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Question Number 85762 by Power last updated on 24/Mar/20

Commented by MJS last updated on 24/Mar/20

≈1.43312742672

1.43312742672

Commented by I want to learn more last updated on 24/Mar/20

If   x   =   ⌈1, 2, 3, 4, 5⌉  ∴        x   =   1  +  (1/(2  +  (1/(3  +  (1/(4  +  (1/(5  + x))))))))  ∴        x   =   1  +  (1/(2  +  (1/(3  +  ((5  +  x)/(21 + 4x))))))  ∴        x   =   1  +  (1/(2  +  (1/((63 + 12x + 5 + x)/(21 + 4x)))))  ∴        x   =   1  +  (1/(2  +  ((21 + 4x)/(68 + 13x))))  ∴        x   =   1  +  (1/((136 + 26x + 21 + 4x)/(68 + 13x)))  ∴        x   =   1  +  ((68 + 13x)/(157 + 30x))  ∴        x   =   ((157  +  30x  +  68  +  13x)/(157  +  30x))  ∴        x   =   ((225  +  43x)/(157  +  30x))  ∴          30x^2   +  157x   =  225  +  43x  ∴          30x^2   +  114x  − 225  =  0  a  =  30,   b  =  114,   c  =  − 225  ∴         x   =   ≈  1.433166662

Ifx=1,2,3,4,5x=1+12+13+14+15+xx=1+12+13+5+x21+4xx=1+12+163+12x+5+x21+4xx=1+12+21+4x68+13xx=1+1136+26x+21+4x68+13xx=1+68+13x157+30xx=157+30x+68+13x157+30xx=225+43x157+30x30x2+157x=225+43x30x2+114x225=0a=30,b=114,c=225x=1.433166662

Commented by Power last updated on 24/Mar/20

thank you sir

thankyousir

Commented by MJS last updated on 24/Mar/20

but the question includes “...”, which means  to infinity. you simply ignored this.

butthequestionincludes...,whichmeanstoinfinity.yousimplyignoredthis.

Commented by I want to learn more last updated on 24/Mar/20

It is a recurring  ⌈1, 2, 3, 4, 5⌉ sir

Itisarecurring1,2,3,4,5sir

Commented by I want to learn more last updated on 24/Mar/20

I think it approximate to your answer sir

Ithinkitapproximatetoyouranswersir

Commented by MJS last updated on 24/Mar/20

I think it′s [1,2,3,4,5,6,...]  my value is an approximation of this

Ithinkits[1,2,3,4,5,6,...]myvalueisanapproximationofthis

Commented by I want to learn more last updated on 24/Mar/20

Oh. I don′t know sir. i used  ⌈1, 2, 3, 4, 5⌉  as shown in the question.

Oh.Idontknowsir.iused1,2,3,4,5asshowninthequestion.

Commented by MJS last updated on 24/Mar/20

if I give you a sequence a_n =⟨1, 2, 3, 4, 5, ...⟩  would you really take it as  a_n =⟨1, 2, 3, 4, 5, 1, 2, 3, 4, 5, ...⟩ ?  I guess no

ifIgiveyouasequencean=1,2,3,4,5,...wouldyoureallytakeitasan=1,2,3,4,5,1,2,3,4,5,...?Iguessno

Commented by I want to learn more last updated on 24/Mar/20

That is infinite continue fraction sir.  It is assume that the sequence repeat in that manner.  As the dot dot dot ...

Thatisinfinitecontinuefractionsir.Itisassumethatthesequencerepeatinthatmanner.Asthedotdotdot...

Commented by MJS last updated on 24/Mar/20

this is just your interpretation  x=2+(1/(1+(1/(2+(1/(1+(1/(1+(1/(4+(1/(1+(1/(1+(1/(6+(1/(1+...))))))))))))))))))  x=e if we continue  2+[1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, ...]  if I want a repetition I must show the first  repeated number(s) like this:  2+[1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 2, ...]

thisisjustyourinterpretationx=2+11+12+11+11+14+11+11+16+11+...x=eifwecontinue2+[1,2,1,1,4,1,1,6,1,1,8,...]ifIwantarepetitionImustshowthefirstrepeatednumber(s)likethis:2+[1,2,1,1,4,1,1,6,1,1,2,...]

Commented by I want to learn more last updated on 25/Mar/20

Ok sir. Thanks.

Oksir.Thanks.

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