微分應用之根與係數關係?

令 f(x) = x⁴+ 7x³ + 5x² + 3x + 1 = (x - a)(x - b)(x - c)(x - d) , 並記 Sᵣ = aʳ + bʳ + cʳ + dʳ .

f '(x)

= (x - b)(x - c)(x - d) + (x - a)(x - c)(x - d) + (x - a)(x - b)(x - d) + (x - a)(x - b)(x - c)

= x³ - (b+c+d)x² + (bc+bd+cd)x - bcd + (x - a)(x - c)(x - d) + (x - a)(x - b)(x - d) + (x - a)(x - b)(x - c)

= x³ - (-7 - a)x² + (5 - a(-7 - a))x - (-3 - a(5 - a(-7 - a)))

+ x³ - (-7 - b)x² + (5 - b(-7 - b))x - (-3 - b(5 - b(-7 - b)))

+ x³ - (-7 - c)x² + (5 - c(-7 - c))x - (-3 - c(5 - c(-7 - c)))

+ x³ - (-7 - d)x² + (5 - d(-7- d))x - (-3 - d(5 - d(-7 - d)))

= x³ + (a + 7)x² + (a² + 7a + 5)x + a³ + 7a² + 5a + 3

+ x³ + (b + 7)x² + (b² + 7b + 5)x + b³ + 7b² + 5b + 3

+ x³ + (c + 7)x² + (c² + 7c + 5)x + c³ + 7c² + 5c + 3

+ x³ + (d + 7)x² + (d² + 7d + 5)x + d³ + 7d² + 5d + 3

= S₀x³ + (S₁+ 7S₀)x² + (S₂+ 7S₁+ 5S₀)x + S₃+ 7S₂+ 5S₁+ 3S₀

另一方面, f '(x) = 4(1)x³ + 3(7)x² + 2(5)x + 1(3) , 比較係數得

S₀ = (4-0) , S₁+ 7S₀ = (4-1)7 , S₂+ 7S₁+ 5S₀ = (4-2)5 , S₃+ 7S₂+ 5S₁+ 3S₀ = (4-3)3

⇒ S₀ = 4 , S₁= - 7 , S₂= 39 , S₃= - 247

又 S₄= - 7S₃- 5S₂- 3S₁- S₀ = 1551

故 S₅ = - 7S₄- 5S₃- 3S₂- S₁= - 9732

別解:

x⁴= - 7x³ - 5x² - 3x - 1

x⁵= - 7x⁴- 5x³ - 3x² - x

∴ a⁵ + b⁵ + c⁵ + d⁵

= - 7(a⁴+ b⁴+ c⁴+ d⁴) - 5(a³ + b³ + c³ + d³) - 3(a² + b² + c² + d²) - (a + b + c + d) ... ①

a⁴+ b⁴+ c⁴+ d⁴

= - 7(a³ + b³ + c³ + d³) - 5(a² + b² + c² + d²) - 3(a + b + c + d) - 4 ... ②

②代入①:

a⁵ + b⁵ + c⁵ + d⁵

= 44(a³ + b³ + c³ + d³) + 32(a² + b² + c² + d²) + 20(a + b + c + d) + 28 ... ③

x³ = - 7x² - 5x - 3 - 1/x

∴ a³ + b³ + c³ + d³

= - 7(a² + b² + c² + d²) - 5(a + b + c + d) - 12 - (1/a + 1/b + 1/c + 1/d)

= - 7(a² + b² + c² + d²) - 5(a + b + c + d) - 12 - (abc + abd + acd + bcd)/(abcd)

= - 7(a² + b² + c² + d²) - 5(a + b + c + d) - 12 - (- 3)/1

= - 7(a² + b² + c² + d²) - 5(a + b + c + d) - 9 ... ④

a² + b² + c² + d² = (a + b + c + d)² - 2(ab + ac + ad + bc + bd + cd) = (- 7)² - 2(5) = 39 ,

代入④: a³ + b³ + c³ + d³ = - 7(39) - 5(- 7) - 9 = - 247

代入③: a⁵ + b⁵ + c⁵ + d⁵ = 44(- 247) + 32(39) + 20(- 7) + 28 = - 9732