[math-fun] Are we simple, yet?
I mentioned f(nth convergent(√3)) = 1/56 (-40 + 2^(3 n/2) (13 + 12 Sqrt[2] + (-1)^n (13 - 12 Sqrt[2]))) where f is Newman's rational counter. Why didn't I say f(1 + (-2 (1 - Sqrt[3])^(-1 + n) + 2 (1 + Sqrt[3])^(-1 + n))/(-(1 - Sqrt[3])^n + (1 + Sqrt[3])^n)) ? Because instead of this formula, I got the puzzling expression In[83]:= FullSimplify[ FindSequenceFunction[Convergents[Sqrt[3], 22]]@n, n \[Element] Integers] Out[83]= (2 Sqrt[3] (-2 - Sqrt[3])^n QPochhammer[2 - Sqrt[3], -2 + Sqrt[3], n])/ ((-1 + (-2 - Sqrt[3])^n) QPochhammer[-1, -2 + Sqrt[3], n]) (Strangely, the supposedly equivalent FullSimplify[FindSequenceFunction[Convergents[Sqrt[3], 22], n], n \[Element] Integers] simply gives up.) OK, let's cut it a break and do the numerators and denoms separately: In[84]:= FullSimplify[ FindSequenceFunction[Numerator /@ #, n]/ FindSequenceFunction[Denominator /@ #, n] &@ Convergents[Sqrt[3], 22], n \[Element] Integers] Out[84]= (3 (2 + Sqrt[ 3]) ((2 - Sqrt[3])^( n/2) (724 Sqrt[2 + Sqrt[3]] + 418 Sqrt[ 3 (2 + Sqrt[3])] + (989 + 571 Sqrt[3]) (-2 + (-1)^n (2 + Sqrt[2]))) + (2 + Sqrt[3])^( n/2) (1978 + 1142 Sqrt[3] + 724 Sqrt[2 + Sqrt[3]] + 418 Sqrt[3 (2 + Sqrt[3])] - (-1)^ n Root[16 + 31648 #1 + 7824944 #1^2 - 7912 #1^3 + #1^4 &, 3])))/((2 + Sqrt[3])^( n/2) ((2 + Sqrt[2]) (6393 + 3691 Sqrt[3]) + (-1)^ n Root[144 + 613728 #1 + 326963664 #1^2 + 51144 #1^3 + #1^4 &, 2]) - 4 I^n (2 - Sqrt[3])^( n/2) (Cos[(n \[Pi])/2] Root[9 - 81740892 #1^2 + #1^4 &, 4] + I (6393 + 3691 Sqrt[3]) Sin[(n \[Pi])/2])) Ingratitude. BtW, In[87]:= Convergents[1/(Sqrt[3] - 1), 9] Out[87]= {1, 3/2, 4/3, 11/8, 15/11, 41/30, 56/41, 153/112, 209/153} I didn't know that In[88]:= NestList[1 + 1/2/# &, 1, 8] Out[88]= {1, 3/2, 4/3, 11/8, 15/11, 41/30, 56/41, 153/112, 209/153} I'm a little surprised that neither this formula nor the first one above requires any odd-even machinery, since the continued fraction of √3 has period 2: 1 1 2 1 2 1 2 ... --rwg
First, an irrelevant puzzle: Where is the "full-size, 170k" picture of "Warren's" Libyan meteorite? http://www.sciencedirect.com/science/article/pii/S0012821X13004998 Spoiler at bottom. On Wed, Oct 9, 2013 at 2:44 AM, Bill Gosper <billgosper@gmail.com> wrote:
I mentioned f(nth convergent(√3)) = 1/56 (-40 + 2^(3 n/2) (13 + 12 Sqrt[2] + (-1)^n (13 - 12 Sqrt[2]))) where f is Newman's rational counter. Why didn't I say f(1 + (-2 (1 - Sqrt[3])^(-1 + n) + 2 (1 + Sqrt[3])^(-1 + n))/(-(1 - Sqrt[3])^n + (1 + Sqrt[3])^n)) ? Because instead of this formula, I got the puzzling expression In[83]:= FullSimplify[ FindSequenceFunction[Convergents[Sqrt[3], 22]]@n, n \[Element] Integers]
Out[83]= (2 Sqrt[3] (-2 - Sqrt[3])^n QPochhammer[2 - Sqrt[3], -2 + Sqrt[3], n])/ ((-1 + (-2 - Sqrt[3])^n) QPochhammer[-1, -2 + Sqrt[3], n])
(Strangely, the supposedly equivalent FullSimplify[FindSequenceFunction[Convergents[Sqrt[3], 22], n], n \[Element] Integers] simply gives up.)
[...]
Simplifying the QPochhammer expression down to the powers of surds requires (((-5 - 3 Sqrt[3])^(1 + n) - (1 - Sqrt[3])^(1 + n)) ((-5 - 3 Sqrt[3])^ n - (1 + Sqrt[3])^n))/ (((-5 - 3 Sqrt[3])^n - (1 - Sqrt[3])^n) ((-5 - 3 Sqrt[3])^(1 + n) - (1 + Sqrt[3])^(1 + n))) -> (1 + (-2 + Sqrt[3])^(1 + n))/(1 + (-2 + Sqrt[3])^n) (integer n). This seems pretty hard. NeilB did it with undetermined "coefficients": Suppose the answer is (a+b s^n)/(c+s^n). Cross multiply. Eliminate a, b, and c from the special cases n=1,2,3,4. Solve for s. Now solve for a,b,c. Cross multiply and verify for general integer n. How long before Mathematica, e.g., can do these? --rwg http://ars.els-cdn.com/content/image/1-s2.0-S0012821X13004998-gr001.jpg
On Wed, Oct 9, 2013 at 5:33 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Wed, Oct 9, 2013 at 2:44 AM, Bill Gosper <billgosper@gmail.com> wrote:
I mentioned f(nth convergent(√3)) = 1/56 (-40 + 2^(3 n/2) (13 + 12 Sqrt[2] + (-1)^n (13 - 12 Sqrt[2]))) where f is Newman's rational counter. Why didn't I say f(1 + (-2 (1 - Sqrt[3])^(-1 + n) + 2 (1 + Sqrt[3])^(-1 + n))/(-(1 - Sqrt[3])^n + (1 + Sqrt[3])^n)) ?
Or better yet, Sqrt[3] (-1 - 2/(-1 + (-2 + Sqrt[3])^n)) .
Because instead of this formula, I got the puzzling expression
In[83]:= FullSimplify[ FindSequenceFunction[Convergents[Sqrt[3], 22]]@n, n \[Element] Integers]
Out[83]= (2 Sqrt[3] (-2 - Sqrt[3])^n QPochhammer[2 - Sqrt[3], -2 + Sqrt[3], n])/ ((-1 + (-2 - Sqrt[3])^n) QPochhammer[-1, -2 + Sqrt[3], n])
(Strangely, the supposedly equivalent FullSimplify[FindSequenceFunction[Convergents[Sqrt[3], 22], n], n \[Element] Integers] simply gives up.)
[...]
Simplifying the QPochhammer expression down to the powers of surds requires
(((-5 - 3 Sqrt[3])^(1 + n) - (1 - Sqrt[3])^(1 + n)) ((-5 - 3 Sqrt[3])^ n - (1 + Sqrt[3])^n))/ (((-5 - 3 Sqrt[3])^n - (1 - Sqrt[3])^n) ((-5 - 3 Sqrt[3])^(1 + n) - (1 + Sqrt[3])^(1 + n)))
-> (1 + (-2 + Sqrt[3])^(1 + n))/(1 + (-2 + Sqrt[3])^n) (integer n).
But only if you foolishly use DiscreteRatio to poof the Pochhammers.
This seems pretty hard. NeilB did it with undetermined "coefficients": Suppose the answer is (a+b s^n)/(c+s^n). Cross multiply. Eliminate a, b, and c from the special cases n=1,2,3,4. Solve for s. Now solve for a,b,c. Cross multiply and verify for general integer n. How long before Mathematica, e.g., can do these? --rwg
Years ago, I wrote for WRI a QFunctionExpand, which they only partially installed. Again, the problem: Out[314]= (2 Sqrt[3] (-2 - Sqrt[3])^n QPochhammer[2 - Sqrt[3], -2 + Sqrt[3], n])/ ((-1 + (-2 - Sqrt[3])^n) QPochhammer[-1, -2 + Sqrt[3], n]) -> Sqrt[3] (-1 - 2/(-1 + (-2 + Sqrt[3])^n)) In[315]:= QFunctionExpand[%] Out[315]= (Sqrt[3] (-2 - Sqrt[3])^ n (1 + (-2 + Sqrt[3])^n))/(-1 + (-2 - Sqrt[3])^n) -> Sqrt[3] (-1 - 2/(-1 + (-2 + Sqrt[3])^n)) In[316]:= # -> 1/FullSimplify[1/#] &[-2 - Sqrt[3]] Out[316]= -2 - Sqrt[3] -> 1/(-2 + Sqrt[3]) In[317]:= FullSimplify[%% /. %, n \[Element] Integers] Out[317]= Sqrt[3] (-1 - 2/(-1 + (-2 + Sqrt[3])^n)) -> Sqrt[3] (-1 - 2/(-1 + (-2 + Sqrt[3])^n)) Let's hope WRI installs more of it. --rwg
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Bill Gosper