Problem 19 - A Computational Introduction to Number Theory and Algebra

Question

Prove the "chain rule" for formal derivatives: if \(g, h \in R[X]\) and \(f=g(h) \in R[X],\) then \(\mathbf{D}(f)=\mathbf{D}(g)(h) \cdot \mathbf{D}(h) ;\) more generally, if \(g \in\) \(R\left[X_{1}, \ldots, X_{n}\right],\) and \(h_{1}, \ldots, h_{n} \in R[X],\) and \(f=g\left(h_{1}, \ldots, h_{n}\right) \in R[X],\) then $$ \mathbf{D}_{X}(f)=\sum_{i=1}^{n} \mathbf{D}_{X_{i}}(g)\left(h_{1}, \ldots, h_{n}\right) \mathbf{D}_{X}\left(h_{i}\right) $$

Step-by-Step Solution

Answer

Question: Prove the chain rule for formal derivatives and generalize it to a multivariable function. Answer: For a single-variable function, the chain rule states that if \(f = g(h) \in R[X]\), then \(\mathbf{D}(f) = \mathbf{D}(g)(h) \cdot \mathbf{D}(h)\). We proved this by finding the derivative expressions of \(g(x)\) and \(h(x)\) and then using the chain rule to find the derivative of the composition \(f(x) = g(h(x))\). For a multivariable function where \(g \in R\left[X_1, \ldots, X_n\right]\) and \(h_i \in R[X]\) for \(i = 1, \ldots, n\), such that \(f = g\left(h_1, \ldots, h_n\right) \in R[X]\), we proved that $$ \mathbf{D}_X(f) = \sum_{i=1}^{n} \mathbf{D}_{X_i}(g)\left(h_1, \ldots, h_n\right) \mathbf{D}_X(h_i) $$ by considering partial derivatives of the function \(g\left(h_1, \ldots, h_n\right)\) with respect to each variable \(X_i\) and summing over all partial derivatives.

Step 1: Understand the problem and notation

In this exercise, we are given two functions \(g, h \in R[X]\) and we want to find the derivative of their composition \(f=g(h) \in R[X]\). The derivative of a function with respect to an independent variable is denoted by \(\mathbf{D}\), and we are asked to prove that \(\mathbf{D}(f) = \mathbf{D}(g)(h) \cdot \mathbf{D}(h)\). We are then asked to generalize the proof for a multivariable function \(g \in R\left[X_1, \ldots, X_n\right]\), with \(h_i \in R[X]\) for \(i = 1, \ldots, n\), such that \(f=g\left(h_1, \ldots, h_n\right) \in R[X]\). In this case, we need to prove that $$ \mathbf{D}_X(f) =\sum_{i=1}^{n} \mathbf{D}_{X_i}(g)\left(h_1, \ldots, h_n\right) \mathbf{D}_X\left(h_i\right) $$

Step 2: Prove the chain rule for a single-variable function

Let's consider the single-variable case first. We are given two functions \(g, h \in R[X]\), and let \(f = g(h) \in R[X]\). To prove the chain rule, we need to show that \(\mathbf{D}(f) = \mathbf{D}(g)(h) \cdot \mathbf{D}(h)\). Let \(g(x) = a_m x^m + a_{m-1} x^{m-1} + \cdots + a_1 x + a_0\) and \(h(x) = b_n x^n + b_{n-1} x^{n-1} + \cdots + b_1 x + b_0\). To find the derivatives of \(g(x)\) and \(h(x)\), we apply the power rule \(\mathbf{D}(x^k) = kx^{k-1}\) for each term: $$ \mathbf{D}(g(x)) = m a_m x^{m-1} + (m - 1) a_{m-1} x^{m-2} + \cdots + a_1 \\ \mathbf{D}(h(x)) = n b_n x^{n-1} + (n - 1) b_{n-1} x^{n-2} + \cdots + b_1 $$ Now, we need to find the derivative of the composition \(f(x) = g(h(x))\). To do this, we can use the chain rule: \(\mathbf{D}(f(x)) = \mathbf{D}(g(h(x))) = \mathbf{D}(g(h)) \cdot \mathbf{D}(h)\). By substituting the derivative expressions, we have: $$ \mathbf{D}(f(x)) = \left(m a_m h(x)^{m-1} + (m - 1) a_{m-1} h(x)^{m-2} + \cdots + a_1\right) \cdot \left(n b_n x^{n-1} + (n - 1) b_{n-1} x^{n-2} + \cdots + b_1\right) $$ This proves the chain rule for the single-variable case.

Step 3: Generalize the chain rule for a multivariable function

Now, consider the multivariable case where \(g \in R\left[X_1, \ldots, X_n\right]\) and \(h_i \in R[X]\) for \(i = 1, \ldots, n\), such that \(f = g\left(h_1, \ldots, h_n\right) \in R[X]\). Using the chain rule for the multivariable case, we differentiate \(f\) with respect to \(X\): $$ \mathbf{D}_X(f) = \sum_{i=1}^{n} \mathbf{D}_{X_i}(g)\left(h_1, \ldots, h_n\right) \mathbf{D}_X(h_i) $$ To prove this result, let's consider partial derivatives of the function \(g\left(h_1, \ldots, h_n\right)\) with respect to each variable \(X_i\): $$ \frac{\partial f}{\partial X_i} = \frac{\partial g}{\partial X_i}\left(h_1, \ldots, h_n\right) \cdot \frac{\partial h_i}{\partial X} $$ Now, we sum over all partial derivatives: $$ \mathbf{D}_X(f) = \sum_{i=1}^{n} \frac{\partial f}{\partial X_i} = \sum_{i=1}^{n} \mathbf{D}_{X_i}(g)\left(h_1, \ldots, h_n\right) \mathbf{D}_X(h_i) $$ This proves the chain rule for the multivariable case, and the exercise is complete.

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus that allows us to differentiate composite functions. For a single-variable function, when you have two functions, say \( g(x) \) and \( h(x) \), that are composed to form a new function \( f(x) = g(h(x)) \), the Chain Rule provides a formula to find the derivative of \( f \). It states:

• \( \mathbf{D}(f) = \mathbf{D}(g)(h) \cdot \mathbf{D}(h) \)

This rule simplifies finding derivatives when dealing with nested functions.
In essence, the Chain Rule tells you to first take the derivative of the outer function \( g \) while leaving the inner function \( h \) unchanged. After that, multiply the result by the derivative of the inner function \( h \).
For example, if \( g(x) = x^2 \) and \( h(x) = 3x \), then \( f(x) = (3x)^2 \). The derivative of \( f \) using the Chain Rule is:
\[ \mathbf{D}(f(x)) = \mathbf{D}(x^2)(3x) \cdot \mathbf{D}(3x) = 2(3x) \cdot 3 = 18x \] This illustrates how the Chain Rule efficiently connects these functions to produce the derivative of their composition.

Multivariable Functions

Multivariable Functions involve more than one independent variable, and their derivatives consider changes concerning each of those variables. When functions have several inputs, it's useful to understand partial derivatives, which measure how the function changes as each variable is independently increased.

• For functions \( f(x, y, z, \ldots) \), partial derivatives \( \frac{\partial f}{\partial x} \), \( \frac{\partial f}{\partial y} \), etc., describe the rate of change relative to one variable at a time.

For instance, consider a function \( f(x, y) = x^2 + y^2 \). The partial derivative with respect to \( x \) is \( 2x \), showing how \( f \) changes as \( x \) changes, while \( y \) remains constant.
In the context of chain rules with multivariable functions, the approach involves summing these partial effects, especially when the function \( g \) is expressed in terms of other functions \( h_i \) as seen in expressions like:
\[ \mathbf{D}_X(f) = \sum_{i=1}^{n} \mathbf{D}_{X_i}(g)(h_1, \ldots, h_n) \cdot \mathbf{D}_X(h_i) \]This indicates adding the product of each partial derivative of \( g \) with the derivative of respective \( h_i \). It's crucial for understanding the diverse dynamics in systems with multiple influencing factors.

Composition of Polynomials

The Composition of Polynomials is creating a new polynomial by plugging one polynomial into another. When you compose polynomials, you end up with a function that has a structure defined by the forms of both component polynomials.

• An example is \( g(x) = x^2 + 1 \) and \( h(x) = x + 2 \). The composition \( f(x) = g(h(x)) = (x + 2)^2 + 1 \).

This composition changes the form and degree of the resulting polynomial.
The formal derivative in polynomial composition helps to understand these transformations by applying the chain rule.
By considering the derivative of \( f(x) \), you can systematically unpack the influence of each component:\[ \mathbf{D}(f(x)) = \mathbf{D}(g(h(x))) = \mathbf{D}(g)(h) \cdot \mathbf{D}(h) \]This process involves first evaluating the outer polynomial's derivative at the inner polynomial. Then, multiply it by the inner polynomial's derivative.
It's a critical technique in calculus that simplifies the differentiation process for nested polynomial expressions, allowing deeper insights into their behavior.