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Chapter 3: Problem 17

Beweisen Sie f眉r \(n \in \mathbb{N}_{\geq 2}:\) $$ \prod_{k=2}^{n}(k-1)=(n-1) ! $$

Short Answer

Expert verified
Question: Prove that for all natural numbers \(n \geq 2\), the following equation holds: \(\prod_{k=2}^{n}(k-1)=(n-1) !\) Answer: By using induction, we showed that the equation holds for the base case \(n=2\). Assuming the equation holds for \(n=m\), we also demonstrated it for \(n=m+1\). Therefore, the formula holds for all natural numbers \(n \geq 2\).

Step by step solution

01

Stating the base case

For the base case, let \(n = 2\). In this case, the left side \(\prod_{k=2}^{n}(k-1) = (2 - 1) = 1\), and the right side is \((n-1)! = (2-1)! = 1! = 1\). Thus, the base case is true.
02

Inductive hypothesis

Suppose the formula holds for some \(n=m\), i.e., $$ \prod_{k=2}^{m}(k-1)=(m-1)! $$
03

Inductive step

We want to show that the formula also holds for \(n=m+1\): $$ \prod_{k=2}^{m+1}(k-1) = (m) ! $$ By the definition of the product notation, we can write the left side as: $$ \prod_{k=2}^{m+1}(k-1) = (m - 1)(\prod_{k=2}^{m}(k-1)) $$ From the inductive hypothesis, we know that \(\prod_{k=2}^{m}(k-1) = (m-1)!\). Thus, we can rewrite the left side: $$ (m - 1)(\prod_{k=2}^{m}(k-1)) = (m - 1) ((m - 1)!) $$ By the definition of the factorial, we have: $$ (m - 1) ((m - 1)!) = m! $$ Which is what we wanted to show. Therefore, by induction, the formula holds for all \(n \in \mathbb{N}_{\geq 2}\).

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Key Concepts

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

Factorials
Factorials are fundamental when working with sequences and series in mathematics. A factorial of a non-negative integer, denoted as \( n! \), represents the product of all positive integers less than or equal to \( n \). This series of multiplicative steps makes factorials key in many branches of mathematics, especially in permutations and combinatorics.

For instance, when you see \( 5! \), this translates to \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \). As the numbers climb, the factorials grow quickly in size. For example, \( 10! \) is a whopping \( 3,628,800 \).

Factorials also play a significant role in defining mathematical expressions and solving proofs, especially in polynomial and algebraic identities. Understanding how they work and their properties helps in tackling more complex mathematical problems and proofs, such as those involving summation and product notations as seen in the provided exercise.
Base Case
In mathematical induction, the base case is the starting point. It acts as the foundation upon which inductive reasoning is built. Essentially, when you begin a proof by induction, you must first show that the proposition holds for the initial value, normally the smallest number in the domain of interest.

For the exercise provided, the base case is when \( n = 2 \). You compute the problem at this starting point and verify it holds true, making sure the left side equals the right. Here, both sides equate to 1, meaning our base case is correct. This verification is a crucial step, because if the base case fails, the entire inductive argument collapses. Think of it like the first domino in a series - if it doesn't fall, none of the others will either.
Inductive Step
The inductive step in mathematical induction is about moving from one case to the next. Once the base case is confirmed, you assume the formula works for some arbitrary case \( n = m \). This assumption is called the inductive hypothesis.

Next, the goal is to prove that if the formula holds true for \( n = m \), then it must also hold for \( n = m+1 \). This is essentially "pushing" the proof one step forward. In the provided solution, we assume the formula \( \prod_{k=2}^{m}(k-1) = (m-1)! \) is true. To prove it for \( n = m+1 \), we expand the product on the left-hand side and simplify it using the inductive hypothesis, finally showing that it equals \( m! \).

Completing this step rigorously means you have extended the truth from the base case, through \( n = m \), to all subsequent integers within the set. This highlights the beauty and power of induction - establishing a truth for infinitely many cases by just a few logical steps.

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Most popular questions from this chapter

F眉r zwei in Serie geschaltete Widerst盲nde \(R_{1}\) und \(R_{2}\) gilt $$ R_{\mathrm{ges}}=R_{1}+R_{2} $$ bei Parallelschaltung erhalt man $$ \frac{1}{R_{\mathrm{gex}}}=\frac{1}{R_{1}}+\frac{1}{R_{2}} $$ Beweisen Sie mittels vollst盲ndiger Induktion, dass f眉r eine beliebige Zahl \(n\) von Widerst盲nden bei serieller Schaltung $$ R_{\mathrm{gex}}=\sum_{k=1}^{n} R_{k} $$ und bei Parallelschaltung $$ \frac{1}{R_{\text {ges }}}=\sum_{k=1}^{n} \frac{1}{R_{k}} $$ gilt.

Finden Sie den Fehler im folgenden.,Beweis" daf眉r, dass der Mars bewohnt ist: Satz: Wenn in einer Menge von \(n\) Planeten einer bewohnt ist, dann sind alle bewohnt. Beweis mittels vollst盲ndiger Induktion: \(n=1:\) trivial \(n \rightarrow n+1\) : Laut Annahme sind von einer Menge von \(n\) Planeten alle bewohnt, sobald nur einer bewohnt ist. Nun betrachten wir eine Menge von \(n+1\) Planeten (die wir willk眉rlich mit \(p_{1}\) bis \(p_{n+1}\) bezeichnen). Von diesen schlie脽en wir vorl盲ufig einen aus unsere Betrachtungen aus, z. B. \(p_{n+1}\). Wenn von der 眉briggebliebenen Menge von \(n\) Planeten nur einer bewohnt ist, sind laut Annahme alle bewohnt. Nun schlie脽en wir von den \(n\) bewohnten Planeten einen aus, z. B. \(p_{1}\), und nehmen \(p_{n+1}\) wieder hinzu. Wir erhalten wieder eine Menge von \(n\) Planeten, die bis auf \(p_{n+1}\) alle bewohnt sind. Auf jeden Fall ist einer bewohnt. demnach alle, also ist auch \(p_{n+1}\) bewohnt. Korollar: Der Mars ist bewohnt. Beweis: Betrachten Sie die \(n\) Planeten des Sonnensystems. Je nach aktueller Meinung zum Status des Pluto ist \(n=8\) oder \(n=9\), doch auf jeden Fall ist \(n\) endlich. Die Erde ist bewohnt, damit sind alle Planeten des Sonnensystems bewohnt - auch der Mars.

Bestimmen Sie die Summe aller nat眉rlichen Zahlen von eins bis tausend.

Beweisen Sie f眉r alle \(n \in \mathbb{N}_{\geq 2}\) : $$ \prod_{k=2}^{n}\left(1-\frac{2}{k(k+1)}\right)=\frac{1}{3}\left(1+\frac{2}{n}\right) $$

K枚nnen Angaben von Werten 眉ber \(100 \%\) sinnvoll sein?
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