Recurrence $k$-th pattern by substitution












2














The recurrence is $T(n) = 9 T(frac{n}{3}) + {n^2}$.
Simplified to ${3^2} T(frac{n}{3}) + {n^2}$. Then



$$begin{align}
T(n)&= 3^4T(frac{n}{3^2}) + 4n\
&= 3^6T(frac{n}{3^3}) + 10n\
&= 3^8T(frac{n}{3^4}) + 28 n.
end{align}$$



I am trying to find the $k$-th pattern of this recurrence.






combinatorics recurrence-relations







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  • 1




    How did you determine these were equivalent expressions?
    – David Peterson
    Nov 24 at 6:33










  • What if $3nmid n$? Or do you need only $T_k=T(3^k)$? (In the latter case, you get a first-order linear recurrence $T_k=9T_{k-1}+9^k$ which is easily solved: $T_k=(T_0+k)9^k$.)
    – metamorphy
    Nov 24 at 6:43
















2














The recurrence is $T(n) = 9 T(frac{n}{3}) + {n^2}$.
Simplified to ${3^2} T(frac{n}{3}) + {n^2}$. Then



$$begin{align}
T(n)&= 3^4T(frac{n}{3^2}) + 4n\
&= 3^6T(frac{n}{3^3}) + 10n\
&= 3^8T(frac{n}{3^4}) + 28 n.
end{align}$$



I am trying to find the $k$-th pattern of this recurrence.






combinatorics recurrence-relations







share|cite|improve this question




















  • 1




    How did you determine these were equivalent expressions?
    – David Peterson
    Nov 24 at 6:33










  • What if $3nmid n$? Or do you need only $T_k=T(3^k)$? (In the latter case, you get a first-order linear recurrence $T_k=9T_{k-1}+9^k$ which is easily solved: $T_k=(T_0+k)9^k$.)
    – metamorphy
    Nov 24 at 6:43














2












2








2







The recurrence is $T(n) = 9 T(frac{n}{3}) + {n^2}$.
Simplified to ${3^2} T(frac{n}{3}) + {n^2}$. Then



$$begin{align}
T(n)&= 3^4T(frac{n}{3^2}) + 4n\
&= 3^6T(frac{n}{3^3}) + 10n\
&= 3^8T(frac{n}{3^4}) + 28 n.
end{align}$$



I am trying to find the $k$-th pattern of this recurrence.






combinatorics recurrence-relations







share|cite|improve this question















The recurrence is $T(n) = 9 T(frac{n}{3}) + {n^2}$.
Simplified to ${3^2} T(frac{n}{3}) + {n^2}$. Then



$$begin{align}
T(n)&= 3^4T(frac{n}{3^2}) + 4n\
&= 3^6T(frac{n}{3^3}) + 10n\
&= 3^8T(frac{n}{3^4}) + 28 n.
end{align}$$



I am trying to find the $k$-th pattern of this recurrence.






combinatorics recurrence-relations




combinatorics recurrence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 7:23









Robert Z

93.2k1061132




93.2k1061132










asked Nov 24 at 6:23









aryamank

466




466








  • 1




    How did you determine these were equivalent expressions?
    – David Peterson
    Nov 24 at 6:33










  • What if $3nmid n$? Or do you need only $T_k=T(3^k)$? (In the latter case, you get a first-order linear recurrence $T_k=9T_{k-1}+9^k$ which is easily solved: $T_k=(T_0+k)9^k$.)
    – metamorphy
    Nov 24 at 6:43














  • 1




    How did you determine these were equivalent expressions?
    – David Peterson
    Nov 24 at 6:33










  • What if $3nmid n$? Or do you need only $T_k=T(3^k)$? (In the latter case, you get a first-order linear recurrence $T_k=9T_{k-1}+9^k$ which is easily solved: $T_k=(T_0+k)9^k$.)
    – metamorphy
    Nov 24 at 6:43








1




1




How did you determine these were equivalent expressions?
– David Peterson
Nov 24 at 6:33




How did you determine these were equivalent expressions?
– David Peterson
Nov 24 at 6:33












What if $3nmid n$? Or do you need only $T_k=T(3^k)$? (In the latter case, you get a first-order linear recurrence $T_k=9T_{k-1}+9^k$ which is easily solved: $T_k=(T_0+k)9^k$.)
– metamorphy
Nov 24 at 6:43




What if $3nmid n$? Or do you need only $T_k=T(3^k)$? (In the latter case, you get a first-order linear recurrence $T_k=9T_{k-1}+9^k$ which is easily solved: $T_k=(T_0+k)9^k$.)
– metamorphy
Nov 24 at 6:43










1 Answer
1






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oldest

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1














In my opinion something is wrong in your approach.



If $T(n)={3^2} T(frac{n}{3}) + {n^2}$, then $T(frac{n}{3^k})={3^2} T(frac{n}{3^{k+1}}) + (frac{n}{3^k})^2$ and
$$begin{align}T(n)&={3^2} T(frac{n}{3}) + {n^2}={3^2} left({3^2} T(frac{n}{3^2}) + frac{n^2}{3^2}right) + n^2\
&={3^4} T(frac{n}{3^2}) + 2n^2
={3^4} left({3^2} T(frac{n}{3^3}) + frac{n^2}{3^4}right) + 2n^2\
&={3^6} T(frac{n}{3^3}) + 3n^2=dots={9^k} T(frac{n}{3^k}) + kn^2.\
end{align}$$



P.S. Do you know the Master Theorem?






share|cite|improve this answer























  • I don't, but I understand where I went wrong - thanks!
    – aryamank
    Nov 24 at 17:19











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1 Answer
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1 Answer
1






active

oldest

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1














In my opinion something is wrong in your approach.



If $T(n)={3^2} T(frac{n}{3}) + {n^2}$, then $T(frac{n}{3^k})={3^2} T(frac{n}{3^{k+1}}) + (frac{n}{3^k})^2$ and
$$begin{align}T(n)&={3^2} T(frac{n}{3}) + {n^2}={3^2} left({3^2} T(frac{n}{3^2}) + frac{n^2}{3^2}right) + n^2\
&={3^4} T(frac{n}{3^2}) + 2n^2
={3^4} left({3^2} T(frac{n}{3^3}) + frac{n^2}{3^4}right) + 2n^2\
&={3^6} T(frac{n}{3^3}) + 3n^2=dots={9^k} T(frac{n}{3^k}) + kn^2.\
end{align}$$



P.S. Do you know the Master Theorem?






share|cite|improve this answer























  • I don't, but I understand where I went wrong - thanks!
    – aryamank
    Nov 24 at 17:19
















1














In my opinion something is wrong in your approach.



If $T(n)={3^2} T(frac{n}{3}) + {n^2}$, then $T(frac{n}{3^k})={3^2} T(frac{n}{3^{k+1}}) + (frac{n}{3^k})^2$ and
$$begin{align}T(n)&={3^2} T(frac{n}{3}) + {n^2}={3^2} left({3^2} T(frac{n}{3^2}) + frac{n^2}{3^2}right) + n^2\
&={3^4} T(frac{n}{3^2}) + 2n^2
={3^4} left({3^2} T(frac{n}{3^3}) + frac{n^2}{3^4}right) + 2n^2\
&={3^6} T(frac{n}{3^3}) + 3n^2=dots={9^k} T(frac{n}{3^k}) + kn^2.\
end{align}$$



P.S. Do you know the Master Theorem?






share|cite|improve this answer























  • I don't, but I understand where I went wrong - thanks!
    – aryamank
    Nov 24 at 17:19














1












1








1






In my opinion something is wrong in your approach.



If $T(n)={3^2} T(frac{n}{3}) + {n^2}$, then $T(frac{n}{3^k})={3^2} T(frac{n}{3^{k+1}}) + (frac{n}{3^k})^2$ and
$$begin{align}T(n)&={3^2} T(frac{n}{3}) + {n^2}={3^2} left({3^2} T(frac{n}{3^2}) + frac{n^2}{3^2}right) + n^2\
&={3^4} T(frac{n}{3^2}) + 2n^2
={3^4} left({3^2} T(frac{n}{3^3}) + frac{n^2}{3^4}right) + 2n^2\
&={3^6} T(frac{n}{3^3}) + 3n^2=dots={9^k} T(frac{n}{3^k}) + kn^2.\
end{align}$$



P.S. Do you know the Master Theorem?






share|cite|improve this answer














In my opinion something is wrong in your approach.



If $T(n)={3^2} T(frac{n}{3}) + {n^2}$, then $T(frac{n}{3^k})={3^2} T(frac{n}{3^{k+1}}) + (frac{n}{3^k})^2$ and
$$begin{align}T(n)&={3^2} T(frac{n}{3}) + {n^2}={3^2} left({3^2} T(frac{n}{3^2}) + frac{n^2}{3^2}right) + n^2\
&={3^4} T(frac{n}{3^2}) + 2n^2
={3^4} left({3^2} T(frac{n}{3^3}) + frac{n^2}{3^4}right) + 2n^2\
&={3^6} T(frac{n}{3^3}) + 3n^2=dots={9^k} T(frac{n}{3^k}) + kn^2.\
end{align}$$



P.S. Do you know the Master Theorem?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 24 at 7:19

























answered Nov 24 at 6:51









Robert Z

93.2k1061132




93.2k1061132












  • I don't, but I understand where I went wrong - thanks!
    – aryamank
    Nov 24 at 17:19


















  • I don't, but I understand where I went wrong - thanks!
    – aryamank
    Nov 24 at 17:19
















I don't, but I understand where I went wrong - thanks!
– aryamank
Nov 24 at 17:19




I don't, but I understand where I went wrong - thanks!
– aryamank
Nov 24 at 17:19


















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