Formula for Multiplying Two Elements in a Polynomial Ring

In summary: Then, the equation becomes$$p(x)q(x) = \left( \sum_{k=0}^{m+n} \left( \sum_{i+j=k} a_i b_j \right) x^k \right)$$The term ##a_0 b_{n+1} x^{n+1}## can be identified with ##k=n+1## and ##a_1 b_{n+2} x^{n+2}## will be the leading term. Hence,$$p(x)q(x) = \left( \sum_{k=0}^{n+1} \left( \sum
  • #1
Bashyboy
1,421
5

Homework Statement


I would like to show that if ##p(x) = \sum_{i=1}^m a_i x^i## and ##q(x) = \sum_{j=1}^n b_j x^j##, then ##p(x)q(x) = \sum_{k=0}^{m+n} \left( \sum_{i+j=k} a_i b_j \right) x^k##, where the polynomial ring is assumed to be commutative.

Homework Equations

The Attempt at a Solution



The base case of ##m=n=1## is trivial; one just simply compares the formula to a "brute force" calculation. So, suppose that the formula holds for polynomials ##p## and ##q## where ##p## has length ##1## and ##q## has length ##n##. Then

$$p(x)q(x) = p(x) \sum_{j=1}^{n+1} b_j x^j = p(x) \sum_{j=1}^n b_j x^j + p(x)b_{n+1}x^{n=1}$$

By the induction, hypothesis, ##p(x) \sum_{j=1}^n b_j x^j = \sum_{k=0}^{n+1} \left( \sum_{i+j=k} a_i b_j \right) x^k##, and so

$$p(x)q(x) = \sum_{k=0}^{n+1} \left( \sum_{i+j=k} a_i b_j \right) x^k + a_0 b_{n+1} x^{n+1} + a_1 b_{n+2} x^{n+2}$$

The term ##a_0 b_{n+1} x^{n+1}## can be identified with ##k=n+1## and ##a_1 b_{n+2} x^{n+2}## will be the leading term. Hence,

$$p(x)q(x) = \sum_{k=0}^{n+1} \left( \sum_{i+j=k} a_i b_j \right) x^k + a_0 b_{n+1} x^{n+1} + a_1 b_{n+2} x^{n+2}$$

I feel that this last point is a bit shaky, but I'll let you be the judge of that. By commutativity and symmetry we get the other induction. Now for the double induction. Assume the formula holds for polynomials of length ##m## and ##n##. Then

$$p(x)q(x) = \sum_{i=1}^{m+1} a_i x^i \sum_{j=1}^{n+1} b_j x^j = \left( \sum_{i=1}^{m} a_i x^i + a_{m+1} x^{m+1} \right) \left(\sum_{j=1}^{n} b_j x^j + b_{n+1} x^{n+1} \right)$$

$$= \sum_{i=1}^{m} a_i x^i \sum_{j=1}^{n} b_j x^j + \sum_{i=1}^m a_i b_{n+1} x^{i+n+1} + \sum_{j=1} a_{m+1} b_j x^{j + m+1} + a_{m+1} b_{n+1} x^{m+n+2}$$

At this point, I can apply the induction hypothesis on the first term, but I am unsure how to properly combine this mess to get the desired formula...
 
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  • #2
A suggestion: don't confuse the length with the degree of a polynomial , a polynomial of length ##1## can have a degree ##m##, so it is of the following monomial form ##P(x)=a_{m}x^{m}##. For the rest the induction is the correct idea ...
 
  • #3
Bashyboy said:

Homework Statement


I would like to show that if ##p(x) = \sum_{i=1}^m a_i x^i## and ##q(x) = \sum_{j=1}^n b_j x^j##, then ##p(x)q(x) = \sum_{k=0}^{m+n} \left( \sum_{i+j=k} a_i b_j \right) x^k##, where the polynomial ring is assumed to be commutative.

...

Fix up the lower summation limits: they should all start at either 0 or at 1 (and, preferably, all at 0).
 

FAQ: Formula for Multiplying Two Elements in a Polynomial Ring

What is the formula for multiplying two elements in a polynomial ring?

The formula for multiplying two elements in a polynomial ring is as follows: (a + bx)(c + dx) = ac + (ad + bc)x + bdx². This formula is also known as the FOIL method.

What does each term in the formula represent?

The terms in the formula represent the coefficients of the polynomial. The first term, ac, is the product of the first coefficients of each element. The second term, ad + bc, is the sum of the product of the first and second coefficients of each element. The third term, bdx², is the product of the second coefficients of each element.

Can this formula be used for polynomials with more than two terms?

Yes, this formula can be extended for polynomials with more than two terms. For example, to multiply (a + bx + cx²)(d + ex + fx²), we would use the same formula and combine like terms to get the final product.

What is the purpose of multiplying two elements in a polynomial ring?

Multiplying two elements in a polynomial ring is used to simplify and combine polynomials. This is useful in various mathematical operations, such as finding the roots of polynomials or solving equations involving polynomials.

Are there any special cases or rules when using this formula?

One special case to note is when multiplying polynomials with the same variable raised to different powers. For example, (a + bx)(c + dx²) would be expanded as ac + adx² + bcx + bdx³. Additionally, it is important to remember to combine like terms and simplify the final product.

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