Question

What is the complexity for the recurrence relation: T(n) = 2T(n/2) + 5n^2

score 1.3k · Answer 1 · 2026-04-15 06:57:19

1.3k

Views

What is the complexity for the recurrence relation: T(n) = 2T(n/2) + 5n^2

Published on 15 Apr 2026 - 6:57

score 107 · Answer 2 · 2026-04-14 04:06:34

107

Views

Maximum row sums after removing column at each step

Published on 14 Apr 2026 - 4:06

score 259 · Answer 3 · 2026-04-10 23:21:46

259

Views

Calculate sum with binomials using generating functions

Published on 10 Apr 2026 - 23:21

score 134 · Answer 4 · 2026-04-15 23:44:22

134

Views

When is it the case that $\sum_{k=0}^nf(n,k) = \int_{\mathbf{R}}f(n,x) \, \mathrm{d}x$?

Published on 15 Apr 2026 - 23:44

score 85 · Answer 5 · 2026-04-17 05:26:44

85

Views

Evaluate $g(n,j)=\sum _{k=1}^n \frac{k^j (-1)^{n-k} \binom{n}{k}}{\frac{1}{2} n (n+1)-k}$

Published on 17 Apr 2026 - 5:26

score 138 · Answer 6 · 2026-04-16 00:13:43

138

Views

Evaluate $\sum _{j=0}^n \frac{\binom{n}{j}^2}{(j+1)^2}$

Published on 16 Apr 2026 - 0:13

score 24 · Answer 7 · 2026-03-28 11:52:53

24

Views

Summation in constraint

Published on 28 Mar 2026 - 11:52

score 74 · Answer 8 · 2026-04-08 23:59:37

74

Views

Finding series from its sum, then finding its reciprocal's sum

Published on 08 Apr 2026 - 23:59

score 95 · Answer 9 · 2026-03-27 02:37:23

95

Views

A summation with Stirling numbers of the Second kind

Published on 27 Mar 2026 - 2:37

score 41 · Answer 10 · 2026-04-11 21:58:34

41

Views

Help calculating series

Published on 11 Apr 2026 - 21:58

score 288 · Answer 11 · 2026-04-13 23:52:24

288

Views

How to find Taylor series for $\ln(1+2x)$ using power series of $\frac{1}{1-x}$

Published on 13 Apr 2026 - 23:52

score 59 · Answer 12 · 2026-03-25 08:09:19

59

Views

Proof of a relation of elements in group rings

Published on 25 Mar 2026 - 8:09

score 211 · Answer 13 · 2026-04-17 09:30:01

211

Views

A binomial summation: $\sum_{k=0}^{n} \frac{{n \choose k}^2}{k+1}$

Published on 17 Apr 2026 - 9:30

score 356 · Answer 14 · 2026-04-12 17:53:49

356

Views

Proving $\sum _{k=1}^n \frac{(-1)^{k-1} 16^k (k-1)! k! (k+n-1)!}{((2 k)!)^2 (n-k)!}=\frac{4}{n}\sum _{k=1}^n \frac{1}{2 k-1}$

Published on 12 Apr 2026 - 17:53

score 65 · Answer 15 · 2026-04-27 19:36:47

65

Views

$\Sigma_{\tau \in S} 2^{-|\tau|} = 2^{-|\sigma|}$, for any prefix-free $S \subseteq 2^{<\omega}$ s.t. $[S] = [\sigma]$.

Published on 27 Apr 2026 - 19:36

List Question

What is the complexity for the recurrence relation: T(n) = 2T(n/2) + 5n^2

Maximum row sums after removing column at each step

Calculate sum with binomials using generating functions

When is it the case that $\sum_{k=0}^nf(n,k) = \int_{\mathbf{R}}f(n,x) \, \mathrm{d}x$?

Evaluate $g(n,j)=\sum _{k=1}^n \frac{k^j (-1)^{n-k} \binom{n}{k}}{\frac{1}{2} n (n+1)-k}$

Evaluate $\sum _{j=0}^n \frac{\binom{n}{j}^2}{(j+1)^2}$

Summation in constraint

Finding series from its sum, then finding its reciprocal's sum

A summation with Stirling numbers of the Second kind

Help calculating series

How to find Taylor series for $\ln(1+2x)$ using power series of $\frac{1}{1-x}$

Proof of a relation of elements in group rings

A binomial summation: $\sum_{k=0}^{n} \frac{{n \choose k}^2}{k+1}$

Proving $\sum _{k=1}^n \frac{(-1)^{k-1} 16^k (k-1)! k! (k+n-1)!}{((2 k)!)^2 (n-k)!}=\frac{4}{n}\sum _{k=1}^n \frac{1}{2 k-1}$

$\Sigma_{\tau \in S} 2^{-|\tau|} = 2^{-|\sigma|}$, for any prefix-free $S \subseteq 2^{<\omega}$ s.t. $[S] = [\sigma]$.

Trending Questions

Popular # Hahtags

Popular Questions