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4 solutions
Thanks for the compliment, although, this problem is inspired by another similar problem or shall I say, it is "stolen". Your solution is absolutely correct and I approve it (if that even means something to you). My way:
x = 1 5 6 + 1 5 6 + . . .
x 2 = 1 5 6 + 1 5 6 + 1 5 6 + . . .
x 2 = 1 5 6 + x
x 2 − x − 1 5 6 = 0
Nulls for this equations are:
2 1 ± 2 5
And because x is obviously positive, then x = − 1 2 .
Only solution, therefore is x = 1 3
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why x is obviously positive? x is a number which is square root of a positive number. sqrt(156 + x) = x can be proved when x is 13 or x is -12.
sqrt(156 + 13) = 13 and also sqrt(156 + (-12)) = sqrt(156 -12) = -12 (sqrt can be negative as well!)
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No, the solution of a square root of a positive number is defined as the positive solution.
x can be -12 and 13. The solution is wrong.
It is known that sqrt(y) can be +x or -x Proof Assume x x = y Multiplying two sides with -1 and -1 again (-1) x (-1) x = y -x -x = y Therefore if x x = y then -x*-x is also equal to y. Since square root of a number is defined as two equal numbers multiplied to get that number and -x = -x, -x is also a root of y.
x = -12 Proof: x = sqrt(156+x) -12 = sqrt(156-12) -12 = sqrt(144) -12 = (+/-)12 -12 = -12
x = 13 Proof: 13 = sqrt(169) 13 = (+/-)13 13 = 13
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After I saw your comment, I did some reading and stumbled across many discussions in which people told each other that only positive values are using because of convention. It makes things a lot easier to work without imaginary numbers and takes away the headache of understanding the real proof of why is it so. (For more reading, check out this discussion on Stack Exchange)
Suppose 1 5 6 + 1 5 6 + 1 5 6 + ⋯ = x
Now,
x = x + 1 5 6
x 2 = x + 1 5 6
x 2 − x − 1 5 6 = 0
We get x = 1 3
Forgive me if I'm being stupid, but isn't this question logically problematic? Essentially the answers given below go "if the limit is finite, then it must be 13", but how do we know that the limit is finite? We are effectively trying to find f(f(f(f(f(..., when we have no initial input value for the function f.
N.B. f(f(f(f(... doesn't make grammatical sense, whereas ...f(f(f(x)))... would. This latter value would indeed be 13.
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Well, only if x = 13! I can't remember my dynamical systems theory to work out what would happen otherwise...not obvious that iterations of f would converge anyway.
solution of the equation y =x^2 - x for every { x (i+1)= x (i) + 2n /n=1, 2, ............}, x(1) =2
Very nice problem of recursion!
To make the process simpler, we can use x , in such way: x = 1 5 6 + 1 5 6 + 1 5 6 + ⋯ .
Now it's easy to notice, that the problem can be re-written to: x = 1 5 6 + x
The number satisfying above equation is 1 3 because:
1 3 = 1 5 6 + 1 3
1 3 = 1 3 2
1 3 = 1 3
Answer: 1 3