Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation + 2* for each integer k z 2 - 1 atisfies the following formula. f- 2" +1- 3 for every integer n21 Proof (by mathematical induction): Suppose f,. f. fg. is a sequence that satisfies the recurrence relation f-f + 2" for each integer k z 2, with initial condition f,- 1. Ne need to show that when the sequence f,, f, fy... is defined in this recursive way, all the terms in the sequence also satisfy the exxplicit formula shown above. So let the property P(n) be the equation f- 2" +1 - 3. We will show that P(n) is true for every integer n2 1. Show that P(1) is true: The left-hand side of P(1) is 22 – 3 , which equals 1 . The right-hand side of P(1) is 1 . Since the left-hand and right-hand sides equal each other, P(1) is true. Show that for each integer k 2 1, if P(k) is true, then P(k + 1) is true: gt+1 - 3 , where f,, f, fa... is a sequence defined by the recurrence relation et k be any integer with k z 1, and suppose that P(k) is true. In other words, suppose that f,- f, -f + 2* for each integer k 2 2, with initial condition f, - 1. This is the inductive hypothesis.] Ne must show that P(k + 1) is true. In other words, we must show that f. 2*=2 – 3 Now the left-hand side of P(k + 1) is f* - f, +2* + 1 by definition of fi, fs, f., + 2* +1 by inductive hypothesis V - 3 by the laws of algebra and this is the right-hand side of P(k + 1). Hence the inductive step is complete. Thus, both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]
Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation + 2* for each integer k z 2 - 1 atisfies the following formula. f- 2" +1- 3 for every integer n21 Proof (by mathematical induction): Suppose f,. f. fg. is a sequence that satisfies the recurrence relation f-f + 2" for each integer k z 2, with initial condition f,- 1. Ne need to show that when the sequence f,, f, fy... is defined in this recursive way, all the terms in the sequence also satisfy the exxplicit formula shown above. So let the property P(n) be the equation f- 2" +1 - 3. We will show that P(n) is true for every integer n2 1. Show that P(1) is true: The left-hand side of P(1) is 22 – 3 , which equals 1 . The right-hand side of P(1) is 1 . Since the left-hand and right-hand sides equal each other, P(1) is true. Show that for each integer k 2 1, if P(k) is true, then P(k + 1) is true: gt+1 - 3 , where f,, f, fa... is a sequence defined by the recurrence relation et k be any integer with k z 1, and suppose that P(k) is true. In other words, suppose that f,- f, -f + 2* for each integer k 2 2, with initial condition f, - 1. This is the inductive hypothesis.] Ne must show that P(k + 1) is true. In other words, we must show that f. 2*=2 – 3 Now the left-hand side of P(k + 1) is f* - f, +2* + 1 by definition of fi, fs, f., + 2* +1 by inductive hypothesis V - 3 by the laws of algebra and this is the right-hand side of P(k + 1). Hence the inductive step is complete. Thus, both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Mathematical Induction Proof Explanation**
This proof demonstrates that the sequence defined by the recurrence relation:
\[ f_k = f_{k-1} + 2^k \; \text{for each integer} \; k \geq 2 \]
with the initial condition:
\[ f_1 = 1 \]
satisfies the formula:
\[ f_n = 2^n + 1 - 3 \; \text{for every integer} \; n \geq 1. \]
**Proof (by mathematical induction):**
*Base Case: Show that \( P(1) \) is true:*
- The left-hand side of \( P(1) \) is \( 2^2 - 3 \), which equals \( 1 \).
- The right-hand side of \( P(1) \) is \( 1 \).
- Since both sides equal each other, \( P(1) \) is true.
*Inductive Step: Show that for each integer \( k \geq 1 \), if \( P(k) \) is true, then \( P(k + 1) \) is true:*
- Assume \( P(k) \) is true, i.e., suppose \( f_k = 2^k + 1 - 3 \).
By the recurrence relation:
\[ f_k = f_{k-1} + 2^k \; \text{with initial condition} \; f_1 = 1. \]
*(This assumption is the inductive hypothesis.)*
- We must show \( P(k + 1) \) is true. Specifically, we need to show:
\[ f_{k+1} = 2^{k+1} + 1 - 3. \]
- The left-hand side of \( P(k + 1) \) is calculated as follows:
\[
\begin{align*}
f_{k+1} &= f_k + 2^{k+1} \quad \text{(by definition of \( f_k \))} \\
&= (2^k + 1 - 3) + 2^{k+1} \quad \text{(by inductive hypothesis)} \\
&= 2 \cdot 2^k + 1 - 3 \quad \text](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2b801ef-259c-473d-9009-1e1b33ddc8b5%2Fe9070604-3b27-45de-9c4f-c31863b8b822%2F4szbey_processed.png&w=3840&q=75)
Transcribed Image Text:**Mathematical Induction Proof Explanation**
This proof demonstrates that the sequence defined by the recurrence relation:
\[ f_k = f_{k-1} + 2^k \; \text{for each integer} \; k \geq 2 \]
with the initial condition:
\[ f_1 = 1 \]
satisfies the formula:
\[ f_n = 2^n + 1 - 3 \; \text{for every integer} \; n \geq 1. \]
**Proof (by mathematical induction):**
*Base Case: Show that \( P(1) \) is true:*
- The left-hand side of \( P(1) \) is \( 2^2 - 3 \), which equals \( 1 \).
- The right-hand side of \( P(1) \) is \( 1 \).
- Since both sides equal each other, \( P(1) \) is true.
*Inductive Step: Show that for each integer \( k \geq 1 \), if \( P(k) \) is true, then \( P(k + 1) \) is true:*
- Assume \( P(k) \) is true, i.e., suppose \( f_k = 2^k + 1 - 3 \).
By the recurrence relation:
\[ f_k = f_{k-1} + 2^k \; \text{with initial condition} \; f_1 = 1. \]
*(This assumption is the inductive hypothesis.)*
- We must show \( P(k + 1) \) is true. Specifically, we need to show:
\[ f_{k+1} = 2^{k+1} + 1 - 3. \]
- The left-hand side of \( P(k + 1) \) is calculated as follows:
\[
\begin{align*}
f_{k+1} &= f_k + 2^{k+1} \quad \text{(by definition of \( f_k \))} \\
&= (2^k + 1 - 3) + 2^{k+1} \quad \text{(by inductive hypothesis)} \\
&= 2 \cdot 2^k + 1 - 3 \quad \text
![If a bank pays interest at a rate of \( i \) compounded \( m \) times a year, then the amount of money \( P_k \) at the end of \( k \) time periods (where one time period = \( 1/m \)th of a year) satisfies the recurrence relation:
\[
P_k = \left( 1 + \frac{i}{m} \right) P_{k-1}
\]
with initial condition \( P_0 = \) the initial amount deposited. Find an explicit formula for \( P_n \).
The given recurrence relation defines ___ with constant ____, which is _____. Therefore, \( P_n = \) _____ for every integer \( n \geq 0 \).
(Note: The blanks in the text are intended for user interaction to select or fill in values based on the context and equation provided.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2b801ef-259c-473d-9009-1e1b33ddc8b5%2Fe9070604-3b27-45de-9c4f-c31863b8b822%2F1t0sfwr_processed.png&w=3840&q=75)
Transcribed Image Text:If a bank pays interest at a rate of \( i \) compounded \( m \) times a year, then the amount of money \( P_k \) at the end of \( k \) time periods (where one time period = \( 1/m \)th of a year) satisfies the recurrence relation:
\[
P_k = \left( 1 + \frac{i}{m} \right) P_{k-1}
\]
with initial condition \( P_0 = \) the initial amount deposited. Find an explicit formula for \( P_n \).
The given recurrence relation defines ___ with constant ____, which is _____. Therefore, \( P_n = \) _____ for every integer \( n \geq 0 \).
(Note: The blanks in the text are intended for user interaction to select or fill in values based on the context and equation provided.)
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