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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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