Testwiki:Reference desk/Archives/Mathematics/2014 September 22

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September 22

Why does this calculates the square root of two?

Why does this calculates the square root of two?

A[0] = Y[0]/X[0]

Where both X[0] and Y[0] are positive real numbers. They can even be equal to each other.

A[k] = Y[k]/X[k]

Y[k+1] = Y[k] + 2*X[k]

X[k+1] = Y[k] + X[k]

We have A[infinity] = Sqrt(2)

But I do not understand why this works for all possible positive real numbers. It looks so deceptively simple.

Also you can extend this to

A[k] = Y[k]/X[k]

Y[k+1] = Y[k] + N*X[k]

X[k+1] = Y[k] + X[k]

We have A[infinity] = Sqrt(N)

202.177.218.59 (talk) 06:05, 22 September 2014 (UTC)

The expression corresponds to:

The formula can be written as:

$\left(\begin{array}{c}Y[k+1]\\ X[k+1]\end{array}\right)=\left(\begin{array}{cc}1& N\\ 1& 1\end{array}\right)\left(\begin{array}{c}Y[k]\\ X[k]\end{array}\right)$

The matrix $\left(\begin{array}{cc}1& N\\ 1& 1\end{array}\right)$ has eigenvalues $1\pm \sqrt{N}$ with eigenvectors $\left(\begin{array}{c}\pm \sqrt{N}\\ 1\end{array}\right).$ The answer follows from elementary matrix manipulation.

You can also see that

$A[k+1]=\frac{A[k]+N}{A[k]+1}$

and the answer follows from properties of Mobius transformations. — Arthur Rubin (talk) 08:03, 22 September 2014 (UTC)

If you write:

$B[k]=\frac{A[k]+\sqrt{N}}{A[k]-\sqrt{N}}$

Then you get:

$B[k+1]=\frac{1+\sqrt{N}}{1-\sqrt{N}}B[k]=-\frac{N+1+2\sqrt{N}}{N-1}B[k]$

— Arthur Rubin (talk) 19:15, 22 September 2014 (UTC)

Real Analysis: bounded set

The question is Let A=U{(1-1/n, 1+1/n) where n is in natural numbers}. I need to find the lower and upper bounds, supA, and infA. I can find those. The problem I am having is proving that A={0,2}.

I know there are 2 parts to this proof. First, showing that A=(0,2). Second, showing that A≠anything out side (0,2). I am not sure how to begin to show that. Please help. — Preceding unsigned comment added by Pinterc (talk • contribs) 23:15, 22 September 2014 (UTC)

If supA=x, then any y>x is not in A, and likewise if y<infA, y is not in A, by definition of the Infimum_and_supremum. I'm confused by your notation though. The way you've written it, A is neither equal to {0,2} or (0,2)...Do you mean intersection, (not union) in the definition of A? SemanticMantis (talk) 17:11, 23 September 2014 (UTC)

What do you mean by 'A={0,2}'—a two-item set or a closed interval...? --CiaPan (talk) 23:04, 27 September 2014 (UTC)