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2^{2}\times 5^{2}
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factor(100)
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The relative error is about 8.9 \cdot 10^{-5}. It is surprising that the relative error is somewhat small because e and \phi are apparently unrelated (*). It is probably a mathematical ...
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Note that the period for a prime p is a factor of \varphi (p)=p-1 but need not be equal to it. This is because 10^{p-1} \equiv 1 \mod p. The period is the least n for which p|(10^n-1). If ...
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Suppose m,n > N. Then f_m(x) = f_n(x) = -1 when x \in [0, \frac{1}{2}-\frac{1}{N}]. Similarly, f_m(x) = f_n(x) = +1 when x \in [\frac{1}{2}+\frac{1}{N},1]. And |f_m(x)-f_n(x)| \leq 1 ...
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