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 [quote="Trojan Horse"][quote="Zag"]For anyone wanting to solve it with trig functions, think of right triangles, and the Indian chief SOHCAHTOA.[/quote] Oh, duh... [spoiler](sec(arctan 2))^2[/spoiler] If you hadn't given us that hint, I don't think I ever would've gotten it. (Perhaps I should turn in my math Ph.D.)[/quote]
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bonanova
Posted: Mon Mar 08, 2010 7:20 am    Post subject: 1

I hadn't, either. But I found this ...

http://en.wikipedia.org/wiki/Binary_logarithm

Different solution from same person: cosh(arcsinh(2))^2
Trojan Horse
Posted: Mon Mar 08, 2010 4:16 am    Post subject: 0

I'm assuming "lb" is not an abbreviation for "pound". I've never seen that symbol used like this before.
bonanova
Posted: Mon Mar 08, 2010 2:09 am    Post subject: -1

Suggested by a colleague: lb(2/(lb(sqrt(sqrt(sqrt(sqrt(2)))))))
I had not heard of the lb notation, but he assures it's of common use in information theory.
Zag
Posted: Thu Feb 04, 2010 12:43 pm    Post subject: -2

PuzzleScot wrote:

But Trojan Horse's answer is the best I've seen yet! Beautiful. Well done.

Oh sure. I had it is before he did, and he got it from my hint. ~pouts~

I assume it was the "expected" answer, since, once you draw the right triangle with sides of 2 and 1 it practically jumps out at you. I started trying to come up with DevJoe's answer (that is, by drawing a triangle with an angle of 2 radians) but couldn't figure out how to do it easily.
PuzzleScot
Posted: Thu Feb 04, 2010 10:09 am    Post subject: -3

Before the ammendum was added regarding variables, I submitted:

5 = sum(n<=2,n € Z) n^2

[Sum of square of integers <=2. ie, 1^2 + 2^2 = 5]

DevJoe told me he'd submitted this:

Code:
arccos(-sin(arcsec(2)))/arccsc(2)

arcsec(2)=pi/3
sin(pi/3)=sqrt(3)/2
arccos(-sqrt(3)/2)=5pi/6
arccsc(2)=pi/6

So this evaluates to (5pi/6)/(pi/6) = 5.

But Trojan Horse's answer is the best I've seen yet! Beautiful. Well done.
Zag
Posted: Mon Feb 01, 2010 9:34 pm    Post subject: -4

Yep. That's what I had, though, in order to be original, I also submitted this:

(sec(arctan(2!!!!!!!!!!)) ^ (2!!!!!!!!!!!!!!!)

Ph.D. in math, eh? I only managed a BSEE, and that was 26 years ago.
Trojan Horse
Posted: Mon Feb 01, 2010 9:23 pm    Post subject: -5

Zag wrote:
For anyone wanting to solve it with trig functions, think of right triangles, and the Indian chief SOHCAHTOA.

Oh, duh...

(sec(arctan 2))^2

If you hadn't given us that hint, I don't think I ever would've gotten it. (Perhaps I should turn in my math Ph.D.)
Chuck
Posted: Sun Jan 31, 2010 10:24 pm    Post subject: -6

In shoe sizes an underline means add ½ to the number so it would be 2 × 2. The underline is a commonly used notation.
milkshake
Posted: Sun Jan 31, 2010 6:24 pm    Post subject: -7

Does T 2 + 2 count? ^^

(Edit: That's T sub 2. It didn't show up right at first.)
Antrax
Posted: Sun Jan 31, 2010 6:47 am    Post subject: -8

I really like lostdummy's answer. In C you could simply do: sqrt('2' / 2) which is similar but feels less cheaty.
JDTAY
Posted: Sun Jan 31, 2010 3:02 am    Post subject: -9

I have an answer, but it involves number 1984.
Zag
Posted: Sun Jan 31, 2010 2:37 am    Post subject: -10

You should submit that. It might be original, unlike mine.

For anyone wanting to solve it with trig functions, think of right triangles, and the Indian chief SOHCAHTOA.
lostdummy
Posted: Sun Jan 31, 2010 2:32 am    Post subject: -11

I'm sure this is not intended answer, but when I saw "hyperbolic arc-tangent..." I couldn't resist:

StrToInt( BinToStr(2) ) /2

this use functions that I consider more "common" that above mentioned hyper-arc-xyz ;)
Zag
Posted: Fri Jan 29, 2010 6:19 pm    Post subject: -12

This is from the IBM Research Puzzle site, the challenge for January 2010. I rather liked it, primarily because I found it challenging but I managed to solve it.

Present a computation whose result is 5, being a composition of commonly used mathematical functions and field operators (anything from simple addition to hyperbolic arc-tangent functions will do), but using only two constants, both of them 2.

It is too easy to do it using round, floor, or ceiling functions, so we do not allow them.

Update 1/11: You can only use 2 instances of the constant 2, so solutions like (2+2+2/2) is illegal. You can not use variables so (x+x+x+x+x)/x is not allowed. You can use square root and power, but squaring will cost you one "2" and other constant powers are not allowed.