Give an integral value equal to √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + .......?

Question

^_^

Answer 1

suppose given equation √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + .... = x
now x^2 = 6+√(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 +.....
x^2 = 6 + x
as the other part except 6 can be written as x again
x^2 - x -6 =0
x^2 -3x+2x-6 =0
x(x-3)+2(x-3)=0
(x-3)(x+2) = 0
so x can be either 3 or -2

Answer 2

Suppose you have a recurrence a[n+1] = sqrt(6 + a[n]) for some starting value a[0].

Remember Newton's method to iterate to a zero of f(x) by starting with an initial estimate a[0] such that a[0] is hopefully near enough to some some b such that f(b) = 0 that iterating...

a[n+1] = a[n] - f(a[n]) / f'(a[n])

...converges to b? So let's try to match these:

a[n] - f(a[n]) / f'(a[n]) = sqrt(6 + a[n])
x - f(x) / f'(x) = sqrt(6 + x)
x - sqrt(6 + x) = f(x) / f'(x)
f'(x) / f(x) = 1 / (x - sqrt(6 + x))
(ln f(x))' = 1 / (x - sqrt(6 + x))
(ln f(x))' = (x + sqrt(6 + x)) / ( (x + sqrt(6 + x)) (x - sqrt(6 + x)) )
(ln f(x))' = (x + sqrt(6 + x)) / (x^2 - (6 + x))
(ln f(x))' = (x + sqrt(6 + x)) / (x^2 - x - 6)
(ln f(x))' = (x + sqrt(6 + x)) / ((x - 3)(x + 2))

Letting u^2 = 6 + x so that dx = 2u du and x = u^2 - 6 the right hand side is
((u^2 - 6) + sqrt(u^2)) / ((u^2 - 6 - 3)(u^2 - 6 + 2))
(u^2 - 6 + |u|) / ((u^2 - 9)(u^2 - 4))
For u >= 0 with |u| = u this is (u^2 + u + 6) / ((u^2 - 9)(u^2 - 4))
= ((u + 3)(u - 2)) / ((u + 3)(u - 3)(u + 2)(u - 2))
= 1 / ((u - 3)(u + 2))
Likewise if u <= 0 so that |u| = - u you get
= 1 / ((u + 3)(u - 2))
Remember the factor of dx = 2u du and integrate:
|u| = u =>
2u du / ((u - 3)(u + 2)) = du ( (A / (u - 3)) + (B / (u + 2)) )
= du ( (6/5)/(u - 3) + (4/5)/(u + 2) )
(ln f(x))' = ((6/5)ln(u - 3) + (4/5)ln(u + 2) + C1)'
ln f(x) = (6/5)ln(u - 3) + (4/5)ln(u + 2) + C1
f(x) = C2 exp( (6/5)ln(u - 3) + (4/5)ln(u + 2) )
= C2 (u - 3)^(6/5) (u + 2)^(4/5)
= 0
So we can have u = 3 or u = - 2, |u| = u so throw out -2 and get u^2 = 6 + x = 3^2 = 9 so x = 3 is a root of f(x) = 0. So start with u = a[0] having square >= 6 and you iterate to 3.

Like they said. :-)

As a check, note that if a[n] = 3 - h for some h near 0, then a[n+1] = sqrt(6 + a[n]) = sqrt(6 + 3 - h) = sqrt(9 - h) = 3 sqrt(1 - h/9)
= 3 (1 - h/9)^(1/2)
= 3 (1 - (1/2)(h/9) + ... terms in h^2 and higher )
= 3 - h/6 + terms in h^2 and higher

so for suffuciently small h near zero, an iteration starting at 3 - h converges to 3

Answer 3

Let √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + .. = x.
therefore, x^2 = 6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + ..
thus x^2 = 6+x
=> x^2 - x - 6 = 0
Thus x = 3 or x = -2.
Bcoz x is a squareroot of some number, it cant be negative. Thus x = 3
Therefore,
√(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + √(6 + .. = 3

Answer 4

Its 6 6 6 what a devil of a question

Answer 5

x=sqrt(6(sqrt6....)
x^2=6+(sqrt 6(square root of 6...)
substitute the x
x^2=6+x
x^2-x-6=0

x^2-x+1/4=1/4+6
(x-1/2)^2=25/4
x-1/2=5/2
x=1/2-2.5
0r1/2+2.5

aprox. to equal to
x=3,-2
-2 is extranous root
so the answer is3

Answer 6

let this be y
since this an infinite series it won't cause anydifference if we take one sqrt 6 out from it.
that is

y=sqrt(6+y)
squaring both sides

y^2=6+y
which is a quadratic whose roots are 3 & -2 .
y can't be -ve so y = 3 .

Answer 7

the answer is 3

Answer 8

wrong answer by him too