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Math Help - Proof of a polynomial function having a solution in R.

  1. #1
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    Proof of a polynomial function having a solution in R.

    Don't really know what I'm doing with this proof here a little help would be amazing right now. Here is the question.

    Let
    f(x)=a0+a1x+...+anx^n (x ϵ ℝ)
    be a polynomial function, an ≠ 0. Suppose n is odd. show that the equation f(x)=0 has at least one solution in ℝ.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ml692787 View Post
    Don't really know what I'm doing with this proof here a little help would be amazing right now. Here is the question.

    Let
    f(x)=a0+a1x+...+anx^n (x ϵ ℝ)
    be a polynomial function, an ≠ 0. Suppose n is odd. show that the equation f(x)=0 has at least one solution in ℝ.
    A proof can go along these lines.

    Without loss of generality, assume a_n > 0

    since n is odd, and f(x) is continuous, we have \lim_{x \to \infty}f(x) > 0 and \lim_{x \to - \infty} f(x) < 0

    thus, by the intermediate value theorem, there exists a real number x \in (-\infty, \infty) such that f(x) = 0
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    Thanks a lot this makes a lot of sense after reading the chapter and putting it together with your thoughts, very much appreciated!
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  4. #4
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    I like to do it with sequences instead. It is more athestically pleasing. (However, the proof using field theory is even more nicer but this is not an algebra thread.)

    Let f(x) be a polynomial in \mathbb{R} of odd degree. Define x_n = f(n). This is a sequence. Now argue that \lim \ x_n = + \infty without lose of generality if the leading coefficient it positive. And define y_n = f(-n) and show \lim \ y_n = -\infty. Thus, this means by definition there is a number n_1<0 so that f(n_1)<0 and there is a number n_2>0 so that f(n_2)>0. Thus, there is a zero by intermediate value theorem. Q.E.D.
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