Find the recurrence reaction between u(n+1) and u(n)

Question

I'm given $$u_{n+1} = \sqrt {\sum_{k=0}^n u_k} $$ and $u(0) > 0.$ I have to find the recurrence relation between $u(n)$ and $u(n+1)$, but I can't manage to find it ... would be glad to have some help ! (btw this is just an intermediate questions for this problem)

Answer 1

$$u_{n+1} = \sqrt {\sum_{k=0}^n u_k} $$ $$\implies u^2_{n+1} = \sum_{k=0}^n u_k \tag1$$ Similarly we have that, $$ u^2_{n} = \sum_{k=0}^{n-1} u_k \tag2$$ So, we can say that $$(1)-(2) \implies u_n=u^2_{n+1}-u^2_{n}$$ $$\implies u^2_{n+1}=u^2_n+u_{n}$$ $$\implies u_{n+1}=\sqrt {u^2_n+u_{n}}$$ This is the required recurrence relation.

Hope this helps you.

Answer 2

$u_{n+1}^2 = \sum_{k=0}^n u_k$ and $u_{n}^2 = \sum_{k=0}^{n-1} u_k$

Then $u_{n+1}^2-u_{n}^2=u_{n} \Rightarrow u_{n+1}^2=u_{n}^2+u_{n} \Rightarrow u_{n+1}=\sqrt{u_{n}^2+u_{n}}$

Answer 3

Let $$S=\sum_{k=0}^{n-1}u_k.$$

then

$$u_{n+1}^2=S+u_n$$

and

$$u_n^2=S.$$

hence

$$u_{n+1}^2=u_n^2+u_n.$$