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Chapter 7: Problem 16

Beweise induktiv die Gleichung $$ (1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \cdots\left(1+x^{2^{n}}\right)=\frac{1-x^{2^{n-1}}}{1-x} \text { f眉r } n=0,1,2, \quad ; x \neq 1 $$

Short Answer

Expert verified
The equation holds for all \( n \geq 0 \) by induction.

Step by step solution

01

Base Case

First, we prove the base case where \( n = 0 \). For \( n = 0 \), our equation becomes \( (1 + x) \). On the right-hand side, it's \( \frac{1 - x^{2^{0-1}}}{1-x} \). Simplifying gives \( \frac{1 - 1}{1 - x} = 1 \). Since both sides equal one, the base case holds true.
02

Inductive Hypothesis

Assume that the equation holds for some \( n = k \). This means we assume: \( (1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^{k}}) = \frac{1-x^{2^{k-1}}}{1-x} \). We need to prove it for \( n = k+1 \).
03

Inductive Step

We need to show that if the statement holds for \( n = k \), it must hold for \( n = k+1 \). The left side of the equation for \( n = k+1 \) is \( (1+x)(1+x^2)\cdots(1+x^{2^k})(1+x^{2^{k+1}}) \). By the inductive hypothesis, the first part equals \( \frac{1-x^{2^{k}}}{1-x} \). Thus, the product becomes \( \frac{1-x^{2^{k}}}{1-x}(1+x^{2^{k+1}}) \).
04

Simplify the Inductive Step

Expand \( \frac{1-x^{2^{k}}}{1-x}(1+x^{2^{k+1}}) \) using distributive property: \( \frac{1-x^{2^{k}} + x^{2^{k+1}} - x^{2^{k}+2^{k+1}}}{1-x} \). Simplify it to: \( \frac{1 - x^{2^{k+1}}}{1-x} \), which matches the right side of the equation for \( n = k+1 \).
05

Conclusion

The base case holds, and if the equation holds for any \( n = k \), then it also holds for \( n = k+1 \). By mathematical induction, the equation holds for all integer \( n \geq 0 \).

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Key Concepts

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

Base Case
The first step in mathematical induction is establishing the base case. This means we need to verify that the statement we're trying to prove is true for the initial value of the integer variable, often denoted as \( n = 0 \). In this exercise, when we plug \( n = 0 \) into the equation, the left-hand side becomes the simple expression \( (1 + x) \). Meanwhile, the right-hand side simplifies to \( \frac{1 - x^{2^{-1}}}{1-x} \). Simplifying further, we get \( \frac{1 - 1}{1-x} = 1 \). It confirms that, at \( n = 0 \), both sides equal 1, making the base case valid. Validating the base case is crucial since it serves as the foundation for the inductive proof.

Once the base case is correctly proven, we can confidently move forward with assuming the truth of the statement for an arbitrary case.
Inductive Hypothesis
The inductive hypothesis is the assumption phase in a proof by induction. Here, we assume that the proposition holds true for some arbitrary integer \( n = k \). This means assuming the given formula:
  • \((1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^{k}}) = \frac{1-x^{2^{k-1}}}{1-x}\)
Our task then, under this assumption, is to prove that the statement also holds true for \( n = k+1 \). This step supports bridging the base case to any desired number by following the sequence of numbers from the base case further. Remembering that the inductive hypothesis is merely an assumption at this stage, is key to understanding how induction interrelates our logical steps.
Inductive Step
The inductive step is the crux of the proof by induction. Here, we demonstrate that if the statement holds for \( n = k \), then it also holds for \( n = k+1 \). To do so, we start by expressing the left-hand side for \( n = k+1 \):
  • \((1+x)(1+x^2)\cdots(1+x^{2^k})(1+x^{2^{k+1}})\)
By leveraging the inductive hypothesis, the first part of this product reflects \( \frac{1-x^{2^k}}{1-x} \). Multiplying this expression by \( (1+x^{2^{k+1}}) \), we apply the distributive property to simplify:
  • \(\frac{1-x^{2^k}}{1-x}(1+x^{2^{k+1}})\)
Expanding and simplifying will eventually confirm that this expression simplifies to match the right-hand side of the equation for \( n = k+1 \). This step demonstrates the efficiency of induction, moving from an assumption to proof incrementally.
Algebraic Simplification
Algebraic simplification plays an essential role, especially in the inductive step. In our exercise, the simplification involves expanding and modifying the expression:
  • \(\frac{1-x^{2^{k}} + x^{2^{k+1}} - x^{2^{k}+2^{k+1}}}{1-x}\)
Each term is carefully managed to ensure meaningful reduction. By applying the distributive property, we reveal the intended algebraic equality. Effective simplification requires attention to algebraic rules, such as combining like terms and factoring, to maintain logical abstraction.Algebraic skills are vital for decomposing complex expressions into simpler counterparts, validating each step through simplicity, and thus confirming the validity of the inductive step. Through thorough simplification, we ensure that our induction hypothesis stands true when extended to \( n = k+1 \).

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Most popular questions from this chapter

F眉r jedes nat眉rliche \(n\) ist \(\left(a_{1}+a_{2}+\cdots+a_{k}\right)^{n}\) gleich der Summe aller Ausdr眉cke $$ \frac{n !}{n_{1} ! n_{2} ! \cdots n_{k} !} a_{1}^{n_{1}} a_{2}^{n_{2}} \cdots a_{k^{k}} $$ wobei f眉r \(n_{1}, n_{2}, \ldots, n_{k}\) alle Kombinationen von Zahlen aus \(\mathbf{N}_{0}\) einzutragen sind, die der Bedingung \(n_{1}+n_{2}+\cdots+n_{k}=n\) gen眉gen. Hinweis: Verfahre 盲hnlich wie beim Beweis des binomischen Satzes.

Sei \(x \neq y .\) Dann ist f眉r jedes nat眉rliche \(n>1\) $$ \frac{x^{n}-y^{n}}{x-y}=x^{n-1}+x^{n-2} y+x^{n-3} y^{2}+\cdots+x y^{n-2}+y^{n-1} $$ F眉r \(n=2\) ist dies nichts anderes als die wohlbekannte Gleichung \(x^{2}-y^{2}=(x+y)(x-y)\).

Versuche, f眉r die folgenden Summen einen, ,geschlossenen Ausdruck", also eine Summenformel zu finden und best盲tige sie induktiv (berechne etwa die Summen f眉r einige \(n\) und versuche, eine Gesetzm盲Bigkeit zu entdecken. Oder benutze geeignete Umformungen bzw. schon bekannte Formeln): a) \(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\cdots+\frac{1}{n(n+1)}\), etwas allgemeiner: $$ \frac{1}{n(n+1)}+\cdots+\frac{1}{(n+k-1)(n+k)} $$ b) \(1+3+5+\cdots+(2 n-1)\). c) \(1-4+9-+\cdots+(-1)^{n+1} n^{2}\) d) \(a_{1}+a_{2}+\cdots+a_{n}\), wobei die Differenz aufeinanderfolgender Glieder konstant \(=\Delta\) sei (also \(a_{2}-a_{1}=a_{3}-a_{2}=\cdots=a_{n}-a_{n-1}=\Delta\) ). Das Ergebnis ist die arithmetische Summenformel. e) \(1 \cdot 2+2 \cdot 3+\cdots+n(n+1)\) f) \(1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+\cdots+n(n+1)(n+2)\).

Es seien \(n\) Objekte gegeben, die aber nicht mehr verschieden zu sein brauchen. Sie seien vielmehr eingeteilt in \(k\) Gruppen \(G_{1}, \ldots, G_{k}\) jeweils gleicher Objekte (Objekte derselben Art), genauer: die Objekte jeder festen Gruppe \(G_{p}\) sind unter sich gleich (sind von derselben Art), die Objekte einer Gruppe \(G_{p}\) sind verschieden von denen der Gruppe \(G_{q}\), falls \(p \neq q\) ist \(G_{p}\) enthalte \(n_{p}\) Objekte, so da \(B\) also \(n_{1}+n_{2}+\cdots+n_{k}=n\) ist. Dann gibt es insgesamt $$ \frac{n !}{n_{1} ! n_{2} ! \cdots n_{k} !} $$ verschiedene Permutationen dieser \(n\) Objekte. Hinweis: Eine Permutation ist bestimmt, wenn die \(n_{1}\) Positionen f眉r die Objekte aus \(G_{1}\), die \(n_{2}\) Positionen f眉r die Objekte aus \(G_{2}, \ldots\) festgelegt sind. Die Herstellung einer Permutation l盲uft also darauf hinaus, die \(n\) Positionen so auf \(k\) K盲stchen zu verteilen, \(\mathrm{da} B\) im ersten K盲stchen \(n_{1}\) Positionen, im zweiten \(n_{2}\) Positionen sind usw.

20\. Auch wenn \(n\) gegebene Objekte \(a, b, \ldots\) nicht alle voneinander verschieden sind, nennt man jede Anordnung derselben (also jede Verteilung auf \(n\) K盲stchen \(\left.K_{1}, \quad, K_{n}\right)\) eine Permutation der \(a, b, \ldots\) Drei verschiedene Buchstaben \(a, b, c\) besitzen \(3 !=6\) Permutationen, die Buchstaben \(a, a, b\) jedoch nur noch drei: \(a a b, a b a, b a a\). Bestimme alle (verschiedenen) Permutationen der Buchstaben \(a, a, a, b\) und ebenso der Buchstaben \(a, a\), \(b, b .\) Z盲hle ab, wie viele es gibt und vgl mit Aufgabe \(21 .\)
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