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Posts posted by DaedalusSFN


You're wrong, the Planck length & Planck time do make a difference, in so that time dilation becomes radical at that level so as to transform the linearity of the temporal dimensions we know & love.
Please provide evidence / reference that "time dilation becomes radical at that level so as to transform the linearity of the temporal dimensions we know & love". As far as I know, relativity makes no assertion about the nature of spacetime at the Planck length / Planck time scale.

But it is limited when your next chronological position is a goolgleplex of a second away.
Why are you incorrectly using the name for an extremely large number to denote a fraction of a second? It lends nothing to your credibility and, to make things worse, it demonstrates how incredibly ignorant you are when it comes to math and physics.
What you are not considering is that I said I would use (tP) because that is a perceivable measurement to explain it easier.
Do you really think that units of Planck time are actually perceivable? The unit of time you use to describe your idea does NOT matter. You keep insisting that there exists some discreet duration of time that somehow disproves time dilation, but you fail to realize that time dilation is dependent on the observer's frame of reference. You refuse to learn the physics that defines what time dilation is and state that you do NOT need to know the mathematics to understand it. How arrogant you must be to think you can understand the laws of nature without having to understand the language used to define such laws.
If I start saying your next chronological position is an instant away, it becomes imperceptible to most people because it is absolute so can only be described by a mathematical expression . (tP) is the smallest expression we have t=Δ(tP)for all observers.
Although Planck time is currently the name for the smallest unit of time, I can easily define units of time that are smaller than that. However, none of that matters. Time dilation is an observable phenomenon that occurs in nature. Not only have we observed it, but we have also measured it and defined mathematical models that predict it. Nothing you say will change that. So, either present us a mathematical model that explains why you are right (units of measurements are not mathematical models) or go back to your cave like a good little troll...

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:
I appreciate your great effort, however again avoidance of the statement that shows time dilation is not true.
I predicted this would happen.
You don't seem to realise that I know what time dilation is and about, yes you are correct in that I could not do the maths required, but the point is I don't require maths for something that doe's not happen.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.

This next song is titled, "G Minor Blues". Although I use blues scales, it's not blues at all. However, it is an interesting piece of music. Enjoy!!!
[video=youtube;N1U5t1P8MXE]

In 2014, my New Year resolutions were to write more music and improve my composition skills. So, to ring in the new year, I wrote this song titled "Resolution". Enjoy!!!
[video=youtube;0eOWi86TFU4]

This next song is titled, "Blue Moon Rising". It's an interesting piece of music that was inspired by all the talk about the blue Moon back in 2015. Enjoy!!!
[video=youtube;CTDpOzcfoWo]

When I write music, it takes the form of my mood at the time and reflects my emotional state of mind. This song, "Endless Night", was written during a period in my life when I was having a tough time. Regardless of how I feel, music has always helped me overcome the hardships that I've faced. As such, this song is one of the most melodramatic pieces that I've written. Enjoy!!!!
[video=youtube;v8WFEpVQS8I]

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;0a79NAa2VE] 
Thanks for the comments everyone. I'm glad you enjoyed my music. :note2: :note: :cool:
This is my favorite so far. I'm no musician, but I do enjoy listening to classical. Your mention of transposing keys, a kind of layering I suppose, caught my interest as I am currently rereading Hofstadter's Gödel, Escher, Bach, and he writes at length in the Introduction on some of the recursive layering techniques Bach used. (In fugues and canons I think?)
Since you have an interest in math as well, may I ask if you have you read Hofstadter? Anyway, thanks for sharing and welcome to the menagerie.I haven't read Hofstadter, but I'll check it out when I have time. As you can tell, my biggest influence is Bach.
Missing you is my fav, mainly because it has a bit of an "epic" touch.
Yeah, it's one of my favorites too!
April was my favourite, even when I can't say exactly why. The others were a tad too restless for my present mood.
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] 

This is a very interesting piece of music. I took the theme from the beginning of the song, transposed it in key, and then varied the music. Enjoy!!!
[video=youtube;nO3tSJ69DtQ]

I wrote this song for my two boys, Maxximus and Adam. At the time, I was extremely busy with work and was constantly on the road. Enjoy!!!
[video=youtube;7a3IUKu1hnc] 
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] 
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^{n1}}[/math]
such that
[math] a^{\left \langle 1 \right \rangle} = a = a^{a^{(11)}} [/math],
[math] a^{\left \langle 2 \right \rangle} = (a)^a = a^{a^{(21)}} [/math],
[math] a^{\left \langle 3 \right \rangle} = ((a)^a)^a = a^{a^{(31)}} [/math],
[math] a^{\left \langle 4 \right \rangle} = (((a)^a)^a)^a = a^{a^{(41)}} [/math],
etc...
Now that we have the generalized form, [math]a^{a^{n1}}[/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 NewtonRhapson 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^{n1}}[/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^{n1}}\right)=n[/math]
We can now use the general form, [math]a^{a^{n1}}[/math], to derive the nested logarithm by using natural logarithms:
[math]\text{ln}\left(a^{a^{n1}}\right)=a^{n1} \, \text{ln}\left(a\right)[/math]
Divide both sides by [math]\text{ln}\left(a\right)[/math]:
[math]\frac{\text{ln}\left(a^{a^{n1}}\right)}{\text{ln}\left(a\right)}=a^{n1}[/math]
Take the natural log of both sides:
[math]\text{ln}\left(\frac{\text{ln}\left(a^{a^{n1}}\right)}{\text{ln}\left(a\right)}\right)=\left(n1\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^{n1}}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n1[/math]
Add one to both sides:
[math]1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{a^{n1}}\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^{n1}}[/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^{n1}}[/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^{n1}}\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]1n=\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n1}}\right)}\right)}{\text{ln}\left(a\right)}[/math]
Multiply both sides by [math]\text{ln}\left(a\right)[/math]:
[math](1n)\, \text{ln}\left(a\right)=\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n1}}\right)}\right)[/math]
Raise [math]e[/math] to the power of each side:
[math]e^{(1n)\, \text{ln}\left(a\right)}=e^{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n1}}\right)}\right)}[/math]
Simplify the result:
[math]a^{1n}=\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n1}}\right)}[/math]
Multiply both sides by [math]\text{ln}\left(a^{a^{n1}}\right)[/math]:
[math]a^{1n}\, \text{ln}\left(a^{a^{n1}}\right)=\text{ln}\left(a\right)[/math]
Divide both sides by [math]a^{1n}[/math]:
[math]\text{ln}\left(a^{a^{n1}}\right)=\frac{\text{ln}\left(a\right)}{a^{1n}}[/math]
Raise [math]e[/math] to the power of each side:
[math]e^{\text{ln}\left(a^{a^{n1}}\right)}=e^{\left(\frac{\text{ln}\left(a\right)}{a^{1n}}\right)}[/math]
Simplify the result:
[math]a^{a^{n1}}=a^{a^{n1}}[/math]
Q.E.D.
Trivial Identities
[math]a^{\left \langle 0 \right \rangle}=a^{a^{(01)}}=a^{a^{1}}=a^{\frac{1}{a}}=\sqrt[a]{a}[/math]
[math]a^{\left \langle 1 \right \rangle}=a^{a^{(11)}}=a^{a^0}=a[/math]
[math]a^{\left \langle 2 \right \rangle}=a^{a^{(21)}}=a^{a^1}=a^a[/math]
[math]\left(a^{\left \langle b \right \rangle}\right)^{a^c}=\left(a^{a^{b1}}\right)^{a^c}=a^{a^{b1} a^{c}}=a^{a^{b+c1}}=a^{\left \langle b+c \right \rangle}[/math]
[math]\left(a^{\left \langle b \right \rangle}\right)^{a^{c}}=\left(a^{a^{b1}}\right)^{a^{c}}=a^{a^{b1} a^{c}}=a^{a^{bc1}}=a^{\left \langle bc \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^{(b1)}}[/math]
Take the natural log of both sides:
[math]\text{ln}\left(a^{a^{(b1)}}\right)=\text{ln}\left(a^{\left \langle b \right \rangle}\right)[/math]
Simplify the result:
[math]a^{(b1)}\, \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^{(b1)}=\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(c1\right)\, a^{(b1)}+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)^{(c1)}}=\left(a^{a^{(b1)}}\right)^{\left(a^{a^{(b1)}}\right)^{(c1)}}=[/math]
[math]\left(a^{a^{(b1)}}\right)^{\left(a^{(c1)\, a^{(b1)}}\right)}=a^{\left(a^{(b1)}\right)\left(a^{(c1)\, a^{(b1)}}\right)}[/math]
Take the natural log of the simplified version:
[math]\text{ln}\left(a^{\left(a^{(b1)}\right)\left(a^{(c1)\, a^{(b1)}}\right)}\right)=\left(a^{(b1)}\right)\left(a^{(c1)\, a^{(b1)}}\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^{(b1)}\right)\left(a^{(c1)\, a^{(b1)}}\right)}\right)}{\text{ln}\left(a\right)}=\left(a^{(b1)}\right)\left(a^{(c1)\, a^{(b1)}}\right)[/math]
Take the natural log of both sides:
[math]\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b1)}\right)\left(a^{(c1)\, a^{(b1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(\left(a^{(b1)}\right)\left(a^{(c1)\, a^{(b1)}}\right)\right)[/math]
Simplify the right hand side:
[math]\text{ln}\left(\left(a^{(b1)}\right)\left(a^{(c1)\, a^{(b1)}}\right)\right)=\text{ln}\left(a^{(b1)}\right) + \text{ln}\left(a^{(c1)\, a^{(b1)}}\right)=[/math]
[math](b1) \, \text{ln}\left(a\right)\, + \, (c1)\, a^{(b1)} \, \text{ln}\left(a\right)=\text{ln}\left(a\right)\, \left((b1)\, +\, (c1)\, a^{(b1)}\right)[/math]
such that
[math]\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b1)}\right)\left(a^{(c1)\, a^{(b1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(a\right)\, \left((b1)\, +\, (c1)\, a^{(b1)}\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^{(b1)}\right)\left(a^{(c1)\, a^{(b1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=(b1) \, + \, (c1)\, a^{(b1)}[/math]
Complete the nested logarithm by adding one to both sides:
[math]1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b1)}\right)\left(a^{(c1)\, a^{(b1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=1+(b1) \, + \, (c1)\, a^{(b1)}[/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^{(b1)}\right)\left(a^{(c1)\, a^{(b1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=(c1)\, a^{(b1)}\, +\, 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^{(a1)}}\left(x^{(a2)}\left(1+\left(a1\right)\, \text{ln}\left(x\right)\right)\right)=x^{\left \langle a \right \rangle}\left(x^{(a2)}\left(1+\left(a1\right)\, \text{ln}\left(x\right)\right)\right)[/math]
[math]\frac{dy}{dx} \ a^{\left \langle x \right \rangle} = a^{a^{(x1)}} \, a^{(x1)} \, \text{ln}\left(a\right)^2=a^{\left \langle x \right \rangle} \left(a^{(x1)} \, \text{ln}\left(a\right)^2\right)[/math]
[math]\frac{dy}{dx} \ x^{\left \langle x \right \rangle} = x^{x^{(x1)}}\left(x^{(x2)}+x^{(x1)}\, \text{ln}\left(x\right)\left(\text{ln}\left(x\right)+\frac{x1}{x}\right)\right)=x^{\left \langle x \right \rangle}\left(x^{(x2)}+x^{(x1)}\, \text{ln}\left(x\right)\left(\text{ln}\left(x\right)+\frac{x1}{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^{(a2)}\left(1+\left(a1\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^{(x1)} \, \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^{(x2)}+x^{(x1)}\, \text{ln}\left(x\right)\left(\text{ln}\left(x\right)+\frac{x1}{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^{a1}} \, dx = \text{unknown}[/math]
[math]\int a^{\left \langle x \right \rangle} \, dx = \int a^{a^{x1}} \, dx = \frac{\text{Ei}\left(a^{x1}\, \text{ln}(a)\right)}{\text{ln}(a)}+C[/math]
[math]\int x^{\left \langle x \right \rangle} \, dx = \int x^{x^{x1}} \, 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 NewtonRhapson Method
First it is important to note that the NewtonRhapson 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)^{(a1)}}b}{\left(x_{n}\right)^{\left(x_{n}\right)^{(a1)}}\left(\left(x_{n}\right)^{(a2)}\left(1+\left(a1\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).


Welcome to Hypography Daedalus! Always nice to see fellow musicians around here.
Enjoy your stay!
Thanks Buffy!

Although I enjoy working on challenging mathematical problems and discussing physics, I also love playing guitar and piano, and writing musical compositions. I've been playing and writing music for over 26 years, but these past three years my skill has finally developed to the point where I'm able to compose symphonies, chamber pieces, and all sorts of musical creations.
I compose music using Pro Tools 10 and render the music using the XPand2 synthesizer and Garritan's orchestral sound packages. However, Pro Tools lacks decent tools to edit the MIDI and add dynamics to the voices. I'm looking at getting a Mac because I've seen how Logic Pro works and it's nice to be able to use Bezier curves instead of having to manually edit the MIDI data in Pro Tools.
Instead of posting all of my compositions at once, I will give people time to comment on each one. My music is classical in nature and I can't afford to pay real musicians to record it. So, it's not music everyone will enjoy. However, if you like classical music, then I hope you'll enjoy my music. On that note, I present you all my song, "Forged by Destiny", which is a song that I wrote that my brother and I used to play on guitar. Unfortunately, he injured himself on the job and can no longer play guitar like he used to. :( So, I turned it into a symphony for him. Please, keep in mind that I copyright all of my music. Enjoy!!!
[video=youtube;l_5i7ACtq5I]

Hello everyone,
I'm Daedalus and I love writing music, working on mathematical problems like the Collatz conjecture, and enjoy theorizing about the nature of the Universe. I'm usually active at scienceforums.net, but thought it would be nice to meet new people, learn from them, and have a new outlet to share my music, math, and ideas.
I Think Time Dilation Semantics Are Incorrect?
in Silly Claims Forum
Posted · Edited by DaedalusSFN
Furthermore, just because a process exhibits Tsymmetry does not actually mean it operates in a true reversal of time. It's just that the process can work in reverse. Of course, I have my own theory of time using a mathematical framework that extends relativity and demonstrates that time is simply a mathematical consequence of space, but I won't go into that here.
I would urge caution before telling someone they are wrong regarding mainstream physics when your intent is to imply they are wrong within your own theories.