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Given polynomials \(g, h \in R[X],\) show how to compute their composition \(g(h) \in R[X]\) using \(O\left(\operatorname{len}(g)^{2} \operatorname{len}(h)^{2}\right)\) operations in \(R\).

Short Answer

Expert verified
The key idea behind finding the composition of two polynomials g(X) and h(X) is to replace the variable X in the polynomial g(X) with the polynomial h(X) and then expand the resulting expression using the binomial theorem and the distributive law. This is called polynomial substitution. Finally, combine the terms with the same powers of X and count the total number of operations in the ring R to ensure they are in the order of O(len(g)² len(h)²).

Step by step solution

01

Write down the given polynomials

Write down the given polynomials \(g(X)\) and \(h(X)\) in the form of summation. \(g(X) = \sum_{i=0}^{n} a_i X^i\) \(h(X) = \sum_{j=0}^{m} b_j X^j\) where \(a_i, b_j \in R\), \(n = \operatorname{len}(g) - 1\), and \(m = \operatorname{len}(h) - 1\).
02

Find the composition of polynomials

Replace the variable \(X\) in \(g(X)\) with the polynomial \(h(X)\) to find the composition \(g(h(X))\). \(g(h(X)) = \sum_{i=0}^{n} a_i \left(\sum_{j=0}^{m} b_j X^j\right)^i\)
03

Expand the composition

Expand the composition \(g(h(X))\) by carefully applying the binomial theorem for each power of \(X\) and the distributive law. \(g(h(X)) = \sum_{i=0}^{n} a_i \sum_{k_1+k_2+\cdots+k_m=i} \binom{i}{k_1,k_2,\cdots,k_m} b_1^{k_1} b_2^{k_2} \cdots b_m^{k_m} X^{(k_1+2k_2+\cdots+mk_m)}\) Here, \(\binom{i}{k_1,k_2,\cdots,k_m}\) represents the multinomial coefficient, which can be calculated as \(\frac{i!}{k_1!k_2!\cdots k_m!}\).
04

Combine terms with the same powers of X

Combine all terms with the same powers of \(X\) in the expanded form of \(g(h(X))\) by summing over the multinomial coefficients. \(g(h(X)) = \sum_{p=0}^{nm} c_p X^p\) where \(c_p = \sum_{i=0}^{n} a_i \sum_{\substack{k_1+k_2+\cdots+k_m=i \\ k_1+2k_2+\cdots+mk_m=p}} \binom{i}{k_1,k_2,\cdots,k_m} b_1^{k_1} b_2^{k_2} \cdots b_m^{k_m}\).
05

Count the total number of operations

Now, we will count the total number of operations in the ring \(R\). For each term \(c_p X^p\), we have used \((n+1)(m+1)\) multiplications, as there are \(n+1\) polynomial coefficients in \(g\) and \(m+1\) in \(h\). For each \(nm\) result, we have performed \((n+1)(m+1)\) multiplications. Therefore, the total number of operations is \(\operatorname{len}(g)\operatorname{len}(h)(\operatorname{len}(g)\operatorname{len}(h)) = \operatorname{len}(g)^{2}\operatorname{len}(h)^{2}\), which is in the order of \(O\left(\operatorname{len}(g)^{2} \operatorname{len}(h)^{2}\right)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Computational Complexity
When evaluating the efficiency of an algorithm, especially in mathematical operations like polynomial composition, the term computational complexity plays a crucial role. This concept measures the amount of resources required for an algorithm to run, most often in terms of time or space. Computational complexity is commonly expressed using Big O notation, which provides an upper limit on the time (or the number of operations) an algorithm will take to complete as a function of the input size.

In the given exercise, the goal is to compute the composition of two polynomials using a certain number of operations within the ring of real numbers, denoted by R. The provided step-by-step solution indicates that the total number of operations is in the order of O(len(g)^2 len(h)^2), which means that the time complexity grows quadratically with the lengths of the polynomials g and h. This estimation helps students and mathematicians alike understand the efficiency and feasibility of performing polynomial composition with large polynomials.
Binomial Theorem
The binomial theorem is a fundamental principle in algebra that describes the algebraic expansion of powers of a binomial. According to this theorem, it is possible to expand the power (x + y)^n into a sum involving terms of the form a*x^b*y^c, where the coefficient a of each term is a specific positive integer known as a binomial coefficient. These coefficients can be found in Pascal's triangle or calculated using the formula:

\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]
where n! denotes the factorial of n, and k is the term's specific exponent in the expansion. The binomial theorem streamlines the process of polynomial expansion, which is crucial when working with polynomial composition as shown in the exercise. By applying the binomial theorem to the polynomial h(X) raised to the power of i, we systematically expand the composition of polynomials into a series of more manageable terms.
Multinomial Coefficients
Building upon the binomial theorem, when expanding polynomials involving more than two terms—like in the composition g(h(X)) from the given exercise—multinomial coefficients come into play. These coefficients are a generalization of binomial coefficients for a sum of variables raised to an integer power.

The multinomial coefficient associated with the sequence (k_1, k_2, ..., k_m) in the expansion of (x_1 + x_2 + ... + x_m)^n is given by:

\[\binom{n}{k_1,k_2,\cdots,k_m} = \frac{n!}{k_1!k_2!\cdots k_m!}\]
Meaning that the coefficient is the ratio of the factorial of the total power n to the product of factorials of individual powers k_1, k_2, ..., k_m. This concept is essential when expanding the inner polynomial h(X) within g(X), as it allows us to find the coefficients for each term in the expansion and thereby ultimately find the coefficients of the composite polynomial.

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Most popular questions from this chapter

Given a polynomial \(g \in R[X]\) and an element \(x \in R,\) a particularly elegant and efficient way of computing \(g(x)\) is called Horner's rule. Suppose \(g=\sum_{i=0}^{k-1} a_{i} X^{i},\) where \(k \geq 0\) and \(a_{i} \in R\) for \(i=0, \ldots, k-1 .\) Horner's rule computes \(g(x)\) as follows: $$ y \leftarrow 0_{R} $$ for \(i \leftarrow k-1\) down to 0 do $$ y \leftarrow y x+a_{i} $$ output \(y\) Show that this algorithm correctly computes \(g(x)\) using \(k\) multiplications in \(R\) and \(k\) additions in \(R\).

Let \(g, h \in R[X, Y]\) with \(g=\sum_{i=0}^{m-1} g_{i} Y^{i}\) and \(h=\sum_{i=0}^{m-1} h_{i} Y^{i},\) where each \(g_{i}\) and \(h_{i}\) is a polynomial in \(X\) of degree less than \(k .\) The product \(f:=g h \in R[X, Y]\) may be written \(f=\sum_{i=0}^{2 m-2} f_{i} Y^{i},\) where each \(f_{i}\) is a polynomial in \(X .\) Show how to compute \(f,\) given \(g\) and \(h,\) using \(O(M(k m))\) operations in \(R\). Hint: for an appropriately chosen integer \(t>0,\) first convert \(g, h\) to \(\tilde{g}, \tilde{h} \in R[X],\) where \(\tilde{g}:=\sum_{i=0}^{m-1} g_{i} X^{t i}\) and \(\tilde{h}:=\sum_{i=0}^{m-1} h_{i} X^{t i} ;\) next, compute \(\tilde{f}:=\tilde{g} \tilde{h} \in R[X] ;\) finally, "read off" the \(f_{i}\) 's from the coefficients of \(\tilde{f}\).

This problem is the analog of Exercise 3.48 for \(R[X] .\) Let us first define the notion of a "floating point" reversed Laurent series \(\hat{z},\) which is represented as a pair \((g, e),\) where \(g \in R[X]\) and \(e \in \mathbb{Z}-\) the value of \(\hat{z}\) is \(g X^{e} \in R\left(\left(X^{-1}\right)\right),\) and we call len \((g)\) the precision of \(\hat{z} .\) We say that \(\hat{z}\) is a length \(k\) approximation of \(z \in R\left(\left(X^{-1}\right)\right)\) if \(\hat{z}\) has precision \(k\) and \(\hat{z}=(1+\varepsilon) z\) for \(\varepsilon \in R\left(\left(X^{-1}\right)\right)\) with \(\operatorname{deg}(\varepsilon) \leq-k,\) which is the same as saying that the high-order \(k\) coefficients of \(\hat{z}\) and \(z\) are equal. Show that given \(h \in R[X]\) with \(\operatorname{lc}(h) \in R^{*},\) and positive integer \(k,\) we can compute a length \(k\) approximation of \(1 / h \in R\left(\left(X^{-1}\right)\right)\) using \(O(M(k))\) operations in \(R\). Hint: using Newton iteration, show how to go from a length \(t\) approximation of \(1 / h\) to a length \(2 t\) approximation, making use of just the high-order \(2 t\) coefficients of \(h,\) and using \(O(M(t))\) operations in \(R\).

Let \(F\) be a field. Let \(z=s / t,\) where \(s, t \in F[X], \operatorname{deg}(s)<\operatorname{deg}(t)\) and \(\operatorname{gcd}(s, t)=1\). (a) Show that if \(F\) is finite, there exist integers \(k, k^{\prime}\) such that \(0 \leq k

Assume \(2_{R} \in R^{*}\). Suppose that we are given two polynomials \(g, h \in R[X]\) of length at most \(\ell,\) along with a primitive \(2^{n}\) th root of unity \(\omega \in R,\) where \(2 \ell \leq 2^{n}<4 \ell .\) Let us "pad" \(g\) and \(h,\) writing \(g=\sum_{i=0}^{2^{n}-1} a_{i} X^{i}\) and \(h=\sum_{i=0}^{2^{n}-1} b_{i} X^{i},\) where \(a_{i}\) and \(b_{i}\) are zero for \(i \geq \ell\). Show that the following algorithm correctly computes the product of \(g\) and \(h\) using \(O(\ell \operatorname{len}(\ell))\) operations

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