Putnam problem of the day (by the HMD dated March 1, 2008)

pdf: included in Caderno (see “caderno” page) 

On March 1st, 2008, the Putnam problem of the day displayed on the  Harvard’s Math Department site was stated as follows:

” Evaluate

\displaystyle\sqrt[8]{2207-\dfrac{1}{2207-\dfrac{1}{2207-\cdots }}}

Express your answer in the form

\dfrac{a+b\sqrt{c}}{d},

where a,b,c,d are integers.  

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Solution

To evaluate the radicand I start by seeing that the continued fraction

x=\dfrac{1}{2207-\dfrac{1}{2207-\cdots}}

satisfies

x=\dfrac{1}{2207-x}.

Thus,  since  \dfrac{1}{2}\left( 2207+\sqrt{2207^2-4}\right) \approx 2207, the only solution left is

x=\dfrac{2207-\sqrt{2207^2-4}}{2}.

A few algebraic manipulations give

2207-x=\dfrac{2207+987\sqrt{5}}{2};

hence

\displaystyle\sqrt[8]{2207-\dfrac{1}{2207-\dfrac{1}{2207-\cdots}}}=\sqrt[8]{\dfrac{2207+987\sqrt{5}}{2}}.

In order to have

\sqrt[8]{\dfrac{2207+987\sqrt{5}}{2}}=\dfrac{a+b\sqrt{c}}{d}

or equivalently,

\dfrac{d^8}{2}\left( 2207+987\sqrt{5}\right) =\left( a+b\sqrt{c}\right) ^8,

with a,b,c integers, d^8/2 should also be an integer; therefore d should be even. I assume that d=2; On the other hand  c should be 5. Thus,

2^7\left( 2207+987\sqrt{5}\right) =126\,336\sqrt{5}+282\,496=\left( a+b\sqrt{5}\right) ^8

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\displaystyle a+b\sqrt{5}=\sqrt[8]{2^{7}\left( 2207+987\sqrt{5}\right) }

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\displaystyle a=\sqrt[8]{2^{7}\left( 2207+987\sqrt{5}\right) }-b\sqrt{5}.

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Since, for b=2

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\displaystyle a=\sqrt[8]{2^{7}\left( 2207+987\sqrt{5}\right) }-2\sqrt{5}<1,

this possibility is excluded. It remains  b=1

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\displaystyle a=\sqrt[8]{2^{7}\left( 2207+987\sqrt{5}\right) }-\sqrt{5}\approx 5,\,236\,1-2,\,236\,1=3,\,000

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Now I confirm

\displaystyle \left( 3+\sqrt{5}\right) ^{8}=126\,336\sqrt{5}+282\,496.

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So, the solution I came was

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\displaystyle\sqrt[8]{2207-\dfrac{1}{2207-\dfrac{1}{2207-\cdots }}}=\dfrac{3+\sqrt{5}}{2}.

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Remark: The calculation of \displaystyle \left( 3+\sqrt{5}\right) ^{8}=126\,336\sqrt{5}+282\,496

can be done by hand as follows

\displaystyle\left( 3+\sqrt{5}\right) ^{2}=6\sqrt{5}+14

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\displaystyle\left( 3+\sqrt{5}\right) ^{4}=\left( 6\sqrt{5}+14\right) ^{2}=168\sqrt{5}+376

\bigskip

\displaystyle\left( 3+\sqrt{5}\right) ^{8}=\left( 168\sqrt{5}+376\right)^{2}=126\,336\sqrt{5}+282\,496

Update March, 20: you can compare with this solution   (Putnam 1995, Problem B5 )

Addendum of March 7, 2009: Comment/Proof of the convergence of the continued fraction by Vishal Lama (comment dated March 7, 2009)

In the solution presented in your post, x denotes an expression that we can’t assume, beforehand, is a finite number. x may perhaps be infinite! Therefore, the way to go about computing the expression (the infinite continued fraction) given in the problem is as follows.

The infinite continued fraction is defined as the limit of the sequence (a_n), where a_0 = 2207 and a_n = 2207 - 1/a_{n-1} for all n \geq 1. Then, we show that the sequence a_n is bounded from below (a_n>2206 for all n \geq 0, which can be shown by a simple induction) and that it is also strictly decreasing (which can be shown using induction, again).  Now, we invoke the Monotone Convergence Theorem to conclude that the sequence does indeed have a (finite) limit, which we can now denote by x. Once we establish that x (which is the infinite continued fraction!) is finite, we can compute x the way you did in your solution. Basically, we have to go through all that trouble just to prove that the given infinite continued fraction is indeed finite! Only after that can the computation begin!

Part of my reply was: “I did not prove the convergence of the continued fraction. Thanks for doing it.
Basically I assumed that convergence based on a certain numerical evidence, but of course this evidence proves nothing.”

ADDENDUM of May 5, 2010: Alternatively we can prove the convergence applying the Śleszýnki-Pringsheim Theorem (a reference here,  wikipedia): if for all naturals j\left\vert b_{j}\right\vert \geq \left\vert a_{j}\right\vert +1, then the continued fraction \mathcal{K}_{1}^{\infty }\left( a_{j}/b_{j}\right) converges. Since a_j=-1 and b_j=2207, the theorem hypothesis is satisfied:

\left\vert b_{j}\right\vert=2207 \geq 1+1=\left\vert a_{j}\right\vert +1.

 

 

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Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Caderno, Exercícios Matemáticos, Exercise, Fracções Contínuas, Matemática, Math, Problem, Problemas, Putnam com as etiquetas , , , , . ligação permanente.

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