 # Discrete Math Question: Induction. Define a sequence recursively as follows. x_1=1 and for n in N, x_{n+1}=sqrt{(x_n)^2+1//(x_n)^2} Mariah Sparks 2022-07-17 Answered
Discrete Math Question: Induction
Define a sequence recursively as follows. ${x}_{1}=1$ and for $n\in N,{x}_{n+1}=\sqrt{\left({x}_{n}{\right)}^{2}+1/\left({x}_{n}{\right)}^{2}}$
Prove using mathematical induction that for all $n\in N$, $1\le {x}_{n}\le \sqrt{n}$
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Step 1
Base Case, clear.
Induction Hypothesis: Assume that $1\le {x}_{n}\le \sqrt{n}$ for all $n\le m$.
Step 2
Induction step: Consider ${x}_{m}$. We note that
${x}_{m}=\sqrt{\left({x}_{m-1}{\right)}^{2}+\frac{1}{\left({x}_{m-1}{\right)}^{2}}}\le \sqrt{\left({x}_{m-1}{\right)}^{2}+1}\le \sqrt{\left(\sqrt{m-1}{\right)}^{2}+1}=\sqrt{m}$
The "$1\le$" portion is easier and is left to you to figure out. Notice that we did not need to find a closed form formula.

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