Joining the nested radicals 1
What is the minimum value x can take in order for f ( x ) : f ( x ) = x + x + x + x + x + . . . to have a real number value assuming that it goes on to infinity?
If the minimum value of x can be expressed as − b a , find a + b . . . Try my Other Problems
The answer is 5.
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3 solutions
Ur solution has the same problems as the rest of the solutions here though (including mine)... See Calvin's comment.
let f(x) = y y=sqrt(x+y) dx/dy = 2y-1 = 0 (min value) y=1/2, x = -1/4 -> a=1, b=4, a+b=5
You can use this to help you with LATEX.
You cannot start off by assuming that f ( x ) = y is a finite value. How do you know that?
Solution without calculus
Let f ( x ) = S for simplicity.
S = x + S S 2 = x + S ( S − 0 . 5 ) 2 = x + 0 . 2 5 S = ± x + 0 . 2 5 + 0 . 5 So, x ≥ 0 . 2 5 in order for S to be a real number. Therefore the answer is − 0 . 2 5 = − 4 1 = − b a a + b = 5
Even though in this question it is not required, the solution S = − x + 0 . 2 5 + 0 . 5 can be discarded because it makes S < x
Not true.
Your solution starts of by assuming that S is a finite value, and starts to solve for S . Note that we do have ∞ = x + ∞ , and so you still need to establish why S = ∞ .
Second, you also need to show that it does converge to a finite value, as opposed to not converging at all.
Your question poses both of these issues, but your solution does not deal with them.
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Im not very sure how to address this. But S would not be infinite unless x is infinite because as in S 0 = x + S 1 , S 1 = x + S 2 . . . , S n = x + S n + 1 . . .
S 0 = S 1 = S 2 . . . = S n = . . . = S N ,
S N would have a less significant impact compared to S n on the value of S 0 given that N > n , as to the "impact" difference between S n and S n + 1 is a square root. I don't know how to explain this clearer though. And im not sure if my thinking is correct.
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The proper way to address this, is to remember that this expression is the limit of the sequence x , x + x , x + x + x . One possible way is to show that each of these values is lesser than some finite number ( x + 1 looks like a good possibility), which then shows that the limit of this sequence is not infinity.
I don't think this is correct
You want to say for minimum
4 x + 1 = 0
Which yields x = 4 − 1
But the solution
y = 2 1 + 4 x + 1
Is valid for x > 0
You can check by substituting x = 0 in formula which gives '1' as answer but we know it is '0'
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Not sure about x = 0 but if you substitute any number smaller than − 0 . 2 5 it would yield an imaginary. This method works for all the other numbers other than 0
EDIT: Assuming x = 0
S = S + 0
S 2 − S = 0
S = 0 or 1
Maybe for the case of 0 , we have to take the solution y = 2 1 − 4 x + 1
Perhaps there are more of this special cases that exists in other numbers (Maybe imaginary numbers?).
I didn't know this was a calculus problem until after I solved it..
Let y = f ( x ) . Square both sides to get y 2 = x + x + x + … and substitute like usual nested radicals to get y 2 = x + y . Rearranging gives a nice quadratic to work with, x = y 2 − y .
The minimum of a quadratic is at the y-value of the vertex, which is found by substituting 2 a − b for y where a and b are the usual coefficients of a quadratic..
x = ( 2 1 ) 2 − 2 1 = 4 − 1