Proving $a_n>2$ where $a_k=sqrt{2a_{k-1}}$
Need to prove that $a_n > 2$ in the sequence: $a_k = sqrt{2a_{k-1}}$ for $k>1$.
$a_1 = 3$.
Not sure how to use the inductive hypothesis: $P(k) = sqrt{2a_{k-1}}$ to find $P(k+1)$.
discrete-mathematics
edited Nov 27 '18 at 4:42
Blue
47.7k870151
asked Nov 27 '18 at 2:20
cap
103
add a comment |
Need to prove that $a_n > 2$ in the sequence: $a_k = sqrt{2a_{k-1}}$ for $k>1$.
$a_1 = 3$.
Not sure how to use the inductive hypothesis: $P(k) = sqrt{2a_{k-1}}$ to find $P(k+1)$.
discrete-mathematics
edited Nov 27 '18 at 4:42
Blue
47.7k870151
asked Nov 27 '18 at 2:20
cap
103
3
Stop trying to delete your questions.
– fleablood
Nov 27 '18 at 3:12
The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
– fleablood
Nov 27 '18 at 5:26
add a comment |
Need to prove that $a_n > 2$ in the sequence: $a_k = sqrt{2a_{k-1}}$ for $k>1$.
$a_1 = 3$.
Not sure how to use the inductive hypothesis: $P(k) = sqrt{2a_{k-1}}$ to find $P(k+1)$.
discrete-mathematics
edited Nov 27 '18 at 4:42
Blue
47.7k870151
asked Nov 27 '18 at 2:20
cap
103
Need to prove that $a_n > 2$ in the sequence: $a_k = sqrt{2a_{k-1}}$ for $k>1$.
$a_1 = 3$.
Not sure how to use the inductive hypothesis: $P(k) = sqrt{2a_{k-1}}$ to find $P(k+1)$.
discrete-mathematics
discrete-mathematics
edited Nov 27 '18 at 4:42
Blue
47.7k870151
asked Nov 27 '18 at 2:20
cap
103
edited Nov 27 '18 at 4:42
Blue
47.7k870151
asked Nov 27 '18 at 2:20
cap
103
edited Nov 27 '18 at 4:42
Blue
47.7k870151
edited Nov 27 '18 at 4:42
Blue
47.7k870151
edited Nov 27 '18 at 4:42
Blue
47.7k870151
47.7k870151
asked Nov 27 '18 at 2:20
cap
103
asked Nov 27 '18 at 2:20
cap
103
asked Nov 27 '18 at 2:20
cap
103
103
3
Stop trying to delete your questions.
– fleablood
Nov 27 '18 at 3:12
The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
– fleablood
Nov 27 '18 at 5:26
add a comment |
3
Stop trying to delete your questions.
– fleablood
Nov 27 '18 at 3:12
The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
– fleablood
Nov 27 '18 at 5:26
3
3
Stop trying to delete your questions.
– fleablood
Nov 27 '18 at 3:12
Stop trying to delete your questions.
– fleablood
Nov 27 '18 at 3:12
The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
– fleablood
Nov 27 '18 at 5:26
The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
– fleablood
Nov 27 '18 at 5:26
add a comment |
1 Answer
1
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You first need to show your base case. For $k = 2$, we have
$$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$
Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.
We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
begin{align}
a_{k+1} &= sqrt{2 a_k}\
&= (sqrt{2})(sqrt{a_k})\
end{align}
But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.
Hence by the principle of mathematical induction, we are done :)
answered Nov 27 '18 at 2:52
AlkaKadri
1,459411
add a comment |
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1 Answer
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1 Answer
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You first need to show your base case. For $k = 2$, we have
$$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$
Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.
We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
begin{align}
a_{k+1} &= sqrt{2 a_k}\
&= (sqrt{2})(sqrt{a_k})\
end{align}
But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.
Hence by the principle of mathematical induction, we are done :)
answered Nov 27 '18 at 2:52
AlkaKadri
1,459411
add a comment |
You first need to show your base case. For $k = 2$, we have
$$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$
Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.
We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
begin{align}
a_{k+1} &= sqrt{2 a_k}\
&= (sqrt{2})(sqrt{a_k})\
end{align}
But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.
Hence by the principle of mathematical induction, we are done :)
answered Nov 27 '18 at 2:52
AlkaKadri
1,459411
add a comment |
You first need to show your base case. For $k = 2$, we have
$$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$
Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.
We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
begin{align}
a_{k+1} &= sqrt{2 a_k}\
&= (sqrt{2})(sqrt{a_k})\
end{align}
But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.
Hence by the principle of mathematical induction, we are done :)
answered Nov 27 '18 at 2:52
AlkaKadri
1,459411
You first need to show your base case. For $k = 2$, we have
$$a_2 = sqrt{2 a_1} = sqrt{2 cdot 3} = (sqrt{2})(sqrt{3}) > (sqrt{2})(sqrt{2}) = 2$$
Now let $P(k)$ be the statement: $a_k > 2$ (notice that we don't set $P(k)$ equal to anything, it just represents the statement). $P(k)$ is usually called the inductive hypothesis.
We need to show that this implies $P(k + 1)$ (which is the statement that $a_{k + 1} > 2$). Notice that we'll go through a nearly identical process as the base case. We have,
begin{align}
a_{k+1} &= sqrt{2 a_k}\
&= (sqrt{2})(sqrt{a_k})\
end{align}
But by the inductive hypothesis, we have that $a_k > 2 Rightarrow sqrt{a_k} > sqrt{2} Rightarrow (sqrt{2})(sqrt{a_k}) > (sqrt{2})(sqrt{2}) = 2$ so that $a_{k+1} > 2$, which shows $P(k+1)$.
Hence by the principle of mathematical induction, we are done :)
answered Nov 27 '18 at 2:52
AlkaKadri
1,459411
answered Nov 27 '18 at 2:52
AlkaKadri
1,459411
answered Nov 27 '18 at 2:52
AlkaKadri
1,459411
answered Nov 27 '18 at 2:52
AlkaKadri
1,459411
1,459411
add a comment |
add a comment |
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Stop trying to delete your questions.
– fleablood
Nov 27 '18 at 3:12
The induction hypothesis isn't $P(k) = sqrt{2a_{k-1}}$. The inductive hypothesis is $P(x) $ is the statement $a_k > 2$ . So to prove $P(k+1)$ we have $a_k = sqrt{2a_k} > sqrt{2cdot 2} = 2$. So $P(k+1)$ is true. That's all.
– fleablood
Nov 27 '18 at 5:26