Question

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

Answer 1

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.