Re: [math-fun] How I miss Mizan Rahman, or, what the heck is this?
George Andrews got it from Bailey's ₂Ψ₂ transformation (rather than finding it scrawled on the wall of an ancient tomb.) Merely with elementary manipulations, starting with partial fractions, I reduced it to n q Sum[---------------, {n, -∞, ∞}] == 1 + 2 n 2 (1 - q ) 2 2 4 QPochhammer[q , q ] -------------------- 2 4 QPochhammer[q, q ] which is presumably known, since it's a simple q-extension of In[1653]:= Sum[(2 n - 1)^-2, {n, -∞, ∞}] Out[1653]= π^2/4. Exercises: Evaluate 3 n + 1 q Sum[---------------, {n, -∞, ∞}] 1 + 2 n 2 (1 - q ) and n q Sum[------------, {n, -∞, ∞}] 1 + 2 n 1 - q Spoilers: 2 2 4 QPochhammer[q , q ] -------------------- 2 4 QPochhammer[q, q ] and 0 (off the unit circle). Problem: n integer, t real, L=Limit[sin nt,n->∞]. Then L=0 or it doesn't exist? --rwg On Sat, Jul 2, 2016 at 11:19 AM, Bill Gosper <billgosper@gmail.com> wrote:
Out[1131]= k q Sum -------------------------------, 1 - 2 k 2 1 + 2 k 2 (1 - q ) (1 - q ) k,-∞,∞
= 1 2 4 (1 + q) QFactorial[-(-), q ] 2 ----------------------------- 2 4 (1 - q )
q-Pochhammers in the sum but not the summand? "q-Mittag Leffler"?
It q-retracts to 16 Sum[-----------, {k,-∞,∞}] == 2 2 (1 - 4 k ) 2 2 π
Mathematica is dumbstruck. For minutes on end. Are these actually common?
Test for q=1/2: In[1132]:= N[Activate[%1131/.q->.5]] Out[1132]= True
Term ratio: In[1180]:= DiscreteRatio[%1131[[1,1]],k] Out[1180]= (q (1-q^(1-2 k))^2 (1-q^(1+2 k))^2)/((1-q^(-1-2 k))^2 (1-q^(3+2 k))^2)
Nine terms of the sum give 27 term agreement with the product, despite the lack of q^k²:
In[1189]:= Series[%1131/.\[Infinity]->9/.-\[Infinity]->-9//Activate,{q,0,29}] Out[1189]= 1+5 q+12 q^2+24 q^3+44 q^4+68 q^5+102 q^6+150 q^7+202 q^8+270 q^9+354 q^10+446 q^11+561 q^12+693 q^13+838 q^14+1002 q^15+1195 q^16+1407 q^17+1638 q^18+1906 q^19+2179 q^20+2491 q^21+2842 q^22+3202 q^23+3610 q^24+4042 q^25+4504 q^26+5008 q^27+5549 q^28+6114 q^29+O[q]^30==1+5 q+12 q^2+24 q^3+44 q^4+68 q^5+102 q^6+150 q^7+202 q^8+270 q^9+354 q^10+446 q^11+561 q^12+693 q^13+838 q^14+1002 q^15+1195 q^16+1407 q^17+1638 q^18+1906 q^19+2179 q^20+2491 q^21+2842 q^22+3202 q^23+3610 q^24+4042 q^25+4504 q^26+5008 q^27+5550 q^28+6114 q^29+O[q]^30
In[1190]:= Subtract@@% Out[1190]= -q^28+O[q]^30
I got it by mildly abusing the (bilateral) "q-Dixon's theorem" in http://math.sun.ac.za/~hproding/pdffiles/Triples-long.pdf which is not the (unilateral) q-Dixon's I get from path invariant matrices. --rwg
Dear all, The right side is \psi(q)^4, where \psi(q) generates the triangular numbers, so the identity gives the number of reps of a number as the sum of 4 triangles, which must be well-known. Mike Hirschhorn. ________________________________ From: mathfuneavesdroppers@googlegroups.com <mathfuneavesdroppers@googlegroups.com> on behalf of Bill Gosper <billgosper@gmail.com> Sent: 07 July 2016 13:08:14 To: math-fun@mailman.xmission.com Subject: Re: How I miss Mizan Rahman, or, what the heck is this? George Andrews got it from Bailey's ??? transformation (rather than finding it scrawled on the wall of an ancient tomb.) Merely with elementary manipulations, starting with partial fractions, I reduced it to n q Sum[---------------, {n, -?, ?}] == 1 + 2 n 2 (1 - q ) 2 2 4 QPochhammer[q , q ] -------------------- 2 4 QPochhammer[q, q ] which is presumably known, since it's a simple q-extension of In[1653]:= Sum[(2 n - 1)^-2, {n, -?, ?}] Out[1653]= ?^2/4. Exercises: Evaluate 3 n + 1 q Sum[---------------, {n, -?, ?}] 1 + 2 n 2 (1 - q ) and n q Sum[------------, {n, -?, ?}] 1 + 2 n 1 - q Spoilers: 2 2 4 QPochhammer[q , q ] -------------------- 2 4 QPochhammer[q, q ] and 0 (off the unit circle). Problem: n integer, t real, L=Limit[sin nt,n->?]. Then L=0 or it doesn't exist? --rwg On Sat, Jul 2, 2016 at 11:19 AM, Bill Gosper <billgosper@gmail.com<mailto:billgosper@gmail.com>> wrote: Out[1131]= k q Sum -------------------------------, 1 - 2 k 2 1 + 2 k 2 (1 - q ) (1 - q ) k,-?,? = 1 2 4 (1 + q) QFactorial[-(-), q ] 2 ----------------------------- 2 4 (1 - q ) q-Pochhammers in the sum but not the summand? "q-Mittag Leffler"? It q-retracts to 16 Sum[-----------, {k,-?,?}] == 2 2 (1 - 4 k ) 2 2 ? Mathematica is dumbstruck. For minutes on end. Are these actually common? Test for q=1/2: In[1132]:= N[Activate[%1131/.q->.5]] Out[1132]= True Term ratio: In[1180]:= DiscreteRatio[%1131[[1,1]],k] Out[1180]= (q (1-q^(1-2 k))^2 (1-q^(1+2 k))^2)/((1-q^(-1-2 k))^2 (1-q^(3+2 k))^2) Nine terms of the sum give 27 term agreement with the product, despite the lack of q^k^2: In[1189]:= Series[%1131/.\[Infinity]->9/.-\[Infinity]->-9//Activate,{q,0,29}] Out[1189]= 1+5 q+12 q^2+24 q^3+44 q^4+68 q^5+102 q^6+150 q^7+202 q^8+270 q^9+354 q^10+446 q^11+561 q^12+693 q^13+838 q^14+1002 q^15+1195 q^16+1407 q^17+1638 q^18+1906 q^19+2179 q^20+2491 q^21+2842 q^22+3202 q^23+3610 q^24+4042 q^25+4504 q^26+5008 q^27+5549 q^28+6114 q^29+O[q]^30==1+5 q+12 q^2+24 q^3+44 q^4+68 q^5+102 q^6+150 q^7+202 q^8+270 q^9+354 q^10+446 q^11+561 q^12+693 q^13+838 q^14+1002 q^15+1195 q^16+1407 q^17+1638 q^18+1906 q^19+2179 q^20+2491 q^21+2842 q^22+3202 q^23+3610 q^24+4042 q^25+4504 q^26+5008 q^27+5550 q^28+6114 q^29+O[q]^30 In[1190]:= Subtract@@% Out[1190]= -q^28+O[q]^30 I got it by mildly abusing the (bilateral) "q-Dixon's theorem" in http://math.sun.ac.za/~hproding/pdffiles/Triples-long.pdf which is not the (unilateral) q-Dixon's I get from path invariant matrices. --rwg -- You received this message because you are subscribed to the Google Groups "mathfuneavesdroppers" group. To unsubscribe from this group and stop receiving emails from it, send an email to mathfuneavesdroppers+unsubscribe@googlegroups.com<mailto:mathfuneavesdroppers+unsubscribe@googlegroups.com>. For more options, visit https://groups.google.com/d/optout.
I haven't been following this discussion, but psi^4 is https://oeis.org/A008438 ("sum of divisors of 2n+1"), which is quite a long entry with many references and comments. If there's something new to add there, please add it (or tell me what to add)! Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Wed, Jul 6, 2016 at 11:37 PM, Mike Hirschhorn <m.hirschhorn@unsw.edu.au> wrote:
Dear all,
The right side is \psi(q)^4, where \psi(q) generates the triangular numbers, so the identity gives the number of reps of a number as the sum of 4 triangles, which must be well-known.
Mike Hirschhorn.
________________________________ From: mathfuneavesdroppers@googlegroups.com < mathfuneavesdroppers@googlegroups.com> on behalf of Bill Gosper < billgosper@gmail.com> Sent: 07 July 2016 13:08:14 To: math-fun@mailman.xmission.com Subject: Re: How I miss Mizan Rahman, or, what the heck is this?
George Andrews got it from Bailey's ??? transformation (rather than finding it scrawled on the wall of an ancient tomb.) Merely with elementary manipulations, starting with partial fractions, I reduced it to n q Sum[---------------, {n, -?, ?}] == 1 + 2 n 2 (1 - q )
2 2 4 QPochhammer[q , q ] -------------------- 2 4 QPochhammer[q, q ]
which is presumably known, since it's a simple q-extension of
In[1653]:= Sum[(2 n - 1)^-2, {n, -?, ?}]
Out[1653]= ?^2/4.
Exercises: Evaluate
3 n + 1 q Sum[---------------, {n, -?, ?}] 1 + 2 n 2 (1 - q )
and n q Sum[------------, {n, -?, ?}] 1 + 2 n 1 - q
Spoilers: 2 2 4 QPochhammer[q , q ] -------------------- 2 4 QPochhammer[q, q ]
and 0 (off the unit circle).
Problem: n integer, t real, L=Limit[sin nt,n->?]. Then L=0 or it doesn't exist? --rwg
On Sat, Jul 2, 2016 at 11:19 AM, Bill Gosper <billgosper@gmail.com<mailto: billgosper@gmail.com>> wrote: Out[1131]= k q Sum -------------------------------, 1 - 2 k 2 1 + 2 k 2 (1 - q ) (1 - q ) k,-?,?
= 1 2 4 (1 + q) QFactorial[-(-), q ] 2 ----------------------------- 2 4 (1 - q )
q-Pochhammers in the sum but not the summand? "q-Mittag Leffler"?
It q-retracts to 16 Sum[-----------, {k,-?,?}] == 2 2 (1 - 4 k ) 2 2 ?
Mathematica is dumbstruck. For minutes on end. Are these actually common?
Test for q=1/2: In[1132]:= N[Activate[%1131/.q->.5]] Out[1132]= True
Term ratio: In[1180]:= DiscreteRatio[%1131[[1,1]],k] Out[1180]= (q (1-q^(1-2 k))^2 (1-q^(1+2 k))^2)/((1-q^(-1-2 k))^2 (1-q^(3+2 k))^2)
Nine terms of the sum give 27 term agreement with the product, despite the lack of q^k^2:
In[1189]:= Series[%1131/.\[Infinity]->9/.-\[Infinity]->-9//Activate,{q,0,29}] Out[1189]= 1+5 q+12 q^2+24 q^3+44 q^4+68 q^5+102 q^6+150 q^7+202 q^8+270 q^9+354 q^10+446 q^11+561 q^12+693 q^13+838 q^14+1002 q^15+1195 q^16+1407 q^17+1638 q^18+1906 q^19+2179 q^20+2491 q^21+2842 q^22+3202 q^23+3610 q^24+4042 q^25+4504 q^26+5008 q^27+5549 q^28+6114 q^29+O[q]^30==1+5 q+12 q^2+24 q^3+44 q^4+68 q^5+102 q^6+150 q^7+202 q^8+270 q^9+354 q^10+446 q^11+561 q^12+693 q^13+838 q^14+1002 q^15+1195 q^16+1407 q^17+1638 q^18+1906 q^19+2179 q^20+2491 q^21+2842 q^22+3202 q^23+3610 q^24+4042 q^25+4504 q^26+5008 q^27+5550 q^28+6114 q^29+O[q]^30
In[1190]:= Subtract@@% Out[1190]= -q^28+O[q]^30
I got it by mildly abusing the (bilateral) "q-Dixon's theorem" in http://math.sun.ac.za/~hproding/pdffiles/Triples-long.pdf which is not the (unilateral) q-Dixon's I get from path invariant matrices. --rwg
-- You received this message because you are subscribed to the Google Groups "mathfuneavesdroppers" group. To unsubscribe from this group and stop receiving emails from it, send an email to mathfuneavesdroppers+unsubscribe@googlegroups.com<mailto: mathfuneavesdroppers+unsubscribe@googlegroups.com>. For more options, visit https://groups.google.com/d/optout. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (3)
-
Bill Gosper -
Mike Hirschhorn -
Neil Sloane