Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 101800 by Quvonchbek last updated on 04/Jul/20

Commented by mr W last updated on 05/Jul/20

$v e r y i n t e r e s t i n g! t h a n k s s i r!$
Commented by mr W last updated on 04/Jul/20

$w h a t i s t h e q u e s t i o n ?$
Commented by prakash 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
	$g(k)=(x/(1−kp))=x(1+kp+k^2 p^2 +....) f(k)=(x/(1−kp))+(y/(1−kq))+(z/(1−kr))+(u/(1−ks))+(v/(1−kt)) =(x+y+z+u+v) +k(px+qy+rz+su+tv) +k^2 (p^2 x+..+t^2 v) +k^9 (p^9 x+q^9 y+..+t^9 v) +higher powers of k f(k)=((A_0 +A_1 k+A_2 k^2 +A_3 k^3 +A_4 k^4 )/(1+B_1 k+B_2 k^2 +B_3 k^3 +B_4 k^4 +B_5 k^5 )) comparing with given equation and equating A_0 +A_1 k+A_2 k^2 +A_3 k^3 +A_4 k^4 =(1+B_1 k+B_2 k^2 +B_3 k^3 +B_4 k^4 +B_5 k^5 ) ×(2+3k+16k^2 +31k^3 +103k^4 +235k^5 +674k^6 +1669k^7 +4326k^8 +11595k^9 +...) A_0 =1 (k^0 ) A_1 =2B_1 +3 (k^1 ) A_2 =3B_1 +2B_2 +16 (k^2 ) A_3 =31+16B_1 +3B_2 +2B_3 (k^3 ) A_4 =103+31B_1 +16B_2 +3B_3 +2B_4 0=235+103B_1 +31B_2 +16B_3 +3B_4 +2B_5 0=674+235B_1 +103B_2 +31B_3 +16B_4 +3B_5 0=1669+674B_1 +235B_2 +103B_3 +31B_4 +16B_5 0=4526+1669B1+674B_2 +235B_3 +103B_4 +31B_5 0=11595+4526B_1 +1669B_2 +674B_3 +235B_4 +103B_5 B_1 =−1 B_2 =−5 B_3 =1 B_4 =3 B_5 =−1 A_1 =1 A_2 =3 A_3 =2 A_4 =1 f(k)=((1+k+3k^2 +2k^3 +k^4 )/(1−k−5k^2 +k^3 +3k^4 −k^5 )) Partial fraction =((29−46k)/(10(k^2 −3k+1)))+((8k+3)/(2(k^2 −k−1)))−(2/(5(k+1))) ((29−46k)/(10(k^2 −3k+1)))=((29−46k)/(10(k−(((3+(√5))/2))(k−((3−(√5))/2)))) Also need to decome((8k+3)/(k^2 −k−1)) into linear partial fraction so we will get −(2/(5(k+1)))+4 more linear part fractions. ((A_0 +A_1 k+A_2 k^2 +A_3 k^3 +A_4 k^4 )/(1+B_1 k+B_2 k^2 +B_3 k^3 +B_4 k^4 +B_5 k^5 )) =(x/(1−kp))+(y/(1−kq))+(z/(1−kr))+(u/(1−ks))+(v/(1−kt)) After comparing partial fraction with brown expression we can see that (x,p)=(−(2/5),−1) remaining pairs (y,q),(z,r),(u,s),(v,t) can be calculated from remaining 4 partial fractions. All pairs are interchangable due to symmetry$
	$g (k) = \frac{x}{1 - k p} = x (1 + k p + k^{2} p^{2} + . . . .)$ $f (k) = \frac{x}{1 - k p} + \frac{y}{1 - k q} + \frac{z}{1 - k r} + \frac{u}{1 - k s} + \frac{v}{1 - k t}$ $= (x + y + z + u + v)$ $+ k (p x + q y + r z + s u + t v)$ $+ k^{2} (p^{2} x + . . + t^{2} v)$ $+ k^{9} (p^{9} x + q^{9} y + . . + t^{9} v)$ $+ h i g h e r p o w e r s o f k$ $f (k) = \frac{A_{0} + A_{1} k + A_{2} k^{2} + A_{3} k^{3} + A_{4} k^{4}}{1 + B_{1} k + B_{2} k^{2} + B_{3} k^{3} + B_{4} k^{4} + B_{5} k^{5}}$ $c o m p a r i n g w i t h g i v e n e q u a t i o n$ $a n d e q u a t i n g$ $A_{0} + A_{1} k + A_{2} k^{2} + A_{3} k^{3} + A_{4} k^{4}$ $= (1 + B_{1} k + B_{2} k^{2} + B_{3} k^{3} + B_{4} k^{4} + B_{5} k^{5})$ $\times (2 + 3 k + 16 k^{2} + 31 k^{3} + 103 k^{4}$ $+ 235 k^{5} + 674 k^{6} + 1669 k^{7} + 4326 k^{8} + 11595 k^{9} + . . .)$ $A_{0} = 1 (k^{0})$ $A_{1} = 2 B_{1} + 3 (k^{1})$ $A_{2} = 3 B_{1} + 2 B_{2} + 16 (k^{2})$ $A_{3} = 31 + 16 B_{1} + 3 B_{2} + 2 B_{3} (k^{3})$ $A_{4} = 103 + 31 B_{1} + 16 B_{2} + 3 B_{3} + 2 B_{4}$ $0 = 235 + 103 B_{1} + 31 B_{2} + 16 B_{3} + 3 B_{4} + 2 B_{5}$ $0 = 674 + 235 B_{1} + 103 B_{2} + 31 B_{3} + 16 B_{4} + 3 B_{5}$ $0 = 1669 + 674 B_{1} + 235 B_{2} + 103 B_{3} + 31 B_{4} + 16 B_{5}$ $0 = 4526 + 1669 B 1 + 674 B_{2} + 235 B_{3} + 103 B_{4} + 31 B_{5}$ $0 = 11595 + 4526 B_{1} + 1669 B_{2} + 674 B_{3} + 235 B_{4} + 103 B_{5}$ $B_{1} = - 1$ $B_{2} = - 5$ $B_{3} = 1$ $B_{4} = 3$ $B_{5} = - 1$ $A_{1} = 1$ $A_{2} = 3$ $A_{3} = 2$ $A_{4} = 1$ $f (k) = \frac{1 + k + 3 k^{2} + 2 k^{3} + k^{4}}{1 - k - 5 k^{2} + k^{3} + 3 k^{4} - k^{5}}$ $Partial fraction$ $= \frac{29 - 46 k}{10 (k^{2} - 3 k + 1)} + \frac{8 k + 3}{2 (k^{2} - k - 1)} - \frac{2}{5 (k + 1)}$ $\frac{29 - 46 k}{10 (k^{2} - 3 k + 1)} = \frac{29 - 46 k}{10 (k - (\frac{3 + \sqrt{5}}{2}) (k - \frac{3 - \sqrt{5}}{2})}$ $Also need to decome \frac{8 k + 3}{k^{2} - k - 1} into$ $linear partial fraction$ $so we will get$ $- \frac{2}{5 (k + 1)} + 4 more linear part fractions .$ $\frac{A_{0} + A_{1} k + A_{2} k^{2} + A_{3} k^{3} + A_{4} k^{4}}{1 + B_{1} k + B_{2} k^{2} + B_{3} k^{3} + B_{4} k^{4} + B_{5} k^{5}}$ $= \frac{x}{1 - k p} + \frac{y}{1 - k q} + \frac{z}{1 - k r} + \frac{u}{1 - k s} + \frac{v}{1 - k t}$ $After comparing partial fraction$ $with brown expression we can$ $see that (x, p) = (- \frac{2}{5}, - 1)$ $remaining pairs (y, q), (z, r), (u, s), (v, t)$ $can be calculated from remaining$ $4 partial fractions .$ $All pairs are interchangable due$ $to symmetry$
	Commented by mr W last updated on 05/Jul/20

	$c a n w e f i n d$ $p^{10} x + q^{10} y + r^{10} z + s^{10} u + t^{10} v = ?$ $w i t h o u t t o g e t x, y, z, u, v, p, q, r, s, t ?$
	Commented by prakash jain last updated on 05/Jul/20

	$A_{0} + A_{1} k + A_{2} k^{2} + A_{3} k^{3} + A_{4} k^{4}$ $= (1 + B_{1} k + B_{2} k^{2} + B_{3} k^{3} + B_{4} k^{4}) \times$ $(\underset{i = 0}{\sum^{9}} a_{i} k^{i} + a_{10} k^{10} + . . .)$ $a_{0} to a_{9} are given values .$ $0 = a_{10} + B_{1} a_{9} + B_{2} a_{8} + B_{3} a_{7} + B_{4} a_{6} + B_{5} a_{5}$ $a_{10} is unknown$ $a_{10} - 1 \cdot 11595 - 5 \cdot 4526 + 1 \cdot 1669 + 3 \cdot 674 - 1 \cdot 235 = 0$ $a_{10} = 30769$
	Commented by prakash jain last updated on 05/Jul/20

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

	$For wolfram alpha calculation . I used$ $answers given in paper .$
	Commented by mr W last updated on 05/Jul/20

	$v e r y n i c e, t h a n k s s i r!$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum