February 2018 - mathfun@mailman.xmission.com

[math-fun] Integer power means -- the reverse problem
by Keith F. Lynch 10 Feb '18

10 Feb '18

I've been discussing when the integer power mean of two positive integers is an integer. When it's *not* an integer, it's still obvious what kind of number it is. The arithmetic or harmonic mean of two integers is always a rational number. The quadratic (2) or citardauq (-2) power mean of two integers is always a quadratic surd, i.e. the square root of a rational number. The cubic (3) or cibuc power mean of two integers is always a the cube root of a rational number. But what about the opposite question? Every number c between the positive integers a and b is the xth power mean of a and b for some number x. It's the unique non-zero solution to (a^x + b^x) / 2 = c^x. (Except when c is the geometric mean of a and b, in which case it *is* zero.) It's negative if c is less than the geometric mean of a and b, i.e. if c^2 < ab. Every x corresponds to a unique number between a and b, and it's monotonic, i.e. increasing x always increases the mean. I'm interested in the cases where c is an integer. The first non-integer example of such an x is (1^x + 4^x) / 2 = 3^x, for which the solution is approximately 2.261405723477886. What kind of number is x, in the general case? Obviously it's sometimes an integer. When it's not an integer, is it ever rational? Is it ever algebraic, and if so, of what order? Is it ever transcendental? Is it in this case? I eventually realized that the formula I found, c = ab/q, applies, not just to the quadratic and citardauq means, but to all power means where c is the xth mean and q is the -xth mean. That explains why the arithmetic-harmonic mean always equals the geometric mean. So do the quadratic-citardauq mean, the cubic-cibuc mean, etc. The arithmetic-harmonic mean is what you get when you separately take the arithmetic mean of two number and the harmonic mean of the same two numbers, then you repeat the process with those two means, then you repeat the process with those two means or means, etc., always getting two new numbers, which rapidly approach each other. Analogously with the better known arithmetic-geometric mean. All power means are, for lack of a better word, slideruleable. For any xth power mean, all numbers can be arranged in order on a line such that the xth power mean of a and b is always halfway between a and b. For instance for the arithmetic mean, the numbers are equally spaced, like on a ruler. For the geometric mean, a log scale is used. This is *not* the case with the arithmetic-geometric mean. Is there an official name for this useful property? (The term "number" in the above should always be read as "real number." It's way too early in the morning (10:42) for me to think about complex numbers, especially complex exponents.)

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[math-fun] thousands of useless formulas (2)
by françois mendzina essomba2 10 Feb '18

10 Feb '18

Some formulas... Formula 1 %pi=2^(n)*sqrt(15-sqrt(15)*sqrt(28*2^(-10)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^5-13))/sqrt(70); Approx : n=9 2^(9)*sqrt(15-sqrt(15)*sqrt(28*2^(-10)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^5-13))/sqrt(70); Formula 2 %pi=2^(n)*sqrt(6-sqrt(6)*sqrt(11*2^(-8)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^4-5))/sqrt(22); Approx : n=9 2^(9)*sqrt(6-sqrt(6)*sqrt(11*2^(-8)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^4-5))/sqrt(22); Formula 3 %pi=2^(n)*sqrt(3-sqrt(16*2^(-6)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^3-7))/2^(3/2); Approx : n=9 2^(9)*sqrt(3-sqrt(16*2^(-6)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^3-7))/2^(3/2); Formula 4 %pi=2^(n)*sqrt(3-sqrt(3)*sqrt(5*2^(-4)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^2-2))/sqrt(5); Approx : n=9 2^(9)*sqrt(3-sqrt(3)*sqrt(5*2^(-4)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^2-2))/sqrt(5); Formula 5 %pi=2^(n)*sqrt(30-sqrt(30)*sqrt(59*2^(-40)*((2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2)))))))))))^20-29))/sqrt(590); Approx : n=11 2^(11)*sqrt(30-sqrt(30)*sqrt(59*2^(-40)*((2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2)))))))))))^20-29))/sqrt(590); Formula 6 %pi=2^(n)*sqrt(15-sqrt(3)*sqrt(149*2^(-100)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^50-74))/sqrt(745); Approx : n=11 2^11*sqrt(15-sqrt(3)*sqrt(149*2^(-100)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))))^50-74))/sqrt(745); Formula 7 %pi=2^(n)*sqrt(150-sqrt(30)*sqrt(1499*2^(-1000)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^500-749))/(5*sqrt(2998)); Approx : n=11 2^(11)*sqrt(150-sqrt(30)*sqrt(1499*2^(-1000)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))))^500-749))/(5*sqrt(2998)); Formula 8 2^(n)*sqrt(sqrt(15)*sqrt(121-61*2^(-80)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))^40)-30*2^(-40)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))^20)/(11*sqrt(10)*2^(-20)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))^10); Approx : n=9 2^9*sqrt(sqrt(15)*sqrt(121-61*2^(-80)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^40)-30*2^(-40)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^20)/(11*sqrt(10)*2^(-20)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^10); Formula 9 %pi=2^(n)*(4*sqrt(10)*sqrt(sqrt(3)*sqrt(317-314*2^(-1/80)*sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^(1/80))-3))/sqrt(157); Formula 10 %pi=2^(n)*(4*sqrt(10)*sqrt(sqrt(3)*sqrt(326*2^(-1/80)*sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^(1/80)-323*2^(-1/40)*sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^(1/40))-3*2^(-1/80)*sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^(1/80)))/(sqrt(163)*2^(-1/160)*sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^(1/160)); Formula 11 %pi=2^(n)*(2*sqrt(5)*sqrt(sqrt(42*2^(-1/30)*sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^(1/30)-41*2^(-1/15)*sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^(1/15))-2^(-1/30)*sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^(1/30)))/(sqrt(7)*2^(-1/60)*sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))))^(1/60));

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[math-fun] thousands of useless formulas
by françois mendzina essomba2 09 Feb '18

09 Feb '18

1) a(n) = (2^n*sqrt((5762440*sqrt(20753571721*u(n)^198-13553934642*u(n)^99+4518487526)-830142868840*u(n)^99+271078692840)^(1/3)- (8584820*11^(2/3))/(5762440*sqrt(20753571721*u(n)^198-13553934642*u(n)^99+4518487526)-830142868840*u(n)^99+271078692840)^(1/3)+2950*11^(1/3)))/(11^(1/6)*sqrt(144061)) ; 2) (2^n*sqrt( (-306267858538119720*u(n)^(1/99)+1990548992520*sqrt(23673207321*u(n)^(2/99)-47605249842*u(n)^(1/99)+23933777126)+307942175442951720)^(1/3)- 15965632980/(-306267858538119720*u(n)^(1/99)+1990548992520*sqrt(23673207321*u(n)^(2/99)-47605249842*u(n)^(1/99)+23933777126)+307942175442951720)^(1/3)-64350))/( sqrt(51287)) ; 3) a(n)=(2^n*sqrt(- (16483125780*u(n)^(1/297))/(-328586152174659720*u(n)^(1/99)+2067396673320*sqrt(25262068046*u(n)^(2/99)-50796633642*u(n)^(1/99)+25536359601)+330372055793209320)^(1/3)+ (-328586152174659720*u(n)^(1/99)+2067396673320*sqrt(25262068046*u(n)^(2/99)-50796633642*u(n)^(1/99)+25536359601)+330372055793209320)^(1/3)/u(n)^(1/297)-66330))/( sqrt(53267)); for each of them u(n+1)=sqrt(2*u(n)+2)/2 u(2) = sqrt(2)/2 n ---------- infinity a(n) ------- Pi

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[math-fun] Draft of my February 2018 blog post
by James Propp 09 Feb '18

09 Feb '18

Hi, My new essay is called "Roasting a Dodo and Biking on Mars: The Magic of Dimensional Analysis" and as usual I plan to publish it on the 17th. If you have comments, please sign them, so that I can ask followup questions. Thanks, Jim Propp Title: Roasting a Dodo and Biking on Mars: The Magic of Dimensional Analysis Beginning: The first time I encountered dimensional analysis, I didn't recognize it for the magic wand it is; it seemed to be just a form of mathematical hygiene --- useful for avoiding mistakes, but not much else. To get a sense of what I mean by hygiene, consider a question that came up in my own household ... Read more: https://mathenchant.wordpress.com?p=2141&shareadraft=5a7dd23703363

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Re: [math-fun] Power-mean conjectures
by Keith F. Lynch 09 Feb '18

09 Feb '18

Earlier this evening I wrote: > The quadratic mean, q, is defined as: (a^2 + b^2) / 2 = q^2 > The citardauq mean, c, is defined as: (1/a^2 + 1/b^2)/2 = 1/c^2 > Do some algebra, and get c^2 = (a^2 b^2)/(q^2), which shows that c > can't be an integer unless q is an integer. No, wait, no it doesn't; > q could be a rational number or the square root of a rational number. > Back to the drawing board. Sigh. That was stupid. Since all numbers are positive, I can simply take the square root of both sides and get c = ab/q. And by construction, q can only be an integer or the square root of an integer. So I've proven that the citardauq mean of two positive integers can only be an integer when the quadratic mean is an integer. QED.

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Re: [math-fun] Power-mean conjectures
by Keith F. Lynch 09 Feb '18

09 Feb '18

"Christian Boyer" <cboyer(a)club-internet.fr> wrote: > About your conjecture 1: > "Euler and Legendre demonstrated that x^3 + y^3 = k * z^3 is > impossible with distinct integers, for k = 1, 2, 3, 4, 5." > About your conjecture 2: > "Legendre showed also that x^4 + y^4 = 2 * z^2 is impossible if x != y." Thanks. What about higher powers? This is smelling a lot like Fermat's Last Theorem, which is way above my pay grade. Gareth McCaughan <gareth.mccaughan(a)pobox.com> wrote: > Keith's conjecture 3 was: "No two positive integers have both a > geometric mean that's an integer and a quadratic mean that's an > integer." .... Thanks for the proof that I guessed right (for values of "guess" that consist of trying all pairs of distinct positive integers less than ten thousand, about 50 million pairs). I've since tested all power means from -5 through +5. It looks like there are no integer means for any power whose absolute value is more than 2. (Though admittedly I haven't tried powers of plus or minus six or more, or pairs of integers where either integer was more than 9999.) So it looks like there are just five kinds of power means that can give integers, the -2, the harmonic (-1), geometric (0), arithmetic (1), and quadratic (2). Is there a name for the -2 power mean? Until someone can give me something better, I'll call it the citardauq mean. Of those five possible means, there are 32 possible combinations of means that are and aren't integers, for each pair of positive integers. But apparently only 12 combinations happen. Here are how many I get for each combination 24981771 00000 none 24954440 00010 arithmetic only 8638 00011 arithmetic and quadratic 8638 00100 geometric only 14118 00110 geometric and arithmetic 4322 01000 harmonic only 11565 01010 harmonic and arithmetic 338 01011 harmonic, arithmetic, and quadratic 269 01100 harmonic and geometric 524 01110 harmonic, geometric, and arithmetic 325 10011 citardauq, arithmetic, and quadratic 53 11011 citardauq, harmonic, arithmetic, and quadratic The first example of the last has the following four means: 28, 35, 80, and 100. I'll leave it as an exercise for the reader to figure out what the integers these are the means of are. I had already noticed that the geometric and quadratic means are never both integers. That was my conjecture 3. I now notice that the same is true of the citardauq and geometric means; they never appear together. Only two other rules are necessary. No quadratic means without arithmetic means. That's easy to explain. Arithmetic means are integers if and only if the numbers they're the means of have the same parity (both odd or both even). Quadratic means also require the parity be the same, since squares share the parity of what they're the square of. The final rule is that there's never a citardauq mean unless there's a quadratic mean. The quadratic mean, q, is defined as: (a^2 + b^2) / 2 = q^2 The citardauq mean, c, is defined as: (1/a^2 + 1/b^2)/2 = 1/c^2 Do some algebra, and get c^2 = (a^2 b^2)/(q^2), which shows that c can't be an integer unless q is an integer. No, wait, no it doesn't; q could be a rational number or the square root of a rational number. Back to the drawing board. Sigh.

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[math-fun] Number Pi
by françois mendzina essomba2 08 Feb '18

08 Feb '18

Hello, two formulas for number Pi Formula 1 Pi=2^(n+1)*sqrt(30)*sqrt(sqrt(11*2^(16/15)-21*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots))))))))^(1/30))-(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots))))))))^(1/60))/(sqrt(11)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots))))))))^(1/120)) ; approximative value for n=8 ; 2^(8+1)*sqrt(30)*sqrt(sqrt(11*2^(16/15)-21*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^(1/30))-(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^(1/60))/(sqrt(11)*(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^(1/120)); Formula 2 a(n)=(2^(n)/sqrt(271))*sqrt(A(n)+B(n)-330); A(n)=378900*(u(n))^(1/45)/(-5743197000*(u(n))^(1/45)+7317000*sqrt(617102*(u(n))^(2/15)-1276266*(u(n))^(1/15)+660969)+5948721000); B(n)=(-5743197000*(u(n))^(1/45)+7317000*sqrt(617102*(u(n))^(2/15)-1276266*(u(n))^(1/15)+660969)+5948721000)/(u(n))^(1/45); u(n+1)=sqrt(2*u(n)+2)/2 ; u(2)= sqrt(2)/2 ; n----- infinity ; a(n) ---- Pi this formula becomes: a(n)=sqrt(-(378900*v(n)^(1/45))/(-5743197000*v(n)^(1/15)+7317000*sqrt(617102*v(n)^(2/15)-1276266*v(n)^(1/15)+660969)+5948721000)^(1/3)+ (-5743197000*v(n)^(1/15)+7317000*sqrt(617102*v(n)^(2/15)-1276266*v(n)^(1/15)+660969)+5948721000)^(1/3)/v(n)^(1/45)-330)/(2^(-n)*sqrt(271)); V(n+1)=sqrt(2*v(n)+2)/2 ; v(2)= sqrt(2)/2 ; n----- infinity ; a(n) ---- Pi Best regards

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[math-fun] Packing points in a lattice
by James Propp 07 Feb '18

07 Feb '18

I just posted a question at MathOverflow ( https://mathoverflow.net/questions/292407/packing-points-in-a-lattice) that may interest some of you: Let $L$ be the square or triangular lattice in the plane, with nearest neighbors having distance 1. Has anyone studied the problem of finding the maximum (okay, supremum) density achieved by a subset of $L$ in which no two points are closer together than distance $d$, for various values of $d$? In some ways this is more natural for the triangular lattice than for the square lattice, since in the off-lattice version of the problem the answer is a triangular lattice of side-length $d$. (Or, turning this around, the problem is more interesting on a square lattice than a triangular lattice because of incommensurability between the lattice $L$ and the subset of $L$ that achieves maximum density, rescaled, when $d$ is large.) As a concrete example, consider the square grid with $d=\sqrt{5}$. A tilted square lattice with density 1/5 avoids distances $<d$. Dually, we can cover $L$ with sets of the form $\{(i,j), (i-1,j), (i+1,j), (i,j-1), (i,j+1)\}$, none of which can contain more than a single point in common with a subset of the grid that avoids distances less than $\sqrt{5}$. So the maximum density is 1/5. It is not clear to me that for every $d$ the supremum is achieved by a periodic packing. Certainly the existence of Wang tiles suggests that a suitably broad version of the packing problem will run afoul of decideability issues. But approximate balls are very different from Wang tiles, so my instinct is that only periodic optima occur. Pointers to the literature would be appreciated. Also, it seems to me that a form of linear programming duality could be used to establish upper bounds on density, generalizing the argument I give above for $d=\sqrt{5}$ on the square lattice. Jim Propp

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[math-fun] "Was sind und was sollen die Zahlen?"
by James Propp 07 Feb '18

07 Feb '18

What does the title "Was sind und was sollen die Zahlen?" mean? I've seen it translated as "What are numbers and what should they be?". But I've also seen it translated as "What are numbers and what are they good for?", which is very different. (I only took one semester of German, but wouldn't "should be" be "sollen sein", rather than "sollen"?) Jim Propp

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[math-fun] Power-mean conjectures
by Keith F. Lynch 06 Feb '18

06 Feb '18

I've been looking at pairs of positive integers which have integer power means, including the harmonic, geometric, arithmetic, quadratic (RMS), cubic, and quartic means. Conjecture 1: No two positive integers have a cubic mean that's an integer. Conjecture 2: No two positive integers have a quartic mean that's an integer. Conjecture 3: No two positive integers have both a geometric mean that's an integer and a quadratic mean that's an integer. All three are true for all possible pairs of positive integers less than ten thousand. Are they true for all larger integers too? Other than that, all possible combinations appear to exist. For instance 5 and 45 have a harmonic mean of 9, a geometric mean of 15, and an arithmetic mean of 25. 4 and 28 have a harmonic mean of 7, an arithmetic mean of 16, and a quadratic mean of 20. And plenty of pairs have just 0, 1, or 2 means that are integers, in all possible combinations that don't violate my conjectures. Does anyone have a proof or counterexample for any of my conjectures? Thanks.

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