Posers and Puzzles

Posers and Puzzles

  1. Joined
    11 Nov '05
    Moves
    43938
    21 Oct '07 15:33
    There are two positive numbers, A and B, which happen to have exactly the same decimals and in the same order as well.
    Another property of these numbers is that when you invert one of them you get the other, and vice versa.

    Is this information enough to decide which are these numbers? And if so, what are they?
  2. SubscriberAttilaTheHorn
    Erro Ergo Sum
    In the Green Room
    Joined
    09 Jul '07
    Moves
    461192
    21 Oct '07 16:07
    (√5 – 1)/2 = ca. 0.618033989. . . and the reciprocal is
    2/(√5 – 1) = ca. 1.618033989. . .
  3. SubscriberAttilaTheHorn
    Erro Ergo Sum
    In the Green Room
    Joined
    09 Jul '07
    Moves
    461192
    21 Oct '07 16:11
    For some reason, these numbers don't print out in the forum. the numbers are
    the square root of 5 minus 1and divide that answer by 2. the reciprocal is 2 divided by (the square root of 5 minus 1).
    these numbers are 0.618033989. . . and 1.618033989. . .
  4. Joined
    30 Oct '04
    Moves
    7797
    21 Oct '07 16:46
    Hm...
    Okay so let's say [A] is the integer part of A and [B] the integer part of B. Then A= [A]+x and B=[B]+x where x is the fractional part of both as the condition states. Then form 1/([A]+x) = [B]+x we get [A][B]+([A]+[B])x+x^2 = 1
    Since the numbers are positive [A].[B] is greater or equal to 1, unless either [A] or [B] is 0. Thus we have without loss of generality 1= [A]x+x^2 or
    x^2+[A]x-1=0. The positive root of this equation is x=(-[A]+sqrt([A]^2+4))/2. Now we need to find for what integers [A] this becomes less than 1 (since x is a fraction). So we need to evaluate how does the function (-n+sqrt(4+n^2))/2 behave with the increasing of n. This would always less than 1 since sqrt(4+n^2) is less then n+2 thus the final result is less than 1 for all n besides 0.
    Does that sound right?
  5. Standard memberwolfgang59
    Mr. Wolf
    at home
    Joined
    09 Jun '07
    Moves
    46122
    21 Oct '07 19:251 edit
    Originally posted by AttilaTheHorn
    For some reason, these numbers don't print out in the forum. the numbers are
    the square root of 5 minus 1and divide that answer by 2. the reciprocal is 2 divided by (the square root of 5 minus 1).
    these numbers are 0.618033989. . . and 1.618033989. . .
    Correct.

    Let the numbers we are looking for be (A+x) & x where A is an integer and x is between 0 and 1.

    Then we have
    A+x = 1/x

    x^2 + Ax = 1

    (x + A/2)^2 = 1 + (A^2)/4

    x = sqrt{1 + (A^2)/4} - A/2

    x = 1/2[sqrt{4 + (A^2)} - A]

    We can see that for A>1 x>1 therefore our only solution is A=1
    substituting for A.

    x=1/2[sqrt{5} -1]

    Nice problem with unexpected solution.
  6. Joined
    30 Oct '04
    Moves
    7797
    21 Oct '07 19:573 edits
    Originally posted by wolfgang59


    x = 1/2[sqrt{4 + (A^2)} - A]

    We can see that for A>1 x>1 therefore our only solution is A=1
    substituting for A.

    How do we see that?
    For A=2 for example we get 1/2 (sqrt (8) -2)) = sqrt (8)/2 - 1.
    Now sqrt (8) < 3 thus sqrt (8) /2 is less than 1,5.
    Then x=sqrt(8)/2 - 1 which is less than 1,5-1 = 0,5 which is less than 1.
    In fact for any integer larger than 0 we get an x smaller than 1.
    EDITED: Grrr something wrong with "less" sign.
  7. Standard memberwolfgang59
    Mr. Wolf
    at home
    Joined
    09 Jun '07
    Moves
    46122
    21 Oct '07 20:181 edit
    You are correct. Bad mistake by me - my scribblings were scrambled! Obviously more than one solution.


    ... and not being able to use 'less than' sign is a PAIN!! 🙁
  8. Standard memberwolfgang59
    Mr. Wolf
    at home
    Joined
    09 Jun '07
    Moves
    46122
    21 Oct '07 20:32
    Originally posted by FabianFnas
    There are two positive numbers, A and B, which happen to have exactly the same decimals and in the same order as well.
    Another property of these numbers is that when you invert one of them you get the other, and vice versa.

    Is this information enough to decide which are these numbers? And if so, what are they?
    Thanks to ilywrin the answer is NO. Insufficient info for a unique solution.
  9. Joined
    11 Nov '05
    Moves
    43938
    22 Oct '07 05:46
    wolfgang59 among others showed nicely the calculations leading to the numerical value by AttilaTheHorn among others. Well done!

    Next question:

    Calculate the value of the expression
    sqrt(1+sqrt(1+sqrt(1+ ... )))
    where ... stands for sqrt(1+ ... ) recursively
    and sqrt stands for square root.
  10. Joined
    30 Oct '04
    Moves
    7797
    22 Oct '07 06:52
    Originally posted by FabianFnas
    wolfgang59 among others showed nicely the calculations leading to the numerical value by AttilaTheHorn among others. Well done!

    Next question:

    Calculate the value of the expression
    sqrt(1+sqrt(1+sqrt(1+ ... )))
    where ... stands for sqrt(1+ ... ) recursively
    and sqrt stands for square root.
    I will try my best - not sure how "legal" is this but as a formal approach it might work. Let us denote x=sqrt(1+sqrt(1+(sqrt(1+...)))
    Now we can write sqrt(1+x)=x since the both continue to infinity (hopeful that this is okay). Thus we get again quadratic equation: x^2-x-1=0 or x=(1+sqrt5)x2^(-1) since the other root is obviously negative and x is strictly positive. Would that work as an answer?
  11. Standard memberwolfgang59
    Mr. Wolf
    at home
    Joined
    09 Jun '07
    Moves
    46122
    22 Oct '07 07:34
    Originally posted by ilywrin
    I will try my best - not sure how "legal" is this but as a formal approach it might work. Let us denote x=sqrt(1+sqrt(1+(sqrt(1+...)))
    Now we can write sqrt(1+x)=x since the both continue to infinity (hopeful that this is okay). Thus we get again quadratic equation: x^2-x-1=0 or x=(1+sqrt5)x2^(-1) since the other root is obviously negative and x is strictly positive. Would that work as an answer?
    ilywrin seems to have got it but one of his logic jumps is not clear (at least not to me!) Also I am unclear as to why he discarded one solution of his quadratic?

    Define x as ilywrin has done; then

    x=sqrt[1+sqrt(..

    square the equation

    x^2 = 1+x

    Solutions of which are x=(1+sqrt5)/2 and (1-sqrt5)/2
  12. Standard memberwolfgang59
    Mr. Wolf
    at home
    Joined
    09 Jun '07
    Moves
    46122
    22 Oct '07 09:46
    Originally posted by wolfgang59
    ilywrin seems to have got it but one of his logic jumps is not clear (at least not to me!) Also I am unclear as to why he discarded one solution of his quadratic?

    Define x as ilywrin has done; then

    x=sqrt[1+sqrt(..

    square the equation

    x^2 = 1+x

    Solutions of which are x=(1+sqrt5)/2 and (1-sqrt5)/2
    This is rubbish. I dont know why its rubbish but it is!

    Firstly by playing around a bit I can come up with many 'solutions'. For example consider squaring the equation x^2 = x + 1 then making substitution for x .... Let y=x^2 and you get a quadratic in y with incompatible results.

    BUT more importantly just LOKING at the original problem;

    Take one term and x = sqrt(1+d) so x is greater than 1.

    Take two terms and x = sqrt[1 + sqrt(1 + e)} so x is greater than 2.

    Take 3 terms and ....

    I can make it as big as I like!

    I therefore believe that Fabian's original problem is invalid .. although I dont understand why!
  13. Joined
    30 Oct '04
    Moves
    7797
    22 Oct '07 11:48
    Pretty much - what this problem is about is to find the limit of the sequence Kn, when n tends to infinity, and Kn = sqrt(1+(sqrt(1+...))) - n 1s total.
    K(n+1)>Kn but with how much? Let us define t(n)=K(n+1) -Kn . If t(n) tends to 0 as n grows to infinity then Kn is a bounded sequence, and we know that it has a limit. Else Kn grows arbitrarily large with n.
    But t(1)=sqrt (2) - 1; t(2)= sqrt(1+sqrt(2)) -sqrt(2); so we see that t(n) is a decreasing sequence. (one can prove this by induction). Since it has positive members it would tend to 0 when n tends to infinity.
    Then Kn is bounded and increasing.
    Thus I believe we can use the formal method. As to wolfgang's remark - we discard the negative root since the sequence has positive members, hence must have a positive limit.
  14. Standard memberPBE6
    Bananarama
    False berry
    Joined
    14 Feb '04
    Moves
    28719
    22 Oct '07 15:16
    Originally posted by wolfgang59
    This is rubbish. I dont know why its rubbish but it is!

    Firstly by playing around a bit I can come up with many 'solutions'. For example consider squaring the equation x^2 = x + 1 then making substitution for x .... Let y=x^2 and you get a quadratic in y with incompatible results.

    BUT more importantly just LOKING at the original problem;

    Take one ...[text shortened]... herefore believe that Fabian's original problem is invalid .. although I dont understand why!
    This is the rubbish post. When you needlessly complicate the problem by squaring both sides of x^2 = x + 1 and substitute in y = x^2, you don't get a quadratic because it contains a square root sign. Secondly, your logic with the successive square roots doesn't follow. It's true that if "d" is a positive real number, then SQRT(1+d) >= 1, but not true that if "e" is a positive real number then SQRT[1 + SQRT(1+e)] >= 2. For example, let "e" be 0.21, then SQRT(1+e) = 1.1, and SQRT(1+1.1) =~ 1.449 which is less than 2.

    The original problem is quite valid, and ilywrin has already posted the correct solution.
  15. Standard memberwolfgang59
    Mr. Wolf
    at home
    Joined
    09 Jun '07
    Moves
    46122
    22 Oct '07 17:45
    Originally posted by PBE6
    This is the rubbish post. When you needlessly complicate the problem by squaring both sides of x^2 = x + 1 and substitute in y = x^2, you don't get a quadratic because it contains a square root sign. Secondly, your logic with the successive square roots doesn't follow. It's true that if "d" is a positive real number, then SQRT(1+d) >= 1, but not true that if ...[text shortened]... e original problem is quite valid, and ilywrin has already posted the correct solution.
    OK! My post rubbishing my post is also rubbish.

    [hands held aloft & waving white flag!]
Back to Top