Algebra Questions

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

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

$${very}\:{interesting}!\:{thanks}\:{sir}! \\$$

Commented by mr W last updated on 04/Jul/20

$${what}\:{is}\:{the}\:{question}? \\$$

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

$${g}\left({k}\right)=\frac{{x}}{\mathrm{1}−{kp}}={x}\left(\mathrm{1}+{kp}+{k}^{\mathrm{2}} {p}^{\mathrm{2}} +....\right) \\$$$${f}\left({k}\right)=\frac{{x}}{\mathrm{1}−{kp}}+\frac{{y}}{\mathrm{1}−{kq}}+\frac{{z}}{\mathrm{1}−{kr}}+\frac{{u}}{\mathrm{1}−{ks}}+\frac{{v}}{\mathrm{1}−{kt}} \\$$$$=\left({x}+{y}+{z}+{u}+{v}\right) \\$$$$\:\:\:\:\:+{k}\left({px}+{qy}+{rz}+{su}+{tv}\right) \\$$$$\:\:\:\:\:\:+{k}^{\mathrm{2}} \left({p}^{\mathrm{2}} {x}+..+{t}^{\mathrm{2}} {v}\right) \\$$$$\:\:\:\:\:\:\:+{k}^{\mathrm{9}} \left({p}^{\mathrm{9}} {x}+{q}^{\mathrm{9}} {y}+..+{t}^{\mathrm{9}} {v}\right) \\$$$$\:\:\:\:\:\:\:\:+{higher}\:{powers}\:{of}\:{k} \\$$$${f}\left({k}\right)=\frac{{A}_{\mathrm{0}} +{A}_{\mathrm{1}} {k}+{A}_{\mathrm{2}} {k}^{\mathrm{2}} +{A}_{\mathrm{3}} {k}^{\mathrm{3}} +{A}_{\mathrm{4}} {k}^{\mathrm{4}} }{\mathrm{1}+{B}_{\mathrm{1}} {k}+{B}_{\mathrm{2}} {k}^{\mathrm{2}} +{B}_{\mathrm{3}} {k}^{\mathrm{3}} +{B}_{\mathrm{4}} {k}^{\mathrm{4}} +{B}_{\mathrm{5}} {k}^{\mathrm{5}} } \\$$$${comparing}\:{with}\:{given}\:{equation} \\$$$${and}\:{equating} \\$$$${A}_{\mathrm{0}} +{A}_{\mathrm{1}} {k}+{A}_{\mathrm{2}} {k}^{\mathrm{2}} +{A}_{\mathrm{3}} {k}^{\mathrm{3}} +{A}_{\mathrm{4}} {k}^{\mathrm{4}} \\$$$$=\left(\mathrm{1}+{B}_{\mathrm{1}} {k}+{B}_{\mathrm{2}} {k}^{\mathrm{2}} +{B}_{\mathrm{3}} {k}^{\mathrm{3}} +{B}_{\mathrm{4}} {k}^{\mathrm{4}} +{B}_{\mathrm{5}} {k}^{\mathrm{5}} \right) \\$$$$\:\:\:\:\:×\left(\mathrm{2}+\mathrm{3}{k}+\mathrm{16}{k}^{\mathrm{2}} +\mathrm{31}{k}^{\mathrm{3}} +\mathrm{103}{k}^{\mathrm{4}} \right. \\$$$$\left.\:\:\:+\mathrm{235}{k}^{\mathrm{5}} +\mathrm{674}{k}^{\mathrm{6}} +\mathrm{1669}{k}^{\mathrm{7}} +\mathrm{4326}{k}^{\mathrm{8}} +\mathrm{11595}{k}^{\mathrm{9}} +...\right) \\$$$${A}_{\mathrm{0}} =\mathrm{1}\:\:\:\left({k}^{\mathrm{0}} \right) \\$$$${A}_{\mathrm{1}} =\mathrm{2}{B}_{\mathrm{1}} +\mathrm{3}\:\:\:\:\left({k}^{\mathrm{1}} \right) \\$$$${A}_{\mathrm{2}} =\mathrm{3}{B}_{\mathrm{1}} +\mathrm{2}{B}_{\mathrm{2}} +\mathrm{16}\:\:\:\left({k}^{\mathrm{2}} \right) \\$$$${A}_{\mathrm{3}} =\mathrm{31}+\mathrm{16}{B}_{\mathrm{1}} +\mathrm{3}{B}_{\mathrm{2}} +\mathrm{2}{B}_{\mathrm{3}} \:\:\:\:\left({k}^{\mathrm{3}} \right) \\$$$${A}_{\mathrm{4}} =\mathrm{103}+\mathrm{31}{B}_{\mathrm{1}} +\mathrm{16}{B}_{\mathrm{2}} +\mathrm{3}{B}_{\mathrm{3}} +\mathrm{2}{B}_{\mathrm{4}} \\$$$$\mathrm{0}=\mathrm{235}+\mathrm{103}{B}_{\mathrm{1}} +\mathrm{31}{B}_{\mathrm{2}} +\mathrm{16}{B}_{\mathrm{3}} +\mathrm{3}{B}_{\mathrm{4}} +\mathrm{2}{B}_{\mathrm{5}} \\$$$$\mathrm{0}=\mathrm{674}+\mathrm{235}{B}_{\mathrm{1}} +\mathrm{103}{B}_{\mathrm{2}} +\mathrm{31}{B}_{\mathrm{3}} +\mathrm{16}{B}_{\mathrm{4}} +\mathrm{3}{B}_{\mathrm{5}} \\$$$$\mathrm{0}=\mathrm{1669}+\mathrm{674}{B}_{\mathrm{1}} +\mathrm{235}{B}_{\mathrm{2}} +\mathrm{103}{B}_{\mathrm{3}} +\mathrm{31}{B}_{\mathrm{4}} +\mathrm{16}{B}_{\mathrm{5}} \\$$$$\mathrm{0}=\mathrm{4526}+\mathrm{1669}{B}\mathrm{1}+\mathrm{674}{B}_{\mathrm{2}} +\mathrm{235}{B}_{\mathrm{3}} +\mathrm{103}{B}_{\mathrm{4}} +\mathrm{31}{B}_{\mathrm{5}} \\$$$$\mathrm{0}=\mathrm{11595}+\mathrm{4526}{B}_{\mathrm{1}} +\mathrm{1669}{B}_{\mathrm{2}} +\mathrm{674}{B}_{\mathrm{3}} +\mathrm{235}{B}_{\mathrm{4}} +\mathrm{103}{B}_{\mathrm{5}} \\$$$${B}_{\mathrm{1}} =−\mathrm{1} \\$$$${B}_{\mathrm{2}} =−\mathrm{5} \\$$$${B}_{\mathrm{3}} =\mathrm{1} \\$$$${B}_{\mathrm{4}} =\mathrm{3} \\$$$${B}_{\mathrm{5}} =−\mathrm{1} \\$$$${A}_{\mathrm{1}} =\mathrm{1} \\$$$${A}_{\mathrm{2}} =\mathrm{3} \\$$$${A}_{\mathrm{3}} =\mathrm{2} \\$$$${A}_{\mathrm{4}} =\mathrm{1} \\$$$${f}\left({k}\right)=\frac{\mathrm{1}+{k}+\mathrm{3}{k}^{\mathrm{2}} +\mathrm{2}{k}^{\mathrm{3}} +{k}^{\mathrm{4}} }{\mathrm{1}−{k}−\mathrm{5}{k}^{\mathrm{2}} +{k}^{\mathrm{3}} +\mathrm{3}{k}^{\mathrm{4}} −{k}^{\mathrm{5}} } \\$$$$\mathrm{Partial}\:\mathrm{fraction} \\$$$$=\frac{\mathrm{29}−\mathrm{46}{k}}{\mathrm{10}\left({k}^{\mathrm{2}} −\mathrm{3}{k}+\mathrm{1}\right)}+\frac{\mathrm{8}{k}+\mathrm{3}}{\mathrm{2}\left({k}^{\mathrm{2}} −{k}−\mathrm{1}\right)}−\frac{\mathrm{2}}{\mathrm{5}\left({k}+\mathrm{1}\right)} \\$$$$\frac{\mathrm{29}−\mathrm{46}{k}}{\mathrm{10}\left({k}^{\mathrm{2}} −\mathrm{3}{k}+\mathrm{1}\right)}=\frac{\mathrm{29}−\mathrm{46}{k}}{\mathrm{10}\left({k}−\left(\frac{\mathrm{3}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\left({k}−\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\right.} \\$$$$\mathrm{Also}\:\mathrm{need}\:\mathrm{to}\:\mathrm{decome}\frac{\mathrm{8}{k}+\mathrm{3}}{{k}^{\mathrm{2}} −{k}−\mathrm{1}}\:\mathrm{into} \\$$$$\mathrm{linear}\:\mathrm{partial}\:\mathrm{fraction} \\$$$$\mathrm{so}\:\mathrm{we}\:\mathrm{will}\:\mathrm{get} \\$$$$−\frac{\mathrm{2}}{\mathrm{5}\left({k}+\mathrm{1}\right)}+\mathrm{4}\:\mathrm{more}\:\mathrm{linear}\:\mathrm{part}\:\mathrm{fractions}. \\$$$$\frac{{A}_{\mathrm{0}} +{A}_{\mathrm{1}} {k}+{A}_{\mathrm{2}} {k}^{\mathrm{2}} +{A}_{\mathrm{3}} {k}^{\mathrm{3}} +{A}_{\mathrm{4}} {k}^{\mathrm{4}} }{\mathrm{1}+{B}_{\mathrm{1}} {k}+{B}_{\mathrm{2}} {k}^{\mathrm{2}} +{B}_{\mathrm{3}} {k}^{\mathrm{3}} +{B}_{\mathrm{4}} {k}^{\mathrm{4}} +{B}_{\mathrm{5}} {k}^{\mathrm{5}} } \\$$$$=\frac{{x}}{\mathrm{1}−{kp}}+\frac{{y}}{\mathrm{1}−{kq}}+\frac{{z}}{\mathrm{1}−{kr}}+\frac{{u}}{\mathrm{1}−{ks}}+\frac{{v}}{\mathrm{1}−{kt}} \\$$$$\mathrm{After}\:\mathrm{comparing}\:\mathrm{partial}\:\mathrm{fraction} \\$$$$\mathrm{with}\:\mathrm{brown}\:\mathrm{expression}\:\mathrm{we}\:\mathrm{can} \\$$$$\mathrm{see}\:\mathrm{that}\:\left({x},{p}\right)=\left(−\frac{\mathrm{2}}{\mathrm{5}},−\mathrm{1}\right) \\$$$$\mathrm{remaining}\:\mathrm{pairs}\:\left({y},{q}\right),\left({z},{r}\right),\left({u},{s}\right),\left({v},{t}\right) \\$$$$\mathrm{can}\:\mathrm{be}\:\mathrm{calculated}\:\mathrm{from}\:\mathrm{remaining} \\$$$$\mathrm{4}\:\mathrm{partial}\:\mathrm{fractions}. \\$$$$\mathrm{All}\:\mathrm{pairs}\:\mathrm{are}\:\mathrm{interchangable}\:\mathrm{due} \\$$$$\mathrm{to}\:\mathrm{symmetry} \\$$

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

$${can}\:{we}\:{find} \\$$$${p}^{\mathrm{10}} {x}+{q}^{\mathrm{10}} {y}+{r}^{\mathrm{10}} {z}+{s}^{\mathrm{10}} {u}+{t}^{\mathrm{10}} {v}=? \\$$$${without}\:{to}\:{get}\:{x},{y},{z},{u},{v},{p},{q},{r},{s},{t}? \\$$

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

$$\\$$$${A}_{\mathrm{0}} +{A}_{\mathrm{1}} {k}+{A}_{\mathrm{2}} {k}^{\mathrm{2}} +{A}_{\mathrm{3}} {k}^{\mathrm{3}} +{A}_{\mathrm{4}} {k}^{\mathrm{4}} \\$$$$=\left(\mathrm{1}+{B}_{\mathrm{1}} {k}+{B}_{\mathrm{2}} {k}^{\mathrm{2}} +{B}_{\mathrm{3}} {k}^{\mathrm{3}} +{B}_{\mathrm{4}} {k}^{\mathrm{4}} \right)× \\$$$$\left(\underset{{i}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}{a}_{{i}} {k}^{{i}} +{a}_{\mathrm{10}} {k}^{\mathrm{10}} +...\right) \\$$$${a}_{\mathrm{0}} \:\mathrm{to}\:{a}_{\mathrm{9}} \:\mathrm{are}\:\mathrm{given}\:\mathrm{values}. \\$$$$\mathrm{0}={a}_{\mathrm{10}} +{B}_{\mathrm{1}} {a}_{\mathrm{9}} +{B}_{\mathrm{2}} {a}_{\mathrm{8}} +{B}_{\mathrm{3}} {a}_{\mathrm{7}} +{B}_{\mathrm{4}} {a}_{\mathrm{6}} +{B}_{\mathrm{5}} {a}_{\mathrm{5}} \\$$$${a}_{\mathrm{10}} \:\mathrm{is}\:\mathrm{unknown} \\$$$${a}_{\mathrm{10}} −\mathrm{1}\centerdot\mathrm{11595}−\mathrm{5}\centerdot\mathrm{4526}+\mathrm{1}\centerdot\mathrm{1669}+\mathrm{3}\centerdot\mathrm{674}−\mathrm{1}\centerdot\mathrm{235}=\mathrm{0} \\$$$${a}_{\mathrm{10}} =\mathrm{30769} \\$$

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

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

$$\mathrm{For}\:\mathrm{wolfram}\:\mathrm{alpha}\:\mathrm{calculation}.\:\mathrm{I}\:\mathrm{used} \\$$$$\mathrm{answers}\:\mathrm{given}\:\mathrm{in}\:\mathrm{paper}. \\$$

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

$${very}\:{nice},\:{thanks}\:{sir}! \\$$