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f(x)+f(x-1)=x^2.If f(55) = 89, what is the value of f(89)? (It is not 55.)

2007-10-24 13:18:13 · 4 answers · asked by yu.s@att.net 2 in Science & Mathematics Mathematics

4 answers

f(x+1) + f(x) = (x+1)²
f(x) + f(x-1) = x²

Subtracting,

f(x+1) - f(x-1) = 2x + 1

So I just considered positive integers for x and treated the cases x odd and x even separately. The result, for x a positive integer, is

f(x) = x(x+1)/2 + 1451 for x even
f(x) = x(x+1)/2 - 1451 for x odd

or more compactly,

f(x) = x(x+1)/2 + 1451*(-1)^x

Then f(89) = 2554

A generalization for x not necessarily an integer (which does satisfy the recursion relation) would be

f(x) = x(x+1)/2 + 1451*cos(π x)

There's probably an elegant way to solve this...

2007-10-24 14:43:49 · answer #1 · answered by Ron W 7 · 2 0

f(x) + f(x-1) = x^2, i.e. f(x) = x^2 - f(x-1).

Using this to compute successive values I get f(89) = 2554. This is actually the fastest way of doing it if you have a spreadsheet handy. ;-)

Otherwise, note that you get
f(89) = 89^2 - f(88)
= 89^2 - (88^2 - f(87))
= 89^2 - 88^2 + 87^2 - f(86)
= ... = 89^2 - 88^2 + 87^2 - 86^2 + ... + 57^2 - 56^2 + f(55).
So we have to evaluate 89^2 - 88^2 + 87^2 - 86^2 + ... + 57^2 - 56^2
= (2(88) + 1) + (2(86) + 1) + ... + (2(56) + 1)
= 2(88 + 86 + ... + 56) + 17 (there are 17 terms)
= 4(44 + 43 + ... + 28) + 17
If you remember 1 + 2 + ... + n = n(n+1)/2 we get
= 4(44(45)/2 - 27(28)/2) + 17
= 4(990 - 378) + 17
= 2465
so f(89) = 2465 + f(55) = 2465 + 89 = 2554.

2007-10-25 00:56:56 · answer #2 · answered by Scarlet Manuka 7 · 0 0

f(x)+f(x-1)=x^2.If f(55) = 89, what is the value of f(89)? (It is not 55.)


f(56) +f(55) = 56^2
f(56) + 89 = 56^2 = 3136

f(56) = 3047

f(89) +f(88) = 89^2 equation (1)
f(88) +f(87) = 88^2 equation (2)
.......................... ............
.......................... .............
f(56) +f(55) =56^2 ..............

by substracting all the equations from equation (1)

we get:
f(89) -f(55) = 89^2 -88^2 - 87^2 -........-56^2

f(89) -f(55) = 89^2 -{ 88^2 +87^2 +.....+56^2}

we use the series formula

let S1 =1^2 +2^2 +3^2 +....+ 55^2 = 55(56)(111)/6 =56980

and
let S2 =1^2 +2^2 +3^2 +.....+88^2 = 88(89)(177)/6 =231044

S2 -S1 = 88^2 +87^2 +...+56^2 = 231044 -56980 =177064

hence ,
f(89) -f(55) = 89^2 -177064 = - 169143
f(89) = -168143 +f(55) = -168143 +89 =-169054

(check for errors)

2007-10-24 20:35:07 · answer #3 · answered by Any day 6 · 0 1

2554

See below for the working:
http://answers.yahoo.com/question/index;_ylt=Aij7GsGiZ_1Stw2G06.9jX7ty6IX;_ylv=3?qid=20071018130016AAHC7ex&show=7#profile-info-u9VwE7FMaa

2007-10-25 00:27:33 · answer #4 · answered by Dr D 7 · 0 0

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