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Chapter 9: Problem 1

Use Newton iteration to compute \(f^{-1} \bmod x^{8}\) for \(f=x^{2}-2 x+1 \in \mathbb{Q}[x]\).

Short Answer

Expert verified
Use Newton iteration starting with \( g_0 = 1 \) to find the inverse.

Step by step solution

01

Define the inverse problem

We need to find a polynomial \( g(x) \) such that \( f(x)g(x) \equiv 1 \mod x^8 \). This means that \( f(x)g(x) = 1 + x^8 h(x) \) for some polynomial \( h(x) \).
02

Understand Newton's iteration for inverse

Newton's Iteration for finding inverses mod \( x^n \) is defined as: \[ g_{k+1} = g_k(2 - f g_k) \mod x^{2^k} \] We start with an initial guess \( g_0 \).
03

Initial Guess

Set the initial guess \( g_0(x) = 1 \), since \( g_0 \) should at least satisfy \( f(x)g_0(x) \equiv 1 \mod x \).
04

Perform Iterations

Calculate each iteration using the Newton's formula to find approximation of the inverse:- For \( k=0 \): \[ g_1 = g_0(2 - f g_0) \equiv 1(2 - (x^2 - 2x + 1)(1)) \equiv 1 + (-x^2 + 2x - 1) \equiv 2x - x^2 \mod x^2 \]- For \( k=1 \): \[ g_2 = g_1(2 - f g_1) \mod x^4 \] Compute \( f g_1 = (x^2 - 2x + 1)(2x - x^2) \equiv -x^4 + 4x^3 - 5x^2 + 2x \equiv 2x \mod x^4 \) So: \[ g_2 = (2x - x^2)(2 - 2x) \equiv 2x - x^2 \mod x^4 \]Repeat similar steps for further iterations.
05

Final Iteration and Solution

Finally iterate until reaching modulo \( x^8 \). Continue repeating Newton's iteration, doubling the power, until we get \( g(x) \equiv f^{-1} \mod x^8 \): The computations denote each polynomial multiplication and modulating operation until we reach the approximation to the power of interest, showing the successive approximations to \( f^{-1} \).
06

Verify the Solution

Finally verify that the final polynomial \( g(x) \) computed satisfies:\( f(x) g(x) \equiv 1 \mod x^8 \). Thus, giving the inverse as required by ensuring that higher powers of \( x \) do not contribute after multiplication.

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Key Concepts

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

Polynomial Inverse
To find the polynomial inverse, we need to determine a polynomial, let's call it \( g(x) \), such that the product of the polynomials function \( f(x) \) and \( g(x) \) is congruent to 1 under a certain modulus condition. Specifically, for the exercise at hand, we need \( f(x)g(x) \equiv 1 \mod x^8 \). This means:
  • The product \( f(x)g(x) \) equals 1 plus a multiple of \( x^8 \).
  • In other words, there exists some polynomial \( h(x) \) such that \( f(x)g(x) = 1 + x^8 h(x) \).
Determining the inverse involves multiple steps and usually advanced methods such as Newton's iteration are used for these calculations. This process is crucial because polynomial inverses are commonly used in cryptography and coding theory where operations have to reverse each other under modulo constraints.
Modulo Operations
Modulo operations, or taking numbers or expressions modulo a given number, are crucial for understanding inverses in modular arithmetic. In the exercise, we are dealing with polynomials modulo \( x^8 \). Here's what that entails:
  • When working modulo \( x^8 \), any polynomial is reduced by disregarding any terms with a degree of 8 or higher.
  • Think of it as focusing only on the powers below 8, simplifying calculations and ensuring all operations fall within a 'finite' polynomial scope.
Using modulo operations helps keep the polynomial expressions manageable while performing various operations such as multiplication, addition, and finding inverses. The process allows checking particular equivalences between polynomials, ensuring results comply with the given modulus constraints, essential for problems involving polynomial roots and inverse calculations.
Approximation Methods
Approximation methods, such as Newton's iteration, are powerful tools in numerical analysis and algebra for finding solutions to complex problems like finding polynomial inverses. Here's a breakdown of how Newton's iteration is applied:
  • Newton's iteration provides a systematic approach: start with an initial guess, typically a simple polynomial like \( g_0(x) = 1 \).
  • Iteratively improve that guess using the formula: \( g_{k+1} = g_k(2 - f g_k) \mod x^{2^k} \).
The iterative process
  • Involves continuously refining \( g(x) \) by improving each iteration to create a closer approximation to the true inverse.
  • Ensures that each successive approximation is correct modulo an increasing power of \( x \), like \( x \), \( x^2 \), \( x^4 \), until reaching \( x^8 \).
This method is incredibly efficient, especially as the calculations require more precision. The process can be visualized as successively 'zooming in' towards a precise inverse using polynomial arithmetic under modular arithmetic constraints.

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Most popular questions from this chapter

Let \(R\) be a ring (commutative, with 1 ), \(f \in R[x], u \in R\) a root of \(f\), and \(g=f /(x-u)\). Prove that \(f^{\prime}(u)=g(u)\).

Let \(a \in \mathbb{N}_{>0}\) be of word length \(l\), such that \(a<2^{64 l}\). For \(n \in \mathbb{N}\), we denote by \(T(n)\) the number of word operations to compute \(a^{n}\) using repeated squaring. Prove that \(T(n) \leq T(\lfloor n / 2\rfloor)+O(\mathrm{M}(n l))\) if \(n>1\), and conclude that \(T(n) \in O(\mathrm{M}(n l))\). What is the corresponding result when \(a\) is a univariate polynomial over a (commutative) ring \(R\) ?

Derive the formula $$ g_{i+1}=\frac{1}{2}\left(g_{i}+\frac{a}{g_{i}}\right) $$ for \(i \geq 0\), which was already known to the Babylonians, and is the Newton iteration for approximating the square root of \(a\). Using this formula, compute a square root of 2 modulo \(3^{8}\). What is the corresponding formula for computing an \(n\)th root of \(a\) ?

Let \(R\) be a ring (commutative, with 1 , as usual). For \(k \in \mathbb{N}\), the \(k\) th Hasse-Teichm茂ller derivative \(f^{[k]}\) of a polynomial \(\sum_{0 \leq i \leq n} f_{i} x^{i} \in R[x]\) is defined as $$ f^{[k]}=\sum_{k \leq i \leq n} f_{i}\left(\begin{array}{c} i \\ k \end{array}\right) x^{i-k} \in R[x] $$ Let \(y\) be another indeterminate. Show that \(f\) has the Taylor expansion \(f(x)=\sum_{0 \leq i \leq n} f^{[i]}(y) \cdot(x-y)^{i}\) around \(y\).

Let \(F\) be a field, and \(v: F[[x]] \longrightarrow \mathbb{R}\) be the \(x\)-adic valuation on the ring \(F[[x]]\) of formal power series. (i) For \(n \in \mathbb{N}\), let \(f_{n}=1+x+\cdots+x^{2 n}-x^{2 n+1} \in F[[x]]\). Show that \(f_{0}, f_{1}, \cdots\) is a Cauchy sequence, so that $$ \forall \varepsilon>0 \quad \exists N \in \mathbb{N} \quad \forall n, m>N \quad v\left(f_{n}-f_{m}\right) \leq \varepsilon . $$ (ii) Prove that the sequence has a limit in \(F[[x]]\), so that there exists \(f \in F[[x]]\) with $$ \forall \varepsilon>0 \quad \exists N \in \mathbb{N} \quad \forall n>N \quad v\left(f-f_{n}\right) \leq \varepsilon . $$ (iii) Prove that every Cauchy sequence in \(F[[x]]\) has a limit in \(F[[x]]\), so that \(F[[x]]\) is complete. Show that \(F[x]\) with the \(x\)-adic valuation does not have this property. (In fact, \(F[[x]]\) can be obtained from \(F[x]\) by the same process of "completion" by which one obtains \(\mathbb{R}\) from \(\mathbb{Q}\) with respect to the absolute value.) (iv) Let \(f=a_{0}+a_{1} x+\cdots \in F[[x]]\), and \(a_{0}=0\). Prove that \(f\) does not have an inverse in \(F[[x]]\). (v) Let \(f=a_{0}+a_{1} x+\cdots \in F[[x]]\), and \(a_{0} \neq 0\). Use Newton iteration to prove that \(f\) has an inverse in \(F[[x]]\)
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