駭客數學第一章五問

[wen1019]

請問老師這五題該怎麼解呢

[thepiano]

第 1 題
3^16 - 1
= (3^8 + 1)(3^8 - 1)
= (3^8 + 1)(3^4 + 1)(3^4 - 1)
= (3^8 + 1)(3^4 + 1)(3^2 + 1)(3^2 - 1)
= (3^8 + 1)(3^4 + 1)(3^2 + 1)(3 + 1)(3 - 1)
= 2 * 4 * 10 * 82 * 6562


第 2 題
f(f(x)) = {a * [(ax + b) / (cx - a)] + b} / {c * [(ax + b) / (cx - a)] - a}
= [(a^2x + ab + bcx - ab) / (cx - a)] / [(acx + bc - acx + a^2) / (cx - a)]
= x


第 3 題
令 y = (x - 1) / (2x + 1)
2xy + y = x - 1
(2y - 1)x = -y - 1
x = (-y - 1) / (2y - 1)

f((x - 1) / (2x + 1)) = f(y) = 5 * [(-y - 1) / (2y - 1)] + 2 = (-y - 7) / (2y - 1)
f(-1) = 2
f(f(-1)) = f(2) = -3


第 4 題
f(-2) = 3 - (-2) = 5
f(2) = 2 * 2 = 4
f(4) = 4 + 3 = 7
......


第 5 題
f(5) = 20
g(f(5)) = g(20) = 61

[wen1019]

謝謝老師
再請教老師3題

1 設f(x)=2f(x-1)+3f(x-2),若f(1)=1且f(2)=2,則f(3)=? ANS:7

2設g(x)為一線型函數,滿足3g(2)=2g(3),5g(4)=4g(5),
且g(5)=5,則g(2)=? ans:2

3 設f(x)=4x+6,g(x)=3x+k, 已知f(g(x))=g(f(x)),則k之值:4

[thepiano]

第 1 題
f(3) = 2f(2) + 3f(1)
......


第 2 題
令 g(x) = ax + b

3g(2) = 2g(3)
3(2a + b) = 2(3a + b)
b = 0

5g(4) = 4g(5)
5(4a + b) = 4 * 5 = 20
a = 1

g(x) = x
g(2) = 2


第 3 題
f(g(x)) = 4(3x + k) + 6
g(f(x)) = 3(4x + 6) + k
......

[Yinong024618]

駭客數學 第八版 第一章 集合與統計
1-3 例題5類題3

第 3 題
令 y = (x - 1) / (2x + 1)
2xy + y = x - 1
(2y - 1)x = -y - 1
x = (-y - 1) / (2y - 1)

f((x - 1) / (2x + 1)) = f(y) = 5 * [(-y - 1) / (2y - 1)] + 2 = (-y - 7) / (2y - 1)
f(-1) = 2
f(f(-1)) = f(2) = -3

請問2xy + y = x - 1到(2y - 1)x = -y - 1是怎麼解的，不明白x - 1如何可以 = -y - 1
謝謝解答者，好人一生平安

[thepiano]

把含有 x 的項移到等號左邊，沒有 x 的項移到等號右邊
左邊提出 x，兩邊再同除以 2y - 1
可得 x = (-y - 1) / (2y - 1)