Help in proving sequence by induction

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Homework Help Overview

The discussion revolves around a sequence defined recursively, where the first term is given as u_1=1 and subsequent terms are calculated using the formula u_{n+1}=-1+sqrt{u_n+7}. Participants are tasked with proving that u_n<2 for all n≥1 using induction and exploring the behavior of the sequence when u_n is close to 2.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the induction proof for the first part of the problem, questioning the assumptions made about the induction variable. There is also a suggestion to manipulate the inequality involving squares to facilitate the proof. For the second part, some participants propose using Taylor series expansion to analyze the behavior of the sequence when u_n is near 2.

Discussion Status

The discussion is active, with participants providing insights and suggestions on how to approach both parts of the problem. There is no explicit consensus yet, but various lines of reasoning are being explored, particularly regarding the use of Taylor series for part b.

Contextual Notes

Participants are navigating the constraints of the problem, particularly in how to correctly apply induction and the implications of approximating u_n as 2-ε. There is also a focus on the assumptions underlying the recursive definition of the sequence.

rock.freak667
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Homework Statement


The sequence u_1,u_2,u_3,... is such that u_1=1 and u_{n+1}=-1+{\sqrt{u_n+7}}

a) Prove by induction that u_n&lt;2 for all n\geq1
b) show that if u_n=2-\epsilon, where \epsilon is small, then u_{n+1}\approx 2-\frac{1}{6}\epsilon



Homework Equations



The Attempt at a Solution


u_{n+1}=-1+sqrt{u_n+7}

\Rightarrow u_n=(u_{n+1}+1)^2-7

Assume statement is true for all k\geq1
then u_k&lt;2
\Rightarrow (u_{k+1}+1)^2-7&lt;2


(u_{k+1}+1)^2-(3)^2&lt;0

((u_{k+1}+1)-3)((u_{k+1}+1)-3)&lt;0

(u_{k+1}+1)-3&gt;0 AND (u_{k+1}+1)-3&lt;0
<br /> u_{k+1}+1&gt;3<br /> <br /> u_{k+1}&gt;2<br />

Thus u_{n+1}&gt;2 is true

<br /> (u_{k+1}+1)-3&lt;0<br /> <br /> u_{k+1}+1&lt;-3<br /> <br /> u_{k+1}&lt;-2<br />
does this affect anything in my proof?

I didn't bother to substitute the values of u_1 and u_2 and so forth as i have already done it and it is so for all n\geq1

but I do not know how to do part b)
 
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rock.freak667 said:

Homework Statement


The sequence u_1,u_2,u_3,... is such that u_1=1 and u_{n+1}=-1+{\sqrt{u_n+7}}

a) Prove by induction that u_n&lt;2 for all n\geq1
b) show that if u_n=2-\epsilon, where \epsilon is small, then u_{n+1}\approx 2-\frac{1}{6}\epsilon<br /> <br /> <br /> <br /> <h2>Homework Equations</h2><br /> <br /> <h2>The Attempt at a Solution</h2><br /> u_{n+1}=-1+sqrt{u_n+7}<br /> <br /> \Rightarrow u_n=(u_{n+1}+1)^2-7<br /> <br /> Assume statement is true for all k\geq1<br /> <br />
<br /> <br /> Didn&#039;t you mean assume it&#039;s true for n\leq{k} ? Actually assuming it is true for n=k is sufficient. But either way is fine.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> then u_k&amp;lt;2<br /> \Rightarrow (u_{k+1}+1)^2-7&amp;lt;2<br /> <br /> <br /> (u_{k+1}+1)^2-(3)^2&amp;lt;0 </div> </div> </blockquote><br /> Good so far... but at this point I&#039;d take the 3^2 to the other side... if x^2&lt;a^2 where a&gt;0, then x&lt;a and x&gt;-a... and then you can use the x&lt;a part to finish...
 
For part b, use the taylor series expansion in terms of \epsilon... you only need the first two terms.
 
So then expand \sqrt{u_n+7} using taylor series and it should work out then?
 
rock.freak667 said:
So then expand \sqrt{u_n+7} using taylor series and it should work out then?

You should substitute in 2-\epsilon for u_n in the expression for u_{n+1}... then get the taylor series of u_{n+1} in terms of epsilon...
 

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