Iteration of $A_{n}(q)=q^nA_{n-1} (q)$

Question

I can't seem to find how $A_{n}(q)=q^nA_{n-1} (q)$ iterates to

$$A_{n}(q) = q^{n+1 \choose 2}A_{0}(q)$$

Where ${a \choose 2} = a(a-1)/2$ and absolute value of q is less than 1.

I understand that I am supposed to find $q^nA_{n-1} (q)$ in terms of $n-2$ and then so on until $n-n$ but I can't seem to figure it out. Thanks!

Answer 1

Consider the sequence $\log A_n(q)$, which satisfies

$$ \log A_n(q) = n \log q + \log A_{n-1}(q). $$

Iteratively, we obtain

$$ \begin{align} \log A_n(q) &= n \log q + \log A_{n-1}(q) \\ &= n \log q + (n-1)\log q + \log A_{n-2}(q) \\ & \qquad \vdots \\ &= n \log q + (n-1)\log q + \dots + \log q + \log A_0(q) \\ &= (n + (n-1) + \dots + 1)\log q + \log A_0(q) \\ &= \frac{n(n+1)}{2}\log q + \log A_0(q) \end{align} $$ and exponentiating gives the result.

Answer 2

Simply look at the first terms: $$A_1(q) = q A_0(q)$$ $$A_2(q) = q^2 A_1(q) = q^2 \cdot q A_0(q)$$ $$A_3(q) = q^3 A_2(q) = q^3 \cdot q^2 \cdot q A_0(q)$$ and so on. The $n$-th term will be (using induction) $$A_n(q)=q^n \cdots q^2 \cdot q A_0(q) = q^{n+\cdots +2+1} A_0(q)$$ Finally, you should know that $n+ (n-1)+ \cdots + 2+1 = \binom{n+1}{2}$, and the result is proved.

Answer 3

$$A_{n}(q)=q^nA_{n-1} (q)$$

Firstly Put $n=1$

$$A_{1}(q)=q^1A_{0} (q)$$

then Put $n=2$

$$A_{2}(q)=q^2A_{1} (q)$$

We got $A_{1}(q)=q^1A_{0} (q)$ in first step $$A_{2}(q)=q^2A_{1} (q)=q^2 .q^1A_{0} (q)=q^{1+2} A_{0} (q)$$

then Put $n=3$ $$A_{3}(q)=q^3A_{2} (q)$$ We got $A_{2}(q)=q^{1+2} A_{0} (q) $ in second step $$A_{3}(q)=q^3A_{2} (q)=q^3 .q^{1+2} A_{0} (q)=q^{1+2+3} A_{0} (q)$$

If you continue $n$ steps in this way , finally you will get:

$$A_{n}(q)=q^{1+2+3+....+n} A_{0} (q)$$