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Re: [math-fun] a^x-b^x=c, solve for x ?
by Bill Gosper 22 Oct '18

22 Oct '18

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(a)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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Re: [math-fun] a^x-b^x=c, solve for x ?
by Axel Vogt 21 Oct '18

21 Oct '18

t^r - t = c; subs(t= exp(y), y= x*ln(b), r=ln(a)/ln(b), %): simplify(%) assuming 0<x, 0<b; x x a - b = c For the example t^r - t - c is (almost) a line at the solution. > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] a^x-b^x=c, solve for x ? > > 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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[math-fun] a^x-b^x=c, solve for x ?
by Henry Baker 21 Oct '18

21 Oct '18

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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Re: [math-fun] a^x-b^x=c, solve for x ?
by Henry Baker 21 Oct '18

21 Oct '18

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. > >Ârwg > >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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Re: [math-fun] a^x-b^x=c, solve for x ?
by Bill Gosper 21 Oct '18

21 Oct '18

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 Baker 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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[math-fun] Twelve Octagon
by Ed Pegg Jr 17 Oct '18

17 Oct '18

There is a twelve octagon toroid. https://math.stackexchange.com/questions/2869725/ --Ed Pegg Jr

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Re: [math-fun] Two ellipse circumferences
by Bill Gosper 17 Oct '18

17 Oct '18

On 2018-10-17 11:20, Henry Baker wrote: > Please stick with ASCII, as it's impossible to read these UTF8 > characters (see below). GAA! What I sent was more legible than ASCII, and pastable into Mathematica to boot! (From which you could extract ASCII if you really wanted it.) Duck-sucking, dog-bleeping xmission.com totally vandalized it. If they allowed, I could have attached Gosper.org/screwxmission.pdf I think I'll stick to mathfuneavesdroppers. Shall I add your name? > > What about time around an elliptical orbit? Any closed form solutions > for -- e.g., approximations to #days from solstice? Yes! How could we possibly have gone this long without a universally recognized Kepler(a,b,t) function? —Bill > > At 11:09 AM 10/17/2018, Bill Gosper wrote: >>(Mail to some kids.) >>> Beginners may find it discouraging that the circumference of an ellipse requires >>> this unfamiliar EllipticE function, but it is actually well worth familiarizing! For >>> example, it provides the world's most rapidly convergent Ï€ formulas, and has fascinating properties. >>> Actually, there are infinitely many eccentricities where the circumference EllipticE is expressible in terms radicals and factorials, but they're factorials of fractions! >> >>You may already know that >> >>In[210]:= (1/2)! >> >>Out[210]= âˆšÏ€/2 >> >>but halves are the only known fractions whose factorials are familiar. >> >>Somewhat amazingly, the 1 Ã— 1/âˆš2 ellipse (bounding box 2Ã—âˆš2) has circumference 9 Ï€^(3/2)/(16 (3/4)!^2) + 32 (3/4)!^2/(9âˆšÏ€) ~ 5.4025755241907 (compared with the perimeter of the bounding box = 6.82842712474619.) >> >>But this is perhaps the nicest case. >> >>The circumference of a 1 by (1/ð“ + 1/âˆšð“)/âˆš2 ellipse is (where ð“ := (1+âˆš5)/2, the Golden Ratio) >> >>ArcLength[Circle[{,}, {(1/ð“ + 1/âˆšð“)/âˆš2, 1}]] == 9 Ï€^(3/2)/(10 âˆš2 5^(7/8) ð“^(1/4) (1/20)! (9/20)!) + 2 âˆš2 5^(3/8) (4 âˆš5 + 10 âˆšð“) (1/20)! (9/20)!/(9 ð“^(1/4) âˆšÏ€) ~ 6.26092807313208, 2Ï€-ish because >> >>In[255]:= N[(1/GoldenRatio + 1/âˆšGoldenRatio)/âˆš2] >> >>Out[255]= 0.992908994700242 >> >>That's rounder than an Indiana circle. >> >>—rwg

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[math-fun] Funny meromorphic function
by Dan Asimov 17 Oct '18

17 Oct '18

This formula f(z) = Product_{1 <= n < oo} (1 + 1/(z-n)^2 - 1/n^2) converges where this sum does: Sum_{1 <= n < oo} (1/(z-n)^2 - 1/n^2) = Sum (n^2 - (z-n)^2) / (n^2 (z-n)^2) = Sum (-z^2 + 2 n z) / (n^2 (z-n)^2) , which clearly converges for z ≠ n in Z+. Does anyone recognize this function, or is it similar to other known functions? —Dan

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Re: [math-fun] Two ellipse circumferences
by Henry Baker 17 Oct '18

17 Oct '18

Please stick with ASCII, as it's impossible to read these UTF8 characters (see below). What about time around an elliptical orbit? Any closed form solutions for -- e.g., approximations to #days from solstice? At 11:09 AM 10/17/2018, Bill Gosper wrote: >(Mail to some kids.) >> Beginners may find it discouraging that the circumference of an ellipse requires >> this unfamiliar EllipticE function, but it is actually well worth familiarizing! For >> example, it provides the world's most rapidly convergent Ï formulas, and has fascinating properties. >> Actually, there are infinitely many eccentricities where the circumference EllipticE is expressible in terms radicals and factorials, but they're factorials of fractions! > >You may already know that > >In[210]:= (1/2)! > >Out[210]= âÏ/2 > >but halves are the only known fractions whose factorials are familiar. > >Somewhat amazingly, the 1 Ã 1/â2 ellipse (bounding box 2Ãâ2) has circumference 9 Ï^(3/2)/(16 (3/4)!^2) + 32 (3/4)!^2/(9âÏ) ~ 5.4025755241907 (compared with the perimeter of the bounding box = 6.82842712474619.) > >But this is perhaps the nicest case. > >The circumference of a 1 by (1/ð + 1/âð)/â2 ellipse is (where ð := (1+â5)/2, the Golden Ratio) > >ArcLength[Circle[{,}, {(1/ð + 1/âð)/â2, 1}]] == 9 Ï^(3/2)/(10 â2 5^(7/8) ð^(1/4) (1/20)! (9/20)!) + 2 â2 5^(3/8) (4 â5 + 10 âð) (1/20)! (9/20)!/(9 ð^(1/4) âÏ) ~ 6.26092807313208, 2Ï-ish because > >In[255]:= N[(1/GoldenRatio + 1/âGoldenRatio)/â2] > >Out[255]= 0.992908994700242 > >That's rounder than an Indiana circle. > >Ârwg

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[math-fun] Two ellipse circumferences
by Bill Gosper 17 Oct '18

17 Oct '18

(Mail to some kids.) > Beginners may find it discouraging that the circumference of an ellipse > requires > this unfamiliar EllipticE function, but it is actually well worth > familiarizing! For > example, it provides the world's most rapidly convergent π formulas, and > has > fascinating properties. > Actually, there are infinitely many eccentricities where the circumference EllipticE is expressible in terms radicals and factorials, but they're factorials of fractions! You may already know that In[210]:= (1/2)! Out[210]= √π/2 but halves are the only known fractions whose factorials are familiar. Somewhat amazingly, the 1 × 1/√2 ellipse (bounding box 2×√2) has circumference 9 π^(3/2)/(16 (3/4)!^2) + 32 (3/4)!^2/(9√π) ~ 5.4025755241907 (compared with the perimeter of the bounding box = 6.82842712474619.) But this is perhaps the nicest case. The circumference of a 1 by (1/𝝓 + 1/√𝝓)/√2 ellipse is (where 𝝓 := (1+√5)/2, the Golden Ratio) ArcLength[Circle[{,}, {(1/𝝓 + 1/√𝝓)/√2, 1}]] == 9 π^(3/2)/(10 √2 5^(7/8) 𝝓^(1/4) (1/20)! (9/20)!) + 2 √2 5^(3/8) (4 √5 + 10 √𝝓) (1/20)! (9/20)!/(9 𝝓^(1/4) √π) ~ 6.26092807313208, 2π-ish because In[255]:= N[(1/GoldenRatio + 1/√GoldenRatio)/√2] Out[255]= 0.992908994700242 That's rounder than an Indiana circle. —rwg

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