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    ''If the reciprocal equation, f(x) = 0, is of odd degree, and therefore has an odd number of roots, one of these must be its own reciprocal.'' How can this be possibly proven?
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    What is a "reciprocal equation"? Answer that, and you should have a good idea of what to do next.
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    (Original post by DFranklin)
    What is a "reciprocal equation"? Answer that, and you should have a good idea of what to do next.
    I'm guessing it's an equation of the form a_0 + a_1x + a_2x^2 + \cdots + a_{n-1}x^{n-1} + a_nx^n = 0, where a_i = a_{n-i}... so for example 3x^5 + 2x^4 + x^3 + x^2 + 2x + 3 = 0 would be one.

    OP: I'm not 100% sure if this is the way to go about it, but if \mbox{f}(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n = 0 for some constants a_i and odd n, then you can make the substitution x = u^2 so that you get a_0 + a_1u^2 + \cdots + a_nu^{2n} and then divide through by u_n so that you get \left( \dfrac{a_0}{u^n} + a_nu^n \right) + \left( \dfrac{a_1}{u^{n-2}} + a_{n-1}u^{n-2} \right) + \cdots + \left( \dfrac{a_{(n-1)/2}}{u} + a_{(n+1)/2}u \right), and then the replace a_{n-i} by a_i (since they're equal), giving you a_0\left( \dfrac{1}{u^n} + u^n \right) + a_1\left( \dfrac{1}{u^{n-2}} + u^{n-2} \right) + \cdots + a_{(n-1)/2}\left( \dfrac{1}{u} + u \right), and then the replace a_{n-i} by a_i. From there the substitution z = \dfrac{1}{u} + u might be of use. This is total guesswork (and may lead to nothing useful).

    EDIT: I'm fairly sure that what I've done will get you absolutely nowhere, but it does lead to some interesting stuff if you follow it far enough (none of it is relevant to what you want though).
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    (Original post by nuodai)
    I'm guessing it's an equation of the form a_0 + a_1x + a_2x^2 + \cdots + a_{n-1}x^{n-1} + a_nx^n = 0, where a_i = a_{n-i}... so for example 3x^5 + 2x^4 + x^3 + x^2 + 2x + 3 = 0 would be one.
    Yes, I can guess too, but (a) what's the point of guessing when the OP knows the asnwer? and (b) I had a reason for wanting the OP to answer in his own words.

    hint
    Why do you think it's an equation of that form? There's nothing in what you say that has anything obvious to do with reciprocals.
    This is total guesswork (and may lead to nothing useful).
    That looks like complete overkill, to be honest.

    Spoiler:
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    You might want to consider the relation between f(x) and f(1/x).
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    (Original post by DFranklin)
    Yes, I can guess too, but (a) what's the point of guessing when the OP knows the asnwer? and (b) I had a reason for wanting the OP to answer in his own words.

    hint
    Why do you think it's an equation of that form? There's nothing in what you say that has anything obvious to do with reciprocals.
    Because Google led me to this :p:
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    (Original post by nuodai)
    Because Google led me to this :p:
    But evidently it didn't lead you to read up to the third sentence in the article... :p:
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    (Original post by DFranklin)
    But evidently it didn't lead you to read up to the third sentence in the article... :p:
    :o:
 
 
 
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