Recurrence relation for the sequence $sqrt{n}, sqrt{sqrt{n}}, sqrt{sqrt{sqrt{sqrt{n}}}}cdots$
To generate a sequence ($A$) like: $n^{frac{1}{2}}, n^{frac{1}{2^2}}, n^{frac{1}{2^3}}, n^{frac{1}{2^4}}, n^{frac{1}{2^5}}, cdots, n^{frac{1}{2^k}}$ we can define a recursive function like:
$$
f(n) =
begin{cases}
1 & text{if $n=2$}\
fleft(sqrt{n}right) & text{otherwise}
end{cases}
$$
For simplicity, we can assume that every $2^i{th}$ root of $n$ is a perfect square of some integral power of $2$, so that any $n^{frac{1}{2^i}}$ gives an integral power of $2$ and we are able to reach $n=2$ eventually (after $log_{_2}(log_{_2}(n))$ steps to be precise).
I was wondering what would be the similar recursive function for a sequence ($B$) like: $n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}} cdots$
$Bequiv n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}}cdotsequiv n^{frac{1}{2^{2^0}}}, n^{frac{1}{2^{2^1}}}, n^{frac{1}{2^{2^2}}}, n^{frac{1}{2^{2^3}}}, n^{frac{1}{2^{2^4}}}cdots $ converges in $log_{_2}(log_{_2}(log_{_2}(n))$ steps.
To visualize, the sequences are:
$A equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{n}}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}tocdots$
and
$B equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}}}}tocdots$
recurrence-relations recursion
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To generate a sequence ($A$) like: $n^{frac{1}{2}}, n^{frac{1}{2^2}}, n^{frac{1}{2^3}}, n^{frac{1}{2^4}}, n^{frac{1}{2^5}}, cdots, n^{frac{1}{2^k}}$ we can define a recursive function like:
$$
f(n) =
begin{cases}
1 & text{if $n=2$}\
fleft(sqrt{n}right) & text{otherwise}
end{cases}
$$
For simplicity, we can assume that every $2^i{th}$ root of $n$ is a perfect square of some integral power of $2$, so that any $n^{frac{1}{2^i}}$ gives an integral power of $2$ and we are able to reach $n=2$ eventually (after $log_{_2}(log_{_2}(n))$ steps to be precise).
I was wondering what would be the similar recursive function for a sequence ($B$) like: $n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}} cdots$
$Bequiv n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}}cdotsequiv n^{frac{1}{2^{2^0}}}, n^{frac{1}{2^{2^1}}}, n^{frac{1}{2^{2^2}}}, n^{frac{1}{2^{2^3}}}, n^{frac{1}{2^{2^4}}}cdots $ converges in $log_{_2}(log_{_2}(log_{_2}(n))$ steps.
To visualize, the sequences are:
$A equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{n}}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}tocdots$
and
$B equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}}}}tocdots$
recurrence-relations recursion
add a comment |
To generate a sequence ($A$) like: $n^{frac{1}{2}}, n^{frac{1}{2^2}}, n^{frac{1}{2^3}}, n^{frac{1}{2^4}}, n^{frac{1}{2^5}}, cdots, n^{frac{1}{2^k}}$ we can define a recursive function like:
$$
f(n) =
begin{cases}
1 & text{if $n=2$}\
fleft(sqrt{n}right) & text{otherwise}
end{cases}
$$
For simplicity, we can assume that every $2^i{th}$ root of $n$ is a perfect square of some integral power of $2$, so that any $n^{frac{1}{2^i}}$ gives an integral power of $2$ and we are able to reach $n=2$ eventually (after $log_{_2}(log_{_2}(n))$ steps to be precise).
I was wondering what would be the similar recursive function for a sequence ($B$) like: $n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}} cdots$
$Bequiv n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}}cdotsequiv n^{frac{1}{2^{2^0}}}, n^{frac{1}{2^{2^1}}}, n^{frac{1}{2^{2^2}}}, n^{frac{1}{2^{2^3}}}, n^{frac{1}{2^{2^4}}}cdots $ converges in $log_{_2}(log_{_2}(log_{_2}(n))$ steps.
To visualize, the sequences are:
$A equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{n}}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}tocdots$
and
$B equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}}}}tocdots$
recurrence-relations recursion
To generate a sequence ($A$) like: $n^{frac{1}{2}}, n^{frac{1}{2^2}}, n^{frac{1}{2^3}}, n^{frac{1}{2^4}}, n^{frac{1}{2^5}}, cdots, n^{frac{1}{2^k}}$ we can define a recursive function like:
$$
f(n) =
begin{cases}
1 & text{if $n=2$}\
fleft(sqrt{n}right) & text{otherwise}
end{cases}
$$
For simplicity, we can assume that every $2^i{th}$ root of $n$ is a perfect square of some integral power of $2$, so that any $n^{frac{1}{2^i}}$ gives an integral power of $2$ and we are able to reach $n=2$ eventually (after $log_{_2}(log_{_2}(n))$ steps to be precise).
I was wondering what would be the similar recursive function for a sequence ($B$) like: $n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}} cdots$
$Bequiv n^{frac{1}{2}}, n^{frac{1}{4}}, n^{frac{1}{16}}, n^{frac{1}{256}}, n^{frac{1}{256^2}}cdotsequiv n^{frac{1}{2^{2^0}}}, n^{frac{1}{2^{2^1}}}, n^{frac{1}{2^{2^2}}}, n^{frac{1}{2^{2^3}}}, n^{frac{1}{2^{2^4}}}cdots $ converges in $log_{_2}(log_{_2}(log_{_2}(n))$ steps.
To visualize, the sequences are:
$A equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{n}}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}tocdots$
and
$B equiv sqrt{n}tosqrt{sqrt{n}}tosqrt{sqrt{sqrt{sqrt{n}}}}tosqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{sqrt{n}}}}}}}}tocdots$
recurrence-relations recursion
recurrence-relations recursion
edited Dec 2 at 16:31
asked Dec 2 at 16:06
dibyendu
356318
356318
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add a comment |
2 Answers
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I'm not sure your recurrence for $A$ is correct. I think what you are looking for is
$f(k) =
begin{cases}
n & text{if }k = 0 \
sqrt{f(k-1)} & text{if }k > 0
end{cases}$
Then $f(0) = n, f(1) = sqrt{n}, f(2) = sqrt{sqrt{n}}$ etc. $f(k) = n^{frac{1}{2^k}}$ as required.
For sequence B, the general term is $f(k) = n^{frac{1}{2^{2^k}}}$. We could write a recurrence
$f(k+1) = n^{frac{1}{2^{2^{k+1}}}} = n^{frac{1}{2^{left(2^k+2^kright)}}} = n^{frac{1}{left(2^{2^k}right)cdot left(2^{2^k}right)}} = left(n^{frac{1}{2^{2^k}}}right)^left(frac{1}{2^{2^k}}right) = f(k)^left(frac{1}{2^{2^k}}right)$.
We can set the base case to be $f(0) = sqrt{n} = n^{frac{1}{2^{2^0}}}$
I guess both are correct. You've defined the recurrence in terms of the number of iterations ($k$) it takes the recurrence to converge, while I tried to define it in terms of the problem size or input size ($n$).
– dibyendu
Dec 4 at 22:30
add a comment |
I've realised that there is no way we can achieve the recurrence (in terms of the input size $n$ alone) with just a single variable $n$, as we did in case of the sequence $A$ [with $f(n)=f(sqrt{n})$].
For sequence $B$, we somehow need to keep tack of the iteration number $i$, like this:
$$
f(n)=f_1(n, 1)
$$
$$
f_1left(n, iright) =
begin{cases}
f_1left(sqrt{n}, 2right) & text{if, $i=1$}\
1 & text{if, $n=2$}\
f_1left(sqrt[i]{n}, i^2right) & text{otherwise}
end{cases}
$$
Any other thought/suggestion is appreciated.
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
I'm not sure your recurrence for $A$ is correct. I think what you are looking for is
$f(k) =
begin{cases}
n & text{if }k = 0 \
sqrt{f(k-1)} & text{if }k > 0
end{cases}$
Then $f(0) = n, f(1) = sqrt{n}, f(2) = sqrt{sqrt{n}}$ etc. $f(k) = n^{frac{1}{2^k}}$ as required.
For sequence B, the general term is $f(k) = n^{frac{1}{2^{2^k}}}$. We could write a recurrence
$f(k+1) = n^{frac{1}{2^{2^{k+1}}}} = n^{frac{1}{2^{left(2^k+2^kright)}}} = n^{frac{1}{left(2^{2^k}right)cdot left(2^{2^k}right)}} = left(n^{frac{1}{2^{2^k}}}right)^left(frac{1}{2^{2^k}}right) = f(k)^left(frac{1}{2^{2^k}}right)$.
We can set the base case to be $f(0) = sqrt{n} = n^{frac{1}{2^{2^0}}}$
I guess both are correct. You've defined the recurrence in terms of the number of iterations ($k$) it takes the recurrence to converge, while I tried to define it in terms of the problem size or input size ($n$).
– dibyendu
Dec 4 at 22:30
add a comment |
I'm not sure your recurrence for $A$ is correct. I think what you are looking for is
$f(k) =
begin{cases}
n & text{if }k = 0 \
sqrt{f(k-1)} & text{if }k > 0
end{cases}$
Then $f(0) = n, f(1) = sqrt{n}, f(2) = sqrt{sqrt{n}}$ etc. $f(k) = n^{frac{1}{2^k}}$ as required.
For sequence B, the general term is $f(k) = n^{frac{1}{2^{2^k}}}$. We could write a recurrence
$f(k+1) = n^{frac{1}{2^{2^{k+1}}}} = n^{frac{1}{2^{left(2^k+2^kright)}}} = n^{frac{1}{left(2^{2^k}right)cdot left(2^{2^k}right)}} = left(n^{frac{1}{2^{2^k}}}right)^left(frac{1}{2^{2^k}}right) = f(k)^left(frac{1}{2^{2^k}}right)$.
We can set the base case to be $f(0) = sqrt{n} = n^{frac{1}{2^{2^0}}}$
I guess both are correct. You've defined the recurrence in terms of the number of iterations ($k$) it takes the recurrence to converge, while I tried to define it in terms of the problem size or input size ($n$).
– dibyendu
Dec 4 at 22:30
add a comment |
I'm not sure your recurrence for $A$ is correct. I think what you are looking for is
$f(k) =
begin{cases}
n & text{if }k = 0 \
sqrt{f(k-1)} & text{if }k > 0
end{cases}$
Then $f(0) = n, f(1) = sqrt{n}, f(2) = sqrt{sqrt{n}}$ etc. $f(k) = n^{frac{1}{2^k}}$ as required.
For sequence B, the general term is $f(k) = n^{frac{1}{2^{2^k}}}$. We could write a recurrence
$f(k+1) = n^{frac{1}{2^{2^{k+1}}}} = n^{frac{1}{2^{left(2^k+2^kright)}}} = n^{frac{1}{left(2^{2^k}right)cdot left(2^{2^k}right)}} = left(n^{frac{1}{2^{2^k}}}right)^left(frac{1}{2^{2^k}}right) = f(k)^left(frac{1}{2^{2^k}}right)$.
We can set the base case to be $f(0) = sqrt{n} = n^{frac{1}{2^{2^0}}}$
I'm not sure your recurrence for $A$ is correct. I think what you are looking for is
$f(k) =
begin{cases}
n & text{if }k = 0 \
sqrt{f(k-1)} & text{if }k > 0
end{cases}$
Then $f(0) = n, f(1) = sqrt{n}, f(2) = sqrt{sqrt{n}}$ etc. $f(k) = n^{frac{1}{2^k}}$ as required.
For sequence B, the general term is $f(k) = n^{frac{1}{2^{2^k}}}$. We could write a recurrence
$f(k+1) = n^{frac{1}{2^{2^{k+1}}}} = n^{frac{1}{2^{left(2^k+2^kright)}}} = n^{frac{1}{left(2^{2^k}right)cdot left(2^{2^k}right)}} = left(n^{frac{1}{2^{2^k}}}right)^left(frac{1}{2^{2^k}}right) = f(k)^left(frac{1}{2^{2^k}}right)$.
We can set the base case to be $f(0) = sqrt{n} = n^{frac{1}{2^{2^0}}}$
answered Dec 4 at 2:49
Akashnil
1
1
I guess both are correct. You've defined the recurrence in terms of the number of iterations ($k$) it takes the recurrence to converge, while I tried to define it in terms of the problem size or input size ($n$).
– dibyendu
Dec 4 at 22:30
add a comment |
I guess both are correct. You've defined the recurrence in terms of the number of iterations ($k$) it takes the recurrence to converge, while I tried to define it in terms of the problem size or input size ($n$).
– dibyendu
Dec 4 at 22:30
I guess both are correct. You've defined the recurrence in terms of the number of iterations ($k$) it takes the recurrence to converge, while I tried to define it in terms of the problem size or input size ($n$).
– dibyendu
Dec 4 at 22:30
I guess both are correct. You've defined the recurrence in terms of the number of iterations ($k$) it takes the recurrence to converge, while I tried to define it in terms of the problem size or input size ($n$).
– dibyendu
Dec 4 at 22:30
add a comment |
I've realised that there is no way we can achieve the recurrence (in terms of the input size $n$ alone) with just a single variable $n$, as we did in case of the sequence $A$ [with $f(n)=f(sqrt{n})$].
For sequence $B$, we somehow need to keep tack of the iteration number $i$, like this:
$$
f(n)=f_1(n, 1)
$$
$$
f_1left(n, iright) =
begin{cases}
f_1left(sqrt{n}, 2right) & text{if, $i=1$}\
1 & text{if, $n=2$}\
f_1left(sqrt[i]{n}, i^2right) & text{otherwise}
end{cases}
$$
Any other thought/suggestion is appreciated.
add a comment |
I've realised that there is no way we can achieve the recurrence (in terms of the input size $n$ alone) with just a single variable $n$, as we did in case of the sequence $A$ [with $f(n)=f(sqrt{n})$].
For sequence $B$, we somehow need to keep tack of the iteration number $i$, like this:
$$
f(n)=f_1(n, 1)
$$
$$
f_1left(n, iright) =
begin{cases}
f_1left(sqrt{n}, 2right) & text{if, $i=1$}\
1 & text{if, $n=2$}\
f_1left(sqrt[i]{n}, i^2right) & text{otherwise}
end{cases}
$$
Any other thought/suggestion is appreciated.
add a comment |
I've realised that there is no way we can achieve the recurrence (in terms of the input size $n$ alone) with just a single variable $n$, as we did in case of the sequence $A$ [with $f(n)=f(sqrt{n})$].
For sequence $B$, we somehow need to keep tack of the iteration number $i$, like this:
$$
f(n)=f_1(n, 1)
$$
$$
f_1left(n, iright) =
begin{cases}
f_1left(sqrt{n}, 2right) & text{if, $i=1$}\
1 & text{if, $n=2$}\
f_1left(sqrt[i]{n}, i^2right) & text{otherwise}
end{cases}
$$
Any other thought/suggestion is appreciated.
I've realised that there is no way we can achieve the recurrence (in terms of the input size $n$ alone) with just a single variable $n$, as we did in case of the sequence $A$ [with $f(n)=f(sqrt{n})$].
For sequence $B$, we somehow need to keep tack of the iteration number $i$, like this:
$$
f(n)=f_1(n, 1)
$$
$$
f_1left(n, iright) =
begin{cases}
f_1left(sqrt{n}, 2right) & text{if, $i=1$}\
1 & text{if, $n=2$}\
f_1left(sqrt[i]{n}, i^2right) & text{otherwise}
end{cases}
$$
Any other thought/suggestion is appreciated.
answered Dec 4 at 22:54
dibyendu
356318
356318
add a comment |
add a comment |
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