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June 25[edit]

Frey curves and Integer Equations[edit]

Could somewhat explain the connection between Frey curves and equations of integers in a somewhat less jargon filled manner than: Frey curve#Application of the epsilon conjecture to Fermat.27s Last Theorem.

I'd like to clarify where the discriminant comes in and how the Modularity theorem allows one to map such equations onto other forms.

Feel free to improve the noted article in the process. Dragons flight (talk) 03:37, 25 June 2008 (UTC)[reply]

The article is on the epsilon conjecture, and that section does a pretty good job of explaining the history of the epsilon conjecture. The point of that section of the article is just that Frey curves are so weird that (now read next section) they do not exist. JackSchmidt (talk) 16:36, 25 June 2008 (UTC)[reply]

A Frey curve is simply the defined form y^2 = x(x-a^n)(x+b^n), which of course exists. One then somehow applies the algebra of curves to arrive at the conclusion that a^n + b^n can not equal c^n for any rational c. I'd like to see the process of reaching that conclusion explained in a way that doesn't require graduate study in number theory. In particular, it would be nice to define terms like conductor and discriminant in this context. Dragons flight (talk) 17:31, 25 June 2008 (UTC)[reply]

I think you misunderstood. A Frey curve is only defined given a solution of a^p + b^p = c^p, for on odd prime p (not for general n, and not for general a,b). The properties discovered by Frey depend essentially on this. I think if you want to understand the proof of Fermat's last theorem, you probably do require some graduate study in number theory. If you only want to see how algebraic curves can be used to answer elementary questions in number theory, then I recommend Murty and Ram's Problems in Algebraic Number Theory, which has a few examples of how projective curves relate to integer equations. This would also be a reasonable beginning at graduate study of number theory. If you are just interested in the development of modern number theory, then Edwards's book on a historical introduction to modern number theory is quite good, but very detailed. You might like [1][2] and Discriminant. In the number field case, the conductor measures a departure from being the maximal order, and the discriminant measures which ideals can ramify (who you've taken n'th roots of). Generically, they just measure which primes are interesting for your particular domain, and the structure of those two sections says that for a Frey curve, no prime is interesting, which means it must be the nicest kind of curve, in particular, it has genus 0. JackSchmidt (talk) 03:59, 26 June 2008 (UTC)[reply]

Logarithm of a sum?[edit]

Can I somehow take the logarithm of a sum?

For example, suppose I have $e^{ax+b}=y$ , and I know a, b and y, and want to figure out x. I do this with $ax+b=\ln y$ , making $x={\frac {\ln y-b}{a}}$ .

But suppose I have $e^{ax}+e^{bx}=y$ , and I again know a, b and y, and want to figure out x. Can this be done? JIP | Talk 17:05, 25 June 2008 (UTC)[reply]

Do you have use of a computer? otherwise stuck?87.102.86.73 (talk) 19:42, 25 June 2008 (UTC)[reply]

I don't think that there is an exact form for the solution to that equation, at least not in terms of elementary functions. As far as I know, there is no simple formula for taking the logarithm of a sum. ln(a + b) = ln(a) + ln(1 + b/a) for all real numbers a and b whenever the equation is defined, but this doesn't solve your problem. 76.238.8.4 (talk) 19:48, 25 June 2008 (UTC)[reply]

Should have found one solution x I could certainly talk about all the other solutions that satisfy the same equation, should you wish to go down that route.. Still no solution though.87.102.86.73 (talk) 19:54, 25 June 2008 (UTC)[reply]

Quick answer - Not easily.
Hard answer - According to Maple (a powerful maths software package) your answer will be of the form

{\frac {\gamma }{b}}

, where

{\gamma }

is a root of

y-e^{z}-e^{\frac {az}{b}}=0

, which doesn't help much. Basically it could be anything, possibly complex, and will be very difficult to find unless a=b, in which case the answer is fairly trivial. Rambo's Revenge (talk) 20:06, 25 June 2008 (UTC)[reply]

Incidentally, you can get an analytic form if you happen to also know

e^{ax}-e^{bx}

, which I suppose you might depending on the context. In general though, there is no simple analytic form except in special cases (e.g. a = 2*b). Dragons flight (talk) 20:22, 25 June 2008 (UTC)[reply]

Question: Would it not be possible to treat the answer as a solution to $f(e^{x})={e^{x}}^{a}+{e^{x}}^{b}-y=0$ ? Then find the derivative with respect to $e^{x}$ and use the Newton-Raphson iteration to find $e^{x}$ (not an analytic solution but the OP didn't specify)? Presumably the OP is allowed a calculator since logs have to be calculated. Zain Ebrahim (talk) 20:43, 25 June 2008 (UTC)[reply]

Yes, you could use Newton's method to approximately determine e^x, from which you could find x. Anyway, if a and b are integers, make the substitution u = e^x and solve the polynomial (or rational expression) u^a + u^b = y (however, this equation can't always be solved exactly, either). Then ln u = x. As others have said, though, there is no general analytic solution to the original equation. As an afterthought, if a is an integral multiple of b, or vise versa, then you could make a substitution like u = e^bx instead. 76.238.8.4 (talk) 20:54, 25 June 2008 (UTC)[reply]

Given the equation $e^{ax}+e^{bx}=c$ and wanting to solve for x given a,b,c. Expand the exponentials into power series: $2-c+\sum _{k=1}^{\infty }{\frac {a^{k}+b^{k}}{k!}}x^{k}=0.$ Truncate at some degree n and solve the polynomial equation $2-c+\sum _{k=1}^{n}{\frac {a^{k}+b^{k}}{k!}}x^{k}=0$ numerically using the Durand-Kerner method. Bo Jacoby (talk) 21:56, 27 June 2008 (UTC).x=y+p+i i=29[reply]