How do you find all the zeros of #f(x) = -7x^9+x^5-x^2+6 #?

1 Answer
May 29, 2016

Answer:

Use Newton's method with suitable first approximations to find approximations of the zeros numerically.

Explanation:

#f(x) = -7x^9+x^5-x^2+6#

This polynomial has #9# zeros near the #9# Complex ninth roots of #6/7#, that is near the zeros of #-7x^9+6#:

#root(9)(6/7)(cos ((2kpi)/9) + i sin((2kpi)/9))# for #k = 0, +-1, +-2, +-3, +-4#

We can use these ninth roots as first approximations for Newton's method:

#f'(x) = -63x^8+5x^4-2x#

Starting with an approximation #a_0#, iterate using the formula:

#a_(i+1) = a_i - (f(a_i))/(f'(a_i))=a_i - (-7a_i^9+a_i^5-a_i^2+6)/(-63a_i^8+5a_i^4-2a_i)#

If your spreadsheet application is anything like mine, it does not handle Complex numbers directly, so expressing this formula requires separate columns for Real and imaginary parts.

I won't bother with that at this time, but I can at least find the Real zero using Real arithmetic:

Putting #a_0 = root(9)(6/7)# we get:

#a_0 ~~ 0.9830179944916754#
#a_1 ~~ 0.9820913414799528#
#a_2 ~~ 0.9820877872098702#
#a_3 ~~ 0.9820877871578138#
#a_4 ~~ 0.9820877871578138#

So it converges quite fast and the first approximation was close to the result.