Menu Close

Question-163812

Tinku Tara 0 Comments
Pin




Question Number 163812 by mathlove last updated on 11/Jan/22
Answered by cortano1 last updated on 11/Jan/22
p(x)= x^3 +1 p(4)= 65
$$\:{p}\left({x}\right)=\:{x}^{\mathrm{3}} +\mathrm{1} \\ $$$$\:{p}\left(\mathrm{4}\right)=\:\mathrm{65} \\ $$
Commented by mathlove last updated on 11/Jan/22
how?
$${how}? \\ $$
Commented by Rasheed.Sindhi last updated on 11/Jan/22
To whom who liked the above post Would you like to tell the reason of liking please?
$$\:\:\:\:\:\:\:\:\:\:\:\:\mathcal{T}{o}\:{whom} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{who}\:{liked}\:{the}\:{above}\:{post} \\ $$$${Would}\:{you}\:{like}\:{to}\:{tell}\:{the}\:{reason}\:{of} \\ $$$${liking}\:{please}? \\ $$
Commented by mr W last updated on 11/Jan/22
maybe he just also wanted to know “how”. in this forum many people use “like” not to show that they really like a post. i think many people give themselves a like such that their post can not be red flagged or for what who knows.
$${maybe}\:{he}\:{just}\:{also}\:{wanted}\:{to}\:{know} \\ $$$$“{how}''.\:{in}\:{this}\:{forum}\:{many}\:{people} \\ $$$${use}\:“{like}''\:{not}\:{to}\:{show}\:{that}\:{they}\:{really} \\ $$$${like}\:{a}\:{post}.\:{i}\:{think}\:{many}\:{people}\:{give} \\ $$$${themselves}\:{a}\:{like}\:{such}\:{that}\:{their} \\ $$$${post}\:{can}\:{not}\:{be}\:{red}\:{flagged}\:{or}\:{for}\:{what} \\ $$$${who}\:{knows}. \\ $$
Commented by Rasheed.Sindhi last updated on 12/Jan/22
Maybe s/he has given a like to increase the importance of the post but ultimately it′ll be the cause to decrease importance of ′likes′ itself!
$$\mathcal{M}{aybe}\:{s}/{he}\:{has}\:{given}\:{a}\:{like}\:{to}\:{increase} \\ $$$${the}\:{importance}\:{of}\:{the}\:{post}\:{but}\:{ultimately} \\ $$$${it}'{ll}\:{be}\:{the}\:{cause}\:{to}\:{decrease}\:{importance} \\ $$$${of}\:'{likes}'\:{itself}! \\ $$
Answered by Ar Brandon last updated on 11/Jan/22
p(3)=28 therefore p(x) must be a non-zero polynomial. Let p(x)=a_0 +a_1 x+a_2 x^2 +∙∙∙+a_n x^n , a_n ≠0 Suppose that p(x)+p((1/x))=p(x)p((1/x)) for all x≠0 Then Σ_(r=0) ^n a_r x^r +Σ_(r=1) ^n (a_r /x^r )=(Σ_(r=0) ^n a_r x^r )(Σ_(r=1) ^n (a_r /x^r )) Multiplying through by x^n , we get that Σ_(r=0) ^n a_r x^(n+r) +Σ_(r=0) ^n a_r x^(n−r) =(Σ_(r=0) ^n a_r x^r )(Σ_(r=0) ^n a_r x^(n−r) ) That is, (a_0 x^n +a_1 x^(n+1) +∙∙∙a_n x^(2n) )+(a_0 x^n +a_1 x^(n−1) +∙∙∙+a_(n−1) x+a_n ) =(a_0 +a_1 x+∙∙∙a_n x^n )(a_0 x^n +a_1 x^(n−1) +∙∙∙+a_(n−1) x+a_n ) Equating the corresponding coefficients of powers of x, we have a_n =^x^0 a_0 a_n , a_(n−1) =^x^1 a_0 a_(n−1) +a_1 a_n a_(n−2) =^x^2 a_2 a_n +a_1 a_(n−1) +a_(n−2) a_0 2a_0 =^x^n a_0 ^2 +a_n ^2 a_n =^x^(2n) a_0 a_n ⇒a_0 =1 (since a_n ≠0) a_(n−1) =a_0 a_(n−1) +a_1 a_n ⇒a_1 a_n =0⇒a_1 =0 a_(n−2) =a_2 a_n +a_1 a_(n−1) +a_(n−2) a_0 ⇒a_(n−2) =a_2 a_n +a_(n−2) ⇒a_2 =0 Continuing this process, we get that a_(n−1) =0 and 2=1+a_n ^2 . Hence a_n =±1. Therefore p(x)=1±x^n Since we are given that p(3)=28 we have 28=1±3^n Therefore p(x) cannot be 1−x^n . Thus, p(x)=1+x^n and 28=1+3^n and hence n=3. So p(x)=1+x^3 and p(4)=65.
$${p}\left(\mathrm{3}\right)=\mathrm{28}\:\mathrm{therefore}\:{p}\left({x}\right)\:\mathrm{must}\:\mathrm{be}\:\mathrm{a}\:\mathrm{non}-\mathrm{zero}\:\mathrm{polynomial}. \\ $$$$\mathrm{Let}\:{p}\left({x}\right)={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +\centerdot\centerdot\centerdot+{a}_{{n}} {x}^{{n}} ,\:{a}_{{n}} \neq\mathrm{0} \\ $$$$\mathrm{Suppose}\:\mathrm{that}\:{p}\left({x}\right)+{p}\left(\frac{\mathrm{1}}{{x}}\right)={p}\left({x}\right){p}\left(\frac{\mathrm{1}}{{x}}\right)\:\mathrm{for}\:\mathrm{all}\:{x}\neq\mathrm{0} \\ $$$$\mathrm{Then}\:\underset{\mathrm{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{\mathrm{r}} {x}^{\mathrm{r}} +\underset{\mathrm{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{a}_{\mathrm{r}} }{{x}^{\mathrm{r}} }=\left(\underset{\mathrm{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{\mathrm{r}} {x}^{\mathrm{r}} \right)\left(\underset{\mathrm{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{a}_{\mathrm{r}} }{{x}^{\mathrm{r}} }\right) \\ $$$$\mathrm{Multiplying}\:\mathrm{through}\:\mathrm{by}\:{x}^{{n}} ,\:\mathrm{we}\:\mathrm{get}\:\mathrm{that} \\ $$$$\underset{\mathrm{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{\mathrm{r}} {x}^{{n}+\mathrm{r}} +\underset{\mathrm{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{\mathrm{r}} {x}^{{n}−\mathrm{r}} =\left(\underset{\mathrm{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{\mathrm{r}} {x}^{\mathrm{r}} \right)\left(\underset{\mathrm{r}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{\mathrm{r}} {x}^{{n}−\mathrm{r}} \right) \\ $$$$\mathrm{That}\:\mathrm{is}, \\ $$$$\left({a}_{\mathrm{0}} {x}^{{n}} +{a}_{\mathrm{1}} {x}^{{n}+\mathrm{1}} +\centerdot\centerdot\centerdot{a}_{{n}} {x}^{\mathrm{2}{n}} \right)+\left({a}_{\mathrm{0}} {x}^{{n}} +{a}_{\mathrm{1}} {x}^{{n}−\mathrm{1}} +\centerdot\centerdot\centerdot+{a}_{{n}−\mathrm{1}} {x}+{a}_{{n}} \right) \\ $$$$\:\:\:\:\:\:=\left({a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+\centerdot\centerdot\centerdot{a}_{{n}} {x}^{{n}} \right)\left({a}_{\mathrm{0}} {x}^{{n}} +{a}_{\mathrm{1}} {x}^{{n}−\mathrm{1}} +\centerdot\centerdot\centerdot+{a}_{{n}−\mathrm{1}} {x}+{a}_{{n}} \right) \\ $$$$\mathrm{Equating}\:\mathrm{the}\:\mathrm{corresponding}\:\mathrm{coefficients}\:\mathrm{of}\:\mathrm{powers}\:\mathrm{of}\:{x}, \\ $$$$\mathrm{we}\:\mathrm{have} \\ $$$$\:\:\:\:\:\:\:\:\:{a}_{{n}} \overset{{x}^{\mathrm{0}} } {=}{a}_{\mathrm{0}} {a}_{{n}} ,\:{a}_{{n}−\mathrm{1}} \overset{{x}^{\mathrm{1}} } {=}{a}_{\mathrm{0}} {a}_{{n}−\mathrm{1}} +{a}_{\mathrm{1}} {a}_{{n}} \\ $$$$\:\:\:\:\:\:\:\:\:{a}_{{n}−\mathrm{2}} \overset{{x}^{\mathrm{2}} } {=}{a}_{\mathrm{2}} {a}_{{n}} +{a}_{\mathrm{1}} {a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} {a}_{\mathrm{0}} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{2}{a}_{\mathrm{0}} \overset{{x}^{{n}} } {=}{a}_{\mathrm{0}} ^{\mathrm{2}} +{a}_{{n}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}_{{n}} \overset{{x}^{\mathrm{2}{n}} } {=}{a}_{\mathrm{0}} {a}_{{n}} \Rightarrow{a}_{\mathrm{0}} =\mathrm{1}\:\left(\mathrm{since}\:{a}_{{n}} \neq\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}_{{n}−\mathrm{1}} ={a}_{\mathrm{0}} {a}_{{n}−\mathrm{1}} +{a}_{\mathrm{1}} {a}_{{n}} \Rightarrow{a}_{\mathrm{1}} {a}_{{n}} =\mathrm{0}\Rightarrow{a}_{\mathrm{1}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}_{{n}−\mathrm{2}} ={a}_{\mathrm{2}} {a}_{{n}} +{a}_{\mathrm{1}} {a}_{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} {a}_{\mathrm{0}} \Rightarrow{a}_{{n}−\mathrm{2}} ={a}_{\mathrm{2}} {a}_{{n}} +{a}_{{n}−\mathrm{2}} \Rightarrow{a}_{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{Continuing}\:\mathrm{this}\:\mathrm{process},\:\mathrm{we}\:\mathrm{get}\:\mathrm{that}\:{a}_{{n}−\mathrm{1}} =\mathrm{0}\:\mathrm{and}\:\mathrm{2}=\mathrm{1}+{a}_{{n}} ^{\mathrm{2}} . \\ $$$$\mathrm{Hence}\:{a}_{{n}} =\pm\mathrm{1}.\:\mathrm{Therefore} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{p}\left({x}\right)=\mathrm{1}\pm{x}^{{n}} \\ $$$$\mathrm{Since}\:\mathrm{we}\:\mathrm{are}\:\mathrm{given}\:\mathrm{that}\:{p}\left(\mathrm{3}\right)=\mathrm{28}\:\mathrm{we}\:\mathrm{have} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{28}=\mathrm{1}\pm\mathrm{3}^{\mathrm{n}} \\ $$$$\mathrm{Therefore}\:{p}\left({x}\right)\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{1}−{x}^{{n}} .\:\mathrm{Thus},\:{p}\left({x}\right)=\mathrm{1}+{x}^{{n}} \:\mathrm{and} \\ $$$$\mathrm{28}=\mathrm{1}+\mathrm{3}^{{n}} \:\mathrm{and}\:\mathrm{hence}\:{n}=\mathrm{3}.\:\mathrm{So}\:{p}\left({x}\right)=\mathrm{1}+{x}^{\mathrm{3}} \:\mathrm{and}\:{p}\left(\mathrm{4}\right)=\mathrm{65}. \\ $$
Commented by mr W last updated on 11/Jan/22
good solution!
$${good}\:{solution}! \\ $$
Commented by Ar Brandon last updated on 11/Jan/22
����
Commented by Tawa11 last updated on 11/Jan/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *