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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)

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  • $\begingroup$ compute $u_{n+1}^2-u_n^2$. $\endgroup$ Dec 6, 2016 at 16:59

3 Answers 3

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

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

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$$u_{n+1} = \sqrt {\sum_{k=0}^n u_k} $$ $$\implies u^2_{n+1} = \sum_{k=0}^n u_k \tag1$$ 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.

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

then

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

and

$$u_n^2=S.$$

hence

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

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