For example, for n = 10, the expression is
(2^10)•√(2-√(2+√(2+
√(2+√(2+√(2+√(2+
√(2+√(2+√(2))))))))))
Note that the first sign is -, the rest being +
What is the value in the limiting case n -> ∞ ?
Update:Once in a while, it's actually worthwhile to check things out with a calculator. The answer is not 0, but a famous constant. What is the constant?
Update 3:zanti3, I can say that with π/2, you're literally halfway there! If you do this for 10 terms, as given in the example, you'll get an excellent approximation of this very famous constant widely known even by schoolchildren.
Update 5:All right, it's time to close this question. The proof involves lots of juggling starting from the trigonometric identity:
Cos(x/2) = √((1/2)(1+Cos(x))))
so that for n -> ∞, the limiting value for this expression is exactly π. For n = 10, you get 3.14159
Answers & Comments
Verified answer
Yes, it can be proven that √(2 + √(2 + √(2 + √(2 + ... = 2. However, 2^n is unbounded. So, this equation comes down to an undefined condition, that being infinity times zero.
My gut feeling is the value is unbounded because exponential functions increase so fast. I'm working on it...
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Geez... if I am following this, each new iteration of the quadratic takes the result half the distance closer to 0. That exactly offsets the doubling of the 2^n component. So, does that mean the limit is 1?? Man, Scythian, you ask some nasty questions.
OK, I tried calculating the first five terms to see where this is going. The series does converge rapidly. To three decimal places, n = 1.571. So, it converges to π/2?
It looks like π/2 is the answer. I did a Web search on formulas for π and found the following:
2/π = (√2 • √(2 + √2) • √((2 + √(2 + √2)) + ...) / 2^n
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Yes, looking at your n=10 example, I can see why you are getting π while I got π/2 -- you have nine √2's after the minus sign, i.e. one less than n. When I did the calculation for a few n values, I entered n √2's after the minus sign. The net effect of this difference is that my 2^n would have been half of your 2^n (your exponent is one bigger), so I'd get half your answer.
let me first compute
y = sqrt(2+sqrt(2+ .....))
then given value is (2^n)*sqrt(2-y)
now y = sqrt(2+y) as there are infinite terms I can do the same
reduction of one y shall not make a difference
so y^2 = 2+y
y^2-y-2 = 0
(y+1)(y-2) = 0
y= -1 or 2
but y cannot be -ve as sqrt is always positive( 2 and -2 both squares are 4 but sqrt(4) = 2)
so given value = 2^n(2-y) = 2^n(2-2) = 0
when n->infinite value is 0
Let x=â(2+â(2+â(2+â(2+...
x^2-2=n
x^2-x-2=0
(x+1)(x-2)=0
Assuming that we take the positive root, x=2.
(2^n)•â(2-â(2+...+â(2)))=(2^n)•â(2-2)
(2^n)•â(2-â(2+...+â(2)))=0