Turn iterative function into polynomial.

Question

So, I have an iterative function that looks something like this.

$$f(x_n) = (x_n + 0.08) \cdot 0.98$$

e.g. So if $n = 2$ and $x$ started at $0$, then the equation would be equal to $(((0 + 0.8) \cdot 0.98) + 0.08) \cdot 0.98$, which is $0.155232$.

Is it possible to create a formula that can find the value of any iteration of the function? e.g being able to rewrite the answer as a polynomial (where $x$ is an integer). If it is, how is it done?

Answer 1

We have $$x_n = 0.98 \cdot x_{n-1} + 0.08 \cdot 0.98$$ $$x_{n+1} = 0.98 \cdot x_n + 0.08 \cdot 0.98$$ Subtract to get $$x_{n+1}=1.98 \cdot x_n-0.98 \cdot x_{n-1}$$ You get the following characteristic polynomial $$p(x)=x^2-1.98x+0.98$$ Whose roots are $x_1=1, \space x_2=0.98$
So the general formula is $x_n=c_1+c_2\cdot 0.98^n$
Using $x_1,x_2$ you can find $c_1=3.92, c_2=-3.92$ so your final equation is: $$x_n=3.92-3.92\cdot 0.98^n$$

Answer 2

Yes, though it wouldn't be a polynomial. In particular, the trick will be to rewrite your function as $$f(x)=m(x-c)+c$$ for some coefficients $m$ and $c$ which you can find algebraically. Then, we can notice that $$f(f(x))=m(m(x-c)+c-c)+c=m^2(x-c)+c$$ that is, the adding and subtracting of $c$ will cancel out, except on the outside (in a sense, we are "conjugating" the function $x\mapsto mx$ by $x\mapsto x+c$ - that is, applying the inverse of the latter, then the former, then the latter again). This allows us to write $$f^n(x)=m^n(x-c)+c.$$

Geometrically, this is noticing that $c$ is the fixed point of $f$ (i.e. the point such that $f(c)=c$) and that in all iterates of $f$ we have $f^n(c)=c$. Then, the slope of the line that is the $n^{th}$ iterate is clearly $m^n$ - and combining these two bits of information yields the closed form.

Answer 3

If the starting value is $t$ and the numbers $0.08,\ 0.98$ are called $a,b$ then your iteration takes $t$ into $(t+a)b.$ Working out the first few iterates we have $$t,\\ bt+ab,\\ b^2t+ab(b+1),\\ b^3t+ab(b^2+b+1).$$ If we apply the sum of a geometric series to the lengthening sequence of powers of $b$ occurring on the right sides, we get for the $n$th iterate the formula $$ b^n t +ab\cdot \frac{b^n-1}{b-1}.$$ One can check that this matches the first few terms above, and it's likely not hard to show it keeps working using induction.