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Question-123265

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Question Number 123265 by Study last updated on 24/Nov/20
Commented by Study last updated on 24/Nov/20
which is a polynomial? why?
$${which}\:{is}\:{a}\:{polynomial}?\:{why}? \\ $$
Commented by mathmax by abdo last updated on 24/Nov/20
3)p(x)=(√3)is apolynom(csnstant)
$$\left.\mathrm{3}\right)\mathrm{p}\left(\mathrm{x}\right)=\sqrt{\mathrm{3}}\mathrm{is}\:\mathrm{apolynom}\left(\mathrm{csnstant}\right) \\ $$
Answered by MJS_new last updated on 24/Nov/20
1) x<0: p(x)=i(√(−x))×i(√(−x))=i^2 (√((−x)(−x)))=−x x≥0: p(x)=x ⇒ p(x)=∣x∣ 2) p(x)=∣x∣ 3) p(x)=(√3) a polynome of degree n usually is defined as p(x)=Σ_(j=0) ^n c_j x^j with c_j ∈C ⇒ only p(x)=(√3) is a polynome; its degree is 0
$$\left.\mathrm{1}\right) \\ $$$${x}<\mathrm{0}:\:{p}\left({x}\right)=\mathrm{i}\sqrt{−{x}}×\mathrm{i}\sqrt{−{x}}=\mathrm{i}^{\mathrm{2}} \sqrt{\left(−{x}\right)\left(−{x}\right)}=−{x} \\ $$$${x}\geqslant\mathrm{0}:\:{p}\left({x}\right)={x} \\ $$$$\Rightarrow\:{p}\left({x}\right)=\mid{x}\mid \\ $$$$\left.\mathrm{2}\right) \\ $$$${p}\left({x}\right)=\mid{x}\mid \\ $$$$\left.\mathrm{3}\right) \\ $$$${p}\left({x}\right)=\sqrt{\mathrm{3}} \\ $$$$\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{degree}\:{n}\:\mathrm{usually}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as} \\ $$$${p}\left({x}\right)=\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}{c}_{{j}} {x}^{{j}} \:\mathrm{with}\:{c}_{{j}} \in\mathbb{C} \\ $$$$\Rightarrow\:\mathrm{only}\:{p}\left({x}\right)=\sqrt{\mathrm{3}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynome};\:\mathrm{its}\:\mathrm{degree}\:\mathrm{is}\:\mathrm{0} \\ $$
Answered by mathmax by abdo last updated on 24/Nov/20
2)p(x)=(√x^2 )=∣x∣ is not polynomial let suppose it p^2 (x)=x^2 ⇒degp=1 ⇒p(x)=ax+b p(1)=1=a+b and p(−1)=1 =−a+b ⇒ { ((a+b=1)),((−a+b=1 ⇒)) :} b=1 ⇒a=0 ⇒p(x)=1 ?impossible p is not constant...!
$$\left.\mathrm{2}\right)\mathrm{p}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}^{\mathrm{2}} }=\mid\mathrm{x}\mid\:\:\mathrm{is}\:\mathrm{not}\:\mathrm{polynomial} \\ $$$$\mathrm{let}\:\mathrm{suppose}\:\mathrm{it}\:\:\mathrm{p}^{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} \:\Rightarrow\mathrm{degp}=\mathrm{1}\:\Rightarrow\mathrm{p}\left(\mathrm{x}\right)=\mathrm{ax}+\mathrm{b} \\ $$$$\mathrm{p}\left(\mathrm{1}\right)=\mathrm{1}=\mathrm{a}+\mathrm{b}\:\mathrm{and}\:\mathrm{p}\left(−\mathrm{1}\right)=\mathrm{1}\:=−\mathrm{a}+\mathrm{b}\:\Rightarrow\begin{cases}{\mathrm{a}+\mathrm{b}=\mathrm{1}}\\{−\mathrm{a}+\mathrm{b}=\mathrm{1}\:\Rightarrow}\end{cases} \\ $$$$\mathrm{b}=\mathrm{1}\:\Rightarrow\mathrm{a}=\mathrm{0}\:\Rightarrow\mathrm{p}\left(\mathrm{x}\right)=\mathrm{1}\:?\mathrm{impossible}\:\:\:\mathrm{p}\:\mathrm{is}\:\mathrm{not}\:\mathrm{constant}…! \\ $$
Commented by mathmax by abdo last updated on 24/Nov/20
1) p(x)=x with x≥0 is not polynom
$$\left.\mathrm{1}\right)\:\mathrm{p}\left(\mathrm{x}\right)=\mathrm{x}\:\:\mathrm{with}\:\mathrm{x}\geqslant\mathrm{0}\:\mathrm{is}\:\mathrm{not}\:\mathrm{polynom} \\ $$

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