Math puzzle
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Check out this gif to know the puzzle: http://gpoulose.home.att.net/untitled.gif George
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Check out this gif to know the puzzle: http://gpoulose.home.att.net/untitled.gif George
I can't even begin to tell you how rusty my math is.... I haven't even had diffy Q's in like 4 years but if you follow the ... logic to its fullest value won't you be takeing some n root of inifinity which is still infinity if memory servers. This should cascade all the way up leaving you with infinity... Please do not shoot the rusty math student! Joseph Dempsey jdempsey@cox.rr.com "Software Engineering is a race between the programmers, trying to make bigger and better fool-proof software, and the universe trying to make bigger fools. So far the Universe in winning." --anonymous
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I can't even begin to tell you how rusty my math is.... I haven't even had diffy Q's in like 4 years but if you follow the ... logic to its fullest value won't you be takeing some n root of inifinity which is still infinity if memory servers. This should cascade all the way up leaving you with infinity... Please do not shoot the rusty math student! Joseph Dempsey jdempsey@cox.rr.com "Software Engineering is a race between the programmers, trying to make bigger and better fool-proof software, and the universe trying to make bigger fools. So far the Universe in winning." --anonymous
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Alright I'll give it a try... (do I get any prize? :) ) Let the original expression be X. Then X > sqrt(1 + 3 * sqrt(1 + 4 * sqrt( 1 + 5... ))) Observe that the RHS can be expressed as (X^2 - 1) / 2 Now, Solve the inequality X > (X^2 - 1)/2 <=> 2X > X^2 - 1 <=> (X - 1)^2 < 2 <=> -sqrt(2) + 1 < X < sqrt(2) + 1 However, also we can see from the original expression that X > 1. Since X is an integer, then X must equals to 2. Eddie!
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Alright I'll give it a try... (do I get any prize? :) ) Let the original expression be X. Then X > sqrt(1 + 3 * sqrt(1 + 4 * sqrt( 1 + 5... ))) Observe that the RHS can be expressed as (X^2 - 1) / 2 Now, Solve the inequality X > (X^2 - 1)/2 <=> 2X > X^2 - 1 <=> (X - 1)^2 < 2 <=> -sqrt(2) + 1 < X < sqrt(2) + 1 However, also we can see from the original expression that X > 1. Since X is an integer, then X must equals to 2. Eddie!
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Check out this gif to know the puzzle: http://gpoulose.home.att.net/untitled.gif George
The puzzle seems to have no solution at all, but... Let's denote the n-th root as x(n): x[n] = sqrt(1+n*x[n+1]), and then x[2] = sqrt(1+2*sqrt(1+3*x[4])), and so on. In other words, x[n+1] = (x[n] - 1)^2/n, and the question of the puzzle is - what is x[2] equal to, if n tends to infinity? It is obvious that the answer depends on how we choose the behavior of x[n]. One possible solution is x[n] = n+1, which gives x[2] = 3, but we can start from any predefined number and construct x[n] so that x[2] will be equal to that number! But if we require that all x[n] are positive, and not growing faster than some power of n, the above "solution" would be the only one. (I guess that if I propose such a solution on a college math exam, they will dismiss me momentarily, but physicists could be satisfied ;) Sorry for long posting - Yuri
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The puzzle seems to have no solution at all, but... Let's denote the n-th root as x(n): x[n] = sqrt(1+n*x[n+1]), and then x[2] = sqrt(1+2*sqrt(1+3*x[4])), and so on. In other words, x[n+1] = (x[n] - 1)^2/n, and the question of the puzzle is - what is x[2] equal to, if n tends to infinity? It is obvious that the answer depends on how we choose the behavior of x[n]. One possible solution is x[n] = n+1, which gives x[2] = 3, but we can start from any predefined number and construct x[n] so that x[2] will be equal to that number! But if we require that all x[n] are positive, and not growing faster than some power of n, the above "solution" would be the only one. (I guess that if I propose such a solution on a college math exam, they will dismiss me momentarily, but physicists could be satisfied ;) Sorry for long posting - Yuri
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Check out this gif to know the puzzle: http://gpoulose.home.att.net/untitled.gif George
Well, I cheated and wrote a little program to figure it out. ;P The answer is 3 according to this:
double recurse(int nValue)
{
const int nMax = 10000;if (nValue == nMax + 1) return sqrt(nMax); return sqrt(1 + nValue \* recurse(nValue + 1));}
void MathPuzzle()
{
cout << "Answer: " << recurse(2) << endl;
}Regards, Alvaro
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Well, I cheated and wrote a little program to figure it out. ;P The answer is 3 according to this:
double recurse(int nValue)
{
const int nMax = 10000;if (nValue == nMax + 1) return sqrt(nMax); return sqrt(1 + nValue \* recurse(nValue + 1));}
void MathPuzzle()
{
cout << "Answer: " << recurse(2) << endl;
}Regards, Alvaro
