Algebra
# Topology

A sequence of functions \(f_n(x)\) is drawn in stages. Each stage in the sequence consists of a connected path of semicircles:

\(f_1(x)\) consists of 1 semicircle of radius 1,

\(f_2(x)\)consists of 2 semicircles of radius \(\frac{1}{2},\)

\(f_3(x)\)consists of 4 semicircles of radius \(\frac{1}{4},\)

and so on.

As \(n\) increases and this process continues indefinitely, what happens to the lengths of \(f_n(x)\) over the interval \([0,2]?\)

Let \(f_n(x)\) be the function on \([0,2]\) whose graph is the Stage \(n\) semicircles. Then \(f_n\left(\frac{1}{3}\right) \rightarrow 0,\) i.e.,

\[\lim_{n \to \infty} f_n\left(\frac{1}{3}\right) = 0.\]

Let \(f_n(x)\) be the function on \([0,2]\) whose graph is the Stage \(n\) semicircles. Then for all \(x\) in \([0,2],\) \(f_n\left(x\right) \rightarrow 0,\) i.e.,

\[\lim_{n \to \infty} f_n\left(x\right) = 0.\]

Note. Another way of phrasing the question is: "Does the sequence \(\{f_n(x)\}\) converge pointwise to \( f(x) = 0\) on the interval \([0,2]?\)

Let \(f_n(x)\) be the function on \([0,2]\) whose graph is the Stage \(n\) semicircles. Then for all \(x\) in \([0,2]\) where they are defined, the **derivatives** \((f_n)'\left(x\right) \rightarrow 0,\) i.e.,

\[\lim_{n \to \infty} (f_n)'\left(x\right) = 0.\]

*Note*. 0 is the derivative of the function \(f(x) = 0\) that the functions \(f_n\) approach.

True or False?

Suppose \(\{f_n\}\) and \(f\) are functions on an interval \(I,\) and \(L(g)\) represents the length of the curve \(g\) on \(I.\)

If \(f_n(x) \to f(x)\) for all \(x \in I,\) then \(L(f_n) \to L(f).\)

Alternate phrasings:

If \(\lim_{n \to \infty} f_n(x) = f(x)\) for all \(x \in I,\) then \(\lim_{n \to \infty}L(f_n) = L(f).\)

If the sequence of curves approach \(f\) at each point, then the lengths of the curves approach the length of \(f.\)

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