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Author SHA1 Message Date
john verzani
83bddc19e3
Merge pull request #153 from fangliu-tju/main
some typos
2025-08-29 15:32:49 -04:00
Fang Liu
bf2d5f6c76 some typos 2025-08-29 13:52:29 +08:00
2 changed files with 2 additions and 2 deletions

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@ -311,7 +311,7 @@ This progression can be seen to be increasing. Cauchy, in his treatise, can see
$$
\begin{align*}
(1 + \frac{1}{m})^n &= 1 + \frac{1}{1} + \frac{1}{1\cdot 2}(1 - \frac{1}{m}) + \\
(1 + \frac{1}{m})^m &= 1 + \frac{1}{1} + \frac{1}{1\cdot 2}(1 - \frac{1}{m}) + \\
& \frac{1}{1\cdot 2\cdot 3}(1 - \frac{1}{m})(1 - \frac{2}{m}) + \cdots \\
&+
\frac{1}{1 \cdot 2 \cdot \cdots \cdot m}(1 - \frac{1}{m}) \cdot \cdots \cdot (1 - \frac{m-1}{m}).

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@ -380,7 +380,7 @@ this by the inverse property. Whereas, by expressing $a=b^{\log_b(a)}$ we have:
$$
a^{(\log_b(x)/\log_b(b))} = (b^{\log_b(a)})^{(\log_b(x)/\log_b(a))} =
a^{(\log_b(x)/\log_b(a))} = (b^{\log_b(a)})^{(\log_b(x)/\log_b(a))} =
b^{\log_b(a) \cdot \log_b(x)/\log_b(a) } = b^{\log_b(x)} = x.
$$