN TRY out EDEXCEL LIVETEXTS......A google search WILl most probably help u finding links M3 M4 AND M5 frther mechanics books...
 
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Re: The Further Maths Thread

hamidali391 said:
@abcde: Can't you just post another question that you incurred problems in? So that I stop undermining my abilities. :mad:
The equation 7x^3 - 4x - 11 = 0 has roots alpha, beta, gamma. Prove that:
(a) 7 Σ α^3 = 33 + 4 Σ α
(b) 7 Σ α^4 = 11 Σ α + 4 Σ α^2
Write down a similar expression for 7 Σα^5 .

The first two parts could be solved. I'm having a problem with bold text.
(undermining your ability is too drastic a step for one problem! :no: ) :)
 
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I am not solving this. I think this is a question from the book, and I hate the book. It will demoralise me to the lowest ebb. :D Chalo, I'll give it a try! Wasay most of hethings in the book are not in the syllabus and in Further Maths, its a universal law, to stick to the syllabus. ;)
 
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Fine, fine. The degree 5 ones aren't in the syllabus but it could be derived I think. :S Anyway, that's A/OK! :D
 
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Sorry, this is in the syllabus but in a totally different way as compared to the book. I have explained it below in the photo. Youwill get it once you solve some past papers. The good thing about our teacher is that he gives questions from the past papers in his worksheet alongwith questions from the book, that he thinks are appropriate, and this allows us to get ready for the real thing.
Don't mind the quality of the picture. I could only find a pencil around me. :S I am kind of short of pens. :p
 

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Re: The Further Maths Thread

@babamama99: This is an exclusive thread for Further Mathematics questions, so I shall move your post to the Mathematics section. Infact, you already posted it there. So no need to post the question at two places.
 
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@hamidali391: An effectively short solution that I had no idea existed. Thank you! :good:
Is this a general sort of formula you used (the one involving the variable n)? What would you do if told to evaluate S 3 and S 4?
 
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Re: The Further Maths Thread

Assalam-o-Alaikum...!

@hamid:

"If the roots of the equation x^4 - px^3 + qx^2 - pq x + 1 = 0 are α, ß, Γ and delta(D) show that:
( α + ß + Γ) ( α + ß + D) ( α + Γ + D) ( ß + Γ + D) = 1"

You can observe that, in the product "( α + ß + Γ) ( α + ß + D) ( α + Γ + D) ( ß + Γ + D)," each of the expression consisting of the sum of the roots of the equation "x^4 - px^3 + qx^2 - pq x + 1 = 0" forms a pattern.
Let's represent the sum "α + ß + Γ + D" by "S" to try to simplify the hairy product.

Then, ( α + ß + Γ) ( α + ß + D) ( α + Γ + D) ( ß + Γ + D) = ( S - D ) ( S - Γ ) ( S - ß ) ( S - α )
= S^4 - DS^3 - ΓS^3 + ΓDS^2 - ßS^3 + ßDS^2 + ßΓS^2 - ßΓDS - αS^3
+ αDS^2 + αΓS^2 - αΓDS + αßS^2 - αßDS - αßΓS + αßΓD
= S^4 - ( α + ß + Γ + D )*S^3 + ( αß + ßΓ + ΓD + Dα + αΓ + ßD )*S^2
- ( αßΓ + ßΓD + ΓDα + αßD)*S + αßΓD
= S^4 - (∑ α)*S^3 + (∑ αß)*S^2 - (∑ αßΓ)*S + αßΓD

Now, from the equation "x^4 - px^3 + qx^2 - pq x + 1 = 0," ∑ α = p = S; ∑ αß = q; ∑ αßΓ = pq; αßΓD = 1

This leads to, ( α + ß + Γ) ( α + ß + D) ( α + Γ + D) ( ß + Γ + D) = p^4 - (p)*p^3 + (q)*p^2 - (pq)*p + 1
= p^4 - p^4 + q*p^2 - q*p^2 + 1
= 1


@abcd:
As regards the method used to find sums of powers of roots of polynomial equations, it's quite simple but requires a clever method. I'll demonstrate the method using a quadratic equation.

Consider the quadratic equation x^2 - 4x + 5 = 0 with roots α and β.

If you multiply the cubic equation with x^n, you get
x^(n+2)-4x^(n+1)+5x^n=0
Since both α and β satisfy this equation,
α^(n+2)-4α^(n+1)+5α^n=0 and
β^(n+2)-4β^(n+1)+5β^n=0
Adding these two equations, we have
∑α^(n+2) - 4 ∑α^(n+1) + 5∑α^n = 0

If you denote ∑α^n by S(n), then this becomes
S(n+2) - 4 S(n+1) + 5 S(n) = 0

We know that α + β = 4 and αβ=5.
So α^2 + β^2 = (α + β)^2 - 2αβ = 4^2 - 2*5 = 6

Putting n = 1, S(3) - 4 S(2) + 5 S(1)=0
We know that S(1) = 4 and S(2) = 6, so S(3) = 4*6 - 5*4=4

I guess you understand the method and can apply it to equations of higher order.


Hope that helps.....:)
 
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Re: The Further Maths Thread

I understand. It really helped. Thank you! :)
However, I'm still having some confusion where S(n) notation is placed (I'm not familiar with it at all except since the last two posts). In the first problem, I realise that S(1) = ∑α = p. But how come S (4) = p^4 , S(3) = p^3 and S(2) = p^2 ?
I know that α + ß + Γ + D = p but S (4) = α^4 + ß^4 + Γ^4 + D^4 which is NOT equal to (α + ß + Γ + D)^4 that can give p^4. :S
Where can I further learn this "S(n+2) - 4 S(n+1) + 5 S(n) = 0" method?
 
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Basically, I need to know how to evaluate S(3) and S(4) in case of cubic and quartic equations.
 
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Oops...;p I guess I had allocated a misleading notation for the sum "α + ß + Γ + D." I should've used the symbol "S(1)" instead of just "S" to represent the sum, to avoid any type of confusion.

So, wherever I wrote "S," I indeed meant S(1). Accordingly, the product "( α + ß + Γ) ( α + ß + D) ( α + Γ + D) ( ß + Γ + D)" in the question lead to the expression "[S(1)]^4 - (∑ α)*[S(1)]^3 + (∑ αß)*[S(1)]^2 - (∑ αßΓ)*S(1) + αßΓD" upon simplification.

Then, from the equation "x^4 - px^3 + qx^2 - pq x + 1 = 0," ∑ α = p = S(1); ∑ αß = q; ∑ αßΓ = pq; αßΓD = 1

Leading to, ( α + ß + Γ) ( α + ß + D) ( α + Γ + D) ( ß + Γ + D) = [p]^4 - (p)*[p]^3 + (q)*[p]^2 - (pq)*p + 1
= p^4 - p^4 + q*p^2 - q*p^2 + 1
= 1

Note that none of S(2), S(3) or S(4) is involved in any of the expressions I wrote.



Hopin' I removed your confusion now....:)


P.S.
To be honest, I don't know of any further maths book covering this "elegant & exceptionally-simple + shortcut" method. However, the past-papers are full of questions on it. Just have a look at some of the questions about polynomials there (along with the corresponding er's/ms's), and you'll master this method very soon....!
 
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Re: The Further Maths Thread

abcde said:
@hamidali391: An effectively short solution that I had no idea existed. Thank you! :good:
Is this a general sort of formula you used (the one involving the variable n)? What would you do if told to evaluate S 3 and S 4?
Yes, it was a general formula that is created once you multiply the whole equation by x^n. You'll find such questions to make the general form in the past papers. So check them out.
 
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