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DaedalusSFN

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  1. Like
    DaedalusSFN reacted to exchemist in What Is Mass?   
    It will go on like this ad infinitum unless everyone stops responding. He will always find a way to give the wheel another spin. Until he is banned.
     
    Maybe now is the time to reveal how the algebra thread went that he started: http://sciforums.com/threads/algebra-help.144627/   This has to be deliberate, I think. 
  2. Like
    DaedalusSFN reacted to Buffy in What Is Mass?   
    Good! You can read a dictionary! :cheer:
     
    Unfortunately you've proven quite well that you do not know how to apply definitions to the real world and use-cases thereof. So:
     

    :umno: Properties are not "attracted."
    This is very confused.
     
    Unfortunately, it's not clear you're even trying to understand here.
     
    That's the thing that annoys people.
     
    So as regards inertia:
     

    Inertia is, of course, a property of mass itself, not "resistance" to the other "physical properties" of mass.
     
    You are obviously not aware of the fact that your statement there is basically gibberish.
     
    And that's annoying, something we have a rule against.
     
     
    You can pretend to be put-upon and treated unfairly here, but you've confused our tolerance of odd ideas as carte blanche to "innocently" "just ask questions" and then insist the answers are wrong and that you know more than Einstein. 
     
    In reality, we allow such behavior to a greater extent than other forums, but only insofar as it serves our purpose to embarrass such trolls in public, make them look foolish and generally put their heads on a pike for others of a similar bent who might wander in here.
     
    You've been warned: DBAA. 
     
     
    “Were you proposing to shoot these people in cold blood, sergeant?" "Nossir. Just a warning shot inna head, sir.” :phones:Buffy
  3. Like
    DaedalusSFN reacted to exchemist in What Is Mass?   
    I see, thanks to Daedelus's response on Antoine's other thread, that Antoine got banned from yet another forum, making at least three now in all, with these words:
     
    "I can't continue to put the rest of the membership through the kind of denial hell you inflict on discussions. You're suspended pending staff review, and I'm going to recommend that, because you can't follow the rules we have, we ban you so you can find a place that appreciates your remolding of science. "
     
    Sounds very rational to me. 
     
    Except that he then came here! 
  4. Like
    DaedalusSFN got a reaction from exchemist in I Think Time Dilation Semantics Are Incorrect?   
    Antoine, also known as JohnLesser at scienceforums.net (SFN), I see that you still haven't learned a single thing since you posted your "Time dilation busted!" thread at SFN. In that thread, I provided you a mathematical break down of how to derive time dilation in special relativity, and showed you that time dilation is not limited to units of Planck length or Planck time. I even used those units in the mathematics to demonstrate why your idea is wrong. Your reply to my post was:
     

    Your thread at SFN got closed because you think you know better than anyone else and now you are posting the same nonsense here at scienceforums.com. The sad thing is that you didn't even try to understand the post where I explained relativity to you because you are firm in the understanding that you must be right and everyone else is wrong. Quite frankly, I could care less if you wish to remain ignorant, at least I tried. There is no amount of nonsense that you can spout to change anyone's mind. To do that, you need to define your ideas with the same rigor that is expected of everyone else.
  5. Like
    DaedalusSFN reacted to DrKrettin in What Is Mass?   
    The problem is that you obviously do not appreciate the presence of an established body of science which involves thousands of professional scientists educated to a PhD level. The idea that you have a new theory without a similar background is frankly absurd. Why should people on this forum waste their time responding to "new theories" from somebody who does not even understand the old ones?
  6. Like
    DaedalusSFN reacted to Buffy in What Is Mass?   
    What will get you into trouble is the combination of these three things:
     
     
    and then:
     
     
    in conjunction with:
     

    You don't get to claim ignorance to justify demanding others accept your novel theories and then come back and say "everything the experts say is wrong."
     
    What will get you banned is violating our rule about "annoying our members," something best summarized by the acronym "DBAA."
     
    As you've discovered we're tolerant of alternative ideas, but what will get you moved to the Silly Claims forum faster than anything is whining that we "do not want to discuss science but rather dictate it."
     
    Remember: you've told us you haven't bothered to actually learn about the topics you're asking about, so if some one simply repeats that back to you, you don't get to say they're being unfair.
     
    We're quite happy to let people make fools of themselves in public and make examples of them.
     
    Welcome to Hypography, antoine!
     
     
    I'm disinclined to acquiesce to your request. Means "no," :phones:
    Buffy
  7. Like
    DaedalusSFN got a reaction from Turtle in Compositions By Daedalus   
    Thanks for the comments everyone. I'm glad you enjoyed my music. :note2: :note: :cool:
     

    I haven't read Hofstadter, but I'll check it out when I have time. As you can tell, my biggest influence is Bach.
     

    Yeah, it's one of my favorites too!
     

    April is also one of my favorites and, if you like the slower paced songs, then you'll like the next two I'm posting. This next song is titled, "Once Upon a Dream". It is my first pipe organ composition, and is heavily influenced by Bach. Enjoy!!!

    [video=youtube;HFCdhwZZ1yA]
  8. Like
    DaedalusSFN got a reaction from Turtle in Compositions By Daedalus   
    The lady who taught me how to read and write music retired last year. Without Mrs. Wood, I wouldn't be able to write the music that I do today. Great teachers are hard to find and, although we are in their lives for a brief period of time, their teachings remain with us forever. This song titled, "Where the Sky is Blue", is dedicated to her. Enjoy!!!

    [video=youtube;0a-79NAa2VE]
  9. Like
    DaedalusSFN got a reaction from DrKrettin in Compositions By Daedalus   
    This next song is titled, "Celeste". I really enjoyed writing the piano part in this one. Enjoy!!!

    [video=youtube;bS8Brxjv5Z4]
  10. Like
    DaedalusSFN got a reaction from DrKrettin in Compositions By Daedalus   
    This next song is titled, "April". I take the theme from the beginning of the song, transpose it in key, and then vary the music. I wrote this song for a friend who's half a world away and it turned out as lovely as her. Enjoy!!!
     
    [video=youtube;mMmNAj0imjc]
  11. Like
    DaedalusSFN got a reaction from sanctus in Compositions By Daedalus   
    Throughout our lives, people come and go. Sometimes they leave such an impression that we cannot help but miss them when they are gone. As I've said many times before, good music comes from the heart, and I wrote this song because there are a few people that I truly miss. Enjoy!!!
     
    [video=youtube;BYwf3Zl9Keg]
  12. Like
    DaedalusSFN got a reaction from Turtle in Compositions By Daedalus   
    This next song is titled, "Celeste". I really enjoyed writing the piano part in this one. Enjoy!!!

    [video=youtube;bS8Brxjv5Z4]
  13. Like
    DaedalusSFN got a reaction from sanctus in Iterated Exponents And Nested Logarithms   
    I was looking into operators beyond exponents when I discovered a few properties of iterated exponents and nested logarithms. I'm not claiming that I am the first to discover these properties. I am well aware of work that has been done on Tetration and the Ackermann function. However, I have not found these properties of iterated exponentiation and nested logarithms anywhere.
     
    Iterated Exponentiation
     
    Most of us are aware of nested exponents and how to simplify the math that uses them:
     
    [math]\left(a^b\right)^c=a^{b\, c}[/math]
     
    So, I began to wonder about iterated exponents and if I could derive any properties associated with them. Now, I do realize that iterated exponentiation is well known. However, let's define iterated exponentiation as
     
    [math]a^{\left \langle n \right \rangle} = \underbrace{(((a)^a)^a)^{\cdot^{\cdot^{a}}}}_n = a^{a^{n-1}}[/math]
     
    such that
     
    [math] a^{\left \langle 1 \right \rangle} = a = a^{a^{(1-1)}} [/math],
     
    [math] a^{\left \langle 2 \right \rangle} = (a)^a = a^{a^{(2-1)}} [/math],
     
    [math] a^{\left \langle 3 \right \rangle} = ((a)^a)^a = a^{a^{(3-1)}} [/math],
     
    [math] a^{\left \langle 4 \right \rangle} = (((a)^a)^a)^a = a^{a^{(4-1)}} [/math],
     
    etc...
     
    Now that we have the generalized form, [math]a^{a^{n-1}}[/math], that predicts the outcome of iterated exponentiation, we can derive the derivatives / integrals using Calculus and define an algorithm for determining iterated roots using the Newton-Rhapson method. However, the operation that I believe is new is what I call the nested logarithm.
     
    Nested Logarithms
     
    The nested log is just like normal logarithms except it determines the value of the iterated exponent:
     
    [math]\text{nLog}_{\, a}\left(a^{\left \langle b \right \rangle}\right)=b[/math]
     
    I've also managed to work out a few generalized forms of the nested logarithm:
     
    [math]\text{nLog}_{\, a}\left(b\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}[/math]
     
    [math]\text{nLog}_{\, a}\left(b\right)=1-\frac{\text{ln}\left(\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)}[/math]
     
     
    Proof that - [math]\text{nLog}_{\, a}\left(b\right)=1+\text{ln}(\text{ln}\left(b\right) / \text{ln}\left(a\right)) / \text{ln}\left(a\right)[/math]
     
    Let [math]a[/math] equal the base of the iterated exponent and let [math]b[/math] equal [math]a^{\left \langle n \right \rangle}[/math], which is also equal to [math]a^{a^{n-1}}[/math].
     
    [math]\text{nLog}_{\, a}\left(b\right)=\text{nLog}_{\, a}\left(a^{\left \langle n \right \rangle}\right)=\text{nLog}_{\, a}\left(a^{a^{n-1}}\right)=n[/math]
     
    We can now use the general form, [math]a^{a^{n-1}}[/math], to derive the nested logarithm by using natural logarithms:
     
    [math]\text{ln}\left(a^{a^{n-1}}\right)=a^{n-1} \, \text{ln}\left(a\right)[/math]
     
    Divide both sides by [math]\text{ln}\left(a\right)[/math]:
     
    [math]\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}=a^{n-1}[/math]
     
    Take the natural log of both sides:
     
    [math]\text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)=\left(n-1\right) \, \text{ln}\left(a\right)[/math]
     
    Again, divide both sides by [math]\text{ln}\left(a\right)[/math]:
     
    [math]\frac{\text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n-1[/math]
     
    Add one to both sides:
     
    [math]1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n[/math]
     
    Substitute [math]a^{\left \langle n \right \rangle}[/math] in place of [math]a^{a^{n-1}}[/math]:
     
    [math]\text{nLog}_{\, a}\left(a^{\left \langle n \right \rangle}\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left \langle n \right \rangle}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n[/math]
     
    Substitute [math]b[/math] in place of [math]a^{\left \langle n \right \rangle}[/math]:
     
    [math]\text{nLog}_{\, a}\left(b\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}[/math]
     
    Q.E.D.
     
     
    Proof that - [math]\text{nLog}_{\, a}\left(b\right)=1-\text{ln}(\text{ln}\left(a\right) / \text{ln}\left(b\right)) / \text{ln}\left(a\right)[/math]
     
    Working backwards, we can prove that the second form is true. Let [math]a[/math] equal the base of the nested exponent and let [math]b[/math] equal [math]a^{\left \langle n \right \rangle}[/math], which is also equal to [math]a^{a^{n-1}}[/math] such that:
     
    [math]1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{\left \langle n \right \rangle}\right)}\right)}{\text{ln}\left(a\right)}=1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}{\text{ln}\left(a\right)}=n[/math]
     
    Move [math]n[/math] and the term using the natural logarithms to the opposite side of the equation:
     
    [math]1-n=\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}{\text{ln}\left(a\right)}[/math]
     
    Multiply both sides by [math]\text{ln}\left(a\right)[/math]:
     
    [math](1-n)\, \text{ln}\left(a\right)=\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)[/math]
     
    Raise [math]e[/math] to the power of each side:
     
    [math]e^{(1-n)\, \text{ln}\left(a\right)}=e^{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}[/math]
     
    Simplify the result:
     
    [math]a^{1-n}=\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}[/math]
     
    Multiply both sides by [math]\text{ln}\left(a^{a^{n-1}}\right)[/math]:
     
    [math]a^{1-n}\, \text{ln}\left(a^{a^{n-1}}\right)=\text{ln}\left(a\right)[/math]
     
    Divide both sides by [math]a^{1-n}[/math]:
     
    [math]\text{ln}\left(a^{a^{n-1}}\right)=\frac{\text{ln}\left(a\right)}{a^{1-n}}[/math]
     
    Raise [math]e[/math] to the power of each side:
     
    [math]e^{\text{ln}\left(a^{a^{n-1}}\right)}=e^{\left(\frac{\text{ln}\left(a\right)}{a^{1-n}}\right)}[/math]
     
    Simplify the result:
     
    [math]a^{a^{n-1}}=a^{a^{n-1}}[/math]
     
    Q.E.D.
     
     
    Trivial Identities
     
    [math]a^{\left \langle 0 \right \rangle}=a^{a^{(0-1)}}=a^{a^{-1}}=a^{\frac{1}{a}}=\sqrt[a]{a}[/math]
     
    [math]a^{\left \langle 1 \right \rangle}=a^{a^{(1-1)}}=a^{a^0}=a[/math]
     
    [math]a^{\left \langle 2 \right \rangle}=a^{a^{(2-1)}}=a^{a^1}=a^a[/math]
     
     
    [math]\left(a^{\left \langle b \right \rangle}\right)^{a^c}=\left(a^{a^{b-1}}\right)^{a^c}=a^{a^{b-1} a^{c}}=a^{a^{b+c-1}}=a^{\left \langle b+c \right \rangle}[/math]
     
    [math]\left(a^{\left \langle b \right \rangle}\right)^{a^{-c}}=\left(a^{a^{b-1}}\right)^{a^{-c}}=a^{a^{b-1} a^{-c}}=a^{a^{b-c-1}}=a^{\left \langle b-c \right \rangle}[/math]
     
     
    [math]\text{nLog}_{\, a}\left(a\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(a\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=1+\frac{\text{ln}\left(1\right)}{\text{ln}\left(a\right)}=1+\frac{0}{\text{ln}\left(a\right)}=1[/math]
     
    Both, [math]\text{nLog}_{\, a}\left(0\right)[/math] and [math]\text{nLog}_{\, a}\left(1\right)[/math], are undefined:
     
    [math]\lim_{b \to 0}\big(\text{nLog}_{\, a}\left(b\right)\big) =\lim_{b \to 0} \left(1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}\right)=\infty[/math]
     
    [math]\lim_{b \to 1}\big(\text{nLog}_{\, a}\left(b\right)\big) =\lim_{b \to 1} \left(1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}\right)=-\infty[/math]
     
     
    Properties of Nested Logarithms
     
    [math]\text{nLog}_{\, a}\left(b\right)=1+\frac{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)} \ \ \ \ \ \ \ \text{nLog}_{\, a}\left(b\right)=1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)}[/math]
     
    [math]\text{nLog}_{\, a}\left(b\right)=1+\text{log}_{a}\left(\text{log}_{a}\left(b\right)\right) \ \ \ \ \ \ \ \text{nLog}_{\, e}\left(b\right)=1+\text{ln}\left(\text{ln}\left(b\right)\right)[/math]
     
    [math]\text{nLog}_{\, a}\left(b\right)=1-\frac{\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)}{\text{ln}\left(a\right)} \ \ \ \ \ \ \ \text{log}_{\, a}\left(b\right)=\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}=a^{\left(\text{nLog}_{\, a}\left(b\right)-1\right)}[/math]
     
    [math]\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(a\right) \left(\text{nLog}_{\, a}\left(b\right) - 1\right) \ \ \ \ \ \ \ \text{ln}\left(\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)=\text{ln}\left(a\right) \left(1-\text{nLog}_{\, a}\left(b\right)\right)[/math]
     
     
    Relationship to Exponentials
     
    [math]a^b=a\ \frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}=a\ \text{log}_{\, a}\left(a^{\left \langle b \right \rangle}\right)[/math]
     
    Proof:
     
    Iterated exponentiation by definition:
     
    [math]a^{\left \langle b \right \rangle}=a^{a^{(b-1)}}[/math]
     
    Take the natural log of both sides:
     
    [math]\text{ln}\left(a^{a^{(b-1)}}\right)=\text{ln}\left(a^{\left \langle b \right \rangle}\right)[/math]
     
    Simplify the result:
     
    [math]a^{(b-1)}\, \text{ln}\left(a\right)=\text{ln}\left(a^{\left \langle b \right \rangle}\right)[/math]
     
    Divide both sides by [math]\text{ln}\left(a\right)[/math]:
     
    [math]a^{(b-1)}=\frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}[/math]
     
    Multiply both sides by [math]a[/math]:
     
    [math]a^{b}=a\ \frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}[/math]
     
    Convert [math]\text{ln}\left(a^{\left \langle b \right \rangle}\right) / \text{ln}\left(a\right)[/math] to a logarithm by the "changing the base" identity:
     
    [math]a^{b}=a\ \text{log}_{\, a}\left(a^{\left \langle b \right \rangle}\right)[/math]
     
    Q.E.D.
     
     
    Relationship between Nested Logarithms and Natural Nested Logarithms
     
    [math]\text{nLog}_{\, a}\left(b\right) = 1+\frac{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}{\text{ln}\left(a\right)}[/math]
     
    Proof:
     
    Natural nested logarithm by definition:
     
    [math]\text{nLn}\left(x\right)=1+\text{ln}\left(\text{ln}\left(x\right)\right)[/math]
     
    Nested logarithm by definition:
     
    [math]\text{nLog}_{\, a}\left(b\right) =1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}= 1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}[/math]
     
    such that
     
    [math]1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}=1+\frac{\left(1+\text{ln}\left(\text{ln}\left(b\right)\right)\right)-\left(1+\text{ln}\left(\text{ln}\left(a\right)\right)\right)}{\text{ln}\left(a\right)}[/math]
     
    Simplify the right side by cancelling the ones in the numerator:
     
    [math]1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}=1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}[/math]
     
    Q.E.D.
     
     
    Relationship between the Difference of Natural Nested Logarithms
     
    [math]\text{nLn}\left(b\right) - \text{nLn}\left(a\right)=\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)[/math]
     
    such that
     
    [math]e^{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}=e^{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}=\text{log}_{\, a}\left(b\right)[/math]
     
    Proof:
     
    [math]1+\frac{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}{\text{ln}\left(a\right)}=1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}[/math]
     
    Subtract one from both sides:
     
    [math]\frac{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}{\text{ln}\left(a\right)}=\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}[/math]
     
    Multiply both sides by [math]\text{ln}\left(a\right)[/math]:
     
    [math]\text{nLn}\left(b\right) - \text{nLn}\left(a\right)=\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)[/math]
     
    Raise [math]e[/math] to the power of both sides:
     
    [math]e^{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}=e^{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}[/math]
     
    Undo the subtraction in the exponent:
     
    [math]\frac{e^{\text{nLn}\left(b\right)}}{e^{\text{nLn}\left(a\right)}}=\frac{e^{\text{ln}\left(\text{ln}\left(b\right)\right)}}{e^{\text{ln}\left(\text{ln}\left(a\right)\right)}}[/math]
     
    Simplify the result:
     
    [math]\frac{e^{1+\text{ln}\left(\text{ln}\left(b\right)\right)}}{e^{1+\text{ln}\left(\text{ln}\left(a\right)\right)}}=\frac{e\ \text{ln}\left(b\right)}{e\ \text{ln}\left(a\right)}=\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}[/math]
     
    Cancel out the [math]e[/math] and change the base:
     
    [math]e^{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}=\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}=\text{log}_{\, a}\left(b\right)[/math]
     
    Q.E.D.
     
     
    The Nested Logarithm of a Doubly Iterated Exponential with the Same Base (or something like that lol)
     
    [math]\text{nLog}_{\, a}\left(\left(a^{\left \langle b \right \rangle}\right)^{\left \langle c \right \rangle}\right)=\left(c-1\right)\, a^{(b-1)}+b[/math]
     
    Proof:
     
    [math]\left(a^{\left \langle b \right \rangle}\right)^{\left \langle c \right \rangle}=\left(a^{\left \langle b \right \rangle}\right)^{\left(a^{\left \langle b \right \rangle}\right)^{(c-1)}}=\left(a^{a^{(b-1)}}\right)^{\left(a^{a^{(b-1)}}\right)^{(c-1)}}=[/math]
     
    [math]\left(a^{a^{(b-1)}}\right)^{\left(a^{(c-1)\, a^{(b-1)}}\right)}=a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}[/math]
     
    Take the natural log of the simplified version:
     
    [math]\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)=\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right) \text{ln}\left(a\right)[/math]
     
    Divide both sides by [math]\text{ln}\left(a\right)[/math]:
     
    [math]\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}=\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)[/math]
     
    Take the natural log of both sides:
     
    [math]\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)\right)[/math]
     
    Simplify the right hand side:
     
    [math]\text{ln}\left(\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)\right)=\text{ln}\left(a^{(b-1)}\right) + \text{ln}\left(a^{(c-1)\, a^{(b-1)}}\right)=[/math]
     
    [math](b-1) \, \text{ln}\left(a\right)\, + \, (c-1)\, a^{(b-1)} \, \text{ln}\left(a\right)=\text{ln}\left(a\right)\, \left((b-1)\, +\, (c-1)\, a^{(b-1)}\right)[/math]
     
    such that
     
    [math]\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(a\right)\, \left((b-1)\, +\, (c-1)\, a^{(b-1)}\right)[/math]
     
    Divide both sides by [math]\text{ln}\left(a\right)[/math]:
     
    [math]\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=(b-1) \, + \, (c-1)\, a^{(b-1)}[/math]
     
    Complete the nested logarithm by adding one to both sides:
     
    [math]1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=1+(b-1) \, + \, (c-1)\, a^{(b-1)}[/math]
     
    Simplify the result:
     
    [math]\text{nLog}_{\, a}\left(\left(a^{\left \langle b \right \rangle}\right)^{\left \langle c \right \rangle}\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=(c-1)\, a^{(b-1)}\, +\, b[/math]
     
    Q.E.D.
     
     
    Derivatives of Iterated Exponents and Nested Logarithms (Found by using Mathematica)
     
    [math]\frac{dy}{dx} \ x^{\left \langle a \right \rangle} = x^{x^{(a-1)}}\left(x^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)=x^{\left \langle a \right \rangle}\left(x^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)[/math]
     
    [math]\frac{dy}{dx} \ a^{\left \langle x \right \rangle} = a^{a^{(x-1)}} \, a^{(x-1)} \, \text{ln}\left(a\right)^2=a^{\left \langle x \right \rangle} \left(a^{(x-1)} \, \text{ln}\left(a\right)^2\right)[/math]
     
    [math]\frac{dy}{dx} \ x^{\left \langle x \right \rangle} = x^{x^{(x-1)}}\left(x^{(x-2)}+x^{(x-1)}\, \text{ln}\left(x\right)\left(\text{ln}\left(x\right)+\frac{x-1}{x}\right)\right)=x^{\left \langle x \right \rangle}\left(x^{(x-2)}+x^{(x-1)}\, \text{ln}\left(x\right)\left(\text{ln}\left(x\right)+\frac{x-1}{x}\right)\right)[/math]
     
    [math]\frac{dy}{dx} \ \text{nLog}_{\, a}\left(x\right) = \frac{1}{x\, \text{ln}\left(x\right)\, \text{ln}\left(a\right)}=\frac{\text{nLog}_{\, a}\left(x\right)}{x\, \text{ln}\left(x\right) \left(\text{ln}\left(a\right) +\text{ln}\left(\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)\right)}[/math]
     
    [math]\frac{dy}{dx} \ \text{nLog}_{\, x}\left(a\right) = \frac{\text{ln}\left(\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)-1}{x\, \text{ln}\left(x\right)^2}=\frac{\text{ln}\left(x\right) \left(1-\text{nLog}_{\, x}\left(a\right)\right)-1}{x\, \text{ln}\left(x\right)^2}[/math]
     
     
    Properties of Derivatives for Iterated Exponents (Found by using Mathematica)
     
    An iterated exponential function that has the constant as the nested exponent, [math]x^{\left \langle a \right \rangle}[/math]:
     
    [math]\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle a \right \rangle}\right) = \left(x^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)[/math]
     
    Which allows us to define the following by substitution:
     
    [math]\frac{dy}{dx} \ x^{\left \langle a \right \rangle} = x^{\left \langle a \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(x^{\left \langle a \right \rangle}\right)[/math]
     
    such that:
     
    [math]x^{\left \langle a \right \rangle}=\frac{\frac{dy}{dx} \ x^{\left \langle a \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle a \right \rangle}\right)}[/math]
     
     
    An iterated exponential function that has the variable as the nested exponent, [math]a^{\left \langle x \right \rangle}[/math]:
     
    [math]\frac{dy}{dx} \ \text{ln}\left(a^{\left \langle x \right \rangle}\right) = \left(a^{(x-1)} \, \text{ln}\left(a\right)^2\right)[/math]
     
    Which allows us to define the following by substitution:
     
    [math]\frac{dy}{dx} \ a^{\left \langle x \right \rangle} = a^{\left \langle x \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(a^{\left \langle x \right \rangle}\right)[/math]
     
    such that:
     
    [math]a^{\left \langle x \right \rangle}=\frac{\frac{dy}{dx} \ a^{\left \langle x \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(a^{\left \langle x \right \rangle}\right)}[/math]
     
     
    An iterated exponential function that has the variable as the base and as the nested exponent, [math]x^{\left \langle x \right \rangle}[/math]:
     
    [math]\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle x \right \rangle}\right) = \left(x^{(x-2)}+x^{(x-1)}\, \text{ln}\left(x\right)\left(\text{ln}\left(x\right)+\frac{x-1}{x}\right)\right)[/math]
     
    Which allows us to define the following by substitution:
     
    [math]\frac{dy}{dx} \ x^{\left \langle x \right \rangle} = x^{\left \langle x \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(x^{\left \langle x \right \rangle}\right)[/math]
     
    such that:
     
    [math]x^{\left \langle x \right \rangle}=\frac{\frac{dy}{dx} \ x^{\left \langle x \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle x \right \rangle}\right)}[/math]
     
     
    Integrals of Iterated Exponents and Nested Logarithms (Found by using Mathematica)
     
    There are no general expressions for the integrals. They are based on the exponential integral, logarithmic integral, or unknown:
     
    [math]\int x^{\left \langle a \right \rangle} \, dx = \int x^{x^{a-1}} \, dx = \text{unknown}[/math]
     
    [math]\int a^{\left \langle x \right \rangle} \, dx = \int a^{a^{x-1}} \, dx = \frac{\text{Ei}\left(a^{x-1}\, \text{ln}(a)\right)}{\text{ln}(a)}+C[/math]
     
    [math]\int x^{\left \langle x \right \rangle} \, dx = \int x^{x^{x-1}} \, dx = \text{unknown}[/math]
     
    [math]\int \text{nLog}_{\, a}\left(x\right) \, dx = \int 1\, +\, \frac{\text{ln}\left (\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)} \, dx = x \, + \, \frac{x \, \text{ln}\left(\frac{\text{ln}(x)}{\text{ln}(a)}\right)}{\text{ln}(a)}\, -\, \frac{\text{li}(x)}{\text{ln}(a)}\, +\, C = x\, \left(\text{nLog}_{\, a}\left(x\right)\right)\, -\, \frac{\text{li}(x)}{\text{ln}(a)}\, +\, C[/math]
     
    [math]\int \text{nLog}_{\, x}\left(a\right) \, dx = \int 1+\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(x\right)}\right)}{\text{ln}\left(x\right)} \, dx = \text{unknown}[/math]
     
    The exponential integral, [math]\text{Ei}\left(x\right)[/math]:
     
    [math]\text{Ei}(x)=\int \limits_{-\infty}^{x} \frac{e^t}{t} \, dt[/math]
     
    The logarithmic integral, [math]\text{li}\left(x\right)[/math]:
     
    [math]\text{li}(x)=\int \limits_{0}^{x} \frac{dt}{\text{ln}(t)}[/math]
     
     
    Finding Iterated Roots and Nested Logarithms using the Newton-Rhapson Method
     
    First it is important to note that the Newton-Rhapson method may fail to converge to the desired result. One must read and understand the proof of quadratic convergence for Newton's iterative method which is described in Wikipedia. With that being said I have not had a problem finding iterated roots or nested logarithms using this approach.
     
    [math]x_{n+1}=x_{n} - \frac{f(x_{n})}{f'(x_{n})}[/math]
     
    Iterated Roots:
     
    To find the iterated root we must use the iterated exponential function that has the constant as the iterated exponent, [math]x^{\left \langle a \right \rangle}[/math]. Using this function with Newton's method yields the following algorithm for iterated roots (Repeat the process until you have obtained the desired result):
     
    [math]\sqrt[\left \langle a \right \rangle]{b}\ \ \ \ \text{approximated by} \ \ \ \ x_{n+1}=x_{n} - \frac{\left(x_{n}\right)^{\left(x_{n}\right)^{(a-1)}}-b}{\left(x_{n}\right)^{\left(x_{n}\right)^{(a-1)}}\left(\left(x_{n}\right)^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)}[/math]
     
    Nested Logarithms:
     
    To find the nested logarithm we must use the iterated exponential function that has the variable as the iterated exponent, [math]a^{\left \langle x \right \rangle}[/math]. Using this function with Newton's method yields the following algorithm for nested logarithms (Repeat the process until you have obtained the desired result):
     
    [math]\text{nLog}_{\, a}\left(b\right)\ \ \ \text{approximated by} \ \ \ x_{n+1}=x_{n} - \frac{a^{a^{\left(x_{n}\right)-1}}-b}{a^{a^{\left(x_{n}\right)-1}} a^{\left(x_{n}\right)-1} \text{ln}\left(a\right)^2}[/math]
     
     
    Graph of [math]x^{\left \langle a \right \rangle}[/math]:
     
    This graph shows both the real part (blue) and the imaginary part (green).
     

     
     
    Graph of [math]a^{\left \langle x \right \rangle}[/math]:
     
    This graph shows both the real part (blue) and the imaginary part (green).
     

     
     
    Graph of [math] x^{\left \langle x \right \rangle}[/math]:
     
    This graph shows both the real part (blue) and the imaginary part (green).
     

     
     
    Graph of [math]\text{nLog}_{\, a} \left(x\right)[/math]:
     
    This graph shows both the real part (blue) and the imaginary part (green).
     

     
     
    Graph of [math]\text{nLog}_{\, x} \left(a\right)[/math]:
     
    This graph shows both the real part (blue) and the imaginary part (green).
     

     
     
    Graph of [math] \text{nLog}_{\, x} \left(x\right)[/math]:
     
    This graph shows both the real part (blue) and the imaginary part (green).
     

  14. Like
    DaedalusSFN got a reaction from Buffy in Iterated Exponents And Nested Logarithms   
    I was looking into operators beyond exponents when I discovered a few properties of iterated exponents and nested logarithms. I'm not claiming that I am the first to discover these properties. I am well aware of work that has been done on Tetration and the Ackermann function. However, I have not found these properties of iterated exponentiation and nested logarithms anywhere.
     
    Iterated Exponentiation
     
    Most of us are aware of nested exponents and how to simplify the math that uses them:
     
    [math]\left(a^b\right)^c=a^{b\, c}[/math]
     
    So, I began to wonder about iterated exponents and if I could derive any properties associated with them. Now, I do realize that iterated exponentiation is well known. However, let's define iterated exponentiation as
     
    [math]a^{\left \langle n \right \rangle} = \underbrace{(((a)^a)^a)^{\cdot^{\cdot^{a}}}}_n = a^{a^{n-1}}[/math]
     
    such that
     
    [math] a^{\left \langle 1 \right \rangle} = a = a^{a^{(1-1)}} [/math],
     
    [math] a^{\left \langle 2 \right \rangle} = (a)^a = a^{a^{(2-1)}} [/math],
     
    [math] a^{\left \langle 3 \right \rangle} = ((a)^a)^a = a^{a^{(3-1)}} [/math],
     
    [math] a^{\left \langle 4 \right \rangle} = (((a)^a)^a)^a = a^{a^{(4-1)}} [/math],
     
    etc...
     
    Now that we have the generalized form, [math]a^{a^{n-1}}[/math], that predicts the outcome of iterated exponentiation, we can derive the derivatives / integrals using Calculus and define an algorithm for determining iterated roots using the Newton-Rhapson method. However, the operation that I believe is new is what I call the nested logarithm.
     
    Nested Logarithms
     
    The nested log is just like normal logarithms except it determines the value of the iterated exponent:
     
    [math]\text{nLog}_{\, a}\left(a^{\left \langle b \right \rangle}\right)=b[/math]
     
    I've also managed to work out a few generalized forms of the nested logarithm:
     
    [math]\text{nLog}_{\, a}\left(b\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}[/math]
     
    [math]\text{nLog}_{\, a}\left(b\right)=1-\frac{\text{ln}\left(\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)}[/math]
     
     
    Proof that - [math]\text{nLog}_{\, a}\left(b\right)=1+\text{ln}(\text{ln}\left(b\right) / \text{ln}\left(a\right)) / \text{ln}\left(a\right)[/math]
     
    Let [math]a[/math] equal the base of the iterated exponent and let [math]b[/math] equal [math]a^{\left \langle n \right \rangle}[/math], which is also equal to [math]a^{a^{n-1}}[/math].
     
    [math]\text{nLog}_{\, a}\left(b\right)=\text{nLog}_{\, a}\left(a^{\left \langle n \right \rangle}\right)=\text{nLog}_{\, a}\left(a^{a^{n-1}}\right)=n[/math]
     
    We can now use the general form, [math]a^{a^{n-1}}[/math], to derive the nested logarithm by using natural logarithms:
     
    [math]\text{ln}\left(a^{a^{n-1}}\right)=a^{n-1} \, \text{ln}\left(a\right)[/math]
     
    Divide both sides by [math]\text{ln}\left(a\right)[/math]:
     
    [math]\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}=a^{n-1}[/math]
     
    Take the natural log of both sides:
     
    [math]\text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)=\left(n-1\right) \, \text{ln}\left(a\right)[/math]
     
    Again, divide both sides by [math]\text{ln}\left(a\right)[/math]:
     
    [math]\frac{\text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n-1[/math]
     
    Add one to both sides:
     
    [math]1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n[/math]
     
    Substitute [math]a^{\left \langle n \right \rangle}[/math] in place of [math]a^{a^{n-1}}[/math]:
     
    [math]\text{nLog}_{\, a}\left(a^{\left \langle n \right \rangle}\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left \langle n \right \rangle}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n[/math]
     
    Substitute [math]b[/math] in place of [math]a^{\left \langle n \right \rangle}[/math]:
     
    [math]\text{nLog}_{\, a}\left(b\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}[/math]
     
    Q.E.D.
     
     
    Proof that - [math]\text{nLog}_{\, a}\left(b\right)=1-\text{ln}(\text{ln}\left(a\right) / \text{ln}\left(b\right)) / \text{ln}\left(a\right)[/math]
     
    Working backwards, we can prove that the second form is true. Let [math]a[/math] equal the base of the nested exponent and let [math]b[/math] equal [math]a^{\left \langle n \right \rangle}[/math], which is also equal to [math]a^{a^{n-1}}[/math] such that:
     
    [math]1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{\left \langle n \right \rangle}\right)}\right)}{\text{ln}\left(a\right)}=1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}{\text{ln}\left(a\right)}=n[/math]
     
    Move [math]n[/math] and the term using the natural logarithms to the opposite side of the equation:
     
    [math]1-n=\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}{\text{ln}\left(a\right)}[/math]
     
    Multiply both sides by [math]\text{ln}\left(a\right)[/math]:
     
    [math](1-n)\, \text{ln}\left(a\right)=\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)[/math]
     
    Raise [math]e[/math] to the power of each side:
     
    [math]e^{(1-n)\, \text{ln}\left(a\right)}=e^{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}[/math]
     
    Simplify the result:
     
    [math]a^{1-n}=\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}[/math]
     
    Multiply both sides by [math]\text{ln}\left(a^{a^{n-1}}\right)[/math]:
     
    [math]a^{1-n}\, \text{ln}\left(a^{a^{n-1}}\right)=\text{ln}\left(a\right)[/math]
     
    Divide both sides by [math]a^{1-n}[/math]:
     
    [math]\text{ln}\left(a^{a^{n-1}}\right)=\frac{\text{ln}\left(a\right)}{a^{1-n}}[/math]
     
    Raise [math]e[/math] to the power of each side:
     
    [math]e^{\text{ln}\left(a^{a^{n-1}}\right)}=e^{\left(\frac{\text{ln}\left(a\right)}{a^{1-n}}\right)}[/math]
     
    Simplify the result:
     
    [math]a^{a^{n-1}}=a^{a^{n-1}}[/math]
     
    Q.E.D.
     
     
    Trivial Identities
     
    [math]a^{\left \langle 0 \right \rangle}=a^{a^{(0-1)}}=a^{a^{-1}}=a^{\frac{1}{a}}=\sqrt[a]{a}[/math]
     
    [math]a^{\left \langle 1 \right \rangle}=a^{a^{(1-1)}}=a^{a^0}=a[/math]
     
    [math]a^{\left \langle 2 \right \rangle}=a^{a^{(2-1)}}=a^{a^1}=a^a[/math]
     
     
    [math]\left(a^{\left \langle b \right \rangle}\right)^{a^c}=\left(a^{a^{b-1}}\right)^{a^c}=a^{a^{b-1} a^{c}}=a^{a^{b+c-1}}=a^{\left \langle b+c \right \rangle}[/math]
     
    [math]\left(a^{\left \langle b \right \rangle}\right)^{a^{-c}}=\left(a^{a^{b-1}}\right)^{a^{-c}}=a^{a^{b-1} a^{-c}}=a^{a^{b-c-1}}=a^{\left \langle b-c \right \rangle}[/math]
     
     
    [math]\text{nLog}_{\, a}\left(a\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(a\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=1+\frac{\text{ln}\left(1\right)}{\text{ln}\left(a\right)}=1+\frac{0}{\text{ln}\left(a\right)}=1[/math]
     
    Both, [math]\text{nLog}_{\, a}\left(0\right)[/math] and [math]\text{nLog}_{\, a}\left(1\right)[/math], are undefined:
     
    [math]\lim_{b \to 0}\big(\text{nLog}_{\, a}\left(b\right)\big) =\lim_{b \to 0} \left(1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}\right)=\infty[/math]
     
    [math]\lim_{b \to 1}\big(\text{nLog}_{\, a}\left(b\right)\big) =\lim_{b \to 1} \left(1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}\right)=-\infty[/math]
     
     
    Properties of Nested Logarithms
     
    [math]\text{nLog}_{\, a}\left(b\right)=1+\frac{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)} \ \ \ \ \ \ \ \text{nLog}_{\, a}\left(b\right)=1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)}[/math]
     
    [math]\text{nLog}_{\, a}\left(b\right)=1+\text{log}_{a}\left(\text{log}_{a}\left(b\right)\right) \ \ \ \ \ \ \ \text{nLog}_{\, e}\left(b\right)=1+\text{ln}\left(\text{ln}\left(b\right)\right)[/math]
     
    [math]\text{nLog}_{\, a}\left(b\right)=1-\frac{\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)}{\text{ln}\left(a\right)} \ \ \ \ \ \ \ \text{log}_{\, a}\left(b\right)=\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}=a^{\left(\text{nLog}_{\, a}\left(b\right)-1\right)}[/math]
     
    [math]\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(a\right) \left(\text{nLog}_{\, a}\left(b\right) - 1\right) \ \ \ \ \ \ \ \text{ln}\left(\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)=\text{ln}\left(a\right) \left(1-\text{nLog}_{\, a}\left(b\right)\right)[/math]
     
     
    Relationship to Exponentials
     
    [math]a^b=a\ \frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}=a\ \text{log}_{\, a}\left(a^{\left \langle b \right \rangle}\right)[/math]
     
    Proof:
     
    Iterated exponentiation by definition:
     
    [math]a^{\left \langle b \right \rangle}=a^{a^{(b-1)}}[/math]
     
    Take the natural log of both sides:
     
    [math]\text{ln}\left(a^{a^{(b-1)}}\right)=\text{ln}\left(a^{\left \langle b \right \rangle}\right)[/math]
     
    Simplify the result:
     
    [math]a^{(b-1)}\, \text{ln}\left(a\right)=\text{ln}\left(a^{\left \langle b \right \rangle}\right)[/math]
     
    Divide both sides by [math]\text{ln}\left(a\right)[/math]:
     
    [math]a^{(b-1)}=\frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}[/math]
     
    Multiply both sides by [math]a[/math]:
     
    [math]a^{b}=a\ \frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}[/math]
     
    Convert [math]\text{ln}\left(a^{\left \langle b \right \rangle}\right) / \text{ln}\left(a\right)[/math] to a logarithm by the "changing the base" identity:
     
    [math]a^{b}=a\ \text{log}_{\, a}\left(a^{\left \langle b \right \rangle}\right)[/math]
     
    Q.E.D.
     
     
    Relationship between Nested Logarithms and Natural Nested Logarithms
     
    [math]\text{nLog}_{\, a}\left(b\right) = 1+\frac{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}{\text{ln}\left(a\right)}[/math]
     
    Proof:
     
    Natural nested logarithm by definition:
     
    [math]\text{nLn}\left(x\right)=1+\text{ln}\left(\text{ln}\left(x\right)\right)[/math]
     
    Nested logarithm by definition:
     
    [math]\text{nLog}_{\, a}\left(b\right) =1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}= 1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}[/math]
     
    such that
     
    [math]1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}=1+\frac{\left(1+\text{ln}\left(\text{ln}\left(b\right)\right)\right)-\left(1+\text{ln}\left(\text{ln}\left(a\right)\right)\right)}{\text{ln}\left(a\right)}[/math]
     
    Simplify the right side by cancelling the ones in the numerator:
     
    [math]1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}=1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}[/math]
     
    Q.E.D.
     
     
    Relationship between the Difference of Natural Nested Logarithms
     
    [math]\text{nLn}\left(b\right) - \text{nLn}\left(a\right)=\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)[/math]
     
    such that
     
    [math]e^{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}=e^{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}=\text{log}_{\, a}\left(b\right)[/math]
     
    Proof:
     
    [math]1+\frac{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}{\text{ln}\left(a\right)}=1+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}[/math]
     
    Subtract one from both sides:
     
    [math]\frac{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}{\text{ln}\left(a\right)}=\frac{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}[/math]
     
    Multiply both sides by [math]\text{ln}\left(a\right)[/math]:
     
    [math]\text{nLn}\left(b\right) - \text{nLn}\left(a\right)=\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)[/math]
     
    Raise [math]e[/math] to the power of both sides:
     
    [math]e^{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}=e^{\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)}[/math]
     
    Undo the subtraction in the exponent:
     
    [math]\frac{e^{\text{nLn}\left(b\right)}}{e^{\text{nLn}\left(a\right)}}=\frac{e^{\text{ln}\left(\text{ln}\left(b\right)\right)}}{e^{\text{ln}\left(\text{ln}\left(a\right)\right)}}[/math]
     
    Simplify the result:
     
    [math]\frac{e^{1+\text{ln}\left(\text{ln}\left(b\right)\right)}}{e^{1+\text{ln}\left(\text{ln}\left(a\right)\right)}}=\frac{e\ \text{ln}\left(b\right)}{e\ \text{ln}\left(a\right)}=\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}[/math]
     
    Cancel out the [math]e[/math] and change the base:
     
    [math]e^{\text{nLn}\left(b\right) - \text{nLn}\left(a\right)}=\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}=\text{log}_{\, a}\left(b\right)[/math]
     
    Q.E.D.
     
     
    The Nested Logarithm of a Doubly Iterated Exponential with the Same Base (or something like that lol)
     
    [math]\text{nLog}_{\, a}\left(\left(a^{\left \langle b \right \rangle}\right)^{\left \langle c \right \rangle}\right)=\left(c-1\right)\, a^{(b-1)}+b[/math]
     
    Proof:
     
    [math]\left(a^{\left \langle b \right \rangle}\right)^{\left \langle c \right \rangle}=\left(a^{\left \langle b \right \rangle}\right)^{\left(a^{\left \langle b \right \rangle}\right)^{(c-1)}}=\left(a^{a^{(b-1)}}\right)^{\left(a^{a^{(b-1)}}\right)^{(c-1)}}=[/math]
     
    [math]\left(a^{a^{(b-1)}}\right)^{\left(a^{(c-1)\, a^{(b-1)}}\right)}=a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}[/math]
     
    Take the natural log of the simplified version:
     
    [math]\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)=\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right) \text{ln}\left(a\right)[/math]
     
    Divide both sides by [math]\text{ln}\left(a\right)[/math]:
     
    [math]\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}=\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)[/math]
     
    Take the natural log of both sides:
     
    [math]\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)\right)[/math]
     
    Simplify the right hand side:
     
    [math]\text{ln}\left(\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)\right)=\text{ln}\left(a^{(b-1)}\right) + \text{ln}\left(a^{(c-1)\, a^{(b-1)}}\right)=[/math]
     
    [math](b-1) \, \text{ln}\left(a\right)\, + \, (c-1)\, a^{(b-1)} \, \text{ln}\left(a\right)=\text{ln}\left(a\right)\, \left((b-1)\, +\, (c-1)\, a^{(b-1)}\right)[/math]
     
    such that
     
    [math]\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(a\right)\, \left((b-1)\, +\, (c-1)\, a^{(b-1)}\right)[/math]
     
    Divide both sides by [math]\text{ln}\left(a\right)[/math]:
     
    [math]\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=(b-1) \, + \, (c-1)\, a^{(b-1)}[/math]
     
    Complete the nested logarithm by adding one to both sides:
     
    [math]1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=1+(b-1) \, + \, (c-1)\, a^{(b-1)}[/math]
     
    Simplify the result:
     
    [math]\text{nLog}_{\, a}\left(\left(a^{\left \langle b \right \rangle}\right)^{\left \langle c \right \rangle}\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=(c-1)\, a^{(b-1)}\, +\, b[/math]
     
    Q.E.D.
     
     
    Derivatives of Iterated Exponents and Nested Logarithms (Found by using Mathematica)
     
    [math]\frac{dy}{dx} \ x^{\left \langle a \right \rangle} = x^{x^{(a-1)}}\left(x^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)=x^{\left \langle a \right \rangle}\left(x^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)[/math]
     
    [math]\frac{dy}{dx} \ a^{\left \langle x \right \rangle} = a^{a^{(x-1)}} \, a^{(x-1)} \, \text{ln}\left(a\right)^2=a^{\left \langle x \right \rangle} \left(a^{(x-1)} \, \text{ln}\left(a\right)^2\right)[/math]
     
    [math]\frac{dy}{dx} \ x^{\left \langle x \right \rangle} = x^{x^{(x-1)}}\left(x^{(x-2)}+x^{(x-1)}\, \text{ln}\left(x\right)\left(\text{ln}\left(x\right)+\frac{x-1}{x}\right)\right)=x^{\left \langle x \right \rangle}\left(x^{(x-2)}+x^{(x-1)}\, \text{ln}\left(x\right)\left(\text{ln}\left(x\right)+\frac{x-1}{x}\right)\right)[/math]
     
    [math]\frac{dy}{dx} \ \text{nLog}_{\, a}\left(x\right) = \frac{1}{x\, \text{ln}\left(x\right)\, \text{ln}\left(a\right)}=\frac{\text{nLog}_{\, a}\left(x\right)}{x\, \text{ln}\left(x\right) \left(\text{ln}\left(a\right) +\text{ln}\left(\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)\right)}[/math]
     
    [math]\frac{dy}{dx} \ \text{nLog}_{\, x}\left(a\right) = \frac{\text{ln}\left(\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)-1}{x\, \text{ln}\left(x\right)^2}=\frac{\text{ln}\left(x\right) \left(1-\text{nLog}_{\, x}\left(a\right)\right)-1}{x\, \text{ln}\left(x\right)^2}[/math]
     
     
    Properties of Derivatives for Iterated Exponents (Found by using Mathematica)
     
    An iterated exponential function that has the constant as the nested exponent, [math]x^{\left \langle a \right \rangle}[/math]:
     
    [math]\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle a \right \rangle}\right) = \left(x^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)[/math]
     
    Which allows us to define the following by substitution:
     
    [math]\frac{dy}{dx} \ x^{\left \langle a \right \rangle} = x^{\left \langle a \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(x^{\left \langle a \right \rangle}\right)[/math]
     
    such that:
     
    [math]x^{\left \langle a \right \rangle}=\frac{\frac{dy}{dx} \ x^{\left \langle a \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle a \right \rangle}\right)}[/math]
     
     
    An iterated exponential function that has the variable as the nested exponent, [math]a^{\left \langle x \right \rangle}[/math]:
     
    [math]\frac{dy}{dx} \ \text{ln}\left(a^{\left \langle x \right \rangle}\right) = \left(a^{(x-1)} \, \text{ln}\left(a\right)^2\right)[/math]
     
    Which allows us to define the following by substitution:
     
    [math]\frac{dy}{dx} \ a^{\left \langle x \right \rangle} = a^{\left \langle x \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(a^{\left \langle x \right \rangle}\right)[/math]
     
    such that:
     
    [math]a^{\left \langle x \right \rangle}=\frac{\frac{dy}{dx} \ a^{\left \langle x \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(a^{\left \langle x \right \rangle}\right)}[/math]
     
     
    An iterated exponential function that has the variable as the base and as the nested exponent, [math]x^{\left \langle x \right \rangle}[/math]:
     
    [math]\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle x \right \rangle}\right) = \left(x^{(x-2)}+x^{(x-1)}\, \text{ln}\left(x\right)\left(\text{ln}\left(x\right)+\frac{x-1}{x}\right)\right)[/math]
     
    Which allows us to define the following by substitution:
     
    [math]\frac{dy}{dx} \ x^{\left \langle x \right \rangle} = x^{\left \langle x \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(x^{\left \langle x \right \rangle}\right)[/math]
     
    such that:
     
    [math]x^{\left \langle x \right \rangle}=\frac{\frac{dy}{dx} \ x^{\left \langle x \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle x \right \rangle}\right)}[/math]
     
     
    Integrals of Iterated Exponents and Nested Logarithms (Found by using Mathematica)
     
    There are no general expressions for the integrals. They are based on the exponential integral, logarithmic integral, or unknown:
     
    [math]\int x^{\left \langle a \right \rangle} \, dx = \int x^{x^{a-1}} \, dx = \text{unknown}[/math]
     
    [math]\int a^{\left \langle x \right \rangle} \, dx = \int a^{a^{x-1}} \, dx = \frac{\text{Ei}\left(a^{x-1}\, \text{ln}(a)\right)}{\text{ln}(a)}+C[/math]
     
    [math]\int x^{\left \langle x \right \rangle} \, dx = \int x^{x^{x-1}} \, dx = \text{unknown}[/math]
     
    [math]\int \text{nLog}_{\, a}\left(x\right) \, dx = \int 1\, +\, \frac{\text{ln}\left (\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)} \, dx = x \, + \, \frac{x \, \text{ln}\left(\frac{\text{ln}(x)}{\text{ln}(a)}\right)}{\text{ln}(a)}\, -\, \frac{\text{li}(x)}{\text{ln}(a)}\, +\, C = x\, \left(\text{nLog}_{\, a}\left(x\right)\right)\, -\, \frac{\text{li}(x)}{\text{ln}(a)}\, +\, C[/math]
     
    [math]\int \text{nLog}_{\, x}\left(a\right) \, dx = \int 1+\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(x\right)}\right)}{\text{ln}\left(x\right)} \, dx = \text{unknown}[/math]
     
    The exponential integral, [math]\text{Ei}\left(x\right)[/math]:
     
    [math]\text{Ei}(x)=\int \limits_{-\infty}^{x} \frac{e^t}{t} \, dt[/math]
     
    The logarithmic integral, [math]\text{li}\left(x\right)[/math]:
     
    [math]\text{li}(x)=\int \limits_{0}^{x} \frac{dt}{\text{ln}(t)}[/math]
     
     
    Finding Iterated Roots and Nested Logarithms using the Newton-Rhapson Method
     
    First it is important to note that the Newton-Rhapson method may fail to converge to the desired result. One must read and understand the proof of quadratic convergence for Newton's iterative method which is described in Wikipedia. With that being said I have not had a problem finding iterated roots or nested logarithms using this approach.
     
    [math]x_{n+1}=x_{n} - \frac{f(x_{n})}{f'(x_{n})}[/math]
     
    Iterated Roots:
     
    To find the iterated root we must use the iterated exponential function that has the constant as the iterated exponent, [math]x^{\left \langle a \right \rangle}[/math]. Using this function with Newton's method yields the following algorithm for iterated roots (Repeat the process until you have obtained the desired result):
     
    [math]\sqrt[\left \langle a \right \rangle]{b}\ \ \ \ \text{approximated by} \ \ \ \ x_{n+1}=x_{n} - \frac{\left(x_{n}\right)^{\left(x_{n}\right)^{(a-1)}}-b}{\left(x_{n}\right)^{\left(x_{n}\right)^{(a-1)}}\left(\left(x_{n}\right)^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)}[/math]
     
    Nested Logarithms:
     
    To find the nested logarithm we must use the iterated exponential function that has the variable as the iterated exponent, [math]a^{\left \langle x \right \rangle}[/math]. Using this function with Newton's method yields the following algorithm for nested logarithms (Repeat the process until you have obtained the desired result):
     
    [math]\text{nLog}_{\, a}\left(b\right)\ \ \ \text{approximated by} \ \ \ x_{n+1}=x_{n} - \frac{a^{a^{\left(x_{n}\right)-1}}-b}{a^{a^{\left(x_{n}\right)-1}} a^{\left(x_{n}\right)-1} \text{ln}\left(a\right)^2}[/math]
     
     
    Graph of [math]x^{\left \langle a \right \rangle}[/math]:
     
    This graph shows both the real part (blue) and the imaginary part (green).
     

     
     
    Graph of [math]a^{\left \langle x \right \rangle}[/math]:
     
    This graph shows both the real part (blue) and the imaginary part (green).
     

     
     
    Graph of [math] x^{\left \langle x \right \rangle}[/math]:
     
    This graph shows both the real part (blue) and the imaginary part (green).
     

     
     
    Graph of [math]\text{nLog}_{\, a} \left(x\right)[/math]:
     
    This graph shows both the real part (blue) and the imaginary part (green).
     

     
     
    Graph of [math]\text{nLog}_{\, x} \left(a\right)[/math]:
     
    This graph shows both the real part (blue) and the imaginary part (green).
     

     
     
    Graph of [math] \text{nLog}_{\, x} \left(x\right)[/math]:
     
    This graph shows both the real part (blue) and the imaginary part (green).
     

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