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$$ 1 x^5-9.067 x^4+24.726 x^3-13.998 x^2-15.278 x+1.014 = 0 $$

Wolfram Result

but i need solve roots using "iteration method" (i don't know how it correctly called in English, but Newton's method is not what I need)

method, what I need, starts from transformation of equation fi(x) = x; if |fi'(x)| < 1 we can find root I used fi(x) = - (1 x^5 - 9.067 x^4 + 24.726 x^3 - 13.998 x^2 + 1.014) / 15.278 but with it I solved just two roots from five

roman
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  • Can you be more specific as to which iteration method? Are you talking some form of Newton's or something else? Regards – Amzoti Apr 15 '13 at 15:02
  • If you like to use WolframAlpha, try prepending "Newton's method" to your query. You can also append "near x=4" to specify the root. You'll get probably more information that you wanted! – Mark McClure Apr 15 '13 at 20:50

1 Answers1

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There are a number of iterative techniques that can be used to solve a equations. Here's one: If you write this equation in the form

$$x = (-1.014 + 13.998 x^2 - 24.726 x^3 + 9.067 x^4 - x^5)/15.278,$$

Then you see that any root of your polynomial is a fixed point of

$$f(x) = (-1.014 + 13.998 x^2 - 24.726 x^3 + 9.067 x^4 - x^5)/15.278.$$

Thus, we might be able to solve your equation by iterating $f$. In fact, the first seven iterates of $f$ starting at zero are:

$$0, 0.066369, 0.062795, 0.0631486, 0.063114, 0.063117, 0.063117, 0.063117$$

In general the derivate at the fixed point should be less than one in absolute value for this to work. I believe that there's one other solution that this technique can find.

If, as you say in your comment, you hope to find other roots, then you might try writing the equation in the form $x=f(x)$ in other ways. One other way, for example is as

$$x = \sqrt[3]{\frac{-1000 x^5+9067 x^4+13998 x^2+15278 x-1014}{24726}}.$$

I obtained this by solving for just the "$x$" under the cube. If we iterate this new function, we might converge to different roots. The obit of zero, for example, is now

$$0, -0.34485, -0.56606, -0.55328, -0.55733, -0.55610, -0.55648, -0.55637$$

Mark McClure
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  • yes, it is method that I need. but with this form of equation I can't get all roots. just two – roman Apr 16 '13 at 06:22
  • @roman That's correct. There might be other ways to write it to get other roots. I don't know if there's any way to re-write the equation to get all the roots. That's rather the point of Newton's method! – Mark McClure Apr 16 '13 at 06:25
  • Newton's method is one of several methods that I must do in my homework – roman Apr 16 '13 at 06:42