Question Number 64903 by naka3546 last updated on 23/Jul/19

$$\mathrm{1},\:\mathrm{3},\:\mathrm{7},\:\mathrm{15},\:\mathrm{30},\:\mathrm{57},\:\mathrm{103},\:{x} \\ $$$${What}'{s}\:\:{x}\:? \\ $$
Commented by Tony Lin last updated on 23/Jul/19

$${f}\left(\mathrm{1}\right)=\mathrm{1},{f}\left(\mathrm{2}\right)=\mathrm{3},{f}\left(\mathrm{3}\right)=\mathrm{7},{f}\left(\mathrm{4}\right)=\mathrm{15} \\ $$$${f}\left(\mathrm{5}\right)=\mathrm{30},{f}\left(\mathrm{6}\right)=\mathrm{57},{f}\left(\mathrm{7}\right)=\mathrm{103},{f}\left(\mathrm{8}\right)={x} \\ $$$${f}\left({x}\right) \\ $$$$={f}\left(\mathrm{1}\right)\frac{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\left({x}−\mathrm{5}\right)\left({x}−\mathrm{6}\right)\left({x}−\mathrm{7}\right)}{\left(\mathrm{1}−\mathrm{2}\right)\left(\mathrm{1}−\mathrm{3}\right)\left(\mathrm{1}−\mathrm{4}\right)\left(\mathrm{1}−\mathrm{5}\right)\left(\mathrm{1}−\mathrm{6}\right)\left(\mathrm{1}−\mathrm{7}\right)} \\ $$$$+{f}\left(\mathrm{2}\right)\frac{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\left({x}−\mathrm{5}\right)\left({x}−\mathrm{6}\right)\left({x}−\mathrm{7}\right)}{\left(\mathrm{2}−\mathrm{1}\right)\left(\mathrm{2}−\mathrm{3}\right)\left(\mathrm{2}−\mathrm{4}\right)\left(\mathrm{2}−\mathrm{5}\right)\left(\mathrm{2}−\mathrm{6}\right)\left(\mathrm{2}−\mathrm{7}\right)} \\ $$$$+{f}\left(\mathrm{3}\right)\frac{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{4}\right)\left({x}−\mathrm{5}\right)\left({x}−\mathrm{6}\right)\left({x}−\mathrm{7}\right)}{\left(\mathrm{3}−\mathrm{1}\right)\left(\mathrm{3}−\mathrm{2}\right)\left(\mathrm{3}−\mathrm{4}\right)\left(\mathrm{3}−\mathrm{5}\right)\left(\mathrm{3}−\mathrm{6}\right)\left(\mathrm{3}−\mathrm{7}\right)} \\ $$$$+{f}\left(\mathrm{4}\right)\frac{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{5}\right)\left({x}−\mathrm{6}\right)\left({x}−\mathrm{7}\right)}{\left(\mathrm{4}−\mathrm{1}\right)\left(\mathrm{4}−\mathrm{2}\right)\left(\mathrm{4}−\mathrm{3}\right)\left(\mathrm{4}−\mathrm{5}\right)\left(\mathrm{4}−\mathrm{6}\right)\left(\mathrm{4}−\mathrm{7}\right)} \\ $$$$+{f}\left(\mathrm{5}\right)\frac{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\left({x}−\mathrm{6}\right)\left({x}−\mathrm{7}\right)}{\left(\mathrm{5}−\mathrm{1}\right)\left(\mathrm{5}−\mathrm{2}\right)\left(\mathrm{5}−\mathrm{3}\right)\left(\mathrm{5}−\mathrm{4}\right)\left(\mathrm{5}−\mathrm{6}\right)\left(\mathrm{5}−\mathrm{7}\right)} \\ $$$$+{f}\left(\mathrm{6}\right)\frac{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\left({x}−\mathrm{5}\right)\left({x}−\mathrm{7}\right)}{\left(\mathrm{6}−\mathrm{1}\right)\left(\mathrm{6}−\mathrm{2}\right)\left(\mathrm{6}−\mathrm{3}\right)\left(\mathrm{6}−\mathrm{4}\right)\left(\mathrm{6}−\mathrm{5}\right)\left(\mathrm{6}−\mathrm{7}\right)} \\ $$$$+{f}\left(\mathrm{7}\right)\frac{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\left({x}−\mathrm{5}\right)\left({x}−\mathrm{6}\right)}{\left(\mathrm{7}−\mathrm{1}\right)\left(\mathrm{7}−\mathrm{2}\right)\left(\mathrm{7}−\mathrm{3}\right)\left(\mathrm{7}−\mathrm{4}\right)\left(\mathrm{7}−\mathrm{5}\right)\left(\mathrm{7}−\mathrm{6}\right)} \\ $$$${x} \\ $$$$={f}\left(\mathrm{8}\right) \\ $$$$=\mathrm{1}×\frac{\mathrm{6}×\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}{\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)\left(−\mathrm{3}\right)\left(−\mathrm{4}\right)\left(−\mathrm{5}\right)\left(−\mathrm{6}\right)} \\ $$$$+\mathrm{3}×\frac{\mathrm{7}×\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}{\mathrm{1}×\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)\left(−\mathrm{3}\right)\left(−\mathrm{4}\right)\left(−\mathrm{5}\right)} \\ $$$$+\mathrm{7}×\frac{\mathrm{7}×\mathrm{6}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}{\mathrm{2}×\mathrm{1}×\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)\left(−\mathrm{3}\right)\left(−\mathrm{4}\right)} \\ $$$$+\mathrm{15}×\frac{\mathrm{7}×\mathrm{6}×\mathrm{5}×\mathrm{3}×\mathrm{2}×\mathrm{1}}{\mathrm{3}×\mathrm{2}×\mathrm{1}×\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)\left(−\mathrm{3}\right)} \\ $$$$+\mathrm{30}×\frac{\mathrm{7}×\mathrm{6}×\mathrm{5}×\mathrm{4}×\mathrm{2}×\mathrm{1}}{\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}×\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)} \\ $$$$+\mathrm{57}×\frac{\mathrm{7}×\mathrm{6}×\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{1}}{\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}×\left(−\mathrm{1}\right)} \\ $$$$+\mathrm{103}×\frac{\mathrm{7}×\mathrm{6}×\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}}{\mathrm{6}×\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}} \\ $$$$=\mathrm{1}−\mathrm{21}+\mathrm{147}−\mathrm{525}+\mathrm{1050}−\mathrm{1197}+\mathrm{721} \\ $$$$=\mathrm{176} \\ $$
Commented by MJS last updated on 23/Jul/19

$$\mathrm{generally}\:\mathrm{it}\:\mathrm{could}\:\mathrm{be}\:\mathrm{any}\:\mathrm{number}. \\ $$$$\mathrm{we}'\mathrm{re}\:\mathrm{interpreting}\:\mathrm{it}\:\mathrm{as} \\ $$$${y}={c}_{\mathrm{6}} {x}^{\mathrm{6}} +{c}_{\mathrm{5}} {x}^{\mathrm{5}} +{c}_{\mathrm{4}} {x}^{\mathrm{4}} +{c}_{\mathrm{3}} {x}^{\mathrm{3}} +{c}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{1}} {x}+{c}_{\mathrm{0}} \\ $$$$\mathrm{with}\:\mathrm{given}\:\mathrm{pairs}\:\left({x},\:{y}\right): \\ $$$$\left(\mathrm{1},\:\mathrm{1}\right)\:\:\left(\mathrm{2},\:\mathrm{3}\right)\:\:\left(\mathrm{3},\:\mathrm{7}\right)\:\:\left(\mathrm{4},\:\mathrm{15}\right)\:\:\left(\mathrm{5},\:\mathrm{30}\right)\:\:\left(\mathrm{6},\:\mathrm{57}\right)\:\:\left(\mathrm{7},\:\mathrm{103}\right) \\ $$$$ \\ $$$$\mathrm{but}\:\mathrm{we}\:\mathrm{can}\:\mathrm{use}\:\mathrm{any}\:\mathrm{function}\:\mathrm{with}\:\mathrm{7}\:\mathrm{unknown} \\ $$$$\mathrm{constants} \\ $$$${y}=\frac{\mathrm{1}}{{c}_{\mathrm{6}} {x}^{\mathrm{6}} +{c}_{\mathrm{5}} {x}^{\mathrm{5}} +{c}_{\mathrm{4}} {x}^{\mathrm{4}} +{c}_{\mathrm{3}} {x}^{\mathrm{3}} +{c}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{1}} {x}+{c}_{\mathrm{0}} } \\ $$$$\mathrm{leads}\:\mathrm{to}\:\frac{\mathrm{3914}}{\mathrm{781}} \\ $$$${y}=\frac{{c}_{\mathrm{6}} }{{x}^{\mathrm{6}} }+\frac{{c}_{\mathrm{5}} }{{x}^{\mathrm{5}} }+\frac{{c}_{\mathrm{4}} }{{x}^{\mathrm{4}} }+\frac{{c}_{\mathrm{3}} }{{x}^{\mathrm{3}} }+\frac{{c}_{\mathrm{2}} }{{x}^{\mathrm{2}} }+\frac{{c}_{\mathrm{1}} }{{x}}+{c}_{\mathrm{0}} \\ $$$$\mathrm{leads}\:\mathrm{to}\:\frac{\mathrm{43339367}}{\mathrm{262144}} \\ $$$${y}=\frac{\mathrm{1}}{\frac{{c}_{\mathrm{6}} }{{x}^{\mathrm{6}} }+\frac{{c}_{\mathrm{5}} }{{x}^{\mathrm{5}} }+\frac{{c}_{\mathrm{4}} }{{x}^{\mathrm{4}} }+\frac{{c}_{\mathrm{3}} }{{x}^{\mathrm{3}} }+\frac{{c}_{\mathrm{2}} }{{x}^{\mathrm{2}} }+\frac{{c}_{\mathrm{1}} }{{x}}+{c}_{\mathrm{0}} } \\ $$$$\mathrm{leads}\:\mathrm{to}\:\frac{\mathrm{1026031616}}{\mathrm{5937579}} \\ $$$$\mathrm{we}\:\mathrm{could}\:\mathrm{also}\:\mathrm{use} \\ $$$${y}={c}_{\mathrm{7}} {x}^{\mathrm{7}} +{c}_{\mathrm{6}} {x}^{\mathrm{6}} +{c}_{\mathrm{5}} {x}^{\mathrm{5}} +{c}_{\mathrm{4}} {x}^{\mathrm{4}} +{c}_{\mathrm{3}} {x}^{\mathrm{3}} +{c}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{1}} {x}+{c}_{\mathrm{0}} \\ $$$$\mathrm{then}\:\mathrm{we}\:\mathrm{have}\:\mathrm{one}\:\mathrm{factor}\:\mathrm{free}\:\mathrm{to}\:\mathrm{choose}: \\ $$$${y}=−\frac{{c}_{\mathrm{0}} +\mathrm{2}}{\mathrm{5040}}{x}^{\mathrm{7}} +\frac{\mathrm{4}{c}_{\mathrm{0}} +\mathrm{7}}{\mathrm{720}}{x}^{\mathrm{6}} −\frac{\mathrm{46}{c}_{\mathrm{0}} +\mathrm{65}}{\mathrm{720}}{x}^{\mathrm{5}} +\frac{\mathrm{56}{c}_{\mathrm{0}} +\mathrm{65}}{\mathrm{144}}{x}^{\mathrm{4}} −\frac{\mathrm{967}{c}_{\mathrm{0}} +\mathrm{749}}{\mathrm{720}}{x}^{\mathrm{3}} +\frac{\mathrm{469}{c}_{\mathrm{0}} +\mathrm{277}}{\mathrm{180}}{x}^{\mathrm{2}} −\frac{\mathrm{11}\left(\mathrm{99}{c}_{\mathrm{0}} −\mathrm{5}\right)}{\mathrm{420}}{x}+{c}_{\mathrm{0}} \\ $$$${x}=\mathrm{8} \\ $$$${y}=\mathrm{174}−{c}_{\mathrm{0}} \\ $$$$\Rightarrow\:\mathrm{any}\:\mathrm{value}\:\mathrm{fits} \\ $$
Answered by MJS last updated on 23/Jul/19

$$\mathrm{1}\:\:\:\mathrm{3}\:\:\:\mathrm{7}\:\:\mathrm{15}\:\:\mathrm{30}\:\:\mathrm{57}\:\mathrm{103}\:\:\:\mathrm{176} \\ $$$$\:\:\mathrm{2}\:\:\:\mathrm{4}\:\:\:\mathrm{8}\:\:\mathrm{15}\:\:\mathrm{27}\:\:\:\mathrm{46}\:\:\:\:\mathrm{73} \\ $$$$\:\:\:\:\:\mathrm{2}\:\:\:\mathrm{4}\:\:\:\mathrm{7}\:\:\mathrm{12}\:\:\:\mathrm{19}\:\:\:\:\mathrm{27} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\mathrm{3}\:\:\:\mathrm{5}\:\:\:\:\mathrm{7}\:\:\:\:\:\:\mathrm{8} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\mathrm{2}\:\:\:\mathrm{2}\:\:\:\:\:\:\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\mathrm{0}\:\:\:−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\:−\mathrm{1} \\ $$
Commented by Tony Lin last updated on 23/Jul/19

$${What}\:{is}\:{the}\:{method}\:? \\ $$
Commented by MJS last updated on 23/Jul/19

$${a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}… \\ $$$$\:\:{b}−{a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{c}−{b}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}−{c}… \\ $$$$\:\:\:\:\:\:\:\:\:\:\left({c}−{b}\right)−\left({b}−{a}\right)\:\:\:\:\:\:\:\:\:\:\:\left({d}−{c}\right)−\left({c}−{b}\right)… \\ $$$$… \\ $$$$\mathrm{until}\:\mathrm{you}\:\mathrm{reach}\:\mathrm{a}\:\mathrm{line}\:\mathrm{with}\:\mathrm{each}\:\mathrm{number}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{value}.\:\mathrm{then}\:\mathrm{start}\:\mathrm{at}\:\mathrm{the}\:\mathrm{bottom} \\ $$$$\mathrm{adding}\:\mathrm{this}\:\mathrm{number}\:\mathrm{once}\:\mathrm{more}\:\mathrm{and}\:\mathrm{go}\:\mathrm{up} \\ $$$$\mathrm{from}\:\mathrm{line}\:\mathrm{to}\:\mathrm{line}.\:\mathrm{this}\:\mathrm{gives}\:\mathrm{the}\:\mathrm{same}\:\mathrm{result} \\ $$$$\mathrm{as}\:\mathrm{the}\:\mathrm{polynomial}\:\mathrm{approach} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{say} \\ $$$$\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{line}\:\mathrm{is}\:{y} \\ $$$$\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{line}\:\mathrm{is}\:\bigtriangleup{y} \\ $$$$\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{line}\:\mathrm{is}\:\bigtriangleup^{\mathrm{2}} {y} \\ $$$$\mathrm{the}\:{n}^{\mathrm{th}} \:\mathrm{line}\:\mathrm{is}\:\bigtriangleup^{{n}−\mathrm{1}} {y} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{similar}\:\mathrm{to}\:\mathrm{the}\:{n}^{\mathrm{th}} \:\mathrm{derivate}\:\mathrm{of}\:\mathrm{a}\:\mathrm{polynome} \\ $$$$\mathrm{of}\:\mathrm{degree}\:{k}.\:\mathrm{you}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{factor} \\ $$$$\mathrm{after}\:{k}\:\mathrm{steps} \\ $$
Commented by Tony Lin last updated on 23/Jul/19

$${is}\:{the}\:{method}\:{only}\:{be}\:{used}\:{on}\: \\ $$$${condition}\:{that}\:{a},{b},{c},{d}\:{are}\:{in}\:{A}.{P}\:? \\ $$
Commented by MJS last updated on 23/Jul/19

$$\mathrm{no}.\:\mathrm{write}\:\mathrm{down}\:\mathrm{any}\:\mathrm{numbers},\:\mathrm{it}\:\mathrm{always}\:\mathrm{works} \\ $$$$\mathrm{but}\:\mathrm{you}'\mathrm{ll}\:\mathrm{only}\:\mathrm{find}\:\mathrm{polynomial}\:\mathrm{solutions} \\ $$$$\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{5}\:\:\:\:\:\mathrm{8}\:\:\:\:\:\mathrm{13}\:\:\:\:\:\mathrm{29} \\ $$$$\:\:\:\:\mathrm{0}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{5}\:\:\:\:\:\mathrm{16} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{0}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{11} \\ $$$$\:\:\:\:\:\:\:\:\:\:−\mathrm{1}\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{0}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:−\mathrm{1}\:\:\:\:\mathrm{1}\:\:\:\:\mathrm{8} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{3}\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{7} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}\:\:\:\:\:\mathrm{5} \\ $$
Commented by Tony Lin last updated on 23/Jul/19

$${thanks}\:{sir}\:.{i}\:{got}\:{it}. \\ $$
Commented by MJS last updated on 23/Jul/19

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$