Five Fives mathematical constants challenge

[DejMar]

There are Mathematical Recreation Puzzles where one uses a given number of one digit and a limited set of mathematical symbols for operations and functions to represent a number. One of the more recognized forms of this puzzle is with five fives. For example, to represent the value 27 one might present 27=5×5+(5+5)/5.
Use of any function that demands one or more letters (to include the Greek and Hebrew letters) in its notation is forbidden, this eliminates using many mathematical numerical constants like i, φ and π.
The challenge here is to present a "Five Fives" expression under the rules for the values of each of these mathematical constants: i, φ and π.

[groza528]

I'll tackle the easy one:
i = sqrt(5+5-55/5)

[Thok]

π = (5-5/5)*arctan(5/5)

[Chuck]

φ = sqrt(5)×5/(5+5)+.5

[jadesmar]

Doesn't sqrt contain 4 letters? And.. arctan is 6 letters. These are both more than one letter.

I guess there could be a symbol for sqrt. But, I think arctan is still out given these rules.

[Chuck]

I could edit in the symbol for square root if necessary.

[Thok]

[Zag]

i without using sqrt:

(- 55/55) ^ .5

or, if you are not even allowing a decimal point:

(-5/5) ^ (5/(5+5))

[DejMar]

Great job, Zag in tackling i without using the decimal point.
Both the decimal point and the radical sign are permitted under the normal rules - thus the notation SQRT( ) is acceptable as a proxy for the radical sign. Functions, operations and other mathematical notations that use letters - to include those of a non-English alphabet - are not otherwise permitted.
Chuck's solution for φ (phi) and groza528's solution for i (the imaginary unit) are both valid. Thok's use of 'arctan' for the trigonometric function and 'zeta' for the Riemann zeta function are invalid solutions for π (pi) . But do not give up hope, there are still solutions for Archimedes constant that are still available.

[Thok]

[DejMar]

A solution for Archimedes' constant: Pi, π: (Spoiler follows)

π = (.5! / .5) ^{((5 + 5) / 5)}

Though it is not standard decimal notation, decimal notation without an integer part is an acceptable form of notation and is also used in the "Five Fives" puzzle. This permits the fractions 1/2 (.5) 11/20 (.55), etc. and even the recurring decimal 5/9 (.5...), etc. with using only one or two fives.
The factorial is one of the principal unary functions that is also used. In some variations of the "Five Fives" puzzle it is permitted to use other factorial-like functions (and in some, to use the Greek letter functions, though I intentionally excluded these in this puzzle) The n!! (double factorial), !n (subfactorial), and n# (primorial) allow many integers to be represented without any additional decimal digits than the fives.

[Thok]

[jadesmar]

[lostdummy]

Actually, we can get pi with less than five 'fives' under those rules.

Best I could get on short notice is 'pi' with only 3 fives.

*EDIT* I found solution for pi with only two fives, which means that now I have solution with any number of fives between two and infinity ;p

[DejMar]

lostdummy,

I am interested in your solution. Though it was not stated, multifactorials of higher than 5th degree should be disallowed. My personal position is that no distinct symbol can be used more than 5 times, i.e. ((5!)/5!!) - 5!! constitutes 5 distinct !, even though there is a separate double exclamation point character in Unicode.

For clarity, not grouping requirements, more than five sets of brackets may be used. [These are personal rules of mine in the "Five Fives" puzzles that I have adopted. With these restrictions I have found solutions for the integers 0 to over 1000.]

[lostdummy]

Well, for 'best' solution with only two fives, I dont use any symbol more than twice:

pi(2*5)= (-.5)!*(-.5)!

Any other combination with more fives than three can use solution with two fives and fact that (5-5)=0, so for example:

pi(4*5) = (-.5)!*(-.5)!+ (5-5)
pi(5*5) = (-.5)!*(-.5)!+ (5-5)*5
pi(6*5) = (-.5)!*(-.5)!+ (5-5)+ (5-5) etc

It gets more tricky for solution with three fives. My initial idea was pi=(-.5)!^(5-5!!!) , but then I found out that !!! does not start from 1 (in which case 5!!! would be 3), but downward from 5 (so 5!!!=10). So, I found alternate way with three fives (which amount to alternate ways to get 2 with two fives):

pi(3*5)= (-.5)!^(5!!!/5)
pi(3*5)= (-.5)!^(5#/5!!)
pi(3*5)= (-.5)!^sqrt(5!/5#)