It is known that:
x+1/x = √(2-√2)
Determine the value of:
x2023 + 1/x2023
I've worked out some details below, but in general use the short cut notation:
f_k(x) = x^k + 1/x^k
f_1(x) is given.
f_n(x) = f_1(x) * f_(n-1)(x) - f_(n-2)(x)
Once you've worked out f_2 and f_3, the rest can be found with this recursive formula.
Putting this into a spreadsheet shows the value has a period of 16, the first 16 being:
f#: value
1: 0.765366864730179
2: -1.4142135623731
3: -1.84775906502257
4: 0
5: 1.84775906502257
6: 1.4142135623731
7: -0.76536686473018
8: -2
9: -0.765366864730179
10: 1.4142135623731
11: 1.84775906502257
12: 0
13: -1.84775906502257
14: -1.41421356237309
15: 0.76536686473018
16: 2
mod(2023,16) = 7
So f_2023 = f_7 = - f_1
= -0.76536686473018
= -√(2-√2)
---- details ----
x+1/x = √(2-√2)
(x+1/x)^2 = 2-√2
x^2+1/x^2 = -√2
(x+1/x)(x^2+1/x^2) = -√2√(2-√2)
x^3+1/x^3 + x+1/x = -√2√(2-√2)
x^3+1/x^3 = -√2√(2-√2) - √(2-√2)
= -(√2 + 1) * √(2-√2)
x^2+1/x^2 = -√2
(x^2+1/x^2)^2 = 2
x^4+1/x^4 + 2 = 2
x^4+1/x^4 = 0
(x^4+1/x^4)(x+1/x) = √(2-√2)*0 = 0
x^5+1/x^5 + x^3+1/x^3 = 0
x^5+1/x^5 = (√2 + 1)√(2-√2)
x^3+1/x^3 = -(√2 + 1) * √(2-√2)
(x^3+1/x^3)^2 = (√2 + 1)^2 * (2-√2)
x^6+1/x^6 + 2 = (3 + 2√2) (2-√2)
x^6+1/x^6 + 2 = (6-4 + 4√2 - 3√2)
x^6+1/x^6 + 2 = 2+√2
x^6+1/x^6 = √2
(x^3+1/x^3)(x^4+1/x^4) = 0 since 2nd term is zero
x^7+1/x^7 + x+1/x = 0
x^7+1/x^7 = -√(2-√2)
(x^4+1/x^4)^2 = 0
x^8+1/x^8 + 2 = 0
x^8+1/x^8 = -2
(x^8+1/x^8)(x+1/x) = -2*√(2-√2)
x^9+1/x^9 + x^7+1/x^7 = -2*√(2-√2)
x^9+1/x^9 = -2*√(2-√2) - (-√(2-√2))
x^9+1/x^9 = -√(2-√2)
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Posted by Larry
on 2024-07-05 22:02:41 |
solution
