Square Root till Infinity
posted by Rajesh Lal at
12:14 AM
Friday, January 9, 2009
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21 Comments:
0
X = 1
I'd have to say 1 too.
I would guess X=1
The only number that I can find that works is 0.
Here is an interesting addition based on the above formula...can you add the next 5 numbers in the sequence?
30, 42, 56, 72, 90
1, obviously
0, obviously
It has to be 0. An Infinite Sqrt series where 'X' is not zero would approach 1, but if that were the case, then the series could be simplified to Sqrt(x + 1) = x so 'x' must = 0
yeah, still zero... as for the random anon with the added question the next five are 110 132 156 182 210 etc...
0 ain't particularly interesting...
If the right-hand side converges, then you can substitute any part of it for the whole.
So, since x = sqrt(x + sqrt(x + ...)),
x = sqrt(x + x)
x = sqrt(2x)
This has two solutions, 0 and 2.
2 is more interesting, than 0, at least to me :)
Here are evaluations of the right-hand side for 1, 2, 3, 4, 5 square root terms with x = 2:
1: 1.41421 = sqrt(2)
2: 1.84776 = sqrt(2 + sqrt(2))
3: 1.96157 = sqrt(2 + sqrt(2 + sqrt(2)))
4: 1.99037 ...
5: 1.99759 ...
This seems to be clearly converging to 2, as the solution indicates.
Both 0 and 2 are solutions and equally uninteresting
i think we have all solved such questions as a part of the maths curriculum (quadratic equations if i remember correctly) in school
This puzzle has the answer as zero, but above was a better puzzle.
The answer to "the next five" in the sequence of...
30, 42, 56, 72, 90,
is
110, 132, 156, 182, 210.
Since we are on to the sequence, let's try to see where if comes from and how it relates to the original question...
Let's look at an expanded form of the original question: suppose the x on the left-hand side is not the same as the number in the nested square root thingy (the SRT)).
x = sqrt(n + sqrt(n + sqrt(n + ... )))
As before, if we assume the series converges, then, since it is infinite, any part of it can be substituted for the whole:
x = sqrt(n + x), then
x^2 = n + x
x^2 - x = n, or
n = x(x-1).
This means SRT(n) converges to x, when n = x(x-1) for any integer x>1.
So, 2 = SRT(2), 3 = SRT(6), 4 = SRT(12), etc.
The sequence of values of n for integer x's is the familiar:
2,6,12,20,30,42,56,72,...
And so, for example, sqrt(42 + sqrt(42 + ...)) = 7, since 42 = 7 * 6.
i say it's 1/infinity
one divided by infinity
which approaches zero infinitely, just as the reciprocal of zero approaches infinity...making your calculator say, 'error'
x+(infinity)=x^(infinity)
i powered the equation to infinity on both sides, canceling the SQRT and powering the other x to infinity, making 1 the only answer
because of one of the properties of radicals:
Proposition (Properties of Radicals) Let be a positive integer greater than 1, and let be a real number.Provided the indicated roots exist.
To simplify the radical means to remove factors from the radical until no factor in the radicand has an exponent greater than or equal to in the index of the radical and the index is as low as possible.
more steps, but i think i got it :D
this is like 0 to the 0 power...
0 to any power is 0..
a power of 0 to any number is 1..
x = the end of infinity.
x=SQRT(x+SQRT(x+SQRT(x+SQRT(x+...))))
x^2=x+SQRT(x+SQRT(x+SQRT(x+SQRT(x+...))))
x^2-x=SQRT(x+SQRT(x+SQRT(x+SQRT(x+...))))=x
x^2-x=x
x^2-2x=0
x(x-2)=0
so x can be 2 or 0 dayum.
Thanks for posting good answer. "guessing" that 1 will work without any proof doesn't help, especially because it doesn't work.
If we take the power 2 of both sides:
we get a 2 + (Same infinity root)=4
so (Infinity root) = 4-2 = 2
let root=y
take power of two sides
x+root=y^2
x=y^2 - root
yet y^2= root^2= x + root
therefore:
x= x + root - root
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