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 Lambert W function and the Super Square Root James Knight Junior Fellow Posts: 6 Threads: 1 Joined: Mar 2008 03/25/2008, 10:09 PM (This post was last modified: 03/25/2008, 10:11 PM by James Knight.) I submitted this proof/research to the University of Waterloo for scholarship consideration about a month ago. Here goes. (note. it's not exactly the same as I am writing it from memory!) Since the beginning of Mathematics, Mathematicians have been trying to provide detailed solutions to complex problems. However, these solutions are limited to what the mathematic community has accepted and can interpret.In addition, there are many other limitations mathematicians may face when providing a solution to a problem. One of these limitations is the lack knowledge or ideas of a certain topic (ie. it's never been thought of before). Tetration is one of these ideas that did not exist because it was still in development. Johann Lambert found an equation that he could not solve using the algebra of his time. The equation looked similar to this; 2^t = 6t. Even though it looks simple, if one uses traditional algebra they will surely become frustrated (**traditional algebra does not include tetration). However, Lambert proposed that there must be a value or values which satisfy the equation. He considered the answer to w e^w = x to be w = W(x). In turn, this would be used to solve equations such as the first one given. Although he answered the problem, he was not able to absolutely deifine his function. That was in the 1700's. It's now 2008 and we have developed a broad range of mathematical knowledge and understanding. We are able now to accurately define the Lambert W function. Let me show you what I mean! Let us take the equation w(e^w) = x and put w in terms of x. Recall that Idea#1 (a^b)^a = a^ba Idea#2 (a^b)#2 = (a^b)^(a^b) = a^(ba^b) Idea#3if (a)^b = a^d b = d and reversely if b = d, a^b = a^d Then I will apply a base e ( make both sides the exponent of e) Idea #3 e^(we^w) = e^x Then I work backwards with idea #2 and #1 to produce: (e^w)#2 = e^x Take the super square root of both sides e^w = w = ln (ssrt (e^x)) Therefore W(x) = ln ssrt (e^X) We could use the New Lambert W function to solve equations like 2^t = 6t but it is truly unnecessary and inefficient. Using this new concept of exponential factoring and super square rooting we can easily solve this equation accurately. 2^t = 6t Divide both sides by 6 (2^t) 1/6 = t(2^(-t)) Multiply both sides by -1 -1/6 = -t (2^-t) Apply base of 2 to both sides. 2 ^(-1/6) = 2^(-t)(2^-t) Factor Exponent 2^(-1/6) = (2^(-t))#2 Super square root each side ssrt (2^(-1/6)) = 2 ^ (-t) Take the log base 2 of each side and multiply by -1 (note lb = log base 2 = binary log) t = - lb (ssrt( 2^(-1/6))) Therefore, because we have new ideas and broader knowledge, we are able to more accurately and precisely define the Lambert W function. In the future, we must always remember to keep an open mind when it comes to mathematics as new ideas and concepts are being created. Hoped you enjoy this!! James Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/25/2008, 11:01 PM Oops! Nice. Why not use new definitions straight! I hope it is correct as well, but I definitely like the approach. By applying new operations one can find out how to define them better. Ivars bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 03/26/2008, 01:05 PM (This post was last modified: 03/26/2008, 01:07 PM by bo198214.) James Knight Wrote:w = ln (ssrt (e^x))This is one of the well known (see wikipedia) formulas re the relation of the functions: $x=W(y)$ defined by $xe^x=y$, $x>-1$ $x=h(y)$ defined by $x^{1/x}=y$, $0 $x=\text{ssqrt}(y)$ defined by $x^x=y$, $x>1/e$ (The range conditions are necessary as neither of the functions is injective.) Well known is: $W(x)=\ln(\text{ssqrt}(e^x))$, $\text{ssqrt}(x)=e^{W(\ln(z))}$ $h(x)=\frac{W(-\ln(x))}{-\ln(x)}$, $W(x)=xh(e^{-x})$ Perhaps not so often mentioned is the derived formula: $h(x)=\frac{W(-\ln(x))}{-\ln(x)}=\frac{\ln(\text{ssqrt}(1/x))}{-\ln(x)}$ However the Lambert W function was not that revolutionary as you describe it. That the function $xe^x$ is injective for $x>-1$ (it has a minimum at -1 as you can easily derive from $0=(xe^x)'=e^x+xe^x=e^x(x+1)$) and its range is $-1/e\dots \infty$, so there *must* be exactly one solution for values in the range! However this solution is not expressible in terms of $+,-,*,/,\exp,\ln,\sin,\cos$ (though the proof such a non-expressability statement should be quite difficult). So its more about giving a name to this solution, i.e. $W$. And it turns out that with this additional function the solution of many other equations can be expressed (for example $ae^x+be^{cx}=d$ as an exercise). andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/29/2009, 06:30 AM (03/25/2008, 10:09 PM)James Knight Wrote: W = ln (ssrt (e^x)) Yes, indeed. I once thought that was unique, but now I have come to realize that it is an example of topological conjugacy. If you would like, there are many more equations like yours, encoded in the diagrams on page 48 of the Tetration Ref.. Topological conjugacy is at the very heart of continuous iteration. It underlies Abel functions, Schroeder functions, regular iteration, intuitive iteration, and the interesting correlation between x^^oo, W(x), and ssqrt(x). « Next Oldest | Next Newest »

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