- #1

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Does that mean f(x)=x^2 has only one root? Even though for every other value of x except zero there are two values for x that you can input to output a particular value for f(x).

What about complex numbers?

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- Thread starter V0ODO0CH1LD
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- #1

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Does that mean f(x)=x^2 has only one root? Even though for every other value of x except zero there are two values for x that you can input to output a particular value for f(x).

What about complex numbers?

- #2

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So, for example, [itex]f(x)=x^2[/itex] has one root, namely 0. But [itex]f(x)=x^2-1[/itex] has two roots, namely 1 and -1. And [itex]f(x)=x^2+1[/itex] has no

As you noticed, this situation is unsatisfactory. Intuitively, we want [itex]x^2[/itex] to have two roots, not one. We want [itex]x^2[/itex] to have two equal roots. We say that the root 0 has multiplicity two.

How is that defined? Well, take an arbitrary number a. We say that a is a root of the polynomial f of multiplicity n if we can write

[tex]f(x)=(x-a)^ng(x)[/tex]

for some polynomial g.

Let's do some examples. Take a=0 and let [itex]f(x)=x+1[/itex]. Then we can only write

[tex]f(x)=(x-0)^0(x+1)[/tex]

so 0 is a root of multiplicity 0. We say that 0 is not a root.

Take a=0 and let [itex]f(x)=x^2+x[/itex], then we can write

[tex]f(x)=(x-0)^1(x+1)[/tex]

so 0 is now a root of multiplicity 1. If we have [itex]f(x)=x^{44}(x+1)[/itex], then 0 is a root of multiplicity 44.

We can see now that [itex]f(x)=x^2-1[/itex] has roots 1 and -1 and both have multiplicity 1. We can also see that [itex]f(x)=x^2[/itex] has only root 0 with multiplicity 2. And [itex]f(x)=x^2+1[/itex] still has no roots.

Now we can easily prove that every polynomial of degree n has at most n roots, if we count with multiplicity. So a quadratic equation [itex]f(x)=x^2+bx+c[/itex] has at most 2 roots, counter with multiplicity. Of course, it can still happen that there are no roots.

Another example: [itex]x^{100}+50x^4+3[/itex] has at most 100 roots, counter with multiplicity. So for example: there might by only 1 root with multiplicity 50, there might be 2 roots with multiplicity 50, there might be 50 roots with multiplicity 2, there might be 100 roots with multiplicity 1, etc.

This was the real case. Now we can extend everything to complex numbers. Now we got something very beautiful. If we allow complex numbers to be roots, then a very famous theorem says: every polynomial of degree n has

For example, [itex]f(x)=x^2+1[/itex] has no real roots, but has two complex roots, namely i and -i. This is the nicest possible result.

- #3

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Are the roots of a polynomial given by the function f(x) defined as the values for x where f(x)=0?

Does that mean f(x)=x^2 has only one root? Even though for every other value of x except zero there are two values for x that you can input to output a particular value for f(x).

Yes, x

What about complex numbers?

Every (nonconstant) polynomial of degree n with complex coefficients has exactly n roots in the complex numbers ... as long as you count multiplicities to account for roots that appear more than once in the linear factorization of the polynomial.

This fact is known as the Fundamental Theorem of Algebra ...

http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

The first widely accepted proof was published by Gauss in 1799 as his Ph.D. thesis when he was 21 years old.

It's a curiosity that the FTA is really a theorem of analysis, not algebra; as its proof depends on the topological completeness of the complex plane.

- #4

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I'm not sure what you mean by "what about complex numbers"; you can define polynomials over the field of complex numbers as well, if you like. It's a rather important theorem that any polynomial with coefficients in the complex numbers (which includes the reals) has a root in the complex numbers. Note that this is not necessarily true of the reals; the polynomial

- #5

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For example, the polynomial x

- #6

Mentallic

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Yes.Are the roots of a polynomial given by the function f(x) defined as the values for x where f(x)=0?

Yes, basically. But there is a particularly important theorem that states that for any polynomial of nDoes that mean f(x)=x^2 has only one root?

The underlined part means that you can have a polynomial of n

It's very common for polynomials to have 2 or more x-values that output to give the same f(x) value. Do you know about inverse functions?Even though for every other value of x except zero there are two values for x that you can input to output a particular value for f(x).

Look back at the theorem and notice the part about complex roots. Every real number is a subset of a complex number (so it can be said that a real number is also complex), it's just that in its complex form z=a+ib, the value of b=0.What about complex numbers?

If we have the function f(x)=x

And again notice that the degree of f(x) is 2, and as it so happens, we have 2 complex roots which in this case are distinct.

EDIT: haha so many replies slipped in before me

- #7

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Graphically, real roots of polynomial is where the graph cuts the x line. :)

- #8

HallsofIvy

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Unfortunately, people are seldom that strict!

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