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## Main Question or Discussion Point

If x^5 has 5 roots, if x^3 has 3 roots and if x^10 has 10 roots, so how many roots has x^3.14 ?

- Thread starter Jhenrique
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If x^5 has 5 roots, if x^3 has 3 roots and if x^10 has 10 roots, so how many roots has x^3.14 ?

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pwsnafu

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It has one root with multiplicity 5.If x^5 has 5 roots,

It has one root with multiplicity 3.if x^3 has 3 roots

It has one root with multiplicity 10.and if x^10 has 10 roots,

One, and the root is also a branch point.so how many roots has x^3.14 ?

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Why one and what's a branch point?One, and the root is also a branch point.

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symbolipoint

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Jhenrique is referring most likely to the degree of a function and not to just specific functions.

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pwsnafu

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The only solution to ##x^{3.14}=0## is ##x=0##. There is no other choice.Why one and what's a branch point?

A branch point is a point where, on the complex plane, traveling around that point is dependent on the direction of travel. Visually, the function ##x^{3.14}## "tears" the complex plane from the point ##x=0##.

Even then the statement "a polynomial of degree 10 has 10 roots" is still wrong. The correct form is "a polynomial with degree 10 has 10 roots counting multiplicity", which is a sloppy way of saying "the sum of the multiplicities of the roots of a polynomial of degree 10 is equal to 10".Jhenrique is referring most likely to the degree of a function and not to just specific functions.

Also Jhenrique, the fundamental theorem of algebra is a theorem about polynomials, ##x^{3.14}## is not a polynomial. You can't expect the pattern to continue.

Last edited:

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pwsnafu

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Suppose we have a polynomial ##p(x)##. Suppose that the degree is ##n##. This means

##p(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0##.

Now suppose that the roots of ##p(x)## are ##x_1, x_2, \ldots, x_j##, each one different. Now the "number of roots" is ##j##. The fundamental theorem does

Each root has what we call "multiplicity", I'm going to denote as ##m_1, m_2, \ldots, m_j##.

This means when we factor

##p(x) = (x-x_1)^{m_1} (x-x_2)^{m_2} \ldots (x-x_j)^{m_j}##

The fundamental theorem states that

- ##j \neq 0##, and
- ##m_1 + m_2 + \ldots+ m_j = n##.

- #7

HallsofIvy

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For n a positive integer, and a non zero, [tex]x^n= a[/tex] has n distinct roots. If n is NOT an integer it has infinitely many roots.If x^5 has 5 roots, if x^3 has 3 roots and if x^10 has 10 roots, so how many roots has x^3.14 ?

(don't know what you mean by "x^5 has 5 roots".

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Are you saying that there are infinitely many complex numbers satisfying ##x^\frac{1}{2}=1##?For n a positive integer, and a non zero, [tex]x^n= a[/tex] has n distinct roots. If n is NOT an integer it has infinitely many roots.

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Mentallic

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Yep!Are you saying that there are infinitely many complex numbers satisfying ##x^\frac{1}{2}=1##?

[tex]x = e^{4i\pi n }[/tex] for all integers n. Granted, they are all the same complex numbers.

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Isn't that rather like saying that there are infinitely many numbers equal to zero, since ##n - n = 0## for all integers ##n##?Yep!

[tex]x = e^{4i\pi n }[/tex] for all integers n. Granted, they are all the same complex numbers.

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Mentallic

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I honestly don't understand it eitherIsn't that rather like saying that there are infinitely many numbers equal to zero, since n−n=0 for all integers n?

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I think it's more like saying that every linear polynomial ##mx+n## with integer coefficients has infinitely many rational roots, ##\{-\frac{nk}{mk}\}_{k\in\mathbb{Z}\setminus\{0\}}##.Isn't that rather like saying that there are infinitely many numbers equal to zero, since ##n - n = 0## for all integers ##n##?

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For example, [itex]x^{√2} = 1[/itex] has infinitely many (complex) solutions.

Indeed, let [itex]x = e^{i\pi√2n}[/itex] for any integer n.

All those solutions are distinct, because otherwise [itex]i\pi√2m = i\pi√2n + 2k\pi i[/itex] or

[itex]√2=\frac{2k}{m-n}[/itex] would be rational, which is false.

Rational exponents have finitely many solutions because then the sequence becomes periodic, when n is a multiple of the denominator.

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HallsofIvy

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