How to solve equation T(n) = 5T(n/5) + sqrt(n), T(1) = 1, T(0) = 0 using recursion tree?











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When I apply master theorem, I get O(n) but when I try to solve it using recursion tree, I get stuck and couldn't make out any solution.



I tried this : 



T(n) = 5T(n/5) + sqrt(n)

T(n) = 5(5T(n/25) + sqrt(n/5)) + sqrt(n) 

     = 25T(n/25) + sqrt(5n) + sqrt(n)

T(n) = 5(5(5T(n/125) + sqrt(n/25)) + sqrt(n/5)) + sqrt(n) 

     = 125T(n/25) + sqrt(25) + sqrt(5n) + sqrt(n)

.

.

.

T(n) = sqrt(n) + sqrt(5n) + sqrt(25n) +  sqrt(125n) + sqrt(625n) + sqrt(3125n) + ...


How do I suppose to solve this GP??






algorithm recursion time-complexity master-theorem







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  • n=5^k; k = log(n) base5; the GP will run long(n) base5 times.
    – roottraveller
    Jul 15 '17 at 16:02















up vote
0
down vote

favorite












When I apply master theorem, I get O(n) but when I try to solve it using recursion tree, I get stuck and couldn't make out any solution.



I tried this : 



T(n) = 5T(n/5) + sqrt(n)

T(n) = 5(5T(n/25) + sqrt(n/5)) + sqrt(n) 

     = 25T(n/25) + sqrt(5n) + sqrt(n)

T(n) = 5(5(5T(n/125) + sqrt(n/25)) + sqrt(n/5)) + sqrt(n) 

     = 125T(n/25) + sqrt(25) + sqrt(5n) + sqrt(n)

.

.

.

T(n) = sqrt(n) + sqrt(5n) + sqrt(25n) +  sqrt(125n) + sqrt(625n) + sqrt(3125n) + ...


How do I suppose to solve this GP??






algorithm recursion time-complexity master-theorem







share|improve this question
























  • n=5^k; k = log(n) base5; the GP will run long(n) base5 times.
    – roottraveller
    Jul 15 '17 at 16:02













up vote
0
down vote

favorite









up vote
0
down vote

favorite











When I apply master theorem, I get O(n) but when I try to solve it using recursion tree, I get stuck and couldn't make out any solution.



I tried this : 



T(n) = 5T(n/5) + sqrt(n)

T(n) = 5(5T(n/25) + sqrt(n/5)) + sqrt(n) 

     = 25T(n/25) + sqrt(5n) + sqrt(n)

T(n) = 5(5(5T(n/125) + sqrt(n/25)) + sqrt(n/5)) + sqrt(n) 

     = 125T(n/25) + sqrt(25) + sqrt(5n) + sqrt(n)

.

.

.

T(n) = sqrt(n) + sqrt(5n) + sqrt(25n) +  sqrt(125n) + sqrt(625n) + sqrt(3125n) + ...


How do I suppose to solve this GP??






algorithm recursion time-complexity master-theorem







share|improve this question















When I apply master theorem, I get O(n) but when I try to solve it using recursion tree, I get stuck and couldn't make out any solution.



I tried this : 



T(n) = 5T(n/5) + sqrt(n)

T(n) = 5(5T(n/25) + sqrt(n/5)) + sqrt(n) 

     = 25T(n/25) + sqrt(5n) + sqrt(n)

T(n) = 5(5(5T(n/125) + sqrt(n/25)) + sqrt(n/5)) + sqrt(n) 

     = 125T(n/25) + sqrt(25) + sqrt(5n) + sqrt(n)

.

.

.

T(n) = sqrt(n) + sqrt(5n) + sqrt(25n) +  sqrt(125n) + sqrt(625n) + sqrt(3125n) + ...


How do I suppose to solve this GP??






algorithm recursion time-complexity master-theorem




algorithm recursion time-complexity master-theorem






share|improve this question















share|improve this question













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edited Jan 26 '17 at 8:28

























asked Jan 25 '17 at 19:43









dj_1993

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445












  • n=5^k; k = log(n) base5; the GP will run long(n) base5 times.
    – roottraveller
    Jul 15 '17 at 16:02


















  • n=5^k; k = log(n) base5; the GP will run long(n) base5 times.
    – roottraveller
    Jul 15 '17 at 16:02
















n=5^k; k = log(n) base5; the GP will run long(n) base5 times.
– roottraveller
Jul 15 '17 at 16:02




n=5^k; k = log(n) base5; the GP will run long(n) base5 times.
– roottraveller
Jul 15 '17 at 16:02












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The final sum has log_5(n) + O(1) terms, since the recursion bottoms out. The largest is sqrt(5^(log_5(n) + O(1)) n) = sqrt(O(n) n) = O(n). The others decrease geometrically, so they don't matter in the big-O accounting (alternatively, divide through by 1 + sqrt(1/5) + sqrt(1/5^2) + ... = Theta(1)).






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    The final sum has log_5(n) + O(1) terms, since the recursion bottoms out. The largest is sqrt(5^(log_5(n) + O(1)) n) = sqrt(O(n) n) = O(n). The others decrease geometrically, so they don't matter in the big-O accounting (alternatively, divide through by 1 + sqrt(1/5) + sqrt(1/5^2) + ... = Theta(1)).






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      The final sum has log_5(n) + O(1) terms, since the recursion bottoms out. The largest is sqrt(5^(log_5(n) + O(1)) n) = sqrt(O(n) n) = O(n). The others decrease geometrically, so they don't matter in the big-O accounting (alternatively, divide through by 1 + sqrt(1/5) + sqrt(1/5^2) + ... = Theta(1)).






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        up vote
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        down vote










        up vote
        1
        down vote









        The final sum has log_5(n) + O(1) terms, since the recursion bottoms out. The largest is sqrt(5^(log_5(n) + O(1)) n) = sqrt(O(n) n) = O(n). The others decrease geometrically, so they don't matter in the big-O accounting (alternatively, divide through by 1 + sqrt(1/5) + sqrt(1/5^2) + ... = Theta(1)).






        share|improve this answer














        The final sum has log_5(n) + O(1) terms, since the recursion bottoms out. The largest is sqrt(5^(log_5(n) + O(1)) n) = sqrt(O(n) n) = O(n). The others decrease geometrically, so they don't matter in the big-O accounting (alternatively, divide through by 1 + sqrt(1/5) + sqrt(1/5^2) + ... = Theta(1)).







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        edited Jul 16 '17 at 11:05

























        answered Jan 25 '17 at 20:05









        David Eisenstat

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