Question Number 101800 by Quvonchbek last updated on 04/Jul/20

Commented bymr W last updated on 05/Jul/20

very interesting! thanks sir!

Commented bymr W last updated on 04/Jul/20

what is the question?

Commented byprakash jain last updated on 05/Jul/20

http://ramanujan.sirinudi.org/Volumes/published/ram03.html Check above link for more theoritical description. in fact same equation is also covered answer that i added is same steps and some help from online simultaneous equation solvers

Answered by prakash jain last updated on 05/Jul/20

g(k)=(x/(1−kp))=x(1+kp+k^2 p^2 +....)  f(k)=(x/(1−kp))+(y/(1−kq))+(z/(1−kr))+(u/(1−ks))+(v/(1−kt))  =(x+y+z+u+v)       +k(px+qy+rz+su+tv)        +k^2 (p^2 x+..+t^2 v)         +k^9 (p^9 x+q^9 y+..+t^9 v)          +higher powers of k  f(k)=((A_0 +A_1 k+A_2 k^2 +A_3 k^3 +A_4 k^4 )/(1+B_1 k+B_2 k^2 +B_3 k^3 +B_4 k^4 +B_5 k^5 ))  comparing with given equation  and equating  A_0 +A_1 k+A_2 k^2 +A_3 k^3 +A_4 k^4   =(1+B_1 k+B_2 k^2 +B_3 k^3 +B_4 k^4 +B_5 k^5 )       ×(2+3k+16k^2 +31k^3 +103k^4      +235k^5 +674k^6 +1669k^7 +4326k^8 +11595k^9 +...)  A_0 =1   (k^0 )  A_1 =2B_1 +3    (k^1 )  A_2 =3B_1 +2B_2 +16   (k^2 )  A_3 =31+16B_1 +3B_2 +2B_3     (k^3 )  A_4 =103+31B_1 +16B_2 +3B_3 +2B_4   0=235+103B_1 +31B_2 +16B_3 +3B_4 +2B_5   0=674+235B_1 +103B_2 +31B_3 +16B_4 +3B_5   0=1669+674B_1 +235B_2 +103B_3 +31B_4 +16B_5   0=4526+1669B1+674B_2 +235B_3 +103B_4 +31B_5   0=11595+4526B_1 +1669B_2 +674B_3 +235B_4 +103B_5   B_1 =−1  B_2 =−5  B_3 =1  B_4 =3  B_5 =−1  A_1 =1  A_2 =3  A_3 =2  A_4 =1  f(k)=((1+k+3k^2 +2k^3 +k^4 )/(1−k−5k^2 +k^3 +3k^4 −k^5 ))  Partial fraction  =((29−46k)/(10(k^2 −3k+1)))+((8k+3)/(2(k^2 −k−1)))−(2/(5(k+1)))  ((29−46k)/(10(k^2 −3k+1)))=((29−46k)/(10(k−(((3+(√5))/2))(k−((3−(√5))/2))))  Also need to decome((8k+3)/(k^2 −k−1)) into  linear partial fraction  so we will get  −(2/(5(k+1)))+4 more linear part fractions.  ((A_0 +A_1 k+A_2 k^2 +A_3 k^3 +A_4 k^4 )/(1+B_1 k+B_2 k^2 +B_3 k^3 +B_4 k^4 +B_5 k^5 ))  =(x/(1−kp))+(y/(1−kq))+(z/(1−kr))+(u/(1−ks))+(v/(1−kt))  After comparing partial fraction  with brown expression we can  see that (x,p)=(−(2/5),−1)  remaining pairs (y,q),(z,r),(u,s),(v,t)  can be calculated from remaining  4 partial fractions.  All pairs are interchangable due  to symmetry

Commented bymr W last updated on 05/Jul/20

can we find  p^(10) x+q^(10) y+r^(10) z+s^(10) u+t^(10) v=?  without to get x,y,z,u,v,p,q,r,s,t?

Commented byprakash jain last updated on 05/Jul/20

  A_0 +A_1 k+A_2 k^2 +A_3 k^3 +A_4 k^4   =(1+B_1 k+B_2 k^2 +B_3 k^3 +B_4 k^4 )×  (Σ_(i=0) ^9 a_i k^i +a_(10) k^(10) +...)  a_0  to a_9  are given values.  0=a_(10) +B_1 a_9 +B_2 a_8 +B_3 a_7 +B_4 a_6 +B_5 a_5   a_(10)  is unknown  a_(10) −1∙11595−5∙4526+1∙1669+3∙674−1∙235=0  a_(10) =30769

Commented byprakash jain last updated on 05/Jul/20

Commented byprakash jain last updated on 05/Jul/20

For wolfram alpha calculation. I used  answers given in paper.

Commented bymr W last updated on 05/Jul/20

very nice, thanks sir!