Continued Surd (1 Viewer)

Tsylana · Oct 5, 2009

BHHS 2009 Ext 2 Trial Question 6 part c

haha a bit of a strange question that i have no idea how to do..

its just pratically find the limitting value of what i posted underneath =)

Tsylana · Oct 5, 2009

kaz1 · Oct 6, 2009

Tsylana · Oct 6, 2009

lol my calculator coulda told me that

Iruka · Oct 6, 2009

Strictly speaking, you probably need to use the Monotone Convergence Theorem to prove that the limit of this thing exists (which means to do this question properly is beyond the 4u syllabus), but if you are willing to accept that it converges on faith, then you let x=that thing with all the surds.

Squaring both sides, we have x^2 = 6 + x, which you solve as a simple quadratic. Clearly x is positive, so you take the positive answer, which, as noted, is 3.

untouchablecuz · Oct 6, 2009

jet · Oct 6, 2009

Tsylana · Oct 6, 2009

I must say that's pretty elegant...

I like it... good question.

shaon0 · Oct 6, 2009

could you post up the BHHS 09 trial?

Drongoski · Oct 6, 2009

Tsylana said:
I must say that's pretty elegant...

I like it... good question.

As Iruka pointed out, this presupposes the existence of a limit which in turn requires the sequence of continued surds is i) monotone increasing and ii) has an upper bound.

untouchablecuz · Oct 6, 2009

could someone post up a formal proof of the said continued fractions convergence

...please

Timothy.Siu · Oct 6, 2009

untouchablecuz said:
could someone post up a formal proof of the said continued fractions convergence

...please

i'm pretty sure 3unitz did it a while back.

Drongoski · Oct 6, 2009

untouchablecuz - I'll attempt proof here.

$\100dpi Consider\ the\ sequence: \ \left \{ a_{n} \right \} \ given\ by:\\ \\ a_{1} =\sqrt{6}\ \ \ a_{2}=\sqrt{6\sqrt{6+\sqrt{6}}\ \ \ a_{3}=\sqrt{6+\sqrt{6+\sqrt{6}}}\\ \\ \cdots \ \ a_{n+1} =\sqrt{6+a_{n}}\\ \\ \\ To\ prove\ sequence\ monotonic\ increasing:\\ \\ Now\ \ a_{n+1}=\sqrt{6+a_{n}}\ =\ \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{\cdots \sqrt{6}}}}}\\ \\ > \sqrt{6+\sqrt{6+\sqrt{6+\sqrt {\cdots \sqrt{0}}}}}\ \ =\ \ a_{n}\\ \\ \therefore \ \ sequence \ \left \{ a_{n} \right \}\ is\ monotonic\ increasing\\ \\ \\ To\ prove\ the \ sequence\ is\ bounded\ above:\\ \\ a_{n}\ =\ \sqrt{6+\sqrt{6+\sqrt{6+\cdots +\sqrt{6+\sqrt{6}}}}}\\ \\ < \ \sqrt{6+\sqrt{6+\sqrt{6+\cdots \sqrt{6+\sqrt{9}}}}}\ =\ \cdots \ =\ \sqrt{6+3}\ =\ 3\\ \\ \therefore \ \ sequence\ is\ bounded\ above.\\ \\ Hence\ sequence\ has\ a\ limit.\\ \\ Then\ work\ out\ limit\ as\ per\ jetblack2007.$

jet · Oct 6, 2009

But you just proved the limit, didn't you?

Drongoski · Oct 6, 2009

jetblack2007 said:
But you just proved the limit, didn't you?

I just established the conditions for existence of a limit. It so happens 3 is an upper bound - in fact the least upper bound (the lub or infimum) and is therefore the limit. But I have not established that 3 is the infimum.

Edit:

I made some typo error and tried to correct and have now ruined my posting; anyway of retrieving it ?

Iruka · Oct 6, 2009

It is not mathematically valid to do ordinary arithmetic with infinite sums and products unless you have established that they converge. There are many examples of paradoxes that arise if you try to do this.

So, as Drogonski has done, you have to establish that the sequence converges to a limit before you can go about finding the limit. So in this case, you use MCT to prove the limit exists. (That is, you need to prove that the sequence is monotonically increasing and bounded above.) You could possibly also prove Cauchy convergence.

Drongoski · Oct 6, 2009

Oh heck! Have to re-construct my "proof" all over again; earler post ruined in a botched editing.

$\100dpi Consider\ th\ sequence\ \left \{ a_{n} \right \}\ given\ by:\\ \\ a_{1}=\sqrt{6}\ \ \ \ a_{2}=\sqrt{6+\sqrt{6}}\ =\ \sqrt{6+a_{1}}\ \ \ \ a_{3}=\sqrt{6+\sqrt{6+\sqrt{6}}}\ =\ \sqrt{6+a_{2}}\\ \\ a_{n+1}=\sqrt{6+\sqrt{6+\sqrt{6+\ \cdots \ +\sqrt{6}}}}\ =\ \sqrt{6+a_{n}}\\ \\ \\ i)\ \ Now\ \ a_{n+1}\ =\ \sqrt{6+\sqrt{6+\sqrt{6+\ \cdots \ +\sqrt{6+\sqrt{6}}}}}\\ \\> \ \ \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{\cdots +\sqrt{6+\sqrt{0}}}}}} \ \ =\ \ a_{n}\\ \\ i.e.\ \ \ \left \{ a_{n} \right \}\ is\ a\ monotonic\ increasing\ sequence\\ \\ \\ ii)\ \ Also:\ \ a_{n}\ =\ \sqrt{6+\sqrt{6+\sqrt{6+\ \cdots \ \sqrt{6+\sqrt{6}}}}}\ \ < \ \ \sqrt{6+\sqrt{6+\sqrt{6+\ \cdots \ \sqrt{6+\sqrt{9}}}}}\\ \\ = \ \ \cdots \ \ =\ \ \sqrt{6+3}\ =\ 3\\ \\ \therefore \ \ sequence\ \left \{ a_{n} \right \}\ is\ monotonic\ increasing\ and\ bounded\ above\ \ and\ \ \therefore \ \ has\ a\ limit.\ \ This\ limit\ can\ then\ be\ derived\ e.g.\ as\ shown\ by\ jetblack2007.$

Continued Surd (1 Viewer)

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