Algebra (spoiler)

[moews]

In article <6...@hoxna.UUCP> t...@hoxna.UUCP ( Tom McGuigan ) writes:
>>A + B + C = 1, A^2 + B^2 + C^2 = 2, A^3 + B^3 + C^3 = 3
>
>(omitting pages and pages of algebraic manipulation)
>
>==> A^4 + B^4 + C^4 = 4.5
>
>Tom McGuigan
>..!ihnp4!homxb!hoxna!tom

This puzzle is more interesting if you let 3=n (i.e., if
a[1]+...+a[n] = 1, a[1]^2 + ... + a[n]^2 = 2, ..., a[1]^n + ... + a[n]^n = n,
what is a[1]^(n+1) + ... + a[n]^(n+1)?) In this case, the answer can be
shown to be the coefficient of x^n in

1 - 1/(1-x)
e - 1 2 3
- ----------------- = x + 5/2 x + 25/6 x + ...,
2
(1-x)

(use Newton's identities),

so the answer is 25/6 (not 9/2; check your algebra.)
--
David Moews mo...@husc4.harvard.edu ...!harvard!husc4!moews

[Tom McGuigan]

Key Words: My Mistake
References: <6...@hoxna.UUCP>, <10...@husc2.UUCP>

At the suggestion of David Moews:

>so ( A^4 + B^4 + C^4 ) = 25/6 (not 9/2; check your algebra.)

I did check my algebra. An incorrect substitution for the value
of A^3 + B^3 + C^3 (I used 2 instead of 3) lead to my answer of
9/2. My apologies, 25/6 is correct.

Tom McGuigan
..!ihnp4!hoxna!tom