An Infinite Continued Fraction
1 + 2 × ⎝ ⎜ ⎜ ⎜ ⎜ ⎛ 1 + 2 × ( 1 + ⋱ 1 ) 1 ⎠ ⎟ ⎟ ⎟ ⎟ ⎞ 1 = ?
The answer is 0.5.
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1 solution
1 + 2 ( 1 + 2 ( 1 + … 1 ) 1 ) 1 = x 1 + 2 x 1 = x ⇒ 2 x 2 + x − 1 = 0
Why does it must be positive?
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Because the numbers you are dividing and adding are all positives, therefore, the result of the combined operations is a positive number.
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That doesn't have to mean it is not a valid solution, just like the sum of all natural numbers is -1/12. Please correct me if i'm wrong.
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@IKstreme 8 – I'm thinking the same thing.
@IKstreme 8 – I have been studying the subject and I've found this.
There's an Euler Theorem stating that the continued fraction solution to the general monic quadratic equation x 2 + b x + c = 0 with real coefficients b , c is given by x = − b − b − b − … c c c and converges or not, depending on both: the coefficient b and the value of the discriminant b 2 − 4 c .
1 ) If b = 0 , the general continued fraction solution is totally divergent; the convergents alternate between 0 and ∞ .
2 ) If b = 0 we distinguish three cases.
a ) If b 2 − 4 c < 0 , the fraction diverges by oscillation, which means that its convergents wander around in a regular or even chaotic fashion, never approaching a finite limit.
b ) If b 2 − 4 c = 0 , the fraction converges to the single root of multiplicity two.
c ) If b 2 − 4 c > 0 , the equation has two real roots, and the continued fraction converges to the largest of these.
@IKstreme 8 – Look at Riemann's reordering theorem. A divergent series, its sum can be whatever you want.
Same question if the function Is right that means it should be having to roots both valid for the function so why isn't -1 a correct answer is this a flaw in mathematics Like the idiotic question 0.9999999999= 1 proof can any one satisfy me with his / her solution
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x can't be -.5 because we can't divide by zero,and this series is convergent so its sum will be the usual sum not the sum for 1+2+3... which is assigning a value to it not actually summing it up ,you will approach ∞ if you do that ,
so the sum must be positive
1 + 2 ( 1 + 2 ( 1 + … 1 ) 1 ) 1 = x ⟹ 1 + 2 x 1 = x ⟹ 2 x 2 + x − 1 = 0 ⟹ ( x + 1 ) ( 2 x − 1 ) = 0 ∴ x = 2 1 , − 1 Because given expression contains only positive terms.Hence,neglecting -1. ∴ x = 0 . 5 0