I noticed that multiplication of polynomial behaves as convolution of their coefficients. And this arises a question for me what is the intuition behind defining multiplication like this? Why we don’t define polynomial multiplication Components-wise instead? Like why not define multiplication like this $$(x^3+2x^2+1)(4x^2+2)=(0)x^3+(8)x^2 + (2)$$?
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@Mariano Suárez-Álvarez Indeed, with combinatorics applications like the ones mentionned in this paper https://www.ams.jhu.edu/~fill/papers/Hadamard.pdf – Jean Marie Dec 18 '22 at 08:30
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Polynomial multiplication is using the usual definition of $\times$ that distributes over $+$. Using distributive property many times over gets the convolution. Component wise multiplication is often useful in data, but it’s just not the usual multiplication that you have from either substituting numbers for variables or thinking of $x$ as a formal variable.
Aaron Goldsmith
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Indeed, it’s just component wise multiplication. You’re right. – Aaron Goldsmith Dec 18 '22 at 04:12
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I see, but why do we define the multiplication of polynomials like this? I mean, for the numbers, we can interpret multiplication as a repetitive sum, but for polynomial multiplication, what is the intuition? @MarianoSuárez-Álvarez – Mubarak Alsaeedi Dec 18 '22 at 04:16
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Still the question hold why we define function multiplication like this? Please forgive me if the question seems silly, I really want to know what is the motivation behind defining the operation like this. – Mubarak Alsaeedi Dec 18 '22 at 04:22
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1@MubarakAlsaeedi If $f, g$ are complex valued functions, we define their product $fg$ by $(fg)(x) = f(x)g(x)$. So once we know how to multiply complex numbers, we know how to multiply complex valued functions, such as polynomials with complex coefficients. – Kakashi Dec 18 '22 at 04:26
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1Say you have sheets of paper, always 2 cm longer than they are wide. You can write the formula for the area of a sheet of paper as $A(w)=w(w+2)$. That would use ordinary multiplication after you plug in a value. So, you can also write $A(w)=w^2+2w$ to get the same area function of width. It’s sometimes because we are pre-performing the arithmetic with variable placeholders. – Aaron Goldsmith Dec 18 '22 at 04:27
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@AaronGoldsmith Aha! And from this, I see why we use the notation R[x] for polynomial ring, it's because x represents an undetermined element of the ring R, and we want the properties of multiplication and addition to hold, so we define the multiplication like this and not in another way. – Mubarak Alsaeedi Dec 18 '22 at 04:59
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1One of the shortest (abstract, I admit) explanation I have found is by Qiao Yuan here : It is due to the isomorphism between the group algebra $K[G]$ of a group $G$ with multiplication and the algebra of finitely supported functions $G \to K$ under convolution, where $K$ is a field. – Jean Marie Dec 18 '22 at 08:19
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1@Mubarak Yes, that's the key idea. We seek a ring $R[x]$ containing $R$ and a new element $x$ that's as general as possible - the only equalities that hold in $R[x]$ are those forced by the ring axioms and $,rx = xr.,$ This implies that equalities true in $R[x]$ map to true equalties when mapped into a ring having an image of $R$ by evaluating at $x,,$ e.g. we can prove identities like the Binomial Theorem. Factor Theorem etc (once and for all) in $\Bbb Z[x,y]$ then specialize them into any ring by an evaluation (ring hom). – Bill Dubuque Dec 18 '22 at 17:43
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For a less trivial example see here for a proof of Sylvester's Determinant Identity using matrices with indeterminate entries ("generic matrices') - a proof so surprising that when first encountered many wrongly believe it is erroneous (even some math professors!). Such is the power of polynomials. See here for further discussion of universal properties of polynomial rings. – Bill Dubuque Dec 18 '22 at 17:44
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With the coefficient-wise product, the evaluation map would not be a ring homomorphism. Many people think of polynomials as functions, so that would be a problem.
I know of an application of coefficient-wise multiplication (Hadamard product) in combinatorics, in the field of partially ordered sets. The paper I would recommend is by Wilhelm Plesken, "Counting with Groups and Rings" 1982, Journal fuel die reine und angewandte Mathematik.
Anton
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Polynomials are functions.
But what are functions? Functions are where, if you input one number, you get a different number.
Consider your example, $(x^3 + 2x^2 +1)(4x^2 + 2)$. Here, we can use the distributive property, which is used for sums of numbers. But remember, polynomials are made of power functions, so polynomials are sums of numbers.
$$(x^3 + 2x^2 + 1)(4x^2) + (x^3 + 2x^2 + 1)(2)$$
Then, since multiplication is communitive, we can do $$((x^3)(4x^2) + (2x^2)(4x^2) + 1(4x^2)) + ((x^3)(2) + (2x^2)(2) + 1(2))$$
Which is what polynomial multiplication is.
You make a very interesting connection between polynomial multiplication and convolutions. In fact, some Python libraries use polynomial multiplication to calculate convolutions. The video below by 3Blue1Brown gives more information about his. Go to 13:40 for that.
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1I see, so the definition of function multiplication arises from the fact that we treat x as undetermined number and we want to preserve the properties of the usual multiplication over numbers. – Mubarak Alsaeedi Dec 18 '22 at 04:53
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Not true. Generally polynomials are not functions, e.g. by little Fermat the function $,f(x) = x^p-x,$ is the zero function on the field $,\Bbb F_p = \Bbb Z_p = $ integers $!\bmod p,,$ but $f(x)$ is not the zero polynomial. See here for more on (formal) polynoimials vs. polynomial functions. @Mubarak – Bill Dubuque Dec 18 '22 at 07:38
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@Mariano Look closer - the question is tagged
abstract-algebraandring-theoryso it is not unreasonable to assume that the OP knows about modular arithmetic or finite fields. – Bill Dubuque Dec 18 '22 at 16:56