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I've seen this formula

$$1RM = \frac{100 \cdot w}{48.8 + 53.8 \cdot e^{-0.075 \cdot r}}$$

but I don't know what does the $e$ means.

The $w$ stands for weight. The $r$ for repetitions but I think the $e$ is from scientific notication but i'm not sure.

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    $\begingroup$ $e$ is the base for the natural logarithm, for the exponential function, $e = \sum_{n = 0}^\infty \frac{1}{n!} \approx 2.7128$. $\endgroup$ – Daniel Fischer Aug 24 '15 at 10:00
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    $\begingroup$ $e$ $\endgroup$ – Surb Aug 24 '15 at 10:01
  • $\begingroup$ I had no idea it was a math constant. Thank you guys $\endgroup$ – Lothre1 Aug 24 '15 at 10:02
  • $\begingroup$ By the way, when I search Google for "e maths", the first hit is mathsisfun.com/numbers/e-eulers-number.html which is correct, and gives you the string "euler's number" on which to search further. $\endgroup$ – Patrick Stevens Aug 24 '15 at 10:50
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    $\begingroup$ @PatrickStevens When I google "e" the second link is the wikipedia one. $\endgroup$ – Surb Aug 24 '15 at 10:52
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$\phantom{1233976327891467283146327864786231476892364983216468236874}$ $$\huge e$$

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    $\begingroup$ Extracted sentences from the link: The number $e$ is an important mathematical constant that is the base of the natural logarithm. We have $$ e= \sum_{n=0}^\infty \frac{1}{n!}=\lim_{n\to \infty}\Big(1+\frac{1}{n}\Big)^n\approx 2.71828$$ Sometimes called Euler's number after the Swiss mathematician Leonhard Euler. Also known as Napier's constant, but Euler's choice of the symbol $e$ is said to have been retained in his honor.The number $e$ is of eminent importance in mathematics. Like the constant $\pi$, $e$ is irrational: it is not a ratio of integers; and it is transcendental. $\endgroup$ – Surb Aug 24 '15 at 10:46
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    $\begingroup$ I think that should probably be in the answer, not a comment. $\endgroup$ – Patrick Stevens Aug 24 '15 at 10:48
  • $\begingroup$ @PatrickStevens I agree with you, but I like this idea to an answer with a single character (not often possible). However, I thought it is worth to extract a few informations from the link in case the link gets broken (although this is a wikipedia link which is unlikely to disappear). To cite Did: "OK, I confess: I always dreamt of reading/posting an answer with only one character... so I could not resist the occasion." $\endgroup$ – Surb Aug 24 '15 at 10:51
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    $\begingroup$ Nicely done. Still, I hope this doesn't get as many upvotes as Did's one-letter answer, that would be a bad incentive for future answerers ;) $\endgroup$ – Daniel Fischer Aug 24 '15 at 10:57
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$e\approx2.7182818284\dots$ is an irrational number. It has several representations. I'll give a few representation, along with a slight generalization.


\begin{align} e=\lim_{n\to\infty}\left(1+\frac1n\right)^n\\ e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n \end{align} This is the usual definition. The $\lim$ sign means, roughly, that you let $n$ get larger and larger. For example, $(1+\frac1{1000})^{1000}=2.7169\dots\approx e$, and you get better and better approximations as $n$ gets bigger. To prove the second thing from the first one, replace $n$ with $\frac nx$, and notice that $\frac nx\to\infty$ means the same thing as $n\to\infty$ (when $x$ is positive, at least. It works for negative $x$, too, though).


\begin{align} e&=\frac1{0!}+\frac1{1!}+\frac1{2!}+\frac1{3!}+\dotsb\\ e^x&=\ 1\ +\ x\ +\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsb \end{align} Remember that $0!=1$. You can try to prove this from the first set of equalities and the binomial formula. That's how Euler did it. (If you want to be rigorous, though, you need more effort. Euler was never very rigorous in this sort of thing.)


\begin{align} \int_1^e\frac1x\operatorname d\!x&=1\\ \int_1^t\frac1x\operatorname d\!x&=\log_e(t) \end{align} I need to explain the notation. The shape surrounded by the curve $y=\frac1x$, the vertical lines $x=1$ and $x=e$, and the $x$-axis has area $1$. More generally, the shape bounded by $\frac1x$, $x=1$, $x=t$, and the $x$-axis has area $\log_e(t)$. I won't prove this here.


And my favorite characterization: $e$ is the unique number that satisfies: $$e^x\ge x+1$$ for all $x$.

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  • $\begingroup$ +1 for your favorite characterization. While $e^x\geq x+1$ for every $x$ is clear to me, I'd be very interested in seeing a proof that $a^x\geq x+1$ for every $x$ implies $a=e$. $\endgroup$ – Surb Aug 25 '15 at 10:26
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    $\begingroup$ @Surb Here's one: Notice that $x^{1/x}$ has a unique maximum at $x=e$ (Steiner's problem). It can be shown using just a few algebraic manipulations that $a^x\ge x+1$ implies that $a$ is a global maximum of $x^{1/x}$ (hint: substitute $x\mapsto\frac xa-1$ into the inequality), which means that $a=e$. $\endgroup$ – Akiva Weinberger Aug 25 '15 at 14:48

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