OT: Interesting math problem...

Will · #1 September 22nd, 2004, 03:58 AM

Narf, the equation means you take the constant e (2.71828183...) to the power of one half times the natural logarithm of x, or log base e of x. It doesn't have to be e though, you can get away with something like 10. So, 10^(0.5 * log(x)). It works because of the laws of logarithms, since a^b=c is equivalent to log(a)(c)=b. Then, it is simple to reconstruct as a^(log(a)(c))=c. Square root is simply a number to the one half power, so 10^(0.5 * log(x)) is the sqrt(x).

Either way, you're left with pretty much imprecise methods. I remember they briefly showed how to do it in high school, and that's probably an algorithm like the one Fyron pointed to, and it would be written in an equation as the Riemann sum of varying powers of 10, which would be a very nasty looking thing. Jack's method is easier to write, so I would go with that one. To write it in terms of an equation for square root, it would be something like:

lim(x->&#8734

R(x,N) = sqrt(N)

--edit: ^ that up there is supposed to be the infinity symbol.

David E. Gervais · #2 September 22nd, 2004, 08:37 AM

Wow, math games,... What's the square root of PI?

Cheers!

geoschmo · #3 September 22nd, 2004, 11:03 AM

At my high school they taught us a way to calculate square roots. It wasn't a simple equation. More of a process that resembled long division. I can't remember it anymore though. It may have been a form of this rod thing Fyron linked to just by a different name. But it didn't look exactly like that as I recall.

narf poit chez BOOM · #4 September 22nd, 2004, 02:06 PM

Quote:

Will said:
Narf, the equation means you take the constant e (2.71828183...) to the power of one half times the natural logarithm of x, or log base e of x. It doesn't have to be e though, you can get away with something like 10. So, 10^(0.5 * log(x)). It works because of the laws of logarithms, since a^b=c is equivalent to log(a)(c)=b. Then, it is simple to reconstruct as a^(log(a)(c))=c. Square root is simply a number to the one half power, so 10^(0.5 * log(x)) is the sqrt(x).

Either way, you're left with pretty much imprecise methods. I remember they briefly showed how to do it in high school, and that's probably an algorithm like the one Fyron pointed to, and it would be written in an equation as the Riemann sum of varying powers of 10, which would be a very nasty looking thing. Jack's method is easier to write, so I would go with that one. To write it in terms of an equation for square root, it would be something like:

lim(x->&#8734

R(x,N) = sqrt(N)

--edit: ^ that up there is supposed to be the infinity symbol.

Ok, sure. I'll pretend I understand all of that, and you can have a nice, shiney medal.

/me gives Will a nice, shiney medal.