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

May 29, 2016

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

#### Explanation:

$f \left(x\right) = - 7 {x}^{9} + {x}^{5} - {x}^{2} + 6$

This polynomial has $9$ zeros near the $9$ Complex ninth roots of $\frac{6}{7}$, that is near the zeros of $- 7 {x}^{9} + 6$:

$\sqrt{\frac{6}{7}} \left(\cos \left(\frac{2 k \pi}{9}\right) + i \sin \left(\frac{2 k \pi}{9}\right)\right)$ for $k = 0 , \pm 1 , \pm 2 , \pm 3 , \pm 4$

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

$f ' \left(x\right) = - 63 {x}^{8} + 5 {x}^{4} - 2 x$

Starting with an approximation ${a}_{0}$, iterate using the formula:

${a}_{i + 1} = {a}_{i} - \frac{f \left({a}_{i}\right)}{f ' \left({a}_{i}\right)} = {a}_{i} - \frac{- 7 {a}_{i}^{9} + {a}_{i}^{5} - {a}_{i}^{2} + 6}{- 63 {a}_{i}^{8} + 5 {a}_{i}^{4} - 2 {a}_{i}}$

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 ${a}_{0} = \sqrt{\frac{6}{7}}$ we get:

${a}_{0} \approx 0.9830179944916754$
${a}_{1} \approx 0.9820913414799528$
${a}_{2} \approx 0.9820877872098702$
${a}_{3} \approx 0.9820877871578138$
${a}_{4} \approx 0.9820877871578138$

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