OT: Interesting math problem...

[narf poit chez BOOM]

Thanks Jack. But, still not an equation.

Kamog, what?

[Instar]

Its a system of equations, linear algebra might get it done

[Will]

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]

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

Cheers!

[geoschmo]

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.

[Timstone]

douglas said:

Took me about 5 minutes
Code:

---

              9 8 8 9

       _______________

1 1 5 | 1 1 3 7 2 3 5 

        1 0 3 5

        ========

          1 0 2 2

            9 2 0 

          ========

            1 0 2 3

              9 2 0

            ========

              1 0 3 5

              1 0 3 5

              ========

                    0

---

Okay, I'm an mechanical engineer and I really don't understand this sollution. How can 1023 minus 920 be 1035??????
Downright stupid if you ask me. But maybe I don't understand how it is put down. Please enlighten me.

[geoschmo]

Will said:
... there is also a sort of mathematical intuition there, that there already is a trivial solution for zeros, and a non-trivial solution most likely will not have the zeros. But, overall, I was just looking to solve it quickly, elegantly, and I determined that rather than proving V != 0, and doing most of the problem twice, I would just ignore that possibility, and go with the more "interesting" one. And, usually, that works

Ok. I guess it is some assumption grounded in years of number theory and probablility or something. Lacking that I went more towards the common sense, brute force method of substition. It was not inherantly obvious to me that because the trivial answer had zeros the non-trivial answer could not. I also don't see how you could make the assumption without proving or disproving posibilities that there isn't more then one correct answer.

[geoschmo]

Timstone said:
Okay, I'm an mechanical engineer and I really don't understand this sollution. How can 1023 minus 920 be 1035??????
Downright stupid if you ask me. But maybe I don't understand how it is put down. Please enlighten me.

Timstone, you've been using computers so long you've forgotten how to do long division.

1023 minus 920 is not 1035, it's 103. The 5 is brought down from the original number so you can do the next step in the process.

[Timstone]

Geo:
Yup, I discovered that too. But still the whole working of this sum eludes me. I can't grasp it. I must say I have only taken a quick look at it. I don't feel much for a thorough study of the phenomena.

[narf poit chez BOOM]

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.