Use neighborhood signs to express e, pi, phi

Question

Is the above description true or false?

$$e=\left(1+\pi-(5^{±})\arccos\left(\fracφ2\right)\right)^{\frac{1}{π-(5^{±})\arccos\left(\fracφ2\right)}}$$

Answer 1

The formula is indeed true, if interpreted appropriately, but you have given it in an entirely unintuitive way. The meaning of the expression $5^{\pm}$ is wholly non-obvious and it could easily have been interpreted as addition, $5{\pm}\ldots$, or exponentiation, $5^{\pm1}$, despite that for the equation to be true, it must mean approaching the number, ($5{\pm}\varepsilon$), where $\varepsilon$ is a small, positive real. In future, please take care to explain your notation in the body of your text, or use notation that is more obvious in its context.

With this in mind, the RHS expression becomes $$\lim_{x\to0}\left(1+\pi-\left(5+x\right)\arccos\left(\frac{\varphi}{2}\right)\right)^{\frac{1}{\pi-\left(5+x\right)\arccos\left(\frac{\varphi}{2}\right)}}$$

The equation is true because $5\arccos\left(\frac{\varphi}{2}\right)=\pi$, so it is directly equivalent to the well-known limit $\lim_{u\to0} \left(1+u\right)^{\frac{1}{u}}$, where $u=-\frac{\pi}{5}x$ and if $x\to0$, then $u\to0$. That is to say, the $\pi$ and $\arccos\ldots$ terms effectively cancel each other out and amount to a scaling of $x$ that is inconsequential to the limit.