math
posted by joe .
The roots of the eqn x^4  3x^2 + 5x  2 = 0 are a, b, c, d, and a^n + b^n + c^n + d^n is denoted by S~n. The equation of the roots a^2, b^2, c^2, d^2 is y^4  6y^3 + 5y^2  13y + 4 = 0. State the value of S~2 and hence show that S~8 = 6S~6  5S~4 + 62.
I answered this question this morning. Check the page previous to this. Let me know if you have questions about it. The post should be near the bottom of the thread begun by jack.
i do not understand the last part with the S~8 = 6S~6  5S~4 + 62.
Ok, that does require a little calculating and observation.
If you take the poly
x^4+px^3+qx^2+rx+s with roots a,b,c,d and multiply it by x^4px^3+qx^2rx+s you get a poly whose terms are x^8,x^6,x^4,x^2,x^0(constant term) and the roots are a^2,b^2,c^2,d^2
The coefficient of the x^6 term is s2.
If you make the sub. y=x^2 you get a poly y^4+p1y^3+p2y^2+p3y+p4. If you repeat this proces that we did above, i.e. negate the coefficients of the odd power terms and multiply them you get a poly whose terms are y^8,,,y^0 all even numbers. The coefficient of the y^6 term is S4. If you do the process once more you get S8.
To evaluate that expression the questioner is asking you to calculate S4, S6 and S8 and verify they satisfy that equation, that's all.
To calculat S6 I observed that in the first y polynomial the coefficient of y^3 is s1, of y^2 is s2 and of y is s3 and that (s1)^33s1s2+3s3=S2 where the lower case s's are the symmetric functions. I'm not sure if you've been asked to calculate that or not. Let me know if this explains matters.
I just noticed a couple typos here. This line;
"To calculat S6 I observed that in the first y polynomial the coefficient of y^3 is s1, of y^2 is s2 and of y is s3 and that (s1)^33s1s2+3s3=S2 "
That should be S6 not S2 at the end. The other typos are minor.
i understand better now, thanks
You're welcome, feel free to post if anything's not clear. I should mention that this would be a fairly challenging problem for college students. Nice job doing this here.
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