Consider the following linear difference equation fk=1+12fk+1+12fk−1, 1≤k≤n−1 with f0=fn=0. How do i find solution?

Apart from trying what Will Jagy said, you can also note that, for n=2,3,4,…, ∑k=2n(n−k+1)⋅(−1)=∑k=2n(n−k+1)(fk−2fk−1+fk−2) =fn+∑k=2n−1((n−k+1−2(n−k)+(n−k−1))fk+((n−2)−2(n−1))f(1)+(n−1)f(0) =fn−nf(1)+(n−1)f(0) Thus, for every n=0,1,2,…, fn(n)=nf(1)−(n−1)f(0)−∑k=2n(n−k+1)=nf(1)−(n−1)f(0)−n(n−1)2

Expert Answer to "Consider the following linear difference equation fk=1+12fk+1+12fk−1, 1≤k≤n−1..."

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gainejavima

Answered question

2021-11-19

Consider the following linear difference equation
$f_{k} = 1 + \frac{1}{2} f_{k + 1} + \frac{1}{2} f_{k - 1}, 1 \leq k \leq n - 1$
with $f_{0} = f_{n} = 0$ . How do i find solution?

Answer & Explanation

Befory

Beginner2021-11-20Added 19 answers

Apart from trying what Will Jagy said, you can also note that, for

n = 2, 3, 4, \dots,

\sum_{k = 2}^{n} (n - k + 1) \cdot (- 1) = \sum_{k = 2}^{n} (n - k + 1) (f_{k} - 2 f_{k - 1} + f_{k - 2})

= f_{n} + \sum_{k = 2}^{n - 1} ((n - k + 1_{- 2} (n - k) + (n - k - 1)) f_{k} + ((n - 2) - 2 (n - 1)) f (1) + (n - 1) f (0)

= f_{n} - n f (1) + (n - 1) f (0)

Thus, for every

n = 0, 1, 2, \dots,

f_{n} (n) = n f (1) - (n - 1) f (0) - \sum_{k = 2}^{n} (n - k + 1) = n f (1) - (n - 1) f (0) - \frac{n (n - 1)}{2}

James Kilian

Beginner2021-11-21Added 20 answers

Easily you can determine that for the homogeneous recurrence equation

f_{k + 1}^{h} - 2_{k}^{h} + f_{k - 1}^{h} = 0

the solution is

f_{k}^{h} = c_{1} + c_{2} k

now, a particular solution which should be polynomial, you can propose

f_{k}^{p} = c_{1} + c_{2} k + c_{3} k^{2}

and the coefficients

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