[math-fun] maximum quadrilateral area
WDS> With given sides occurs when the quad is inscribable in (i.e. inscribed in) a circle (and this indeed does not depend on the order of the sides due to Brahmagupta's area formula). It is also interesting that the "max area happens when quad inscribable" theorem is an "absolute" theorem, i.e. it works in both euclidean and nonEuclidean geometries. And the fact the area then is independent of the side-ordering also is "absolute" (which should be obvious). (I'm recounting all this from memory, since I recall proving this theorem a long time ago. I was going to put it in a paper I never finished.) ---------------------------- So, is the solid angle formed by four dihedrals maximal when it inscribes in a circular cone? (Actually, I should still have the max constraints. But I think the cone part was way messy.) --rwg 3 Sep 2008: "Everybody" knows that, given any triangle, the sum of the angle tangents equals their product. This "Brahmagupts" into Given any quadrilateral, the sum of the angle tangents equals their product times the sum of the cotangents. Can't be new. But probably newer than Brahmagupta, since, even for trapezoids, it just says 0=0. --rwg Am I the only one who hasn't seen In[650]:= FullSimplify[Integrate[Log[1/z]^n, {z, 0, 1}], n \[Element] Integers && n > 0] Out[650]= n! until today? It's just a change of variables in Euler's integral, but why isn't it preferred, or even mentioned, in e.g., A&S? Dicey for complex n?
49 years ago, I used that very (math, not Mma) fact to prove (1/Gamma(s)) Integral_{0,1} ((1-x^r)/(1-x)) ln(1/x)^(s-1) dx = 1 + 1/2^s + . . . + 1/r^s for r a positive integer (and thus a way to interpolate between these partial Zeta sums, by letting r be a non-integer). In the limit this becomes Gamma(s) Zeta(s) = Integral_{0,1} (ln(1/x)^(s-1) / (1-x)) dx which I thought was really cool, until entering college in September and telling this to my freshman advisor (Henry McKean), upon which he reached over and picked up a Whittaker and Watson on his shelf, flipped through it, and showed me essentially the identical formula there. (I was devastated.) --Dan On 2013-08-12, at 11:07 PM, Bill Gosper wrote:
Am I the only one who hasn't seen
In[650]:= FullSimplify[Integrate[Log[1/z]^n, {z, 0, 1}], n \[Element] Integers && n > 0]
Out[650]= n!
until today?
[...] rwg>So, is the solid angle formed by four dihedrals maximal when it inscribes
in a circular cone? (Actually, I should still have the max constraints. But I think the cone part was way messy.) --rwg 3 Sep 2008: "Everybody" knows that, given any triangle, the sum of the angle tangents equals their product. This "Brahmagupts" into Given any quadrilateral, the sum of the angle tangents equals their product times the sum of the cotangents. Can't be new. But probably newer than Brahmagupta, since, even for trapezoids, it just says 0=0.
[...] maxSolidAngle[a_, b_, c_, d_] := 2*ArcCos[((Cos[d] + Cos[c] + Cos[b] + Cos[a])/(4*Cos[a/2]*Cos[b/2]* Cos[c/2]*Cos[d/2])) - Tan[a/2]*Tan[b/2]*Tan[c/2]*Tan[d/2]]
Browsing Carr's synopsis http://archive.org/stream/asynopsiselemen00carrgoog#page/n216/mode/2up to see if that's where Ramanujan got his Bernoulli notation, I found two more formulæ (appended) for the solid angle given the vertex angles: {ArcCos[-1+(1+Cos[a]+Cos[b]+Cos[c])^2/((1+Cos[a]) (1+Cos[b]) (1+Cos[c]))], 2 ArcCos[1/4 (1+Cos[a]+Cos[b]+Cos[c]) Sec[a/2] Sec[b/2] Sec[c/2]], 2 ArcSin[(Sqrt[-1-Cos[2 a]-Cos[2 b]+4 Cos[a] Cos[b] Cos[c]-Cos[2 c]] Sec[a/2] Sec[b/2] Sec[c/2])/(4 Sqrt[2])], 4 ArcTan[Sqrt[Tan[1/4 (a+b-c)] Tan[1/4 (a-b+c)] Tan[1/4 (-a+b+c)] Tan[1/4 (a+b+c)]]]} The equivalence of these gives FullSimplify fits, even for {a,b,c}=π/{2,3,4}. The last one is http://mathworld.wolfram.com/LHuiliersTheorem.html, misprinted in Carr as Llhuillier's by an obviously Welsh typesetter. Amazingly, "huilier" means oiler. (Or cruet.) --rwg
I'm afraid I missed the original mention of this. Because: What are the constraints on the four dihedrals? (Maybe the sum of their vertex angles?) --Dan On 2013-09-16, at 3:09 AM, Bill Gosper wrote:
rwg>So, is the solid angle formed by four dihedrals maximal when it inscribes
in a circular cone? (Actually, I should still have the max constraints. But I think the cone part was way messy.)
DanAsimov> I'm afraid I missed the original mention of this. Because: What are the constraints on the four dihedrals? (Maybe the sum of their vertex angles?) --Dan Sorry, my bad -- terminology. I've been saying dihedral and meaning vertex angle! On Mon, Sep 16, 2013 at 3:09 AM, Bill Gosper <billgosper@gmail.com> wrote:
[...] rwg>So, is the solid angle formed by four [vertex angle]s maximal when it inscribes
in a circular cone? (Actually, I should still have the max constraints.
But I think the cone part was way messy.) --rwg
[...]
maxSolidAngle[a_, b_, c_, d_] := 2*ArcCos[((Cos[d] + Cos[c] + Cos[b] + Cos[a])/(4*Cos[a/2]*Cos[b/2]* Cos[c/2]*Cos[d/2])) - Tan[a/2]*Tan[b/2]*Tan[c/2]*Tan[d/2]]
Browsing Carr's synopsis http://archive.org/stream/asynopsiselemen00carrgoog#page/n216/mode/2up to see if that's where Ramanujan got his Bernoulli notation, I found two more formulæ (appended) for the solid angle given the vertex angles: {ArcCos[-1+(1+Cos[a]+Cos[b]+Cos[c])^2/((1+Cos[a]) (1+Cos[b]) (1+Cos[c]))], 2 ArcCos[1/4 (1+Cos[a]+Cos[b]+Cos[c]) Sec[a/2] Sec[b/2] Sec[c/2]], 2 ArcSin[(Sqrt[-1-Cos[2 a]-Cos[2 b]+4 Cos[a] Cos[b] Cos[c]-Cos[2 c]] Sec[a/2] Sec[b/2] Sec[c/2])/(4 Sqrt[2])], 4 ArcTan[Sqrt[Tan[1/4 (a+b-c)] Tan[1/4 (a-b+c)] Tan[1/4 (-a+b+c)] Tan[1/4 (a+b+c)]]]}
The equivalence of these gives FullSimplify fits, even for {a,b,c}=π/{2,3,4}.
E.g., In[72]:= FullSimplify[% /. {a -> π/2, b -> π/3, c -> π/4}] Out[72]= {ArcCos[1/6 (4 + Sqrt[2])], 2 ArcCos[((3 + Sqrt[2]) Sin[π/8])/Sqrt[3]], 2 ArcSin[Sin[π/8]/Sqrt[3]], 4 ArcTan[Sqrt[Cot[11 π/48] Tan[π/48] Tan[5 π/48] Tan[7 π/48]]]} In[73]:= N[%] Out[73]= {0.445561, 0.445561, 0.445561, 0.445561} In[74]:= FullSimplify[Rest[%%] - First[%%]] Still running since yesterday. --rwg The last one is http://mathworld.wolfram.com/LHuiliersTheorem.html,
misprinted in Carr as Llhuillier's by an obviously Welsh typesetter. Amazingly, "huilier" means oiler. (Or cruet.) --rwg
Luckily, David Fowler's ancestors didn't drop the F. I didn't find any explicit listing of Bernoulli-like numbers or polynomials in Carr's, e.g., as Faulhaber polynomials or Taylor coefficients of tan. "This is a digital copy of a book that was preserved for generations on library shelves before it was carefully scanned by Google as part of a project to make the world’s books discoverable online."
The first page of the Preface to Part I is significantly obscured by at least two other apparently damaged pages folded over it.
participants (2)
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Bill Gosper -
Dan Asimov