moews@husc2.UUCP (moews) (11/16/86)
In article <660@hoxna.UUCP> tom@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 moews@husc4.harvard.edu ...!harvard!husc4!moews
tom@hoxna.UUCP ( Tom McGuigan ) (11/17/86)
Key Words: My Mistake
References: <660@hoxna.UUCP>, <1029@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