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

Question

Liam Brennan · Accepted Answer

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

#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.

Sadie Ackerman · Answer

undefined To find the zeros of the function $ f(x) = -7x^9 + x^5 - x^2 + 6 $, we set the function equal to zero and solve for $ x $:

$$ -7x^9 + x^5 - x^2 + 6 = 0 $$

Unfortunately, there's no general formula to solve polynomial equations of degree higher than 4. However, we can use numerical methods or approximation techniques to estimate the roots. These methods include graphical analysis, Newton's method, or numerical approximation algorithms like the bisection method or Newton-Raphson method.