Continued fractions:-
x = 2 + 2 + 2 + 2 + 2 + ⋱ 1 1 1 1
Given the above, what is the value of x ?
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10 solutions
reurirng
The reason we exclude the negative solution for x , is because we expect x to be the limiting value of the sequence below, for which all terms are positive.
2 , 2 + 2 1 , 2 + 2 + 2 1 1 , …
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Or simply it's 2 plus something positive.
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Is there a way to prove that this "something" is > 0 ?
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@A Former Brilliant Member – x > 0 because if x < 0, then 1/x is undefined.
Indeed! ^_^
why did you wrote it as X=2+1/x? please!
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Look at the denominator of the fraction on the RHS. It's essentially exactly the same as your original expression for x since the fraction continues indefinitely.
nice question ..............
This is best solution
some sort of pattern O O . But anyways, I got it right without guessing. ^ ^
can be easily understood ....thanx
don't you think before substituting that expression as x you require some justification like convergence or bounded ..etc
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To be equal to anything yes, this is to find what the value is assuming convergence.
Firstly, it is obvious that answer is positive and secondly it is converging as denominator is continuously adding the same thing!! But according to me there is no need any justification to solve it with above methodology.
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serious? okk then its fine to me...I hope you can prove it by some rigorously by using definition of sequential limit ....I have no problem ..i just expected to mention you that we really substitute that by x
this eqn becomes x= 2+1/x
now x= (2x+1)/x
then x^2=2x+1
this gives x^2-2x-1=0
this gives roots 1+ sqrt2
and 1- sqrt2
this eqn becomes x= 2+1/x now x= (2x+1)/x then x^2=2x+1 this gives x^2-2x-1=0 this gives roots 1+ sqrt2 and 1- sqrt2
equation is expressed as x = 2 +1/x that is x^2 -2x-1=0 and therefore value of x =1 + underroot 2 Ans K.K.GARG,India
Repeated continued fractions are always a memeber of the field Q [ 2 ] The only one of these options was 1 + 2
We can see that the fraction continues indefinitely. The only answer which continues indefinitely is 1 + sqrt(2), as sqrt(2) is irrational.
squar root of 2 is equl to 1.4142... so,ans is 2.4142...
Option 1: The methodical way... As the fraction continues infinitely many times, we can as well write the denominator as x. So, x = 2 + 1/x. Solving we get exactly x = 1 + /2. (Ignoring the negative root as RHS is a +ve quantity.) .. Option 2: The unorthodox way... The denominator is a +ve quantity > 2. Therefore, x < 2 + 1/2 = 2.5. Only option that satidfies this is 1+ /2 (Ignoring "none of these" option!)
here x=2+1/x it implies x*x=2x+1 ,so x=1+_root of 2
sqrt(N) = 1 + (N-1)/(2+(N-1)/(2+(N-1))...). So 1 + sqrt(N) = 2 + (N-1)/(2+(N-1)/(2+(N-1))...). When N=2, we get the equation presented in this problem. Therefore, the solution is 1 + sqrt(2). See Ben Paul Thurston Blog for more on the generalized continued fraction for square roots.
Just want to add another 'brute force' approach
2 + 1 / (2 + 1 / (2 + 1)) = 2.42857
2 + 1 / (2 + 1 / (2 + 1 / (2 + 1))) = 2.41176
2 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / (2 + 1)))) = 2.41463
1+\sqrt{2} = 2.41421
How can you guarantee that your terms will eventually converge to 1 + 2 ? (The same principle can be applied to iterations when trying to find roots of a polynomial.)