/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 In Evercises \(2.5 .19-2.5 .22, ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Evercises \(2.5 .19-2.5 .22, \Delta\) is the forwand difference operator with spacing \(h>0\). Let \(m\) and \(n\) be nonnegative integers, and let \(x_{0}\) be any real number. Prove by induction on \(n\) that $$ \Delta^{n}\left(x-x_{0}\right)^{m}=\left\\{\begin{array}{ll} 0 & \text { if } 0 \leq m \leq n_{+} \\ n ! h^{n} & \text { if } m=n . \end{array}\right. $$ Does this suggest an analogy between "differencing" and differentiation?

Short Answer

Expert verified
Question: Prove by induction on \(n\) that for all nonnegative integers \(m\) and \(n\), $$ \Delta^n (x-x_0)^m = \left\\{ \begin{array}{ll} 0 & \text{if } 0 \leq m \leq n \\ n! h^{n} & \text{if } m = n \end{array} \right. $$ where \(\Delta\) is the forward difference operator with step size \(h\), and discuss if this suggests an analogy between "differencing" and differentiation.

Step by step solution

01

Base Case (n=0)

For the base case, we need to show that the expression holds true for \(n=0\). When \(n=0\), \(\Delta^n (x-x_0)^m = (x-x_0)^m\), and the right side of the expression becomes: $$ \left\\{ \begin{array}{ll} 0 & \text{if } 0 \leq m \leq 0 \\ 0! h^{0} = 1 & \text{if } m = 0 \end{array} \right. $$ Since both cases give us the same result \((x-x_0)^m = (x-x_0)^0 = 1\), the base case is true.
02

Inductive Hypothesis (n=k)

Now, we assume that the expression is true for some integer \(n=k\), which means: $$ \Delta^k (x-x_0)^m = \left\\{ \begin{array}{ll} 0 & \text{if } 0 \leq m \leq k \\ k! h^{k} & \text{if } m = k \end{array} \right. $$
03

Inductive Step (n=k+1)

We now need to prove that the expression is true for \(n=k+1\). We will do this by calculating \(\Delta^{k+1} (x-x_0)^m\): $$ \Delta^{k+1} (x-x_0)^m = \Delta (\Delta^k (x-x_0)^m) $$ Now, based on the inductive hypothesis, we have two cases for the value of \(\Delta^k (x-x_0)^m\):
04

Case 1: \(0 \leq m \leq k\)

In this case, \(\Delta^k (x-x_0)^m = 0\), so: $$ \Delta (\Delta^k (x-x_0)^m) = \Delta(0) = 0 $$ This satisfies the condition in the expression for \(0 \leq m \leq k+1\), since \(0 \leq m \leq k < k+1\).
05

Case 2: \(m = k\)

In this case, \(\Delta^k (x-x_0)^m = k!h^k\), so: $$ \Delta (\Delta^k (x-x_0)^m) = \Delta (k!h^k) $$ Now, applying the forward difference operator on a constant value (in this case, \(k!h^k\)), we get: $$ \Delta (k!h^k) = 0 $$ This satisfies the condition in the expression for \(0 \leq m \leq k+1\), since \(m = k < k+1\).
06

Case 3: \(m = k+1\)

We need to find the value of \(\Delta^{k+1} (x-x_0)^{k+1}\). Observe that the forward difference operator \(\Delta\) is analogous to differentiation. Applying \(\Delta\) successively to \((x - x_0)^{k+1}\) is like differentiating \((x - x_0)^{k+1}\) with respect to \(x\) for \((k+1)\) times. When we differentiate a polynomial \(k+1\) times, where the degree of the polynomial is \(k+1\), we get the factorial of the degree multiplied by the constant term. Therefore, we can conclude that: $$ \Delta^{k+1} (x-x_0)^{k+1} = (k+1)!h^{k+1} $$ This satisfies the condition in the expression for \(m = k+1\).
07

Conclusion

By proving the base case and using induction, we have shown that the given expression is true for all nonnegative integers \(m\) and \(n\). This proof also suggests an analogy between "differencing" and differentiation, as the forward difference operator behaves similarly to the differentiation operation on polynomials.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Mathematical Induction
Understanding mathematical induction is foundational for proving propositions and statements, particularly those that are defined recursively or have a natural progression. The power of induction lies in its simplicity; it breaks down complex problems into a manageable two-step process. First, one must prove the 'base case,' which typically involves verifying the truthfulness of the statement for an initial value, often the smallest or first in the sequence. In our exercise, the base case was established for a nonnegative integer where n=0.

After affirming the base case, we move to the 'inductive step,' where we assume the statement holds true for an arbitrary case 'k' and then prove it for 'k+1'. This leap from 'k' to 'k+1' relies on the principle that if you can reach the next step from any given step, you can walk the entire staircase step by step. By showing these two parts, we can confidently assert that the proposition is true for all nonnegative integers following the base case. In our problem, by proving the statement for n=k and then for n=k+1, we employ mathematical induction to demonstrate the behavior of the forward difference operator on powers of (x - x_0).
Polynomial Differentiation
Polynomial differentiation is a core concept within calculus, providing a cornerstone for analyzing the behavior of polynomial expressions. Differentiation helps determine the rate at which a function's output changes with respect to its input. When we differentiate a polynomial of degree m, the derivative is a polynomial of degree m-1, where the coefficient of each term is multiplied by its original exponent, reflecting the power rule of differentiation.

For illustration, differentiating x^3 yields 3x^2, and so on—until eventually, after m differentiations, a constant term remains, leading to the derivative collapsing to zero. This process connects intimately with the concept at hand: The forward difference operator can be seen as a discrete equivalent to polynomial differentiation. As with continuous differentiation, applying the forward difference operator n times on a polynomial of degree m where n <= m, results in a polynomial of degree m-n. Conversely, if n > m, you're left with zero, mirroring the outcome of repeated differentiation beyond the order of the polynomial.
Real Analysis
Real analysis is a branch of mathematical analysis dealing with the real numbers and real-valued functions. It provides a rigorous framework for understanding concepts like convergence, limits, continuity, and as pertinent to our task, differentiation among others. The field extends the intuitive notions of calculus to more precise and abstract levels, allowing mathematicians to tackle complex problems and prove fundamental theorems.

In the context of our exercise, real analysis comes into play as we explore the nature of the forward difference operator. This operator can be likened to the derivative in real analysis, but applied to discrete functions or sequences. Real analysis typically deals with continuous functions, but it also sheds light on their discrete counterparts, such as sequences and series. Understanding the deep connections between continuous and discrete analysis not only enriches the study of mathematics but also broadens the tools available for tackling scientific and engineering challenges where data is often quantized rather than continuous.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Prove: \(\lim _{x \rightarrow x_{0}-} f(x)\) exists (finite) if and only if for each \(\epsilon>0\) there is a \(\delta>0\) such that \(\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|<\epsilon\) if \(x_{0}-\delta

Give an example of a function that has zero derivatives of all orders at a local minimum point.

A function \(f\) has a simple zero (or a zero of multiplicity 1) at \(x_{0}\) if \(f\) is differentiable in a neighborhood of \(x_{0}\) and \(f\left(x_{0}\right)=0,\) while \(f^{\prime}\left(x_{0}\right) \neq 0\). (a) Prove that \(f\) has a simple zero at \(x_{0}\) if and only if $$ f(x)=g(x)\left(x-x_{0}\right) $$ where \(g\) is continuous at \(x_{0}\) and differentiable on a deleted neighborhood of \(x_{0}\), and \(g\left(x_{0}\right) \neq 0 .\) (b) Give an example showing that \(g\) in(a) need not be differentiable at \(x_{0}\) -

Assume that \(f\) is differentiable on \((-\infty, \infty)\) and \(x_{0}\) is a critical point of \(f\). (a) Let \(h(x)=f(x) g(x),\) where \(g\) is differentiable on \((-\infty, \infty)\) and $$ f\left(x_{0}\right) g^{\prime}\left(x_{0}\right) \neq 0 $$ Show that the tangent line to the curve \(y=h(x)\) at \(\left(x_{0}, h\left(x_{0}\right)\right)\) and the tangent line to the curve \(y=g(x)\) at \(\left(x_{0}, g\left(x_{0}\right)\right.\) intersect on the \(x\) -axis. (b) Suppose that \(f\left(x_{0}\right) \neq 0 .\) Let \(h(x)=f(x)\left(x-x_{1}\right),\) where \(x_{1}\) is arbitrary. Show that the tangent line to the curve \(y=h(x)\) at \(\left(x_{0}, h\left(x_{0}\right)\right)\) intersects the \(x\) -axis at \(\bar{x}=x_{1}\) (c) Suppose that \(f\left(x_{0}\right) \neq 0 .\) Let \(h(x)=f(x)\left(x-x_{1}\right)^{2},\) where \(x_{1} \neq x_{0} .\) Show that the tangent line to the curve \(y=h(x)\) at \(\left(x_{0}, h\left(x_{0}\right)\right)\) intersects the \(x\) -axis at the midpoint of the interval with endpoints \(x_{0}\) and \(x_{1}\). (d) Let \(h(x)=\left(a x^{2}+b x+c\right)\left(x-x_{1}\right),\) where \(a \neq 0\) and \(b^{2}-4 a c \neq 0 .\) Let \(x_{0}=-\frac{b}{2 a} .\) Show that the tangent line to the curve \(y=h(x)\) at \(\left(x_{0}, h\left(x_{0}\right)\right)\) intersects the \(x\) -axis at \(\bar{x}=x_{1}\). (e) Let \(h\) be a cubic polynomial with zeros \(\alpha, \beta,\) and \(\gamma,\) where \(\alpha\) and \(\beta\) are distinct and \(\gamma\) is real. Let \(x_{0}=\frac{\alpha+\beta}{2}\). Show that the tangent line to the curve \(y=h(x)\) at \(\left(x_{0}, h\left(x_{0}\right)\right)\) intersects the axis at \(\bar{x}=\gamma\)

In Evercises \(2.5 .19-2.5 .22, \Delta\) is the forwand difference operator with spacing \(h>0\). Suppose that \(f^{(n+1)}\) exists on \((a, b), x_{0}, \ldots, x_{n}\) are in \((a, b),\) and \(p\) is the polynomial of degree \(\leq n\) such that \(p\left(x_{i}\right)=f\left(x_{i}\right), 0 \leq i \leq n .\) Prove: If \(x \in(a, b),\) then $$ f(x)=p(x)+\frac{f^{(n+1)}(c)}{(n+1) !}\left(x-x_{0}\right)\left(x-x_{1}\right) \cdots\left(x-x_{n}\right) $$ where \(c,\) which depends on \(x,\) is in \((a, b) .\) HiNT: Let \(x\) be fixed, distinct from \(x_{0}\) \(x_{1}, \ldots . x_{n},\) and consider the function $$ g(y)=f(y)-p(y)-\frac{K}{(n+1) !}\left(y-x_{0}\right)\left(y-x_{1}\right) \cdots\left(y-x_{n}\right) $$ where \(K\) is chosen so that \(g(x)=0 .\) Use Rolle's theorem to show that \(K=\) \(f^{(n+1)}(c)\) for some c in \((a, b)\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.