   Fun math puzzle
  (I use "fun" in the broadest possible sense.)Let S1 = sqrt[2], S2 = sqrt[2]sqrt[2], S3 = sqrt[2]sqrt[2]sqrt[2], and so on. Prove whether the limit as n approaches infinity of Sn exists; if it does, what is its value? If not, why doesn't it exist?(Note: nothing beyond precalculus is required, just common sense and an "aha!" moment.)(Note #2: Remember, exponentiation proceeds from the outside in, not the inside out. That is, 444 is 4256 [about 10^38], not 2564 [about 10^9].)(Note #3: This is not my homework. Besides, my final exams were three weeks ago!)   
  i smell homework....   ;  It looks like it tends to 2. I can prove by induction that it cannot exceed 2. Still thinking...   ;  Given -sqrt(2)2 = 2;sqrt(2)x < 2 iff x < 2;It can be shown numerically thatsqrt(2) ~= 1.414 < 2Therefore the expression does not approach infinity, since <insert technical-stuff here>. I'm too tired to figure out the rest.EDIT - :(sqrt(2)x == 2x/2I need some paper.hmm.222222222222222x/2/2/2/2/2/2/2/2/2/2/2/2/2/2/2Now, given -A DOT B = |A||B|cosCTime to swizzle!!A DOT B = A.x * B.x + A.y * B.y + A.z * B.z;|A| = sqrt( A.x2 + A.y2 + A.z2 );|B| = sqrt( B.x2 + B.y2 + B.z2 );LOOK AT ALL THOSE 2's!!! I THINK WE'RE ONTO SOMETHING.Now, if we do some funky math combinant string theory we get -cosC = A.x * B.x + A.y * B.y + A.z * B.z/( sqrt( A.x2 + A.y2 + A.z2 )*( sqrt( B.x2 + B.y2 + B.z2 ) );Starting to look familiar?We then need to use the quadrilateral divergence theorem to prove the limit of PI as x approaches infinity is equal to the fourth side of the triangle constructed from the initial expression, projected into Eulicidian space. Which is where things start to get a little fuzzy...ds2 = (dz1)2 + ... + (dzP)2 - (dzP+1)2) - ... - (dzP+Q)2;We now need to solve for the quantity s. While this may at first seem difficult, once you realize that s = S - 0;andsn = Sn - 0;The solution becomes somewhat more distinct.Therefore, the answer can only be -sn = Sn = (z1)2 + ... + (zP)2 - (zP+1)2) - ... - (zP+Q)2;Algebraic simplicification and substitution of the values of x, y and z (as shown above) give us the final answer -Sn = 2;QED.[Edited by - Mushu on May 30, 2006 11:40:26 PM];  It might be easier to figure it out if you write 2 to the 1/2 power rather than sqrt(2).EDIT - Mushu beat me to it :(   ;  By the logic mushu explained (which can more formally be proved with induction), 2 is an upper bound for S_{n}.  Now we may assume that lim S_{n} exists.  We can recursively define S_{n} as S_{n+1}=sqrt(2)^S_{n}.  Now, to solve for the limit, we can use the fact that as n approaches infinity, S_{n} tends to be constant.  Thus, we simply need to solve the equation S=sqrt(2)^S, for which 2 is clearly the solution.   ;  S_{1} < 2If S_{n} < 2, then S_{n+1} = sqrt[2]^S_{n} < sqrt[2]^2 = 2.  By induction S_{n} is bounded by 2.The function x^(1/x) is increasing for x < e, so S_{n}^S_{n} < 2^2.  Taking logarithms, multiplying through, and exponentiating, 2*ln(S_{n}) < S_{n}*ln(2), S_n < sqrt[2]^S_{n} = S_{n+1}.  Thus S_{n} is monotone increasing and bounded, so it converges.  I don't know how to show that the limit is two though without saying something like "as n approaches infinity, S_{n} tends to be constant."In the same spirit consider the sequence a_{0}=sqrt[2], a_{n)=sqrt[a_{n-1}+2].   ;Quote:Original post by SheepI don't know how to show that the limit is two though without saying something like "as n approaches infinity, S_{n} tends to be constant."Well, this is a perfectly valid argument.  Informally, you can say that since for all epsilon > 0, and some N, for all n,m > N, |S_{n}-S_{m}| < epsilon, there is no number that is smaller than the difference between any pair of elements in S_{n}, for all n > N.Another way of looking at it is to consider the limit of S_{n} as the "last" element of the sequence.  Then obviously if S_{n} = x, then S_{n+1} = x also.   ;Quote:Original post by eleusiveQuote:Original post by SheepI don't know how to show that the limit is two though without saying something like "as n approaches infinity, S_{n} tends to be constant."Well, this is a perfectly valid argument.  Informally, you can say that since for all epsilon > 0, and some N, for all n,m > N, |S_{n}-S_{m}| < epsilon, there is no number that is smaller than the difference between any pair of elements in S_{n}, for all n > N.That would only prove that S_{n} converges, which is trivial to prove other ways ie all the posts above, but says nothing of the fact that the limit is 2 (in fact, I dare you to explicitly prove convergence of this particular series via Cauchy-ness, ie find me a formula for N corresponding to an epsilon).  I actually want to know, how would you find an N such that |S_{n}-2| < epsilon for n > N?  I have no idea how to do this.Quote:Another way of looking at it is to consider the limit of S_{n} as the "last" element of the sequence.  Then obviously if S_{n} = x, then S_{n+1} = x also.Could you explain this further?  I don't understand what you mean.;Quote:Original post by SheepQuote:Original post by eleusiveQuote:Original post by SheepI don't know how to show that the limit is two though without saying something like "as n approaches infinity, S_{n} tends to be constant."Well, this is a perfectly valid argument.  Informally, you can say that since for all epsilon > 0, and some N, for all n,m > N, |S_{n}-S_{m}| < epsilon, there is no number that is smaller than the difference between any pair of elements in S_{n}, for all n > N.That would only prove that S_{n} converges, which is trivial to prove other ways ie all the posts above, but says nothing of the fact that the limit is 2 (in fact, I dare you to explicitly prove convergence of this particular series via Cauchy-ness, ie find me a formula for N corresponding to an epsilon).  I actually want to know, how would you find an N such that |S_{n}-2| < epsilon for n > N?  I have no idea how to do this.Quote:Another way of looking at it is to consider the limit of S_{n} as the "last" element of the sequence.  Then obviously if S_{n} = x, then S_{n+1} = x also.Could you explain this further?  I don't understand what you mean.For your first point, I also have no idea how to explicitly find a valid N given an epsilon.  It's tedious enough to do that given non-exponential functions, much less a hyper-exponential function.  But that's why we study real analysis [grin]For your second question, I'm just using the fact that S_n gets infinitely close to some value S.  If you consult a real analysis textbook, you can probrably find more formal information on how this kind of argument works (I was taught this technique in a real analysis course).   