Using the backward substitution method, solve the following recurrence relations: a.T(n)= T(n−1)+3forn>1 ,T(1)=0 b.T(n)=3T(n−1) forn>1 ,T(1)=7...

Using the backward substitution method, solve the following recurrence relations: a.T(n)= T(n−1)+3forn>1 ,T(1)=0 b.T(n)=3T(n−1) forn>1 ,T(1)=7 c.T(n)= T(n−1)+n for n>0 ,T(0)=0 d.T(n)= T(n/2)+n for n>1 ,T(1)=1(solve for n=2k) e.T(n)= T(n/3)+1forn>1 ,T(1)=1(solve for n=3k)

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We have put all the question in the above images.

Some formula are use like sum of A.P.(arithmetic progression) in question d.

venereology answered 1 year ago

1. Using domain and range transformations, solve the following recurrence relations: a) T(1) = 1, T(n)...

1. Using domain and range transformations, solve the following recurrence relations: a) T(1) = 1, T(n) = 2T(n/2) + 6n - 1 b) T(1) = 1, T(n) = 3T(n/2) + n^2 - n

Solve the recurrence equation. T(n) = 3T (n/3) + Cn T(1) = C

Solve the recurrence equations by Substitution a) T(n) = 4T (n/2) + n, T (1) =...

Solve the recurrence equations by Substitution a) T(n) = 4T (n/2) + n, T (1) = 1 b) T(n) = 4T (n/2) + n2 , T (1) = 1 c) T(n) = 4T (n/2) + n3 , T (1) = 1

Solve the following recurrence relations. a. x(n) = x(n − 1) + 3 for n >...

Solve the following recurrence relations. a. x(n) = x(n − 1) + 3 for n > 1, x(1) = 0 b. x(n) = 5x(n − 1) for n > 1, x(1) = 6 c. x(n) = x(n/5) + 1 for n > 1, x(1) = 1 (solve for n = 5k )

Give upper and lower bounds for T(n) in the following recurrence: T(n) = 3T(n/4) + n

- Solve the following recurrence relation : T(n) = T(αn) + T((1 − α)n) + n

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Solve the given non-homogeneous recurrence relations: an = an-1 + 6an-2 + f(n) a) an = an-1 + 6an-2 - 2n+1 with a0 = -4, a1= 5 b) an = an-1 + 6an-2 + 5 x 3n with a0 = 2, a1 = 5 c) an = an-1 + 6an-2 - 36n with a0 = 10, a1= 40

Show for the following recurrence that T(n) = T(n/3) + nlog(n) is O(nlog(n)) (not using the...

Show for the following recurrence that T(n) = T(n/3) + n*log(n) is O(n*log(n)) (not using the Master theorem please)

Solve the following recurrence relation for the given initial conditions. y(n+2) - 0.3y(n + 1) + 0.02y(n) = 10 y(0) = 2; y(1) = 0

Solve the following recurrence relation for the given initial conditions.y(n+2) - 0.3y(n + 1) + 0.02y(n) = 10 y(0) = 2; y(1) = 0

Solve the following system of equations using the Substitution Method. 1.x + 3y = – 2...

Solve the following system of equations using the Substitution Method. 1.x + 3y = – 2 5x + 15y = 0 2. x – 4y = 10 3x – 2y = 10 3. 4a + 7b = 54 2a – 3b = 14 4. 2x – 3y = 1 8x – 12y = 4 5. 3x + 4y = 12 6x + 8y = 24 6. 2a – 5b = 10 3a – b = 2

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