polynomial Q (1 Viewer)

[mojako]

Let a, b and c be the roots of x^3 + px + q = 0, and define s_n by
s_n = a^n + b^n + c^n , for n = 1, 2, 3, ...

so, s_1 = 0 , s2 = -2p , s_3 = -3q

Prove that for n > 3,
s_n = - p s_(n-2) - q s_(n-3)

This is from 2003 HSC BTW.

Thanks.

 

[Archman]

by inspection:
a^n + b^n + c^n = (a+b+c)(a^n-1 + b^n-1 + c^n-1) - (ab + bc + ca)(a^n-2 + b^n-2 + c^n-2) + abc(a^n-3 + b^n-3 + c^n-3)

ok fine, this is probably a bit hard to just get by inspection. but the approach is to get everything from s_n = - p s_(n-2) - q s_(n-3) in terms a, b and c. and try to prove the equation.

 

[mojako]

Thanks Archman.
Do you know of any other way though?

The marking guidelines says:
correct solution -> 2 marks
recognition that a^3 = -pa - q in this context or equivalent -> 1 mark
(it's out of 2 marks)

The comment says:
A number attempted a proof by induction but were ultimately unsuccessful, although a few did prove the result required within their attempt at induction without realising that they had done so.
Those who observed that a^3 = -pa - q were generally successful in gaining some marks.

I wish they gave more info
They have written a long document so they should make it more useful for students, IMHO.

 

[Rorix]

It's just
a^n = -pa^(n-2) - q^(n-3)
b^n = ..
c^n = ...
summation?

I tried to do induction when I first saw this question. It wasn't pretty.
Man, I really didn't want to answer this question because of the way I pwned myself in the other polynomial thread

 

[mojako]

I think that's it, haha..
how easy.

by the way, if someone knows,
"although a few did prove the result required within their attempt at induction without realising that they had done so."
will they get the marks?

 

[Rorix]

probably get 1 for successful working and lose 1 for not making the conclusion

 

[Jase]

"They proved it by induction without knowing they had done so?" what the.

 

[mojako]

I think they didn't prove it by induction...
Within their attempt at induction, they did something that Rorix said.
But they didn't mean it as the reason why s_1 = 0 , s2 = -2p , s_3 = -3q

 

[Archman]

It's just
a^n = -pa^(n-2) - q^(n-3)
b^n = ..
c^n = ...
summation?

I tried to do induction when I first saw this question. It wasn't pretty.
Man, I really didn't want to answer this question because of the way I pwned myself in the other polynomial thread

meh, as the saying goes, the most obvious way works most of the time in HSC.

 

[mojako]

It's just
a^n = -pa^(n-2) - q^(n-3)
b^n = ..
c^n = ...
summation?

I tried to do induction when I first saw this question. It wasn't pretty.
Man, I really didn't want to answer this question because of the way I pwned myself in the other polynomial thread

umm... how do you know that?
[ how do you know that a^n = -pa^(n-2) - qa^(n-3) ]

I can easily prove it by induction, but is it naturally true already?

 

[ngai]

umm... how do you know that?
[ how do you know that a^n = -pa^(n-2) - qa^(n-3) ]

I can easily prove it by induction, but is it naturally true already?

umm..by inspection?
heheh : D

 

[Rorix]

if a^3 = -pa - q
then a^n = -pa^(n-2) - qa^(n-3)
it's Rorix's first therom which states: answers can be often found by multiplying by a^(n-3)

 

[mojako]

if a^3 = -pa - q
then a^n = -pa^(n-2) - qa^(n-3)
it's Rorix's first therom which states: answers can be often found by multiplying by a^(n-3)

Oh I see.
The thing is my teacher never told me about Rorix's theorem, so it's understandable that I didn't know it ^_^
Thanks.