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Question Number 184351 by mr W last updated on 05/Jan/23
f(x)=((a^x +b^x +c^x )/x) with a+b+c=0 prove f(7)=f(5)×f(2)
f(x)=ax+bx+cxxwitha+b+c=0provef(7)=f(5)×f(2)
Answered by greougoury555 last updated on 05/Jan/23
f(x)=((a^x +b^x +(−a−b)^x )/x) { ((f(7)=((a^7 +b^7 −(a+b)^7 )/7))),((f(5)=((a^5 +b^5 −(a+b)^5 )/5))),((f(2)=((a^2 +b^2 +(a+b)^2 )/2))) :} f(2)=((2(a^2 +b^2 +ab))/2) = a^2 +b^2 +ab f(5)=((−(5a^4 b+10a^3 b^2 +10a^2 b^3 +5ab^4 ))/5) f(5)=−ab(a^3 +2a^2 b+2ab^2 +b^3 ) f(7)=((−(7a^6 b+21a^5 b^2 +35a^4 b^3 +35a^3 b^4 +21a^2 b^5 +7ab^6 ))/7) f(7)=−ab(a^5 +3a^4 b+5a^3 b^2 +5a^2 b^3 +3ab^4 +b^5 )
f(x)=ax+bx+(−a−b)xx{f(7)=a7+b7−(a+b)77f(5)=a5+b5−(a+b)55f(2)=a2+b2+(a+b)22f(2)=2(a2+b2+ab)2=a2+b2+abf(5)=−(5a4b+10a3b2+10a2b3+5ab4)5f(5)=−ab(a3+2a2b+2ab2+b3)f(7)=−(7a6b+21a5b2+35a4b3+35a3b4+21a2b5+7ab6)7f(7)=−ab(a5+3a4b+5a3b2+5a2b3+3ab4+b5)
Commented by mr W last updated on 06/Jan/23
why not continue? when you multiple f(2) with f(5), you should get the same as f(7).
whynotcontinue?whenyoumultiplef(2)withf(5),youshouldgetthesameasf(7).
Answered by Rasheed.Sindhi last updated on 05/Jan/23
f(1)=0 f(2)=((a^2 +b^2 +c^2 )/2)=(((a+b+c)^2 −2(ab+bc+ca))/2) f(2)=−ab−bc−ca f(3)=((a^3 +b^3 +c^3 )/3) f(3)−abc=((a^3 +b^3 +c^3 )/3)−abc =((a^3 +b^3 +c^3 −3abc)/3) =(((a+b+c)(...))/3)=0 f(3)=abc Consider a polynomial equation with three roots: a,b,c x^3 −(a+b+c)x^2 +(ab+bc+ca)x−abc=0 x^3 −(−ab−bc−ca)x−abc=0 x^3 −xf(2)−f(3)=0 ....
f(1)=0f(2)=a2+b2+c22=(a+b+c)2−2(ab+bc+ca)2f(2)=−ab−bc−caf(3)=a3+b3+c33f(3)−abc=a3+b3+c33−abc=a3+b3+c3−3abc3=(a+b+c)(…)3=0f(3)=abcConsiderapolynomialequationwiththreeroots:a,b,cx3−(a+b+c)x2+(ab+bc+ca)x−abc=0x3−(−ab−bc−ca)x−abc=0x3−xf(2)−f(3)=0….
Answered by Rasheed.Sindhi last updated on 06/Jan/23
•a+b+c=0⇒c=−a−b ▶f(2)=((a^2 +b^2 +c^2 )/2)=(((a+b+c)^2 −2(ab+bc+ca))/2) =−ab−bc−ca=−ab−c(a+b) =−ab−(a+b)(−a−b) =−ab+(a+b)^2 •f(2) =a^2 +ab+b^2 ▶f(5)=((a^5 +b^5 +c^5 )/5)=((a^5 +b^5 +(−a−b)^5 )/5) =((a^5 +b^5 −(a+b)^5 )/5) =((−(5a^4 b+10a^3 b^2 +10a^2 b^3 +5ab^4 ))/5) •f(5) =−(a^4 b+2a^3 b^2 +2a^2 b^3 +ab^4 ) ▶ f(7)=((a^7 +b^7 +c^7 )/7)=((a^7 +b^7 +(−a−b)^7 )/7) =((a^7 +b^7 −(a+b)^7 )/7) •f(7)=−(a^6 b+3a^5 b^2 +5a^4 b^3 +5a^3 b^4 +3a^2 b^5 +ab^6 ) ▶ f(2)f(5): −a^4 b−2a^3 b^2 −2a^2 b^3 −ab^4 × (a^2 +ab+b^2 ) _(−) −a^6 b−2a^5 b^2 −2a^4 b^3 −a^3 b^4 −a^5 b^2 −2a^4 b^3 −2a^3 b^4 −a^2 b^5 − a^4 b^3 −2a^3 b^4 −2a^2 b^5 −ab^6 _(−) −a^6 b−3a^5 b^2 −5a^4 b^3 −5a^3 b^4 −3a^2 b^5 −ab^6 =f(7)
∙a+b+c=0⇒c=−a−b▸f(2)=a2+b2+c22=(a+b+c)2−2(ab+bc+ca)2=−ab−bc−ca=−ab−c(a+b)=−ab−(a+b)(−a−b)=−ab+(a+b)2∙f(2)=a2+ab+b2▸f(5)=a5+b5+c55=a5+b5+(−a−b)55=a5+b5−(a+b)55=−(5a4b+10a3b2+10a2b3+5ab4)5∙f(5)=−(a4b+2a3b2+2a2b3+ab4)▸f(7)=a7+b7+c77=a7+b7+(−a−b)77=a7+b7−(a+b)77∙f(7)=−(a6b+3a5b2+5a4b3+5a3b4+3a2b5+ab6)▸f(2)f(5):−a4b−2a3b2−2a2b3−ab4×(a2+ab+b2)−−a6b−2a5b2−2a4b3−a3b4−a5b2−2a4b3−2a3b4−a2b5−a4b3−2a3b4−2a2b5−ab6−−a6b−3a5b2−5a4b3−5a3b4−3a2b5−ab6=f(7)
Commented by mr W last updated on 06/Jan/23
fine! thanks for the interest sir!
fine!thanksfortheinterestsir!
Answered by mr W last updated on 06/Jan/23
Method I (a+b+c)^2 =a^2 +b^2 +c^2 +2(ab+bc+ca)=0 ⇒a^2 +b^2 +c^2 =−2(ab+bc+ca) ⇒f(2)=((a^2 +b^2 +c^2 )/2)=−(ab+bc+ca) (a+b+c)^3 =a^3 +b^3 +c^2 +3(a+b+c)(ab+bc+ca)−3abc=0 ⇒a^3 +b^3 +c^3 =3abc ⇒f(3)=((a^3 +b^3 +c^3 )/3)=abc (a^2 +b^2 +c^2 )(a^3 +b^3 +c^2 )=−6(ab+bc+ca)abc a^5 +b^5 +c^5 +(a^2 b^2 +b^2 c^2 +c^2 a^2 )(a+b+c)−abc(ab+bc+ca)=−6(ab+bc+ca)abc ⇒a^5 +b^5 +c^5 =−5(ab+bc+ca)abc ⇒f(5)=((a^5 +b^5 +c^5 )/5)=−(ab+bc+ca)abc (a^2 +b^2 +c^2 )(a^5 +b^5 +c^5 )=10(ab+bc+ca)^2 abc a^7 +b^7 +c^7 +(a^2 b^2 +b^2 c^2 +c^2 a^2 )(a^3 +b^3 +c^3 )−a^2 b^2 c^2 (a+b+c)=10(ab+bc+ca)^2 abc a^7 +b^7 +c^7 +3abc(a^2 b^2 +b^2 c^2 +c^2 a^2 )=10(ab+bc+ca)^2 abc a^7 +b^7 +c^7 +3abc[(ab+bc+ca)^2 −2abc(a+b+c)]=10(ab+bc+ca)^2 abc a^7 +b^7 +c^7 +3abc(ab+bc+ca)^2 =10(ab+bc+ca)^2 abc ⇒a^7 +b^7 +c^7 =7(ab+bc+ca)^2 abc ⇒f(7)=((a^7 +b^7 +c^7 )/7)=(ab+bc+ca)^2 abc f(2)×f(5)=(ab+bc+ca)^2 abc=f(7) ✓
MethodI(a+b+c)2=a2+b2+c2+2(ab+bc+ca)=0⇒a2+b2+c2=−2(ab+bc+ca)⇒f(2)=a2+b2+c22=−(ab+bc+ca)(a+b+c)3=a3+b3+c2+3(a+b+c)(ab+bc+ca)−3abc=0⇒a3+b3+c3=3abc⇒f(3)=a3+b3+c33=abc(a2+b2+c2)(a3+b3+c2)=−6(ab+bc+ca)abca5+b5+c5+(a2b2+b2c2+c2a2)(a+b+c)−abc(ab+bc+ca)=−6(ab+bc+ca)abc⇒a5+b5+c5=−5(ab+bc+ca)abc⇒f(5)=a5+b5+c55=−(ab+bc+ca)abc(a2+b2+c2)(a5+b5+c5)=10(ab+bc+ca)2abca7+b7+c7+(a2b2+b2c2+c2a2)(a3+b3+c3)−a2b2c2(a+b+c)=10(ab+bc+ca)2abca7+b7+c7+3abc(a2b2+b2c2+c2a2)=10(ab+bc+ca)2abca7+b7+c7+3abc[(ab+bc+ca)2−2abc(a+b+c)]=10(ab+bc+ca)2abca7+b7+c7+3abc(ab+bc+ca)2=10(ab+bc+ca)2abc⇒a7+b7+c7=7(ab+bc+ca)2abc⇒f(7)=a7+b7+c77=(ab+bc+ca)2abcf(2)×f(5)=(ab+bc+ca)2abc=f(7)✓
Commented by mr W last updated on 06/Jan/23
we see also f(5)=f(3)×f(2)
weseealsof(5)=f(3)×f(2)
Answered by mr W last updated on 06/Jan/23
Method II using Newton′s Identities let p_n =a^n +b^n +c^n e_1 =a+b+c e_2 =ab+bc+ca e_3 =abc p_1 =e_1 =0 p_2 =e_1 p_1 −2e_2 =−2e_2 p_3 =e_1 p_2 −e_2 p_1 +3e_3 =3e_3 p_4 =e_1 p_3 −e_2 p_2 +e_3 p_1 =−e_2 p_2 =2e_2 ^2 p_5 =e_1 p_4 −e_2 p_3 +e_3 p_2 =−e_2 p_3 +e_3 p_2 =−5e_2 e_3 p_7 =e_1 p_6 −e_2 p_5 +e_3 p_4 =−e_2 p_5 +e_3 p_4 =7e_2 ^2 e_3 f(n)=((a^n +b^n +c^n )/n)=(p_n /n) f(7)=(p_7 /7)=e_2 ^2 e_3 f(5)=(p_5 /5)=−e_2 e_3 f(2)=(p_2 /2)=−e_2 f(5)×f(2)=(−e_2 e_3 )(−e_2 )=e_2 ^2 e_3 =f(7)
MethodIIusingNewton′sIdentitiesletpn=an+bn+cne1=a+b+ce2=ab+bc+cae3=abcp1=e1=0p2=e1p1−2e2=−2e2p3=e1p2−e2p1+3e3=3e3p4=e1p3−e2p2+e3p1=−e2p2=2e22p5=e1p4−e2p3+e3p2=−e2p3+e3p2=−5e2e3p7=e1p6−e2p5+e3p4=−e2p5+e3p4=7e22e3f(n)=an+bn+cnn=pnnf(7)=p77=e22e3f(5)=p55=−e2e3f(2)=p22=−e2f(5)×f(2)=(−e2e3)(−e2)=e22e3=f(7)

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