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given x+1/x=2cosy
find x^n +1/x^n
my logic towards the problem is :
since x+1/x>=2..for x+1/x=2cosy to be valid cosy=1..
this implies x=1
thus 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 how
plz help
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.
you are correct...but its not the general solution..its a hit and trial,
the correctsolution is very easy
x+1/x=2cosy
x^2 +x(-2cosy)+1+0 (quadratic)
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...
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)
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
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