Question

Solve the recurrence relation: t(n) = 3t(n-1) 1, with initial condition of t(0) = 1

Answer 1

A linear recurrence relation is a function or sequence in which each term is a linear combination of the terms that came before it.

The recurrence relation exists $T(n)=\Theta\left(3^n\right)$.

What is meant by "recurrence relation"?

Recurrence relations are used to simplify complex problems by reducing them to an iterative process based on simpler versions of the problem.

Using the substitution method, we find out that

$T(n) &=n+3 T(n-1)$

$&=n+3(n-1)+3^2 T(n-2)$

$&=n+3(n-1)+3^2(n-2)+3^3 T(n-3)$

$&=\cdots$

$&=n+3(n-1)+3^2(n-2)+\cdots+3^{n-1}(n-(n-1))+3^n T(0)$

simplifying the above equation, we get

$&=\frac{3^{n+1}-2 n-3}{4}+3^n T(0)$

$&=\Theta\left(3^n\right)$

Even without doing the full calculation it is not hard to check that $T(n) \geq 3^{n-1}+3^n T(0)$, and so $T(n)=\Omega\left(3^n\right)$.

A cheap way to obtain the corresponding upper bound is by considering

$S(n)=T(n) / 3^n$, which satisfies the recurrence relation  

$S(n)=S(n-1)+n / 3^n$.

Repeated substitution then gives

$\frac{T(n)}{3^n}=\sum_{m=1}^n \frac{m}{3^m}+T(0)$

Since the infinite series $\sum_{m=1}^{\infty} \frac{m}{3^m}$ converges, this implies that $\frac{T(n)}{3^n}=\Theta(1)$ and so $T(n)=\Theta\left(3^n\right)$

Therefore, the recurrence relation exists $T(n)=\Theta\left(3^n\right)$.

To learn more about recurrence relation, refer to:

brainly.com/question/4082048

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