I am currently writing a college essay that is utilizing the engineering procedure. I am going to answer my question with the ways I know. Are there any other ways to write mathematically write 2 besides the infinite series?
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$(1+1)^1$
Largest even prime
number of last names of the presidents with the same last name
marty cohen
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@TheGreatDuck it would appear mister Cohen referred to US presidents, however, there are more than two pairs of presidents who share the same last name. Those last names are: Adams (2nd and 6th), Harrison (9th and 23rd), Roosevelt (26th 32nd) and Bush (41st and 43th). If mister Cohen was instead referring to living US presidents who share the same last name he would be correct, at the time of answering both presidents W. Bush and H.W. Bush were still alive and had previously served as president. – JJJ Mar 03 '19 at 21:57
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$$2=\pi\prod_{i=1}^{\infty}\left(1-\frac{1}{4n^2}\right)$$ $$2=\sqrt{\frac{\pi}{\arctan(1)}}$$ It is also the smallest even prime.
Siong Thye Goh
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$$2=\left(\sum_{k\ge 1}\frac{k}{\left(\left(\ln\left(\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n\right)+\sin^2\theta+\cos^2\theta\right)\sum_{n\ge 0}\frac{\cosh y\sqrt{1-\tanh^2y}}{2^n}\right)^{\frac1{k+1}\binom{k+1}2}}\right)!$$
Brian M. Scott
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\begin{align} 2 &= \sqrt{\frac1\pi}\int_{-\infty}^\infty e^{-x^2}\ \mathsf dx\\ &= \int_0^\infty x^2 e^{-x}\ \mathsf dx\\ &=\sum_{n=0}^\infty \frac{(ix)^n +(-ix)^n}{n!} + \frac1i\sum_{n=0}^\infty \frac{(ix)^n -(-ix)^n}{n!}\\ &= \lim_{n\to\infty}\left(1 + \int_1^2 \frac1{nx}\ \mathsf dx \right)^n\\ &= \frac4\pi \prod_{n=1}^\infty\left(\frac{4n^2}{4n^2-1}\right) \end{align}
Math1000
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According to John Von Neumann this is actually the definition of $2$:
$\{\emptyset, \{\emptyset\} \}$
Granted, it is possible that the definition might have changed since then but the way I've understood that literally is two.
user64742
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$$2=1+\left(\frac {1}{1\cdot 2}+\frac {1}{2\cdot 3}+\frac {1}{3\cdot 4}+...\right)$$ because of the telescoping series phenomenom:$$\sum_{j=1}^{\infty}\frac {1}{j(j+1)}=\lim_{n\to \infty}\sum_{j=1}^n\frac {1}{j(j+1)}=\lim_{n\to \infty}\sum_{j=1}^n\left(\frac {1}{j}-\frac {1}{j+1}\right)=$$ $$=\lim_{n\to \infty}\left(1-\frac {1}{n+1}\right)=1.$$
If you are a computer then $2$ is just a display character, but you know a number $10$ in what humans call base $2.$
DanielWainfleet
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After reading your link, I suggest $\frac {1}{\sqrt {\pi}}\int_{-\infty}^{\infty}e^{-x^2}dx+ \lim_{y\to \infty}\frac {\pi (y)\ln y}{y}$ where $\pi (y)$ is the number of primes less than $y$. Or if you feel that "the number of (anything)" is in poor taste, if $P(y)$ is the set of primes less than $y$ then $\pi(y)$ is the $n\in \Bbb Z^+\cup {0}$ such that $P(y)$ is a bijective image of ${j\in \Bbb Z^+: j\le n}$........LOL – DanielWainfleet Mar 30 '19 at 20:16