Question

solve the following recurrence equation using generating functions . G(K)- 7G(K-1)+10G(K-2)=8K+6​

Answer 1

Answer:

Step-by-step explanation:

G(K) - 7G(K-1) + 10G(K-2) = 8K + 6

or, GK - 7GK + 7G + 10GK - 20G = 8K + 6

or, 4GK - 13G = 8K + 6

or, G(4K - 13) = 8K + 6

Answer 2

The solution of the recurrence relation is given by:  $G(k) = \frac{1}{4}(2^k) + \frac{3}{4}(5^k)$ .

To solve the given recurrence relation using generating functions, we define the generating function as:

$G(x) = G(0) + G(1)x + G(2)x^2 + G(3)x^3 + ...$

We multiply both sides of the recurrence relation by x^k and sum overall k:

$\sum[k\ge0] (G(k) - 7G(k-1) + 10G(k-2)) x^k = \sum[k\ge0] (8k + 6) x^k$

Using the linearity of the sum and the fact that the sum of a shifted sequence is equal to the original series multiplied by x, we get:

$G(x) - 7xG(x) + 10x^2G(x) = \sum[k\ge0] (8k + 6) x^k$

Simplifying and solving for G(x), we get:

$G(x) =\frac{[6 + 8x + (7x-1)G(x)]}{(1 - 7x + 10x^2)}$

Now, we use partial fraction decomposition to write the right-hand side as a sum of simpler fractions:

$G(x) = \frac{(\frac{1}{4})}{1-2x} + \frac{(\frac{3}{4})}{(1-5x)}$

We can then use the formula for the geometric series to get:

$G(x) = \frac{1}{4} \sum[k\ge0] (2^k) x^k + \frac{3}{4} \sum[k\ge0] (5^k) x^k$

Simplifying, we get:

$\frac{\frac{1}{4}}{(1-2x)} + \frac{\frac{3}{4}}{(1-5x)}\\\\= \sum[k\ge0] [(\frac{1}{4})2^k + (\frac{3}{4})5^k] x^k$

Therefore, the solution to the given recurrence relation is:

$G(k) = \frac{1}{4}(2^k) + \frac{3}{4}(5^k)$

To learn more about recurrence relation visit:

https://brainly.in/question/42809329

https://brainly.in/question/32502804

#SPJ3