Question

The forward difference operators with spacing \(h>0\) are defined by $$ \begin{array}{c} \Delta^{0} f(x)=f(x), \quad \Delta f(x)=f(x+h)-f(x) . \\ \Delta^{n+1} f(x)=\Delta\left[\Delta^{n} f(x)\right], \quad n \geq 1 . \end{array} $$ (a) Prove by induction on \(n:\) If \(k \geq 2, c_{1}, \ldots, c_{k}\) are constants, and \(n \geq 1,\) then $$ \Delta^{n}\left[c_{1} f_{1}(x)+\cdots+c_{k} f_{k}(x)\right]=c_{1} \Delta^{n} f_{1}(x)+\cdots+c_{k} \Delta^{n} f_{k}(x) $$ (b) Prove by induction: If \(n \geq 1\), then $$ \Delta^{n} f(x)=\sum_{m=0}^{n}(-1)^{n-m}\left(\begin{array}{c} n \\ m \end{array}\right) f(x+m h) $$

Short answer

Question: Prove the following properties for the forward difference operator \(\Delta\) using induction for \(n\in\mathbb{N}\): (a) \(\Delta^n[c_1f_1(x)+\cdots + c_kf_k(x)] = c_1\Delta^n f_1(x) + \cdots + c_k \Delta^n f_k(x)\) (b) \(\Delta^n f(x) = \sum_{m=0}^{n}(-1)^{n-m}\binom{n}{m}f(x+mh)\) Answer: For part (a), we proved the linearity property of the forward difference operator using induction. We started by proving the base case for n=1, then assumed that the property holds true for n=p, and showed that it also holds true for n=p+1. For part (b), we proved the relationship between \(\Delta^n\) and the original functions using induction. We started by proving the base case for n=1, then assumed that the property holds true for n=p, and showed that it also holds true for n=p+1, using the linearity property that we proved in part (a).

Step-by-step solution

Part (a) Base Case (n=1)

We'll begin by proving the base case for n=1: $$ \Delta\left[c_1f_1(x)+\cdots + c_kf_k(x)\right] = c_1\Delta f_1(x) + \cdots + c_k \Delta f_k(x) $$ Let's expand the left-hand side of the equation, using the definition of the forward difference operator: $$ \begin{aligned} \Delta\left[c_1f_1(x)+\cdots + c_kf_k(x)\right] &= \left[c_1f_1(x)+\cdots + c_kf_k(x)+h\right] - \left[c_1f_1(x)+\cdots + c_kf_k(x)\right] \end{aligned} $$ Next, we'll distribute the summation in the expression: $$ \begin{aligned} &= \left[c_1(f_1(x+h)-f_1(x)) + \cdots + c_k(f_k(x+h) - f_k(x))\right] \end{aligned} $$ Now, we see that the expression is equal to the sum of forward differences of the individual functions multiplied by the corresponding constants: $$ \begin{aligned} &= c_1\Delta f_1(x) + \cdots + c_k \Delta f_k(x) \end{aligned} $$ The base case holds true for n=1.

Part (a) Induction Step

Now, we'll assume that the property holds true for n = p: $$ \Delta^{p}\left[c_1f_1(x)+\cdots + c_kf_k(x)\right] = c_1\Delta^{p} f_1(x) + \cdots + c_k \Delta^{p} f_k(x) $$ We want to prove the statement for n=p+1: $$ \Delta^{p+1}\left[c_1f_1(x)+\cdots + c_kf_k(x)\right] = c_1\Delta^{p+1} f_1(x) + \cdots + c_k \Delta^{p+1} f_k(x) $$ By definition of the forward difference operator: $$ \Delta^{p+1}\left[c_1f_1(x)+\cdots + c_kf_k(x)\right] = \Delta\left[\Delta^{p}\left[c_1f_1(x) + \cdots + c_kf_k(x)\right]\right] $$ By the induction hypothesis, substitute the expression for \(\Delta^{p}\left[c_1f_1(x) + \cdots + c_kf_k(x)\right]\): $$ \begin{aligned} &= \Delta\left[c_1\Delta^{p}f_1(x) + \cdots + c_k\Delta^{p}f_k(x)\right] \end{aligned} $$ Now, using the linearity property we've already proved for n=1: $$ \begin{aligned} &= c_1\Delta\left[\Delta^{p}f_1(x)\right] + \cdots + c_k\Delta\left[\Delta^{p}f_k(x)\right] \end{aligned} $$ This is equal to the right-hand side of the statement we wanted to prove for n=p+1, which completes the induction step, and therefore, part (a) is proved.

Part (b) Base Case (n=1)

We'll now move on to part (b) and prove the base case for n=1: $$ \Delta f(x)=\sum_{m=0}^1 (-1)^{1-m}\binom{1}{m}f(x+mh) $$ We'll expand the summands on the right-hand side: $$ \begin{aligned} &= (-1)^0 \binom{1}{0}f(x+0h) + (-1)^1 \binom{1}{1}f(x+1h) \\ &= f(x) - f(x+h) \end{aligned} $$ This matches the definition of the forward difference operator for n=1, so the base case holds true.

Part (b) Induction Step

Now, we'll assume that the property holds true for n=p: $$ \Delta^p f(x) = \sum_{m=0}^{p}(-1)^{p-m}\binom{p}{m}f(x+mh) $$ We want to prove the statement for n=p+1: $$ \Delta^{p+1} f(x) =\sum_{m=0}^{p+1}(-1)^{(p+1)-m}\binom{p+1}{m}f(x+mh) $$ By definition of the forward difference operator: $$ \Delta^{p+1} f(x) = \Delta\left[\Delta^{p}f(x)\right] $$ Substitute the expression for \(\Delta^{p}f(x)\) using the induction hypothesis: $$ \begin{aligned} &= \Delta\left[\sum_{m=0}^{p}(-1)^{p-m}\binom{p}{m}f(x+mh)\right] \end{aligned} $$ We can now use the linearity of the forward difference operator, which we proved in part (a): $$ \begin{aligned} &=\sum_{m=0}^{p}(-1)^{p-m}\binom{p}{m} \Delta f(x+mh) \end{aligned} $$ Now, we'll replace \(\Delta f(x+mh)\) using the base case equation for n=1 that we proved earlier: $$ \begin{aligned} &=\sum_{m=0}^{p}(-1)^{p-m}\binom{p}{m} (f(x+(m+1)h) - f(x+mh)) \end{aligned} $$ Rearrange the terms of the summation: $$ \begin{aligned} &=-f(x) + \sum_{m=0}^{p}((-1)^{p-m+1}\binom{p}{m} - (-1)^{p-m}\binom{p}{m-1})f(x+(m+1)h) \end{aligned} $$ Now, we can simplify the coefficients in the summation using Pascal's rule: \(\binom{p}{k} + \binom{p}{k-1} = \binom{p+1}{k}\): $$ \begin{aligned} &=(-1)^{p+1}f(x+(p+1)h)+\sum_{m=0}^{p}(-1)^{(p+1)-m}\binom{p+1}{m}f(x+mh) \end{aligned} $$ This is equal to the right-hand side of the statement we wanted to prove for n=p+1, which completes the induction step. As a result, part (b) is also proved.

Key concepts

Mathematical Induction

Understanding mathematical induction involves familiarizing oneself with a systematic way of proving statements regarding natural numbers. Although the concept might seem daunting at first, it's similar to climbing a staircase: you take one step at a time, ensuring each step is secure.

Let's begin by looking at the basics of induction. The process consists of two main steps:

• Base Case: First, prove that a statement holds for the initial value, often for the smallest natural number, typically zero or one.
• Inductive Step: Then, assume the statement is true for an arbitrary natural number, let's say 'n'. Next, show that the statement must also be true for 'n+1'. This acts like a domino effect; if it's true for one number, it must be true for the next.

Applying this to the exercise, we prove properties of difference operators by anchoring our proof with a base case and then ensuring the logical progression to an arbitrary step, solidifying the argument with a strong foundation — the inductive step.

In part (a) and (b) of the exercise, the base case is confirmed first for the simplicity of 'n=1'. Once this is secure, we extend the assumptions for a general 'n' (which we assume to hold true), and our goal is to prove for 'n+1', effectively setting off the 'domino effect'. As evidenced in the solution, following the steps carefully is crucial for a watertight mathematical induction proof.

Linearity of Difference Operators

Exploring the concept of linearity in difference operators is akin to uncovering the flexible, distributive nature of these mathematical tools. Just as a beam of white light passes through a prism and distributes into a spectrum, the forward difference operator, when applied to a linear combination of functions, distributes over each function scaled by its respective coefficient.

In mathematical language, if we have a linear combination of functions — summing several functions each multiplied by a constant factor — the forward difference operator can be distributed across each term. This principle, which resembles that of linear functions where the sum of inputs is equal to the sum of their outputs, is crucial for proving part (a) of the exercise. Based on the step-by-step solution, we see that by applying the operator to each function individually and then multiplying by the constants, we arrive at the same result as applying the operator once to the entire combination. In essence, this property simplifies complex expressions, making it easier to discern patterns and establish proofs within the realm of difference equations.

Binomial Coefficients

The world of binomial coefficients is much like the beads on an abacus, each representing a combination of elements that can be repositioned in a limited number of ways. Delving into the properties of binomial coefficients can provide a deeper understanding of various mathematical concepts, including the binomial theorem and combinatorics.

Binomial coefficients, symbolized by \( \binom{n}{k} \), answer the question: 'In how many ways can we choose 'k' elements from a set of 'n' elements?' When 'n' is a natural number, and 'k' ranges from zero to 'n', these coefficients form the entries of Pascal's triangle. Each entry is the sum of the two directly above it, representing Pascal's rule: \( \binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k} \).

In part (b) of the exercise, binomial coefficients play a central role in describing the expansion of powers of the forward difference operator. The step-by-step solution demonstrates how the summands, involving the alternation of signs and the binomial coefficients, precisely capture the increments in the function's inputs, reflecting the underlying combinatorial structure. This showcases the intimate link between algebraic structures and combinatorial concepts, accentuating the importance of binomial coefficients in developing formulas for difference operations.