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

= 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:
 * $$\begin{pmatrix}

Y[k+1] \\ X[k+1] \end{pmatrix}= \begin{pmatrix} 1 & N \\ 1 & 1 \end{pmatrix} \begin{pmatrix} Y[k] \\ X[k] \end{pmatrix}$$
 * The matrix $$\begin{pmatrix}

1 & N \\ 1 & 1 \end{pmatrix}$$ has eigenvalues $$1 \pm \sqrt N$$ with eigenvectors $$\begin{pmatrix} \pm \sqrt N \\ 1 \end{pmatrix}.$$ The answer follows from elementary matrix manipulation.
 * You can also see that
 * $$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)