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Tower of powers

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  • #16
    Originally posted by Leonhard View Post
    And both you and Pixie are confusing this for merely.

    y = z^z^z

    Its rather

    y = z^z^z^z^z^z^z^z^...

    Meaning the process goes on forever. If look close at the power tower, you'll notice a small ellipsis at the top. I should have specified this verbally though, but I thought it was obvious when I wrote it in up-arrow notation.
    Is .999... = 1?

    Comment


    • #17
      Originally posted by lao tzu View Post
      It's a natural error. Everything else in PEMDAS evaluates "ties" left to right, except exponents, which evaluate right to left.
      Interesting. I've never really thought about it before, since Commutativity very often makes the question moot. However, 1/2/2=x would seem a good example of this discrepancy. Are there any others?
      "[Mathematics] is the revealer of every genuine truth, for it knows every hidden secret, and bears the key to every subtlety of letters; whoever, then, has the effrontery to pursue physics while neglecting mathematics should know from the start he will never make his entry through the portals of wisdom."
      --Thomas Bradwardine, De Continuo (c. 1325)

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      • #18
        Originally posted by Boxing Pythagoras View Post
        Interesting. I've never really thought about it before, since Commutativity very often makes the question moot. However, 1/2/2=x would seem a good example of this discrepancy. Are there any others?
        Like exponentiation, subtraction and division are neither associative nor commutative.

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        • #19
          Originally posted by lao tzu View Post
          Is .999... = 1?
          Yes. ".999...." is simply a way of expressing the limit as n->oo of the sum of 9x10-i as i ranges fom 1 to n. That limit is, in fact, 1.


          Jim

          PS Another simple way of validating .999...=1 is:

          x = .9999....
          10x = 9.9999......

          9.9999.... - .99999.... = 9

          thus (10x-x)=9 -> 9x=9 -> x=1
          Last edited by oxmixmudd; 01-30-2015, 08:57 AM.
          My brethren, do not hold your faith in our glorious Lord Jesus Christ with an attitude of personal favoritism. James 2:1

          If anyone thinks himself to be religious, and yet does not  bridle his tongue but deceives his own heart, this man’s religion is worthless James 1:26

          This you know, my beloved brethren. But everyone must be quick to hear, slow to speak and slow to anger; James 1:19

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          • #20
            Originally posted by Leonhard View Post
            Like so.

            And while we're looking at the order of operations, we need mo' parentheses here.

            As written, the LHS simplifies to W(2 ln(z))

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            • #21
              Originally posted by oxmixmudd View Post
              Yes. ".999...." is
              Sorry Jim, didn't mean to punk you on that one. It was an in-joke.

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              • #22
                Np - It's been that kind of day

                Jim
                My brethren, do not hold your faith in our glorious Lord Jesus Christ with an attitude of personal favoritism. James 2:1

                If anyone thinks himself to be religious, and yet does not  bridle his tongue but deceives his own heart, this man’s religion is worthless James 1:26

                This you know, my beloved brethren. But everyone must be quick to hear, slow to speak and slow to anger; James 1:19

                Comment


                • #23
                  Fixed another error,



                  Is wrong, but



                  Is correct.

                  Comment


                  • #24
                    sqrt(2)^sqrt(2)^sqrt(2) = 2

                    The tower grows from there.

                    Confused...

                    K54

                    Comment


                    • #25
                      Originally posted by klaus54 View Post
                      sqrt(2)^sqrt(2)^sqrt(2) = 2
                      This is incorrect. CodeCogsEqn (4).gif

                      You are confusing this Power Tower with a very different case, which is CodeCogsEqn (5).gif

                      These are two entirely different expressions.
                      "[Mathematics] is the revealer of every genuine truth, for it knows every hidden secret, and bears the key to every subtlety of letters; whoever, then, has the effrontery to pursue physics while neglecting mathematics should know from the start he will never make his entry through the portals of wisdom."
                      --Thomas Bradwardine, De Continuo (c. 1325)

                      Comment


                      • #26
                        Originally posted by Boxing Pythagoras View Post
                        This is incorrect. [ATTACH=CONFIG]3817[/ATTACH]

                        You are confusing this Power Tower with a very different case, which is [ATTACH=CONFIG]3818[/ATTACH]

                        These are two entirely different expressions.
                        You keep using [ATTACH=CONFIG ] expressions. There's really no need to actually downloading the gifs of these equations, and then to include them. That's just more trouble for you.

                        Simple use [img ][/img ], and use the handy URL encoding that the CodeCog's Equation Editor uses. Its located down under the editor. This loads faster, and uses less memory on the tweb server, as you offload the fetch to the CodeCog server which seems to be optimized to handle cases like this.

                        Here as example I wrote that equation below as [img ]http://latex.codecogs.com/png.latex?%5Cdpi%7B200%7D%20f%28z%29%20%3D%20z%5E% 7Bz%5E%7Bz%5E%7B...%7D%7D%7D[/img ]

                        Also... please... please please please, stop writing the tower power like



                        Ugly ugly ugly! You might as well just write it in ascii. This implies that the tower power is merely, z to the power of z which has been raised to the power z. Instead write it like this


                        (latex code: f(z)=z^{z^{z^{...}}})

                        So that people can see that this process goes on forever.
                        Last edited by Leonhard; 02-04-2015, 03:16 AM.

                        Comment


                        • #27
                          Originally posted by klaus54 View Post
                          sqrt(2)^sqrt(2)^sqrt(2) = 2

                          The tower grows from there.

                          Confused...

                          K54
                          First of all (and I'll write it in ascii this time), it's not,

                          f(z) = z^z^z

                          But

                          f(z) = z^z^z^z^z^z^z^...

                          Secondly I assume of course the ordinary order right right to left, but to make it clearer this is what it looks like with parenthesees.

                          f(z) = z^(z^(z^(z^(z^(z^(z^(z^(...))))))))

                          Comment


                          • #28
                            Originally posted by Leonhard View Post
                            First of all (and I'll write it in ascii this time), it's not,

                            f(z) = z^z^z

                            But

                            f(z) = z^z^z^z^z^z^z^...

                            Secondly I assume of course the ordinary order right right to left, but to make it clearer this is what it looks like with parenthesees.

                            f(z) = z^(z^(z^(z^(z^(z^(z^(z^(...))))))))
                            Ah, exponentiation is not associative.

                            How does one start at the "end" of an infinite sequence in order to work right to left??

                            K54

                            P.S. E.g., sqrt(2)^((sqrt(2))^(sqrt(2)) is approximately 1.76

                            This demonstrates where I was wrong.

                            P.P.S. And I had forgotten the right to left convention for evaluating exponents. Sorry...
                            Last edited by klaus54; 02-04-2015, 12:58 PM. Reason: ppps

                            Comment


                            • #29
                              Originally posted by klaus54 View Post
                              ...
                              How does one start at the "end" of an infinite sequence in order to work right to left?? (ETA: Or top to bottom when there's no "top"?)
                              ...
                              Did someone answer this? If so, what post number?

                              Thanks!

                              K54

                              P.S. ETA: I get it now... (took me a hour or so of cogitation.) Look at a sequence of the values of n towers as n increases without bound.
                              Last edited by klaus54; 02-04-2015, 10:18 PM.

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                              • #30
                                Infinite_power_tower.svg.png

                                Comment

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