Asymptotic behaviour of recurrence












3















In DNA chain, there are four types of bases: A, C, G, T. Let $g(n)$ be the number of configurations of a DNA chain of length $n$, in which the sequences TT and TG never appear. Write a recurrence for $g(n)$. Determine the asymptotic behavior of the recurrence. Also, prove whether this rate of growth is $o(n^{ n^{1/2}}), Theta(n^{ n^{1/2}}),$ or $Omega(n^{ n^{1/2}})$.




My answer:
$g(n)=A(n)+C(n)+G(n)+T(n)$



$g(n)=g(n-1)+g(n-1)+g(n-1)+A(n-1)+C(n-1)$



$g(n)=3g(n-1)+2g(n-2)$



After solving this recurrence, we get



$$begin{align}
g(n) &= (frac{(4)(17)^{1/2}+16}{(3)(17)^{1/2}+17})(frac{3+(17)^{1/2})}{2})^n \
&+(frac{(17)^{1/2}-5)}{2(17)^{1/2}})(frac{3-(17)^{1/2}}{2})^n
end{align}$$



Therefore when $n$ is large, $(frac{3-(17)^{1/2})}{2})^n$ become smaller and smaller
So we can write $$g(n) = Theta((frac{3+(17)^{1/2})}{2})^n).$$



(Do I determine the asymptotic behavior of the recurrence correctly?)



But I do not know what is the rate of growth, if we take the limit of $frac{g(n)}{n^{ n^{1/2}}}$,where n tend to infinity, it seems impossible to calculate it. How could I determine the rate of growth?






discrete-mathematics asymptotics biology







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  • Write, for example, $Theta$ for $Theta$; this is what's known as MathJax, a type of $LaTeX$.
    – Shaun
    Nov 26 at 15:56












  • Use $frac{a}{b}$ for $frac{a}{b}$.
    – Shaun
    Nov 26 at 15:58










  • Please edit the question.
    – Shaun
    Nov 26 at 15:59










  • The exercise is strange because it has a much simpler solution. Namely, note that the total number of configurations is $4^n$ hence $g(n)leqslant4^n$. Since $4^n=o(n^{sqrt n})$, you are done.
    – Did
    Dec 12 at 9:18


















3















In DNA chain, there are four types of bases: A, C, G, T. Let $g(n)$ be the number of configurations of a DNA chain of length $n$, in which the sequences TT and TG never appear. Write a recurrence for $g(n)$. Determine the asymptotic behavior of the recurrence. Also, prove whether this rate of growth is $o(n^{ n^{1/2}}), Theta(n^{ n^{1/2}}),$ or $Omega(n^{ n^{1/2}})$.




My answer:
$g(n)=A(n)+C(n)+G(n)+T(n)$



$g(n)=g(n-1)+g(n-1)+g(n-1)+A(n-1)+C(n-1)$



$g(n)=3g(n-1)+2g(n-2)$



After solving this recurrence, we get



$$begin{align}
g(n) &= (frac{(4)(17)^{1/2}+16}{(3)(17)^{1/2}+17})(frac{3+(17)^{1/2})}{2})^n \
&+(frac{(17)^{1/2}-5)}{2(17)^{1/2}})(frac{3-(17)^{1/2}}{2})^n
end{align}$$



Therefore when $n$ is large, $(frac{3-(17)^{1/2})}{2})^n$ become smaller and smaller
So we can write $$g(n) = Theta((frac{3+(17)^{1/2})}{2})^n).$$



(Do I determine the asymptotic behavior of the recurrence correctly?)



But I do not know what is the rate of growth, if we take the limit of $frac{g(n)}{n^{ n^{1/2}}}$,where n tend to infinity, it seems impossible to calculate it. How could I determine the rate of growth?






discrete-mathematics asymptotics biology







share|cite|improve this question
























  • Write, for example, $Theta$ for $Theta$; this is what's known as MathJax, a type of $LaTeX$.
    – Shaun
    Nov 26 at 15:56












  • Use $frac{a}{b}$ for $frac{a}{b}$.
    – Shaun
    Nov 26 at 15:58










  • Please edit the question.
    – Shaun
    Nov 26 at 15:59










  • The exercise is strange because it has a much simpler solution. Namely, note that the total number of configurations is $4^n$ hence $g(n)leqslant4^n$. Since $4^n=o(n^{sqrt n})$, you are done.
    – Did
    Dec 12 at 9:18
















3












3








3


1






In DNA chain, there are four types of bases: A, C, G, T. Let $g(n)$ be the number of configurations of a DNA chain of length $n$, in which the sequences TT and TG never appear. Write a recurrence for $g(n)$. Determine the asymptotic behavior of the recurrence. Also, prove whether this rate of growth is $o(n^{ n^{1/2}}), Theta(n^{ n^{1/2}}),$ or $Omega(n^{ n^{1/2}})$.




My answer:
$g(n)=A(n)+C(n)+G(n)+T(n)$



$g(n)=g(n-1)+g(n-1)+g(n-1)+A(n-1)+C(n-1)$



$g(n)=3g(n-1)+2g(n-2)$



After solving this recurrence, we get



$$begin{align}
g(n) &= (frac{(4)(17)^{1/2}+16}{(3)(17)^{1/2}+17})(frac{3+(17)^{1/2})}{2})^n \
&+(frac{(17)^{1/2}-5)}{2(17)^{1/2}})(frac{3-(17)^{1/2}}{2})^n
end{align}$$



Therefore when $n$ is large, $(frac{3-(17)^{1/2})}{2})^n$ become smaller and smaller
So we can write $$g(n) = Theta((frac{3+(17)^{1/2})}{2})^n).$$



(Do I determine the asymptotic behavior of the recurrence correctly?)



But I do not know what is the rate of growth, if we take the limit of $frac{g(n)}{n^{ n^{1/2}}}$,where n tend to infinity, it seems impossible to calculate it. How could I determine the rate of growth?






discrete-mathematics asymptotics biology







share|cite|improve this question
















In DNA chain, there are four types of bases: A, C, G, T. Let $g(n)$ be the number of configurations of a DNA chain of length $n$, in which the sequences TT and TG never appear. Write a recurrence for $g(n)$. Determine the asymptotic behavior of the recurrence. Also, prove whether this rate of growth is $o(n^{ n^{1/2}}), Theta(n^{ n^{1/2}}),$ or $Omega(n^{ n^{1/2}})$.




My answer:
$g(n)=A(n)+C(n)+G(n)+T(n)$



$g(n)=g(n-1)+g(n-1)+g(n-1)+A(n-1)+C(n-1)$



$g(n)=3g(n-1)+2g(n-2)$



After solving this recurrence, we get



$$begin{align}
g(n) &= (frac{(4)(17)^{1/2}+16}{(3)(17)^{1/2}+17})(frac{3+(17)^{1/2})}{2})^n \
&+(frac{(17)^{1/2}-5)}{2(17)^{1/2}})(frac{3-(17)^{1/2}}{2})^n
end{align}$$



Therefore when $n$ is large, $(frac{3-(17)^{1/2})}{2})^n$ become smaller and smaller
So we can write $$g(n) = Theta((frac{3+(17)^{1/2})}{2})^n).$$



(Do I determine the asymptotic behavior of the recurrence correctly?)



But I do not know what is the rate of growth, if we take the limit of $frac{g(n)}{n^{ n^{1/2}}}$,where n tend to infinity, it seems impossible to calculate it. How could I determine the rate of growth?






discrete-mathematics asymptotics biology




discrete-mathematics asymptotics biology






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edited Nov 26 at 16:06

























asked Nov 26 at 15:51









user614642

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  • Write, for example, $Theta$ for $Theta$; this is what's known as MathJax, a type of $LaTeX$.
    – Shaun
    Nov 26 at 15:56












  • Use $frac{a}{b}$ for $frac{a}{b}$.
    – Shaun
    Nov 26 at 15:58










  • Please edit the question.
    – Shaun
    Nov 26 at 15:59










  • The exercise is strange because it has a much simpler solution. Namely, note that the total number of configurations is $4^n$ hence $g(n)leqslant4^n$. Since $4^n=o(n^{sqrt n})$, you are done.
    – Did
    Dec 12 at 9:18




















  • Write, for example, $Theta$ for $Theta$; this is what's known as MathJax, a type of $LaTeX$.
    – Shaun
    Nov 26 at 15:56












  • Use $frac{a}{b}$ for $frac{a}{b}$.
    – Shaun
    Nov 26 at 15:58










  • Please edit the question.
    – Shaun
    Nov 26 at 15:59










  • The exercise is strange because it has a much simpler solution. Namely, note that the total number of configurations is $4^n$ hence $g(n)leqslant4^n$. Since $4^n=o(n^{sqrt n})$, you are done.
    – Did
    Dec 12 at 9:18


















Write, for example, $Theta$ for $Theta$; this is what's known as MathJax, a type of $LaTeX$.
– Shaun
Nov 26 at 15:56






Write, for example, $Theta$ for $Theta$; this is what's known as MathJax, a type of $LaTeX$.
– Shaun
Nov 26 at 15:56














Use $frac{a}{b}$ for $frac{a}{b}$.
– Shaun
Nov 26 at 15:58




Use $frac{a}{b}$ for $frac{a}{b}$.
– Shaun
Nov 26 at 15:58












Please edit the question.
– Shaun
Nov 26 at 15:59




Please edit the question.
– Shaun
Nov 26 at 15:59












The exercise is strange because it has a much simpler solution. Namely, note that the total number of configurations is $4^n$ hence $g(n)leqslant4^n$. Since $4^n=o(n^{sqrt n})$, you are done.
– Did
Dec 12 at 9:18






The exercise is strange because it has a much simpler solution. Namely, note that the total number of configurations is $4^n$ hence $g(n)leqslant4^n$. Since $4^n=o(n^{sqrt n})$, you are done.
– Did
Dec 12 at 9:18












1 Answer
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You solved the recurrence and found its asymptotic behavior correctly. The growth is $c^n$, where $c=frac12(3+sqrt{17})$, and you are asked how this compares with $n^{n^{1/2}}$. The latter is faster: to see this, take logs of both, and note that
$$
(log c)n=oBig(n^{1/2}log nBig)
$$






share|cite|improve this answer





















  • Thank you very much
    – user614642
    Nov 27 at 3:53











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1 Answer
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active

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1














You solved the recurrence and found its asymptotic behavior correctly. The growth is $c^n$, where $c=frac12(3+sqrt{17})$, and you are asked how this compares with $n^{n^{1/2}}$. The latter is faster: to see this, take logs of both, and note that
$$
(log c)n=oBig(n^{1/2}log nBig)
$$






share|cite|improve this answer





















  • Thank you very much
    – user614642
    Nov 27 at 3:53
















1














You solved the recurrence and found its asymptotic behavior correctly. The growth is $c^n$, where $c=frac12(3+sqrt{17})$, and you are asked how this compares with $n^{n^{1/2}}$. The latter is faster: to see this, take logs of both, and note that
$$
(log c)n=oBig(n^{1/2}log nBig)
$$






share|cite|improve this answer





















  • Thank you very much
    – user614642
    Nov 27 at 3:53














1












1








1






You solved the recurrence and found its asymptotic behavior correctly. The growth is $c^n$, where $c=frac12(3+sqrt{17})$, and you are asked how this compares with $n^{n^{1/2}}$. The latter is faster: to see this, take logs of both, and note that
$$
(log c)n=oBig(n^{1/2}log nBig)
$$






share|cite|improve this answer












You solved the recurrence and found its asymptotic behavior correctly. The growth is $c^n$, where $c=frac12(3+sqrt{17})$, and you are asked how this compares with $n^{n^{1/2}}$. The latter is faster: to see this, take logs of both, and note that
$$
(log c)n=oBig(n^{1/2}log nBig)
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 at 16:45









Mike Earnest

20.2k11950




20.2k11950












  • Thank you very much
    – user614642
    Nov 27 at 3:53


















  • Thank you very much
    – user614642
    Nov 27 at 3:53
















Thank you very much
– user614642
Nov 27 at 3:53




Thank you very much
– user614642
Nov 27 at 3:53


















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