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How to do alpha^4 + Beta^4

I know alpha^2+beta^2 and ^3 but i dont know how to expand 4 brackets :s-smilie:
it becomes (a^2+b^2)^2-2a^2b^2, which is the difference of two squares
I didn't miss anything
(α+β)2−2αβ=α2+β2\displaystyle (\alpha + \beta)^2-2\alpha \beta = \alpha^2 + \beta^2

α↦α2 \displaystyle \alpha \mapsto \alpha^2 and β↦β2\displaystyle \beta \mapsto \beta^2

So α4+β4=(α2+β2)2−2α2β2=[(α+β)2−2αβ]2−2α2β2\displaystyle \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2-2\alpha^2 \beta^2=[(\alpha+\beta)^2-2\alpha \beta]^2-2\alpha^2 \beta^2

and so on.

I wouldn't bother remembering it.
(edited 7 years ago)
Well I assume Alpha and a Beta make the average male?
number to the power of 4 male
Well there you go then, I literally derived it in like 2 lines for you. It's easier to derive it than to remember it.
Original post by MathMoFarah
it becomes (a^2+b^2)^2-2a^2b^2, which is the difference of two squares


Difference of two squares??? Err, no. Unless you want to be bringing 2\sqrt{2} into it which is a mess.



Well if you are just keep in mind that the last two terms should be squared. I fixed that in my post. :smile:
Original post by RDKGames
Difference of two squares??? Err, no. Unless you want to be bringing 2\sqrt{2} into it which is a mess.


OP wasn't very clear and I thought he meant factorise a^4+b^4 - no other way to do so.
Original post by MathMoFarah
OP wasn't very clear and I thought he meant factorise a^4+b^4 - no other way to do so.


Even then it wouldn't factorise due to a non-integer.
Original post by RDKGames
Even then it wouldn't factorise due to a non-integer.


It still factorises - just not into integers, you can play around with a and b (multiplying by constants, adding/subtracting) to make it integers if that's important for some reason.
Reply 10
Avatar for Zacken
Zacken
22
Original post by RDKGames
Even then it wouldn't factorise due to a non-integer.


Yes it does? Factorisation doesn't have to be solely integers.
Original post by MathMoFarah
It still factorises - just not into integers, you can play around with a and b (multiplying by constants, adding/subtracting) to make it integers if that's important for some reason.


I meant to say non-rationals, but okay. Just that irrationals in a factorised form are awkward to work with, and I doubt this type of question with irrationals in factorised form would be relevant to anything in FP1's roots of quadratics.
(edited 7 years ago)
Original post by RDKGames
Even then it wouldn't factorise due to a non-integer.


Yeh it does, that factorisation is very common.


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Reply 13
Avatar for RichE
RichE
15
A more helpful way to handle these might be the following. If
[br]α+β=A,αβ=B[br][br]\alpha + \beta = A, \quad \alpha \beta = B[br]
and
[br]Sn=αn+βn[br][br]S_n = \alpha^n + \beta^n[br]
then
[br]Sn+2=ASn+1−BSn[br][br]S_{n+2} = A S_{n+1} - B S_n[br]
And +rep / PRSOM to the first one to derive it (above).
Original post by RichE
A more helpful way to handle these might be the following. If
[br]α+β=A,αβ=B[br][br]\alpha + \beta = A, \quad \alpha \beta = B[br]
and
[br]Sn=αn+βn[br][br]S_n = \alpha^n + \beta^n[br]
then
[br]Sn+2=ASn+1−BSn[br][br]S_{n+2} = A S_{n+1} - B S_n[br]


We could solve this reccurance then in terms of A and B then?


Posted from TSR Mobile
Reply 16
Avatar for RichE
RichE
15
Original post by physicsmaths
We could solve this reccurance then in terms of A and B then?


Posted from TSR Mobile


Of course. But my point was if you just wanted S4 or S5 then you could quickly use the recurrence to get them without having to figure out any factorisation. *

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