Ways to write "50" [closed]
A really good friend of mine is an elementary school math teacher. He is turning 50, and we want to put a mathematical expression that equals 50 on his birthday cake but goes beyond the typical "order of operations" problems. Some simple examples are $$e^{\ln{50}}$$ $$100\sin{\frac{\pi}{6}}$$ $$25\sum_{k=0}^\infty \frac{1}{2^k}$$ $$\frac{300}{\pi^2}\sum_{k\in \mathbb{N}}\frac{1}{k^2}$$
What are some other creative ways I can top his cake?
I should note that he is an elementary school teacher. Now he LOVES math, and I can certainly show him a lot of expressions. I don't want them so difficult that it takes a masters degree to solve, but they should certainly be interesting enough to cause him to be wowed. Elementary functions are good, summations are also good, integrals can be explained, so this is the type of expression I'm looking for...
EDIT:: I would make a note that we are talking about a cake here, so use your judgement from here on out. Think of a normal rectangular cake and how big it is. Hence, long strings of numbers, complex integrals, and long summations are not going to work. I appreciate the answers but I need more compact expressions.
$$\frac{2^{\frac{(2\cdot2)!}{2+2}}+22+2^{2^2}-\sqrt 2^2}{2^{2-\frac{2}{2}}}$$
We can use only two famous numbers in mathematics, $\large\pi$ and $\large e$, to produce number $50$.
Click the box to see Wolfram Alpha's output to confirm the result
$$50 = 2\cdot(2\varphi - 1)^4$$ where $\varphi$ is the Golden Ratio.
$$50 = \sum_{i=0}^{+\infty} (0.98)^i$$
(Geometric series)
$$50 = \left(\left(\frac{5^5-5}{5}+5^0\right)\cdot\left(5-5^0\right)\right)^{0.5}$$
$$50 = 0.5 \cdot (5+5)^{\frac{5^0}{0.5}}$$
$$50 = 5\cdot\left(\frac{5}{0.5}+5^0\right)-5$$
(Using only the digits "5" and "0")
$$50 = \frac{3^{3!}-3^{3-3^0}}{3^{3-3^0}}-30$$
(Using only the digits "3" and "0")
$$50 = \frac{(10i)^2\log(i^i)}{\pi}$$
(Using imaginary unit $i$)
$$50 = 3 + 47$$ $$50 = 7 + 43$$ $$50 = 13 + 37$$ $$50 = 19 + 31$$
(As sum of two prime numbers)
$$50 = (7+11)\frac{11}{11-7}+\frac{7+11}{11-7}-11+7$$
(Using only the two next prime numbers of $5$)
$$ 50 = 7+3+ (7-3)\cdot(7+3)$$
(Using only the previous and next prime numbers of $5$)
$$50 = 3\cdot(2^3+3^2)-(2\cdot 3)^{3-2}+3+2$$
(Using only the two previous prime numbers of $5$)
$$ 50 = (1^6-2^5+3^4-4^3+5^2-6^1)^2\cdot(4^1 - 3^2 + 2^3 - 1^4)$$ $$ 50 = 3 - (1^9-2^8+3^7-4^6+5^5-6^4+7^3-8^2+9^1)$$ (Using bases/powers in reverse order)
No fancy math here, but if you want to emphasize how old your friend is getting, nothing says it better than implying he's halfway to the century mark:
$$100\over2$$