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https://gateoverflow.in/33989/how-to-solve-below-recurrence-relation
[closed]
mitesh kumar
asked
in
Combinatory
Dec 23, 2018
closed
Dec 23, 2018
by
mitesh kumar
158
views
1
vote
1
vote
closed with the note:
my mistake
https://gateoverflow.in/33989/how-to-solve-below-recurrence-relation
recurrence-relation
discrete-mathematics
mitesh kumar
asked
in
Combinatory
Dec 23, 2018
closed
Dec 23, 2018
by
mitesh kumar
by
mitesh kumar
158
views
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How to solve below recurrence relation ?
T(n)=2T(n-1)+n-1, T(1)=1 , n>=2 T(n)=2kT(n-k)+2(k-1)(n-(k-1))+2(k-2)(n-(k-2))+.......+n Now k=n-1 T(n)=2(n-1)(1)+2(n-2)(2)+2(n-3)(3)+.......+2(n-n)(n) T(n)=2(n)[ 1/1 + 2/2(2) +3/2 (3) + 2/ 2(4) +....] + n Now I am struck at this point how to proceed from here ?
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radha gogia
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How to solve below recurrence relation ?
T(n)=2√nT(√n)+n In this If we take n=2m then and I divide the entire equation by n so I will get T(2m)/2m=2T(2m/2) + 1 Now T(2m)/2m=S(m), so equation becomes S(m)=2S(m/2)+1 therefore S(m)=⊝( ... #8861;(m) , T(2m)=2m⊝(logn)=⊝(nlogn) Is this correct approach or not ,since I am unable to do it with tree method .
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How to solve below recurrence relation using subtitution method ?
T(n)=T(n-3)+cn2 T(n-3)=T(n-6)+c(n-3)2 T(n-6)=T(n-9)+c(n-6)2 Continuing like this I am getting T(n)=T(n-3k)+cn2+c(n-3k)2+c(n-(3k+3))2+c(n-(3k+6))2+c(n-(3k+9))2+...... Now let k=(n-1)/3 ,I am only able to get terms like cn2+c+4c+25c +64c, with which I am unable to reach any conclusion , so how to proceed through this .
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sampad
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343
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How to solve this recurrence relation?
Solve the equation: $a_n = 5a_{n/3} + 7, \;\; a_1 = 5$ Note: $a_0$ is not given.
sampad
asked
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Combinatory
Nov 23, 2015
by
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343
views
recurrence-relation
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