 # given x+1/x=2cosyfind x^n +1/x^nmy logic towards the problem is :since x+1/x>=2..for x+1/x=2cosy to be valid cosy=1..this implies x=1thus x^n+1/x^n==1+1=2..this is for +ve x..so if x is negative answer is -2..therefore final ans is:2*sgn(x)but i am said that i am wrong...i dnt know where and howplz help

10 years ago

here we know x can only be 1 or -1. so if it is 1 ans is 2.

if x is -1, for n even ans will be 2

if x is -1 and n is odd ans will ne -2.

so we can see evenfor negative x also ans is 2 while sgn you were using would give ans -2 so you can see 2 sgn(x) is wrong.

so ans will be 2 ecxept when x is -1 and n is odd then it will be -2.

10 years ago

you are correct...but its not the general solution..its a hit and trial,

the correctsolution is very easy

x+1/x=2cosy

hence,

x=2cosy +  *sgn(4cosy^2 - 4)^1/2      /  2

x= 2cosy + 2*sgn(-siny^2)^1/2      /2

x= cosy + *sgn i.(siny)

hence x=            cosy + i.siny        or             cosy - i.siny

x^n=          cos ny + i.sin ny       or           cos ny - i.sin ny (de moivres theorem)

and 1/x^n =         cos ny - i.sin ny       or        cos ny + i.sin ny (de moivres theorem for (cis y)^(-n)   )

hence, x^n + 1/x^n = 2cosny (ANS)

you put x=1 and y=0,which is one set of answers for the question but does not relate x and y in amy manner.

hope i helped you...

10 years ago

they havent said that x is a real number. If you cast the equation and solve for x you will get x = cis(y) or cis(-y)

Hence the required answer is 2 cos (ny)

10 years ago

there is a small mistake... in your statement..that u gave

it is not for +ve x the answer is +2

and the answer is not-2 for -ve x

......the answer is,,, for x=+1 the function holds and the answer is+2

and for x=-1 the function holds and the answer is +2

so the function only holds true for x=+1or-1  not for all +ve and -ve x