newton's formula for the sums of powers of roots?

2016-08-29 6:28 am
Please take a look of the photo. In the middle part, it says For each I, by division and gets the following results. Please further explain to me how to get the result by division.


photo: http://s613.photobucket.com/user/Yan_Wa_Chung/media/img026_zps1jv84kiw.jpg.html

回答 (1)

2016-08-29 3:27 pm
✔ 最佳答案
For example ,
let f(x) = a0 x⁵ + a1 x⁴+ a2 x³ + a3 x² + a4 x + a5 = a0 (x - α1) (x - α2) (x - α3) (x - α4) (x - α5) ,

Note that
- a1/a0 = α1 + ... + α5 ,
a2/a0 = α1 α2 + ... + α4 α5 ,
- a3/a0 = α1 α2 α3 + ... + α3 α4 α5 ,
a4/a0 = α1 α2 α3 α4 + ... + α2 α3 α4 α5 ,
- a5/a0 = α1 α2 α3 α4 α5 , then

f(x) / (x - α1)
= a0 (x - α2) (x - α3) (x - α4) (x - α5)

= a0 x⁴+ a0(- α2 - α3 - α4 - α5)x³ + a0(α2 α3 + ... + α4 α5)x² + a0(- α2 α3 α4 - ... - α3 α4 α5)x
+ a0(α2 α3 α4 α5)

= a0 x⁴+ a0(a1/a0 + α1)x³ + a0(a2/a0 - α1(α2 + α3 + α4 + α5))x² + a0(a3/a0 + α1(α2 α3 + ... + α4 α5))x
+ a0(a4/a0 - α1(α2 α3 α4 + ... + α3 α4 α5))

= a0 x⁴+ (a0α1 + a1)x³ + (a2 - α1 a0(α2 + α3 + α4 + α5))x² + (a3 + α1 a0(α2 α3 + ... + α4 α5))x
+ (a4 - α1 (-a3 - α1 a0(α2 α3 + ... + α4 α5))

= a0 x⁴+ (a0α1 + a1)x³ + (a2 - α1 (- a0α1 - a1) )x² + (a3 + α1 a0(α2 α3 + ... + α4 α5))x
+ (a4 - α1 a0(α2 α3 α4 + ... + α3 α4 α5))

= a0 x⁴+ (a0α1 + a1)x³ + (a0α1² + a1α1 + a2)x² + (a3 + α1 (a0α1² + a1α1 + a2))x
+ (a4 - α1 a0(α2 α3 α4 + ... + α3 α4 α5))

= a0 x⁴+ (a0α1 + a1)x³ + (a0α1² + a1α1 + a2)x² + (a0α1³ + a1α1² + a2α1 + a3)x
+ (a4 - α1 (- a0α1³ - a1α1² - a2α1 - a3)

= a0 x⁴+ (a0α1 + a1)x³ + (a0α1² + a1α1 + a2)x² + (a0α1³ + a1α1² + a2α1 + a3)x
+ (a0α1⁴+ a1α1³ + a2α1² + a3α1 + a4)


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