How can I find a root of a polynomial, say, x101+x7−1, without any approximation methods?

Question

paralovut91 · Accepted Answer

If you mean that you wish to express the roots using only the coefficients of the polynomial, the four basic arithmetic operators and radicals, if the polynomial is a quintic or higher, you cannot in general.
There are, of course, many specific cases when you can. For instance, the polynomial in your question factors into

(x^{2} - x + 1) P_{99} (x) where P_{99} (x)

is a lengthy 99-th power polynomial that (as far as I can tell) cannot be factorized any further. The quadratic equation

x^{2} - x + 1 = 0

can be easily solved yielding the complex roots

x = \frac{1 \pm \sqrt{- 3}}{2}

. But the one real root of your polynomial, at

x \tilde{=} 0.980097

, cannot be found this way.
Having said that, once you reduced a more complex problem to finding the roots of a rational polynomial, it is often viewed as good as being solved, with the roots having been found for all practical intents and purposes. That is because the values of the roots can always be calculated easily and efficiently. In a sense it's no different from finding the root of the equation

x^{2} - 2 = 0

; we cannot do that using the four arithmetic operators only, so we invent the radical symbol which allows us to write

x = \pm \sqrt{2}

.

Chettaf04 · Accepted Answer

Applying the formula for real root of an odd degree polonomial , say,

x^{101} + x^{7} = 1

,
The general solution of real root for a polonomial

X^{n} + x^{m} = 1

, where (n and m) are odd positive integers and (n is greater than m)

x = [1 - \frac{1}{n} + \frac{2 m - n + 1}{2! \cdot n^{2}} - (3 m - 2 n + 1) \frac{3 m - n + 1}{3! \cdot n^{3}} + (4 m - 3 n + 1) (4 m - 2 n + 1) \frac{4 m - n + 1}{4! \cdot n^{4}}

- (5 m - 4 n + 1) (5 m - 3 n + 1) (5 m - 2 n + 1) \frac{4 m - n + 1}{4! \cdot n^{4}} + \dots \infty]

You may notice that this series solution is always a convergent series (very similar to sin(t) or cos(t) or

e^{x}

…etc), which we consider exact values in our calculations
The general term is so obvious that is

\frac{{(- 1)}^{i}}{i! \cdot n^{i}} \cdot [Product (j = 1, to j = i - 1) {i \cdot n - j \cdot m + 1}

X = 1 - \frac{1}{n} + \sum [i = 2 to i = \infty] \frac{{(- 1)}^{i}}{i! \cdot n^{i}} \cdot [Product (j = 1, to j = i - 1) {i \cdot n - j \cdot m + 1}

Applying this formula to the above mentioned example we get:

X = [1 - \frac{1}{101} - \frac{86}{2!} \cdot 101^{2} - 79 \cdot \frac{180}{3!} \cdot 101^{3} - 72 \cdot 173 \cdot \frac{274}{4!} \cdot 101^{4}

65 \cdot 166 \cdot 267 \cdot \frac{368}{5}! \cdot 101^{5} - 58 \cdot 159 \cdot 260 \cdot 361 \cdot \frac{462}{6}! \cdot 101^{6}

51 \cdot 152 \cdot 253 \cdot 354 \cdot 455 \cdot \frac{556}{7}! \cdot 101^{7}

44 \cdot 145 \cdot 246 \cdot 374 \cdot 448 \cdot 549 \cdot \frac{650}{8}! \cdot 101^{8}

37 \cdot 138 \cdot 239 \cdot 340 \cdot 441 \cdot 542 \cdot 643 \cdot \frac{744}{9}! \cdot 101^{9}

30 \cdot 131 \cdot 232 \cdot 333 \cdot 434 \cdot 535 \cdot 636 \cdot 737 \cdot \frac{838}{10}! \cdot 101^{10}

How can I find a root of a polynomial, say,

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