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Given-x-p-x-1-x-5-p-x-and-p-1-1-Find-p-1-2-

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Question Number 162823 by john_santu last updated on 01/Jan/22
Given: x.p(x−1)=(x−5).p(x) and p(−1)=1. Find p((1/2)).
$$\:\:{Given}:\:\:{x}.{p}\left({x}−\mathrm{1}\right)=\left({x}−\mathrm{5}\right).{p}\left({x}\right) \\ $$$$\:\:{and}\:{p}\left(−\mathrm{1}\right)=\mathrm{1}.\: \\ $$$$\:\:{Find}\:{p}\left(\frac{\mathrm{1}}{\mathrm{2}}\right). \\ $$
Answered by mr W last updated on 02/Jan/22
p(x)=(x/(x−5))×p(x−1) p(x)=((x(x−1))/((x−5)(x−6)))×p(x−2) ... p(x)=((x(x−1)(x−2)...×7×6)/((x−5)(x−6)(x−7)...×2×1))×p(5) p(x)=((x(x−1)(x−2)(x−3)(x−4)(x−5)(x−6)...×7×6×5×4×3×2×1)/(5×4×3×2×1×(x−5)(x−6)(x−7)...×2×1))×p(5) p(x)=((x(x−1)(x−2)(x−3)(x−4))/(5×4×3×2×1))×p(5) ⇒p(x)=((x(x−1)(x−2)(x−3)(x−4))/(120))×p(5) p(−1)=(((−1)(−2)(−3)(−4)(−5))/(120))×p(5)=1 ⇒p(5)=−1 ⇒p(x)=−((x(x−1)(x−2)(x−3)(x−4))/(120)) ⇒p((1/2))=−(((1/2)((1/2)−1)((1/2)−2)((1/2)−3)((1/2)−4))/(120))=−(7/2^8 )=−(7/(256))
$${p}\left({x}\right)=\frac{{x}}{{x}−\mathrm{5}}×{p}\left({x}−\mathrm{1}\right) \\ $$$${p}\left({x}\right)=\frac{{x}\left({x}−\mathrm{1}\right)}{\left({x}−\mathrm{5}\right)\left({x}−\mathrm{6}\right)}×{p}\left({x}−\mathrm{2}\right) \\ $$$$… \\ $$$${p}\left({x}\right)=\frac{{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)…×\mathrm{7}×\mathrm{6}}{\left({x}−\mathrm{5}\right)\left({x}−\mathrm{6}\right)\left({x}−\mathrm{7}\right)…×\mathrm{2}×\mathrm{1}}×{p}\left(\mathrm{5}\right) \\ $$$${p}\left({x}\right)=\frac{{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)\left({x}−\mathrm{5}\right)\left({x}−\mathrm{6}\right)…×\mathrm{7}×\mathrm{6}×\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}{\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}×\left({x}−\mathrm{5}\right)\left({x}−\mathrm{6}\right)\left({x}−\mathrm{7}\right)…×\mathrm{2}×\mathrm{1}}×{p}\left(\mathrm{5}\right) \\ $$$${p}\left({x}\right)=\frac{{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)}{\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}×{p}\left(\mathrm{5}\right) \\ $$$$\Rightarrow{p}\left({x}\right)=\frac{{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)}{\mathrm{120}}×{p}\left(\mathrm{5}\right) \\ $$$${p}\left(−\mathrm{1}\right)=\frac{\left(−\mathrm{1}\right)\left(−\mathrm{2}\right)\left(−\mathrm{3}\right)\left(−\mathrm{4}\right)\left(−\mathrm{5}\right)}{\mathrm{120}}×{p}\left(\mathrm{5}\right)=\mathrm{1} \\ $$$$\Rightarrow{p}\left(\mathrm{5}\right)=−\mathrm{1} \\ $$$$\Rightarrow{p}\left({x}\right)=−\frac{{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)}{\mathrm{120}} \\ $$$$\Rightarrow{p}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{3}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{4}\right)}{\mathrm{120}}=−\frac{\mathrm{7}}{\mathrm{2}^{\mathrm{8}} }=−\frac{\mathrm{7}}{\mathrm{256}} \\ $$
Commented by Tawa11 last updated on 02/Jan/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Commented by Rasheed.Sindhi last updated on 02/Jan/22
Xcellent Sir! 🏆🏅🎖️🥇
$$\mathcal{X}{cellent}\:\mathcal{S}{ir}! \\ $$🏆🏅🎖️🥇
Answered by Rasheed.Sindhi last updated on 02/Jan/22
x.p(x−1)=(x−5).p(x)....(i) p(−1)=1 ; p((1/2))=? x−1→x (x+1).p(x)=(x−4).p(x−1)...(ii) (i)/(ii) (((x−5).p(x))/((x+1).p(x)))=((x.p(x−1))/((x−4).p(x−1))) (x−5)(x−4)=x(x+1) x^2 −9x+20=x^2 +x x=2 ?? −−−−−−−−−−−−−−−− x.p(x−1)=(x−5).p(x) x=0: 0.p(0−1)=(0−5).p(0) ⇒p(0)=0 x=1: 1.p(1−1)=(1−5).p(1) ⇒p(0)=−4.p(1) ⇒p(1)=0 x=−1: −1.p(−1−1)=(−1−5).p(−1) −p(−2)=−6.1 ⇒p(−2)=6 x.p(x−1)=(x−5).p(x) p(x)=(x/(x−5)).p(x−1) p(x)=(x/(x−5)).(x/(x−5)).p(x−2) =((x/(x−5)))^n p(x−n) x−n=−1 x=n−1 =(((n−1)/(n−6)))^n p(n−1−n) =(((n−1)/(n−6)))^n p(−1) =(((n−1)/(n−6)))^n (1) p(n−1)=(((n−1)/(n−6)))^n p(x)=((x/(x−5)))^(x+1) p((1/2))=(((1/2)/((1/2)−5)))^((1/2)+1) =(−(1/9))^(3/2)
$$\:\:{x}.{p}\left({x}−\mathrm{1}\right)=\left({x}−\mathrm{5}\right).{p}\left({x}\right)….\left({i}\right) \\ $$$$\:\:\:{p}\left(−\mathrm{1}\right)=\mathrm{1}\:\:;\:{p}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=? \\ $$$${x}−\mathrm{1}\rightarrow{x} \\ $$$$\:\:\left({x}+\mathrm{1}\right).{p}\left({x}\right)=\left({x}−\mathrm{4}\right).{p}\left({x}−\mathrm{1}\right)…\left({ii}\right) \\ $$$$\left({i}\right)/\left({ii}\right) \\ $$$$\frac{\left({x}−\mathrm{5}\right).\cancel{{p}\left({x}\right)}}{\left({x}+\mathrm{1}\right).\cancel{{p}\left({x}\right)}}=\frac{{x}.\cancel{{p}\left({x}−\mathrm{1}\right)}}{\left({x}−\mathrm{4}\right).\cancel{{p}\left({x}−\mathrm{1}\right)}} \\ $$$$\left({x}−\mathrm{5}\right)\left({x}−\mathrm{4}\right)={x}\left({x}+\mathrm{1}\right) \\ $$$${x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{20}={x}^{\mathrm{2}} +{x} \\ $$$${x}=\mathrm{2} \\ $$$$?? \\ $$$$−−−−−−−−−−−−−−−− \\ $$$$\:\:{x}.{p}\left({x}−\mathrm{1}\right)=\left({x}−\mathrm{5}\right).{p}\left({x}\right) \\ $$$${x}=\mathrm{0}: \\ $$$$\:\:\mathrm{0}.{p}\left(\mathrm{0}−\mathrm{1}\right)=\left(\mathrm{0}−\mathrm{5}\right).{p}\left(\mathrm{0}\right) \\ $$$$\:\:\:\:\Rightarrow{p}\left(\mathrm{0}\right)=\mathrm{0}\:\:\:\: \\ $$$$\:\:{x}=\mathrm{1}: \\ $$$$\:\:\mathrm{1}.{p}\left(\mathrm{1}−\mathrm{1}\right)=\left(\mathrm{1}−\mathrm{5}\right).{p}\left(\mathrm{1}\right) \\ $$$$\:\:\:\Rightarrow{p}\left(\mathrm{0}\right)=−\mathrm{4}.{p}\left(\mathrm{1}\right) \\ $$$$\:\:\:\Rightarrow{p}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{1}: \\ $$$$\:\:−\mathrm{1}.{p}\left(−\mathrm{1}−\mathrm{1}\right)=\left(−\mathrm{1}−\mathrm{5}\right).{p}\left(−\mathrm{1}\right) \\ $$$$−{p}\left(−\mathrm{2}\right)=−\mathrm{6}.\mathrm{1} \\ $$$$\Rightarrow{p}\left(−\mathrm{2}\right)=\mathrm{6} \\ $$$$\:\:\:\:{x}.{p}\left({x}−\mathrm{1}\right)=\left({x}−\mathrm{5}\right).{p}\left({x}\right) \\ $$$$\:\:{p}\left({x}\right)=\frac{{x}}{{x}−\mathrm{5}}.{p}\left({x}−\mathrm{1}\right) \\ $$$$\:\:{p}\left({x}\right)=\frac{{x}}{{x}−\mathrm{5}}.\frac{{x}}{{x}−\mathrm{5}}.{p}\left({x}−\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:=\left(\frac{{x}}{{x}−\mathrm{5}}\right)^{{n}} {p}\left({x}−{n}\right) \\ $$$${x}−{n}=−\mathrm{1} \\ $$$${x}={n}−\mathrm{1} \\ $$$$\:\:\:\:\:\:=\left(\frac{{n}−\mathrm{1}}{{n}−\mathrm{6}}\right)^{{n}} {p}\left({n}−\mathrm{1}−{n}\right) \\ $$$$\:\:\:\:\:\:=\left(\frac{{n}−\mathrm{1}}{{n}−\mathrm{6}}\right)^{{n}} {p}\left(−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:=\left(\frac{{n}−\mathrm{1}}{{n}−\mathrm{6}}\right)^{{n}} \left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{p}\left({n}−\mathrm{1}\right)=\left(\frac{{n}−\mathrm{1}}{{n}−\mathrm{6}}\right)^{{n}} \\ $$$$\:\:\:\:\:\:\:\:{p}\left({x}\right)=\left(\frac{{x}}{{x}−\mathrm{5}}\right)^{{x}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:{p}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{5}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}} =\left(−\frac{\mathrm{1}}{\mathrm{9}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$ \\ $$
Commented by mr W last updated on 02/Jan/22
you got p(0)=0 ⇒ p(1)=0 ⇒ p(2)=0... but with p(x)=((x/(x−5)))^(x+1) you′ll get p(0)=0 ✓ p(1)=(−(1/4))^2 =(1/(16)) ≠0 ??? p(2)=(−(2/3))^3 =−(8/(27)) ≠0 ??? p(−2)=(((−2)/(−7)))^(−1) =(7/2)≠6 ???
$${you}\:{got}\:{p}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow\:{p}\left(\mathrm{1}\right)=\mathrm{0}\:\Rightarrow\:{p}\left(\mathrm{2}\right)=\mathrm{0}… \\ $$$${but}\:{with}\:{p}\left({x}\right)=\left(\frac{{x}}{{x}−\mathrm{5}}\right)^{{x}+\mathrm{1}} \:{you}'{ll}\:{get} \\ $$$${p}\left(\mathrm{0}\right)=\mathrm{0}\:\checkmark \\ $$$${p}\left(\mathrm{1}\right)=\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{16}}\:\neq\mathrm{0}\:\:\:??? \\ $$$${p}\left(\mathrm{2}\right)=\left(−\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{3}} =−\frac{\mathrm{8}}{\mathrm{27}}\:\neq\mathrm{0}\:\:\:??? \\ $$$${p}\left(−\mathrm{2}\right)=\left(\frac{−\mathrm{2}}{−\mathrm{7}}\right)^{−\mathrm{1}} =\frac{\mathrm{7}}{\mathrm{2}}\neq\mathrm{6}\:\:\:??? \\ $$
Commented by Rasheed.Sindhi last updated on 02/Jan/22
Sir,I think there′s a serious error either in my answer or in the question. Actually I′ve no foundation regarding these kind of questions.So I′m not much confident. (I encountered them only in this forum only). Special THANKS to you to see my answer/comments in this critic way by which I could learn.I have feelings of your busyness in more important questions/work!
$$\mathcal{S}{ir},{I}\:{think}\:{there}'{s}\:{a}\:{serious}\:{error} \\ $$$${either}\:{in}\:{my}\:{answer}\:{or}\:{in}\:{the}\:{question}. \\ $$$${Actually}\:{I}'{ve}\:{no}\:{foundation}\:{regarding} \\ $$$${these}\:{kind}\:{of}\:{questions}.{So}\:{I}'{m}\:{not} \\ $$$${much}\:{confident}.\:\left({I}\:{encountered}\right. \\ $$$$\left.\:{them}\:{only}\:{in}\:{this}\:{forum}\:{only}\right). \\ $$$${Special}\:\mathcal{THANKS}\:{to}\:{you}\:{to}\:{see}\:{my} \\ $$$$\:{answer}/{comments}\:{in}\:{this}\:{critic} \\ $$$$\:{way}\:{by}\:{which}\:{I}\:{could}\:\boldsymbol{{learn}}.{I}\:{have} \\ $$$${feelings}\:{of}\:{your}\:{busyness}\:{in}\:{more} \\ $$$${important}\:{questions}/{work}! \\ $$
Commented by Rasheed.Sindhi last updated on 02/Jan/22
Sir,can we use newton′s identities approach in q#162649?
$$\mathcal{S}{ir},{can}\:{we}\:{use}\:{newton}'{s}\:{identities} \\ $$$${approach}\:{in}\:{q}#\mathrm{162649}? \\ $$
Commented by mr W last updated on 02/Jan/22
i also think something is wrong in the question.
$${i}\:{also}\:{think}\:{something}\:{is}\:{wrong}\:{in} \\ $$$${the}\:{question}. \\ $$
Commented by mr W last updated on 02/Jan/22
in question 162649 basically the newton′s identities are also applicable, but using it directly doesn′t help much.
$${in}\:{question}\:\mathrm{162649}\:{basically}\:{the} \\ $$$${newton}'{s}\:{identities}\:{are}\:{also}\: \\ $$$${applicable},\:{but}\:{using}\:{it}\:{directly} \\ $$$${doesn}'{t}\:{help}\:{much}. \\ $$
Commented by Rasheed.Sindhi last updated on 02/Jan/22
THANKS a lot SIR!
$$\mathbb{THANKS}\:\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{lot}}\:\mathcal{SIR}! \\ $$
Commented by mr W last updated on 02/Jan/22
i could get p(x)=−((x(x−1)(x−2)(x−3)(x−4))/(120)) therefore p((1/2))=−(7/(256)) see my working below.
$${i}\:{could}\:{get} \\ $$$${p}\left({x}\right)=−\frac{{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)\left({x}−\mathrm{4}\right)}{\mathrm{120}} \\ $$$${therefore}\:{p}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\frac{\mathrm{7}}{\mathrm{256}} \\ $$$${see}\:{my}\:{working}\:{below}. \\ $$

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