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I'm hoping that this isn't such a basic question that it gets completely laughed off the site, but why would I want to multiply two polynomials together?

I flipped through some algebra books and have googled around a bit, and whenever they introduce polynomial multiplication they just say 'Suppose you have two polynomials you wish to multiply', or sometimes it's just as simple as 'find the product'. I even looked for some example story problems, hoping that might let me in on the secret, but no dice.

I understand that a polynomial is basically a set of numbers (or, if you'd rather, a mapping of one set of numbers to another), or, in another way of thinking about it, two polynomials are functions, and the product of the two functions is a new function that lets you apply the function once, provided you were planning on applying the original functions to the number and then multiplying the result together.

Elementary multiplication can be described as 'add $X$ to itself $Y$ times', where $Y$ is a nice integer number of times. When $Y$ is not a whole number, it doesn't seem to make as much sense.

Any ideas?

  • 18
    I just want to comment on one subtlety: Actually, a polynomial *is* not a function even though it induces a function from the underlying field to itself. This is particularly apparent when the field over which you are working is finite: then there are only finitely many functions from the field to itself, but still infinitely many polynomials.2010-11-22
  • 3
    Since you mentioned dice: http://math.stackexchange.com/questions/4632/how-can-i-algorithmically-count-the-number-of-ways-n-m-sided-dice-can-add-up-to/4652#46522010-11-22
  • 2
    Isn't multiplying two integers just like multiplying two polynomials and taking the coefficients? ;)2010-11-22
  • 20
    One reason to multiply polynomials is that there are points on a test at stake.2010-11-22
  • 0
    See http://mathoverflow.net/questions/41310/any-sum-of-2-dice-with-equal-probability/41311#413112010-11-22
  • 0
    @Rasmus: I'm sure the op only cares about polynomials on fields of char 0 (where two polynomials are equivalent iff their functions are).2010-11-22
  • 26
    Don't be shy. This is a fantastic question: it is of obvious and wide interest, and moreover challenging: off the top of my head, I'm not sure how to answer it!2010-11-22
  • 0
    Sometimes for questions like these, I'm tempted to just use Lockhart's answer: "because we can". Admittedly the path towards applications is long and tenuous, and in practice (well, my practice at least), one more often factors 'em than multiply 'em...2010-11-22
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    @J.M.: Yes; but you have absolutely no hope of factoring if you don't know how to multiply in the first place!2010-11-23
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    @Arturo: Sure, but I meant it in the vein of "$f(x)$ gets used more often than $f^{(-1)}(x)$ in applications"...2010-11-23
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    For some reason I still remain unconvinced by the attempts at showing the practicality of multiplying two polynomials (a grasping at straws, if you will)... I'll stick with Lockhart's answer, then.2010-11-23
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    @Eric: 1)Also for fields of characteristic zero, polynomials or not the same as polynomial functions. 2)Comments and answers need not address solely the OP.2010-11-23
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    @J.M. - I think that's a misunderstanding of Lockhart's view. He strongly rejects the reduction of mathematics to applications of mathematics to other areas (hence "why think about these things? Because we can..."), but he is very interested pedagogically in the motivation of ideas within math itself. I think he would always try to come up with a substantive answer to a question about motivation.2012-11-12
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    ]If you think so, please reconsider the problem several years from now2017-04-14

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