Re: [math-fun] a^x-b^x=c, solve for x ?
I mean I haven't found a decently convergent series of exact terms, as in gosper.org/newsrope.pdf. Newton's method works great numerically, but soon bogs down using an expansion at x = 1 or 1 + 1/(50 Log[50] - 49 Log[49]). It feels like I'm missing something. Like there's a pFq for these things. —rwg On 2018-10-21 10:00, Allan Wechsler wrote:
I think I'm not following this discussion very well. Just stabbing around with a Haskell window gave me x = 1.14331628080... in just a couple of minutes. The function isn't doing anything exotic in this range, so I would think that a Newton iteration would converge on the answer very fast. What does "abysmal" convergence mean in this context?
On Sun, Oct 21, 2018 at 12:43 PM Henry Baker <hbaker1@pipeline.com> wrote:
Substituting x=log(y), we get
y^log(a)-y^log(b) = c
i.e., y^log(50)-y^log(49)=2
But this "polynomial" may not be any easier to solve.
We could approximate log(a)/log(b) with a continued fraction and solve a trinomial, but it might have a quite high degree.
For example, if we let y = z^49.33508819661897 in y^log(50)-y^log(49)=2,
we get z^193-z^192=2, which doesn't factor over Q, and Maxima can't solve it.
At 07:26 AM 10/21/2018, Bill Gosper wrote:
I usually suggest the techniques in gosper.org/newsrope.pdf , but everything I've tried for 50^x - 49^x = 2 converges abysmally.
—rwg
On 2018-10-19 12:41, Henry Bakerr wrote:
I can solve this equation with Newton's iterations, but I was curious if there might be an elegant closed form solution.
I tried fiddling with hyperbolic trig functions, but nothing seemed any simpler.
real a>b>0,c>0
E.g., a ~ 50, b ~ 49, c ~ 2 => x ~ 1.1438929
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Bill Gosper