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I understand all the algebra, but not "let $e = 2^c$". This feels like the key step. I've been staring at this the whole day, and I would've never been able to prognosticate this substitution.

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James Stewart, Calculus 7th ed. 2011. Not the Early Transcendentals version. Pp 37-8. This isn't the original screenshot. Stewart defines $f'(0)$ as $k$, but I deleted his $k$ and restored $f'(0)$. I see no point to defining $f'(0)$ as $k$.

  • I'm not sure I quite understand your question, but: when Stewart writes "let $e=2^c$," what he's doing is defining $c$: there is some $x$ such that $2^x=e$, and he's using "$c$" to refer to that $x$. – Noah Schweber Jan 11 '21 at 06:25

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If $f(x) = 2^x$, then we can estimate $f'(0)$ by doing a calculation like $$ f'(0) \approx \frac{2^{1/N} - 1}{1/N} $$ for very large $N$, which only requires taking the $N^{\text{th}}$ root of something. If we take $N = 2^k$, then we can compute $2^{1/2^k}$ by taking $k$ iterated square roots. So we can really compute an estimate such as $f'(0) \approx 0.693147$, from which we then estimate $e$ as $\approx 2^{-0.693147}$.

Picking $2^x$ in particular is not crucial, we could take $3^x$ or $10^x$ or anything else just as easily.

Misha Lavrov
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