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Math Help - Purely Transcendental

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    Purely Transcendental

    Given an element x\in \mathbb{R} having a property that no non-constant polynomial in \mathbb{Q} has x as a zero, we call it "transcendental". For example, \pi \in \mathbb{R} is an example.

    However, if we allow "infinite polynomials", i.e. power series then it is no longer transcendental. Because \pi is a zero of f(x)=x - \frac{x^3}{3!}+\frac{x^5}{5!}-....


    My question is: Is there an element in \mathbb{R} which is purely transcendental? Meaning it has no rational power series even.
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    Super Member Rebesques's Avatar
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    What do you mean by "rational power series"? Rational coefficients?
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    Quote Originally Posted by Rebesques View Post
    What do you mean by "rational power series"? Rational coefficients?
    Yes
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    Super Member Rebesques's Avatar
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    Like e^{1/\pi}=\sum_n \frac{1}{n!\pi^n} ?
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    Quote Originally Posted by Rebesques View Post
    Like e^{1/\pi}=\sum_n \frac{1}{n!\pi^n} ?
    He is interested in those numbers, if such exist, which are not the root of any
    rational power series:

    f(x)=\sum_{n=0}^{\infty} a_n x^n,\ a_n \in \bold{Q}

    RonL
    Last edited by CaptainBlack; August 12th 2007 at 12:49 PM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    He is interested in those numbers, if such exist, which are not the root of any
    rational power series:

    f(x)=\sum_{n=0}^{\infty} a_n x^n,\ n \in \bold{Q}

    RonL
    Correct me if I'm wrong, but we should be able to represent any real number as a decimal, yes? Which means that all real numbers, including the transcendentals can be expressed as \sum_{n = 0}^{\infty}a_n(10)^n for some choice of \{ a_n \}. (This argument would seem to depend on the Axiom of Choice?) So my stab at the answer is that there is no such transcendental number.

    -Dan
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    Quote Originally Posted by topsquark View Post
    Correct me if I'm wrong, but we should be able to represent any real number as a decimal, yes? Which means that all real numbers, including the transcendentals can be expressed as \sum_{n = 0}^{\infty}a_n(10)^n for some choice of \{ a_n \}. (This argument would seem to depend on the Axiom of Choice?) So my stab at the answer is that there is no such transcendental number.

    -Dan
    x=\sum_{n = 0}^{\infty}a_n(10)^n does not guarantee that

    <br />
\sum_{n=0}^{\infty} a_n x^n=0,\ a_n \in \bold{Q}<br />

    for some a_0, a_1, ..  \in \bold{Q}

    RonL
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    Super Member Rebesques's Avatar
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    I wonder where he comes up with those

    Ok, suppose such a number x exists -- not a root of any power series with rational coefficients. We will prove Hacker's questions are transcedental.


    The set S={1,x,x^2,...} is then linearly independent over Q, and thus constitutes a transcedence basis for R (over Q). But S is countable, while the degree of the extension of R over Q has the power of the continuum.


    "Quod erat demonstratum"
    Last edited by Rebesques; August 12th 2007 at 05:32 PM. Reason: tried to remember the proof of [R:Q]=c and failed.
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    Quote Originally Posted by Rebesques View Post
    The set S={1,x,x^2,...} is then linearly independent over Q, and thus constitutes a transcedence basis for R (over Q). But S is countable, while the degree of the extension of R over Q has the power of the continuum.
    How do we know that given any r\in \mathbb{R} there exists q_i\in \mathbb{Q} so that r=q_0+q_1x+q_2x^2+..., i.e. is spams?
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    Quote Originally Posted by Rebesques View Post
    I wonder where he comes up with those

    Ok, suppose such a number x exists -- not a root of any power series with rational coefficients. We will prove Hacker's questions are transcedental.


    The set S={1,x,x^2,...} is then linearly independent over Q, and thus constitutes a transcedence basis for R (over Q). But S is countable, while the degree of the extension of R over Q has the power of the continuum.


    "Quod erat demonstratum"
    It is obvious that the set of numbers which are roots of some rational power series has the same cardinality as \bold{R}, but that does not prove tht the set exhausts \bold{R}.

    It is also obvious tha the set is dense in \bold{R}, as it inherits this propety from the set of algebraic numbers which it contains.

    RonL
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    x=\sum_{n = 0}^{\infty}a_n(10)^n does not guarantee that

    <br />
\sum_{n=0}^{\infty} a_n x^n=0,\ a_n \in \bold{Q}<br />

    for some a_0, a_1, ..  \in \bold{Q}

    RonL
    Ah well. I guess when I'm off, I'm way off.

    -Dan
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    Super Member Rebesques's Avatar
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    How do we know (...) spans?

    It is also obvious tha the set is dense in , as it inherits this propety from the set of algebraic numbers which it contains.

    You are both right, I mistakingly took it for a maximal set.
    Last edited by Rebesques; August 13th 2007 at 05:36 PM.
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