polynomialsrecursive algorithms
What theorem can be used to recursively multiply two polynomials together?
Is there another theorem that uses recursion to add together two polynomials recursively?
I'm looking for something that accomplishes these things in much the same way that Horner's method can be used to evaluate a polynomial.
Thank You!
You have two questions, the explicit one about why you would want to multiply polynomials, and an implicit one in your final paragraph about what multiplication by a non-integer might mean or why we would care to multiply by a non-integer in the first place.
To address the last one first: once you have multiplication by integers, multiplication by fractions will very quickly rear its head. What does multiplying by "one and a half" mean, if multiplying by 2 means "add to itself", etc? Well, imagine you have a chocolate bar, those that are made up of smaller squares. You can imagine breaking the bar in half, and then figuring out what three times that half will be; that will be multiplying by "three halves" (a.k.a one and a half). You are really multiplying by an integer, after suitably modifying $X$.
In general, if you need to multiply $X$ by a fraction, $\frac{p}{q}$, imagine dividing $X$ into $q$ equal parts, and then multiplying such a $q$th part of $X$ by $p$ in the sense you have above. That is the same as "multiplying by $\frac{p}{q}$". So multiplying by a fraction is like "abbreviated addition": it means "break up into $q$ equal parts, and then add a $q$th part repeatedly $p$ times."
So at least, multiplying by fractions makes just as much "natural sense" as multiplying by integers does.
Why bother with numbers other than fractions then? Well, in one sense you don't have to: you can try to stick to fractions and nothing more complicated than that, and you can go very far. But as the Greeks discovered a long time ago, you also run into very big walls very quickly. For instance, if you draw a square which is $1$ foot long on each side, and you try to measure how long its diagonal is (say, for construction purposes), then it turns out that the diagonal is not a number that can be expressed as a fraction; it is an irrational number. So very soon you end up having to consider numbers that are not fractions, and if they are lying around sooner or later you are going to have to multiply them to compute stuff.
So you end up having to find some way of multiplying irrationals as well, even though they no longer seem to fit with that same "natural" meaning they had back when we started with integers. One solution is that every irrational can be approximated by a suitable sequence of fractions (think about computing the decimals one at a time; every time you stop, what you have so far as a rational; for example, $\sqrt{2} = 1.4142\ldots$, and you get that $1.4 = \frac{14}{10}$, $1.41=\frac{141}{100}$, $1.414=\frac{1414}{1000}$, etc.) We know what it means to multiply $X$ by each of those fractions in a sensible way, so we say that multiplying $X$ by $\sqrt{2}$ is the number you get by doing the successive multiplications, just like $\sqrt{2}$ is the number you get by doing the successive fractional approximations.
This no longer makes sense as "abbreviated addition", but it turns out that it is very, very necessary and very, very useful, in order to make sense of things and be able to compute things that we need to be able to compute (areas, productivity, interest, etc).
As for multiplying polynomials...
One answer: multiplying functions lets you construct more complicated functions out of simpler ones. Or more to the point, it lets you express more complicated functions in terms of simpler ones. This is particularly important if you want to perform complex computations, as then you my be able to "get away" with performing much simpler computations and then multiplying the results, rather than do the really complicated expression instead.
For instance, say you have a single polynomial like $p(x) = x^2-7x+10$. If you realize that $p(x)$ is the result of multiplying the simpler polynomial $x-2$ by the (also simpler) polynomial $x-5$, then whenever you need to evaluate $p(x)$ at a number, say $17$, instead of having to square $17$, then multiply it by $7$, subtract that form the square you computed, and then adding $10$ (three multiplications and two additions/subtractions), you can instead take $17$, subtract $2$ to get $15$; then take $17$, subtract $5$ to get $12$; and then multiply $15$ by $12$ (one multiplication and two additions/subtractions), because $x^2-7x+10 = (x-2)(x-5)$, so $(17)^2 - 7(17) + 10 = (17-2)(17-5)$. Much simpler to do.
Another: it is usually very hard to find a value $x$ for which the result of doing some complex series of operations will be a desired quantity, $d$. For example, you want to know how much money to put in the bank so that, at the end of five months at a particular interest rate, you will have exactly the amount of money you need to buy that new wide-screen TV. This involves solving equations. Many natural equations can be written down in the form $p(x)=c$ where $p(x)$ is a polynomial expression in the unknown quantity $x$, and $c$ is the desired value. Solving such equations can be dificult in general. If you don't know the quadratic formula, then figuring out the values of $x$ for which the polynomial above $x^2-7x+10$ is equal to zero can be pretty difficult. Or think about something like $x^4 + x^3 - 120x^2 - 121x = 121$.
On the other hand, figuring out when a product is equal to $0$ is very easy, because the only way for a product to be zero is if one of the two factors is equal to zero. So if you take the equation above and you write it as $x^4+x^3-120x^2-121x-121 =0$, then you are trying to find when a certain polynomial is equal to $0$. If you can write $q(x)=x^4+x^3-120x^2-121x-121$ as a product, $q(x) = p(x)r(x)$, then you have that $q(x)=0$ if and only if either $p(x)=0$ or $r(x)=0$. With some luck, $p$ and $r$ will be "easier" than $q$, so you can solve them. (In the above case, $q(x) = (x^2-121)(x^2+x+1)=(x-11)(x+11)(x^2+x+1)$, so the only way you can get $q(x)=0$ is if $x=11$ or $x=-11$).
In fact, this is one way to figure out the quadratic formula (did you ever wonder where it came from?). Why are the solutions to $ax^2 + bx+c = 0$ given by $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$? You can factor out $a$ and get $a(x^2 + Bx + C) = 0$, with $B=\frac{b}{a}$ and $C=\frac{c}{a}$. For this to be zero, you need $x^2+Bx+C=0$. Now, imagine you could write it as a product, $$x^2 + Bx+C = (x-r_1)(x-r_2).$$ What would $r_1$ and $r_2$ be? If you know how to multiply polynomials, you get that $(x-r_1)(x-r_2)=x^2 - (r_1+r_2)x + r_1r_2$, so you need $r_1r_2=C$ and $r_1+r_2 = -B$. Squaring the latter you get $(r_1+r_2)^2 = B^2$; but $(r_1+r_2)^2 = r_1^2 +2r_1r_2 + r_2^2$. On the other hand, $$(r_1-r_2)^2 = r_1^2 - 2r_1r_2 + r_2^2 = (r_1^2+2r_1r_2+r_2^2) - 4r_1r_2 = B^2 - 4C.$$ So $(r_1-r_2)^2 =B^2-4C$. Taking square roots, you have that $r_1-r_2 = \pm \sqrt{B^2-4C}$. And you already know that $r_1+r_2 = -B$. Adding them you get $$2r_1 = -B\pm\sqrt{B^2-4C}\qquad\text{or}\qquad r_1 = \frac{-B\pm\sqrt{B^2-4C}}{2}$$ and taking the difference between $r_1+r_2 = -B$ and $r_1-r_2 = \pm\sqrt{B^2-4C}$ you get $$2r_2 = -B\mp\sqrt{B^2-4C}\qquad\text{or}\qquad r_2 = \frac{-B\mp\sqrt{B^2-4C}}{2}.$$ So you get that $r_1 = \frac{-B+\sqrt{B^2-4C}}{2}$ and $r_2 = \frac{-B-\sqrt{B^2-4C}}{2}$, and plugging in $B=\frac{b}{a}$ and $C=\frac{c}{a}$ gives the usual quadratic formula. No way to find it without knowing how to multiply polynomials!
When you get to Calculus (added: I'm assuming you will "get to Calculus" because you tagged the question as being (algebra-precalculus), so presumably you are taking a course labeled as 'precalculus'; but this may not be the case. If you are not going to "get to Calculus", then this paragraph will not tell you anything useful), you will find that there is a particular operation (differentiation, taking derivatives) which is very useful and very important. It tells you how fast a certain quantity is changing, and it can be used to find all sorts of useful things, like what production level will maximize profit in a factory, how big a dose of medicine and how often you should give to a patient based on how fast they metabolize it, and many other applications. Computing derivatives from first principles with an arbitrary function is pretty labor-intensive; but by reconizing a function as being "made up" (through sums, products, quotients, and compositions) of other, simpler, functions, makes it a very straighforward and easy job.
But in order to be able to recognize that a function is a product of two other functions, you first need to know how to multiply two functions together. Polynomials are one case.
Another situation occurs when the polynomials are measuring different things, and their product is somehow meaningful; maybe one polynomial gives you the length and the other polynomial gives you the width of a certain figure? Their product will be the area, which may be something you need to compute.
And more generally, you can think of polynomials as "abbreviations" for more complicated operations that you are doing with numbers, just like you are thinking of multiplication as "abbreviated addition". In that case, multiplying the two polynomials represents another complicated operation that you need to express in terms of the two simpler ones (addition and multiplication).
$6x^7-7x^5+2x^4-10x^2+20x-6$ =
$6x^7+0x^6-7x^5+2x^4+0x^3-10x^2+20x-6$
The method is essentially start with the coefficient of the highest power, multiply by $x$ and add the next coefficient. Stop when you add the constant coefficient.
So steps in the iteration go:
$6$
$6x[+0]$
$6x^2-7$
$6x^3-7x+2$
$6x^4-7x^2+2x[+0]$
$6x^5-7x^3+2x^2-10$
$6x^6-7x^4+2x^3-10x+20$
$6x^7-7x^5+2x^4-10x^2+20x-6$
Trust this helps
Best Answer
For multiplication, maybe you are looking for Karatsuba's algorithm? See section 1.3 of this book:
http://www.loria.fr/~zimmerma/mca/pub226.html
(Addition is also mentioned, but I'm not sure there's much useful that can be said about polynomial addition).