Use complete induction to prove that $a_n < 2^n$ for every integer $n geq 2$
$begingroup$
Define the sequence of integers $a_0, a_1, a_2, cdots$ as follows
$$ a_i =
begin{cases}
i+1 & 0 leq i leq 2 \
a_{i-1} + a_{i-2} + 2a_{i-3} & i > 2 \
end{cases}
$$
Use complete induction to prove that $a_n < 2^n$ for every integer $n geq 2$
I will prove $a_n < 2^n, forall n geq 2$ by using complete induction
Base Case: Three cases $n = 2, 3, 4$
let $n = 2$
$a_{n} = a_{2} = 2 + 1 = 3 < 4 = 2^2 = 2^n$ By definition, and holds
let $n = 3$
$a_{n} = a_{3} = 3 + 2 + 1 = 6 < 8 = 2^3 = 2^n$ By definition, and holds
let $n = 4$
$a_{n} = a_{4} = 4 + 3 + 2 = 9 < 16 = 2^4 = 2^n$ By definition, and holds
Inductive step: let $n > 4$. Suppose $a_j < 2^j$ whenever $2 leq j < n$. [Inductive hypothesis]
What to prove: $a_n < 2^n$
$a_{n} = a_{i-1} + a_{i-2} + 2a_{i-3}$ [By definition]
$< 2^{n-1} + 2^{n-2} + 2^{n-3+1}$ [By Inductive hypothesis]
$= 2^{n-1} + 2^{n-2} + 2^{n-2}$
$= 2^{n-1} + 2^{n-2 + 1}$
$= 2^{n-1 + 1}$
$= 2^n$
as wanted.
Would this be correct?
induction
asked Dec 1 '18 at 7:53
Tree GarenTree Garen
36219
$endgroup$
add a comment |
$begingroup$
Define the sequence of integers $a_0, a_1, a_2, cdots$ as follows
$$ a_i =
begin{cases}
i+1 & 0 leq i leq 2 \
a_{i-1} + a_{i-2} + 2a_{i-3} & i > 2 \
end{cases}
$$
Use complete induction to prove that $a_n < 2^n$ for every integer $n geq 2$
I will prove $a_n < 2^n, forall n geq 2$ by using complete induction
Base Case: Three cases $n = 2, 3, 4$
let $n = 2$
$a_{n} = a_{2} = 2 + 1 = 3 < 4 = 2^2 = 2^n$ By definition, and holds
let $n = 3$
$a_{n} = a_{3} = 3 + 2 + 1 = 6 < 8 = 2^3 = 2^n$ By definition, and holds
let $n = 4$
$a_{n} = a_{4} = 4 + 3 + 2 = 9 < 16 = 2^4 = 2^n$ By definition, and holds
Inductive step: let $n > 4$. Suppose $a_j < 2^j$ whenever $2 leq j < n$. [Inductive hypothesis]
What to prove: $a_n < 2^n$
$a_{n} = a_{i-1} + a_{i-2} + 2a_{i-3}$ [By definition]
$< 2^{n-1} + 2^{n-2} + 2^{n-3+1}$ [By Inductive hypothesis]
$= 2^{n-1} + 2^{n-2} + 2^{n-2}$
$= 2^{n-1} + 2^{n-2 + 1}$
$= 2^{n-1 + 1}$
$= 2^n$
as wanted.
Would this be correct?
induction
asked Dec 1 '18 at 7:53
Tree GarenTree Garen
36219
$endgroup$
add a comment |
$begingroup$
Define the sequence of integers $a_0, a_1, a_2, cdots$ as follows
$$ a_i =
begin{cases}
i+1 & 0 leq i leq 2 \
a_{i-1} + a_{i-2} + 2a_{i-3} & i > 2 \
end{cases}
$$
Use complete induction to prove that $a_n < 2^n$ for every integer $n geq 2$
I will prove $a_n < 2^n, forall n geq 2$ by using complete induction
Base Case: Three cases $n = 2, 3, 4$
let $n = 2$
$a_{n} = a_{2} = 2 + 1 = 3 < 4 = 2^2 = 2^n$ By definition, and holds
let $n = 3$
$a_{n} = a_{3} = 3 + 2 + 1 = 6 < 8 = 2^3 = 2^n$ By definition, and holds
let $n = 4$
$a_{n} = a_{4} = 4 + 3 + 2 = 9 < 16 = 2^4 = 2^n$ By definition, and holds
Inductive step: let $n > 4$. Suppose $a_j < 2^j$ whenever $2 leq j < n$. [Inductive hypothesis]
What to prove: $a_n < 2^n$
$a_{n} = a_{i-1} + a_{i-2} + 2a_{i-3}$ [By definition]
$< 2^{n-1} + 2^{n-2} + 2^{n-3+1}$ [By Inductive hypothesis]
$= 2^{n-1} + 2^{n-2} + 2^{n-2}$
$= 2^{n-1} + 2^{n-2 + 1}$
$= 2^{n-1 + 1}$
$= 2^n$
as wanted.
Would this be correct?
induction
asked Dec 1 '18 at 7:53
Tree GarenTree Garen
36219
$endgroup$
Define the sequence of integers $a_0, a_1, a_2, cdots$ as follows
$$ a_i =
begin{cases}
i+1 & 0 leq i leq 2 \
a_{i-1} + a_{i-2} + 2a_{i-3} & i > 2 \
end{cases}
$$
Use complete induction to prove that $a_n < 2^n$ for every integer $n geq 2$
I will prove $a_n < 2^n, forall n geq 2$ by using complete induction
Base Case: Three cases $n = 2, 3, 4$
let $n = 2$
$a_{n} = a_{2} = 2 + 1 = 3 < 4 = 2^2 = 2^n$ By definition, and holds
let $n = 3$
$a_{n} = a_{3} = 3 + 2 + 1 = 6 < 8 = 2^3 = 2^n$ By definition, and holds
let $n = 4$
$a_{n} = a_{4} = 4 + 3 + 2 = 9 < 16 = 2^4 = 2^n$ By definition, and holds
Inductive step: let $n > 4$. Suppose $a_j < 2^j$ whenever $2 leq j < n$. [Inductive hypothesis]
What to prove: $a_n < 2^n$
$a_{n} = a_{i-1} + a_{i-2} + 2a_{i-3}$ [By definition]
$< 2^{n-1} + 2^{n-2} + 2^{n-3+1}$ [By Inductive hypothesis]
$= 2^{n-1} + 2^{n-2} + 2^{n-2}$
$= 2^{n-1} + 2^{n-2 + 1}$
$= 2^{n-1 + 1}$
$= 2^n$
as wanted.
Would this be correct?
induction
induction
asked Dec 1 '18 at 7:53
Tree GarenTree Garen
36219
asked Dec 1 '18 at 7:53
Tree GarenTree Garen
36219
edited Dec 1 '18 at 8:45
Tree Garen
asked Dec 1 '18 at 7:53
Tree GarenTree Garen
36219
asked Dec 1 '18 at 7:53
Tree GarenTree Garen
36219
asked Dec 1 '18 at 7:53
Tree GarenTree Garen
36219
36219
add a comment |
add a comment |
1 Answer
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$begingroup$
The second part is fine. But the base case is not just the $n=2$ case. You should have checked it for $n=0$, and $n=1$ too.
answered Dec 1 '18 at 7:59
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
But it doesn't hold for $n = 0$ or $n = 1$?
$endgroup$
– Tree Garen
Dec 1 '18 at 8:01
$begingroup$
You are right. My mistake. But the you should have checked it for $nin{2,3,4}$. Because only then will your inductive proof shows that it works for $n=5$.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 8:06
$begingroup$
I'm confused why. Why three numbers? And I thought the base case for recurrence functions was just so we can use "$a_{i-1} + a_{i-2} + 2a_{i-3}$" part but that part holds for all $i > 2$ so wouldn't going all the way to $n = 5$ be redundant?
$endgroup$
– Tinler
Dec 1 '18 at 8:14
$begingroup$
Three numbers because each $a_i$ is defined from the three previous terms. So, in order to prove it it holds for $a_5$, you must know that it holds for $a_2$, $a_3$, and $a_4$.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 8:26
add a comment |
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$begingroup$
The second part is fine. But the base case is not just the $n=2$ case. You should have checked it for $n=0$, and $n=1$ too.
answered Dec 1 '18 at 7:59
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
But it doesn't hold for $n = 0$ or $n = 1$?
$endgroup$
– Tree Garen
Dec 1 '18 at 8:01
$begingroup$
You are right. My mistake. But the you should have checked it for $nin{2,3,4}$. Because only then will your inductive proof shows that it works for $n=5$.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 8:06
$begingroup$
I'm confused why. Why three numbers? And I thought the base case for recurrence functions was just so we can use "$a_{i-1} + a_{i-2} + 2a_{i-3}$" part but that part holds for all $i > 2$ so wouldn't going all the way to $n = 5$ be redundant?
$endgroup$
– Tinler
Dec 1 '18 at 8:14
$begingroup$
Three numbers because each $a_i$ is defined from the three previous terms. So, in order to prove it it holds for $a_5$, you must know that it holds for $a_2$, $a_3$, and $a_4$.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 8:26
add a comment |
$begingroup$
The second part is fine. But the base case is not just the $n=2$ case. You should have checked it for $n=0$, and $n=1$ too.
answered Dec 1 '18 at 7:59
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
$begingroup$
But it doesn't hold for $n = 0$ or $n = 1$?
$endgroup$
– Tree Garen
Dec 1 '18 at 8:01
$begingroup$
You are right. My mistake. But the you should have checked it for $nin{2,3,4}$. Because only then will your inductive proof shows that it works for $n=5$.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 8:06
$begingroup$
I'm confused why. Why three numbers? And I thought the base case for recurrence functions was just so we can use "$a_{i-1} + a_{i-2} + 2a_{i-3}$" part but that part holds for all $i > 2$ so wouldn't going all the way to $n = 5$ be redundant?
$endgroup$
– Tinler
Dec 1 '18 at 8:14
$begingroup$
Three numbers because each $a_i$ is defined from the three previous terms. So, in order to prove it it holds for $a_5$, you must know that it holds for $a_2$, $a_3$, and $a_4$.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 8:26
add a comment |
$begingroup$
The second part is fine. But the base case is not just the $n=2$ case. You should have checked it for $n=0$, and $n=1$ too.
answered Dec 1 '18 at 7:59
José Carlos SantosJosé Carlos Santos
161k22127232
$endgroup$
The second part is fine. But the base case is not just the $n=2$ case. You should have checked it for $n=0$, and $n=1$ too.
answered Dec 1 '18 at 7:59
José Carlos SantosJosé Carlos Santos
161k22127232
answered Dec 1 '18 at 7:59
José Carlos SantosJosé Carlos Santos
161k22127232
answered Dec 1 '18 at 7:59
José Carlos SantosJosé Carlos Santos
161k22127232
answered Dec 1 '18 at 7:59
José Carlos SantosJosé Carlos Santos
161k22127232
161k22127232
$begingroup$
But it doesn't hold for $n = 0$ or $n = 1$?
$endgroup$
– Tree Garen
Dec 1 '18 at 8:01
$begingroup$
You are right. My mistake. But the you should have checked it for $nin{2,3,4}$. Because only then will your inductive proof shows that it works for $n=5$.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 8:06
$begingroup$
I'm confused why. Why three numbers? And I thought the base case for recurrence functions was just so we can use "$a_{i-1} + a_{i-2} + 2a_{i-3}$" part but that part holds for all $i > 2$ so wouldn't going all the way to $n = 5$ be redundant?
$endgroup$
– Tinler
Dec 1 '18 at 8:14
$begingroup$
Three numbers because each $a_i$ is defined from the three previous terms. So, in order to prove it it holds for $a_5$, you must know that it holds for $a_2$, $a_3$, and $a_4$.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 8:26
add a comment |
$begingroup$
But it doesn't hold for $n = 0$ or $n = 1$?
$endgroup$
– Tree Garen
Dec 1 '18 at 8:01
$begingroup$
You are right. My mistake. But the you should have checked it for $nin{2,3,4}$. Because only then will your inductive proof shows that it works for $n=5$.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 8:06
$begingroup$
I'm confused why. Why three numbers? And I thought the base case for recurrence functions was just so we can use "$a_{i-1} + a_{i-2} + 2a_{i-3}$" part but that part holds for all $i > 2$ so wouldn't going all the way to $n = 5$ be redundant?
$endgroup$
– Tinler
Dec 1 '18 at 8:14
$begingroup$
Three numbers because each $a_i$ is defined from the three previous terms. So, in order to prove it it holds for $a_5$, you must know that it holds for $a_2$, $a_3$, and $a_4$.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 8:26
$begingroup$
But it doesn't hold for $n = 0$ or $n = 1$?
$endgroup$
– Tree Garen
Dec 1 '18 at 8:01
$begingroup$
But it doesn't hold for $n = 0$ or $n = 1$?
$endgroup$
– Tree Garen
Dec 1 '18 at 8:01
$begingroup$
You are right. My mistake. But the you should have checked it for $nin{2,3,4}$. Because only then will your inductive proof shows that it works for $n=5$.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 8:06
$begingroup$
You are right. My mistake. But the you should have checked it for $nin{2,3,4}$. Because only then will your inductive proof shows that it works for $n=5$.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 8:06
$begingroup$
I'm confused why. Why three numbers? And I thought the base case for recurrence functions was just so we can use "$a_{i-1} + a_{i-2} + 2a_{i-3}$" part but that part holds for all $i > 2$ so wouldn't going all the way to $n = 5$ be redundant?
$endgroup$
– Tinler
Dec 1 '18 at 8:14
$begingroup$
I'm confused why. Why three numbers? And I thought the base case for recurrence functions was just so we can use "$a_{i-1} + a_{i-2} + 2a_{i-3}$" part but that part holds for all $i > 2$ so wouldn't going all the way to $n = 5$ be redundant?
$endgroup$
– Tinler
Dec 1 '18 at 8:14
$begingroup$
Three numbers because each $a_i$ is defined from the three previous terms. So, in order to prove it it holds for $a_5$, you must know that it holds for $a_2$, $a_3$, and $a_4$.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 8:26
$begingroup$
Three numbers because each $a_i$ is defined from the three previous terms. So, in order to prove it it holds for $a_5$, you must know that it holds for $a_2$, $a_3$, and $a_4$.
$endgroup$
– José Carlos Santos
Dec 1 '18 at 8:26
add a comment |
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