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$$ y_{k+1}=(1-h)y_{k}-h~~\text{where} ~~k\in\mathbb{N}_{\geq0},~~y_0=0,~~h=\text{non-zero constant} $$

I want to evaluate the general term of the sequence $~\{y\}~$

$$\begin{align} y_0&=0\\ y_1&=(1-h)y_0-h=-h\\ y_2&=(1-h)y_1-h\\&=(1-h)(-h)+(-h)\\ &=(-h)(1-h+1)\\ &=(-h)(2-h)\\ y_3&=(1-h)h_2-h\\ &=(1-h)(-h)(2-h)-h\\ &=(-h)\left\{ (1-h)(2-h)+1 \right\}\\ &=(-h) \left\{ 2-3h+h^2+1 \right\}\\ &=(-h)\left(h^2-3h+3 \right)\\ y_4&=(-h)\left\{ (1-h) \left(h^2-3h+3 \right) +1\right\}\\ &=(-h) \left\{ -h^3+3h^2-6h+4 \right\} \end{align}$$

$$ \begin{cases} y_0=0\\ y_1=-h\\ y_2=(-h)(2-h)\\ y_3=(-h)\left(h^2-3h+3 \right)\\ y_4=(-h)\left\{ -h^3+3h^2-6h+4 \right\} \end{cases} $$

I've failed to estimate the general term $~ y_k ~$

I need your help.

BTW the problem statement precedes the following problem statement(I've already solved this problem(though I haven't checked that my soln for it is correct)).

$$ \text{Evaluate}~~y(x)~~\text{of}~~y'=-y-1~~\wedge~~y(0)=0 $$

Can this help to solve the initially given problem?

TG24
  • 443

1 Answers1

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Note that $$ y_{k+1}+1=(1-h)(y_k+1). $$

We obtain $y_k=(1-h)^k-1$ (when $h\ne1$). If $h=1$ then $y_0=0$, $y_k=-1(k\ge1)$.

Saunders
  • 656