Into the Studio

As some of you already know, I am producing professional recordings of my music at Evermind studio for my first demo / album. We are in the process of selecting qualified student musicians from the University of Oklahoma, University of Central Oklahoma, and Oklahoma City Community College so that we can capture musical qualities that only a musician can provide.

Once I have the album finished, I will upload my music to Spinnup.com. They will then distribute my music to “Spotify, iTunes, Google Play, Amazon, Rhapsody, and WiMPSo”. The coolest thing about Spinnup.com is that not only do I get to keep all of the rights to my music and royalties from the sales, but also that Universal has talent scouts that search the website daily looking for new talent. In addition, I get to send my music to these scouts 3 times per month for consideration. So, as you can tell, this is a very exciting opportunity for me.

As for my family, friends, and members at scienceforums.net, I will be making a special edition of the album where I will personally autograph each copy. I realize that my autograph isn’t worth much, and that my music might never get the recognition that I am hoping to achieve. However, you never know what the future holds, and if I do get signed by Universal, then an autographed CD might be worth something one day 🙂 This special edition will only be available to my family, friends, and members of SFN. So, let me know if you want a CD of my music because I need to know how many special editions I will need to make.

I am also looking for an artist to help design the album’s cover. I was thinking about holding a competition where you can submit artwork to be considered. Of course, the artist would be given a reasonable cash prize and would receive a special edition of the album. However, I haven’t worked out all the details to how this would work, but the final artwork chosen would be considered as a “Work for hire“. So, instead of holding a competition, I might have to hire an artist to create the album cover. Either way, all new artwork will be considered as a “Work for Hire”, and I will also consider using existing artwork if I can pay a reasonable one-time fee for the right to use it as my album cover. If you are interested in designing or providing art for my album cover, please contact me by replying to this post or by sending me an e-mail to clevelandraymond@hotmail.com.

Once this project is completed, I might have to take down the music I have posted on youtube. Although I get to maintain the rights to my music by using Spinnup.com, I am not sure what Universal will require if they decide to bring me on board. It’s a long shot that they will even consider my music for publication, but it’s a chance that I am willing to take. After all, if I do nothing, then nothing will happen. So, for now, you can listen to my music at youtube, or right here on my blog at my post, A Window Into The Soul.

Please note that all of my music is copyrighted. If you wish to use my music for personal or commercial purposes, I require you to obtain my permission.

Unlocked Cage

I ended my marriage of ten years back in 2010. We had grown apart over the course of several years and even though we didn’t fight like cats and dogs, we decided it was best to part ways. There were other issues that played their part, but I am not going to dish out all of the he said / she did stuff because marriage is a two way street and it really does take quite an effort on both parties to make it work or consequently fall apart. “Unlocked Cage” is about the end of my marriage and my comming to terms with my divorce.

Unlocked Cage

Throughout the years I have been cruel
I locked you up just like a fool

Within my heart I caged you in
A love so true should never end

It is hard for me to let you go
It feels like losing part of my soul

I know you just need time to see
That you can share a life with me

I should have seen your colors fade
I should have heard your love song wane

I should have been a better friend
To be with you until the end

But now I must set you free
To release you from your misery

I unlocked the cage I put you in
So spread your wings and fly again

A Window Into The Soul

Music has always been a huge part of my family’s heritage. My grandfather was a violinist in the Detroit Philharmonic and I grew up watching my father plays bluegrass on his fiddle. I even had the privilege of seeing my father perform at the Grand Ole Opry in Nashville, Tennessee when I was five. Most of my aunts, uncles, and cousins on my father’s side of the family play some form of instrument. My brother plays the guitar and I play the guitar and the piano.

I’ve been playing guitar for twenty years and I have played piano for fifteen years. I don’t play music everyday like I am suppose to, but I do practice at least three to four days a week. Music is an integral part of my being no different than the love I have for mathematics. The music I write is an expression of emotion that manifests itself into song which is truly the window into my soul and a reflection of how I feel.

I have decided to create this category in my blog so that I can share with you all the music that I love and the songs that I have written. I play classical music on the piano and metal on the guitar. The type of metal that I play is not what most people relate to the genre. Most of my influences comes from great guitarist such as Michael Angelo Batio, Yngwie Malmsteen, Joe Satriani, Steve Vai, and Eric Johnson. My favorite style of classical music is from the Baroque period and I am heavily influenced by the music of Johann Sebastian Bach.

So without further ado, I present to you all a few of my own compositions. Please be aware that I do not have the best recording software and even though I have tried my best to render my compositions using instruments that sound good, I am restricted by the quality of sounds that are available to me.

Arpeggios In E Minor

This is one of my warm-up riffs that I play on guitar. I wrote this song to help me practice my arpeggios.

Bound By Fate

This is a song that my brother and I wrote. My son, Maxx, who was five years old at the time is playing the drums : )

A Father’s Love

I wrote this lullaby for my two sons, Maxx and Adam. This is my one of my favorites.

In The Dark of Night

This is one of my romantic compositions. I wrote it for my ex-wife when we got married.

Bewtiched

I wrote this song for a friend that helped me get through some of the hard times I faced after my divorce. Her name is Samantha. So I decided to name this one “Bewitched” ; )

A casa d’un drapaire

It’s been a long time since I’ve posted to my blog, but I do have a new arrangement that I would like to share with everyone. I arranged the guitar and violin music in response to a request from a member at scienceforums.net. Here is a link to the thread:

I will post more of my music once I have rendered the compositions or have the time to record my jam sessions. I also plan to use the category to post tips on playing guitar and piano as well as lessons into music theory and composition. I hope you all have enjoyed the music : )

Resolution (A minor)

As promised, I am posting new songs that I have written. This song is named “Resolution” because I made a New Years resolution to write more music this year, and so far I have kept to the promise I made to myself:

Once Upon A Dream

This is my first composition for the pipe organ, and I must say that it turned out far better than I expected. Of course, Bach’s music has influenced me the most, and I would like to think that he would approve of this song. I hope you all enjoy it as much as I do : )

G Minor Blues

As a way of expanding my horizons, I am experimenting with different styles of music. So, this is my first composition to use blues scales. The song begins with a classical style and transitions to blues.

I Will See You Again

I haven’t been able to spend much time with my boys, and I’ve been missing them a lot here lately. So, I wrote this song for them.

This is a really beautiful song that was inspired by a good friend of mine.

Variations in C Minor

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.

Please note that all music and articles are copyrighted.

Nested Exponentiation

The Next Operation Beyond Exponents

I was looking into operations greater than exponents when I discovered a few properties of nested exponentiation. 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 nested exponentiation anywhere. Also, according to Wikipedia, nested exponentation is not even listed as a hyperoperation.

Nested Exponentiation:

Most of us are aware of nested exponents and how to simplify the mathematics associated with them:

$latex \left(a^b\right)^c=a^{b\, c}$

This led me to inquire about series of nested exponents and if I could derive any properties of continued nested exponentiation. I developed the following notation to simplify the respresentation of this operation:

$latex a^{\left \langle n \right \rangle} = \underbrace{((((a)^a)^a)^{\cdots})^{a}}_n = a^{a^{n-1}}$

We can demonstrate this as follows:

$latex a^{\left \langle 1 \right \rangle} = a = a^{a^{(1-1)}}$

$latex a^{\left \langle 2 \right \rangle} = (a)^a = a^{a^{(2-1)}}$

$latex a^{\left \langle 3 \right \rangle} = ((a)^a)^a = a^{a^{(3-1)}}$

$latex a^{\left \langle 4 \right \rangle} = (((a)^a)^a)^a = a^{a^{(4-1)}}$

Now that we have a generalized equation that represents nested exponentiation, we can derive the nested logarithm. We can also use Calculus to derive the derivatives / integrals and define nested roots using the Newton-Rhapson method.

Nested Logarithms:

The nested logarithm is similar to normal logarithms in that it returns the value of the exponent. However, the similarities end there and nested logarithms define a whole new suit of identities that are different from normal logarithms. The following notation is used to denote nested logarithms and shows their relationship to nested exponentiation:

$latex \text{nLog}_{\, a}\left(a^{\left \langle n \right \rangle}\right)=n$

The natural nested logarithm is notated as follows:

$latex \text{nLog}_{\, e}\left(e^{\left \langle n \right \rangle}\right)=\text{nLn}\left(e^{\left \langle n \right \rangle}\right)=n$

The above definitions keeps things simple, but they do not allow us to do calculations involving nested logarithms. We will need to use the definition for nested exponentiation to derive an equation that represents the nested logarithm. This can be accomplished as follows:

Definition for nested exponentiation as previously defined:

$latex a^{\left \langle n \right \rangle} = a^{a^{n-1}}$

Let $latex a$ equal the base of the nested exponential and let $latex b$ equal $latex a^{\left \langle n \right \rangle}$ such that:

$latex \text{nLog}_{\, a}\left(b\right)=\text{nLog}_{\, a}\left(a^{\left \langle n \right \rangle}\right)=\text{nLog}_{\, a}\left(a^{a^{n-1}}\right)=n$

Take the natural log of the generalized equation:

$latex \text{ln}\left(a^{a^{n-1}}\right)=a^{n-1} \, \text{ln}\left(a\right)$

Divide both sides by ln(a):

$latex \frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}=a^{n-1}$

Take the natural logarithm of both sides:

$latex \text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)=\left(n-1\right) \, \text{ln}\left(a\right)$

Again, divide both sides by ln(a):

$latex \frac{\text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n-1$

Add one to both sides:

$latex 1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{a^{n-1}}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n$

Substitute $latex a^{\left \langle n \right \rangle}$ in place of $latex a^{a^{n-1}}$:

$latex \text{nLog}_{\, a}\left(a^{\left \langle n \right \rangle}\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left \langle n \right \rangle}\right)}{\text{ln}\left(a\right)}\right)}{ \text{ln}\left(a\right)}=n$

Substitute $latex b$ in place of $latex a^{\left \langle n \right \rangle}$:

$latex \text{nLog}_{\, a}\left(b\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}$

As we can see from the above proof, nested logarithms are derived from natural logarithms. If we analyze the equation that we just derived for nested logarithms, we can use the “changing base” identity for normal logarithms and see that the nested logarithm is truly nesting normal logarithms.

Changing the base of normal exponentials:

$latex \text{log}_{\, a}\left(b\right)=\frac{\text{log}_{\, d}\left(b\right)}{\text{log}_{\, d}\left(a\right)}=\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}$

Such that:
$latex \text{nLog}_{\, a}\left(b\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}$

$latex \text{nLog}_{\, a}\left(b\right)=1+\frac{\text{ln}\left(\text{log}_{\, a}\left(b\right)\right)}{\text{ln}\left(a\right)}$

$latex \text{nLog}_{\, a}\left(b\right)=1+\text{log}_{\, a}\left(\text{log}_{\, a}\left(b\right)\right)$

There is also another form that the equation for nested logarithms can take. The equation seems counterintuitive, but we will see that this second form allows us to yield some interesting identities for nested logarithms and normal logarithms. To prove the definition, we must first define the equation and work backwards to yield the nested exponential. This second equation for nested logarithms is as follows:

$latex \text{nLog}_{\, a}\left(b\right)=1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)}$

We can see that the fraction in the numerator has been inverted such that the natural log of $latex a$ is now being divided by the natural log of $latex b$. Now that we have defined the second equation for nested logarithms, we can work backwards from this form and derive nested exponentiation such that:

$latex 1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}{\text{ln}\left(a\right)}=n$

Swap the variable $latex n$ and the term using the natural logarithms:

$latex 1-n=\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}{\text{ln}\left(a\right)}$

Multiply both sides by ln(a):

$latex (1-n)\, \text{ln}\left(a\right)=\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)$

Raise $latex e$ to the power of each side:

$latex e^{(1-n)\, \text{ln}\left(a\right)}=e^{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}\right)}$

Simplify the result:

$latex a^{1-n}=\frac{\text{ln}\left(a\right)}{\text{ln}\left(a^{a^{n-1}}\right)}$

Multiply both sides by $latex \text{ln}\left(a^{a^{n-1}}\right)$:

$latex a^{1-n}\, \text{ln}\left(a^{a^{n-1}}\right)=\text{ln}\left(a\right)$

Raise $latex e$ to the power of each side:

$latex e^{\text{ln}\left(a^{a^{n-1}}\right)}=e^{\frac{\text{ln}\left(a\right)}{a^{1-n}}}$

Simplify the result:

$latex a^{a^{n-1}}=a^{a^{n-1}}$

Now that we have the definitions for nested exponentiation and nested logarithms, we can define identities associated with these operations.

Identities for Nested Exponentiation:

$latex a^{\left \langle 0 \right \rangle}=a^{a^{(0-1)}}=a^{a^{-1}}=a^{\frac{1}{a}}=\sqrt[a]{a}$

$latex a^{\left \langle 1 \right \rangle}=a^{a^{(1-1)}}=a^{a^0}=a$

$latex a^{\left \langle 2 \right \rangle}=a^{a^{(2-1)}}=a^{a^1}=a^a$

$latex \left(a^{\left \langle b \right \rangle}\right)^{a^c}=\left(a^{a^{b-1}}\right)^{a^c}=a^{a^{b-1} a^{c}}=a^{a^{b+c-1}}=a^{\left \langle b+c \right \rangle}$

$latex \left(a^{\left \langle b \right \rangle}\right)^{a^{-c}}=\left(a^{a^{b-1}}\right)^{a^{-c}}=a^{a^{b-1} a^{-c}}=a^{a^{b-c-1}}=a^{\left \langle b-c \right \rangle}$

Derivatives for Nested Exponentiation (Derived using Mathematica):

Nested exponential function that has a variable as the base and a constant as the nested exponent:

$latex \frac{dy}{dx} \ x^{\left \langle a \right \rangle} = x^{\left \langle a \right \rangle} \left(x^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)$

The derivative of the natural log of the nested exponential:

$latex \frac{dy}{dx} \ \text{ln}\left(x^{\left \langle a \right \rangle}\right) = \left(x^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)$

Which allows us to define the following by substitution:

$latex \frac{dy}{dx} \ x^{\left \langle a \right \rangle} = x^{\left \langle a \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(x^{\left \langle a \right \rangle}\right)$

Such that:
$latex x^{\left \langle a \right \rangle}=\frac{\frac{dy}{dx} \ x^{\left \langle a \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle a \right \rangle}\right)}$

Nested exponential function that has a constant as the base and a variable as the nested exponent:

$latex \frac{dy}{dx} \ a^{\left \langle x \right \rangle} = a^{\left \langle x \right \rangle} \left(a^{(x-1)} \, \text{ln}\left(a\right)^2\right)$

The derivative of the natural log of the nested exponential:

$latex \frac{dy}{dx} \ \text{ln}\left(a^{\left \langle x \right \rangle}\right) = \left(a^{(x-1)} \, \text{ln}\left(a\right)^2\right)$

Which allows us to define the following by substitution:

$latex \frac{dy}{dx} \ a^{\left \langle x \right \rangle} = a^{\left \langle x \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(a^{\left \langle x \right \rangle}\right)$

Such that:
$latex a^{\left \langle x \right \rangle}=\frac{\frac{dy}{dx} \ a^{\left \langle x \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(a^{\left \langle x \right \rangle}\right)}$

Nested exponential function that has a variable as the base and as the nested exponent:

$latex \frac{dy}{dx} \ x^{\left \langle x \right \rangle} = x^{\left \langle x \right \rangle} \left(x^{(x-2)}+x^{(x-1)}\, \text{ln}\left(x\right)\left(\text{ln}\left(x\right)+\frac{x-1}{x}\right)\right)$

The derivative of the natural log of the nested exponential:

$latex \frac{dy}{dx} \ \text{ln}\left(x^{\left \langle x \right \rangle}\right) = \left(x^{(x-2)}+x^{(x-1)}\, \text{ln}\left(x\right)\left(\text{ln}\left(x\right)+\frac{x-1}{x}\right)\right)$

Which allows us to define the following by substitution:

$latex \frac{dy}{dx} \ x^{\left \langle x \right \rangle} = x^{\left \langle x \right \rangle} \ \frac{dy}{dx} \ \text{ln}\left(x^{\left \langle x \right \rangle}\right)$

Such that:
$latex x^{\left \langle x \right \rangle}=\frac{\frac{dy}{dx} \ x^{\left \langle x \right \rangle}}{\frac{dy}{dx} \ \text{ln}\left(x^{\left \langle x \right \rangle}\right)}$

Integrals for Nested Exponentiation (Derived using Mathematica):

The integrals for nested exponentiation are based on the exponential integral or they are unkown.

$latex \int x^{\left \langle a \right \rangle} \, dx = \int x^{x^{a-1}} \, dx$

$latex \int a^{\left \langle x \right \rangle} \, dx = \int a^{a^{x-1}} \, dx = \frac{\text{Ei}\left(a^{x-1}\, \text{ln}(a)\right)}{\text{ln}(a)}+C$

$latex \int x^{\left \langle x \right \rangle} \, dx = \int x^{x^{x-1}} \, dx$

The exponential integral:

$latex \text{Ei}(x)=\int \limits_{-\infty}^{x} \frac{e^t}{t} \, dt$

Identities for Nested Logarithms:

$latex \text{nLog}_{\, a}\left(b\right)=1+\frac{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}$

$latex \text{nLog}_{\, a}\left(b\right)=1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)}$

$latex \text{nLog}_{\, a}\left(b\right)=1+\text{log}_{a}\left(\text{log}_{a}\left(b\right)\right)$

$latex \text{nLn}\left(b\right)=1+\text{ln}\left(\text{ln}\left(b\right)\right)$

$latex \text{nLog}_{\, a}\left(b\right)=1-\frac{\text{ln}\left(\text{ln}\left(a\right)\right)}{\text{ln}\left(a\right)}+\frac{\text{ln}\left(\text{ln}\left(b\right)\right)}{\text{ln}\left(a\right)}$

$latex \text{log}_{\, a}\left(b\right)=\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}=a^{\left(\text{nLog}_{\, a}\left(b\right)-1\right)}$

$latex \text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(a\right)\left(\text{nLog}_{\, a}\left(b\right)-1\right)$

$latex \text{ln}\left(\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)=\text{ln}\left(a\right)\left(1-\text{nLog}_{\, a}\left(b\right)\right)$

$latex \text{nLog}_{\, a}\left(a\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(a\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=1+\frac{\text{ln}\left(1\right)}{\text{ln}\left(a\right)}=1+\frac{0}{\text{ln}\left(a\right)}=1$

$latex \lim_{b \to 0}\big(\text{nLog}_{\, a}\left(b\right)\big) =\lim_{b \to 0} \left(1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}\right)=\infty$

$latex \lim_{b \to 1}\big(\text{nLog}_{\, a}\left(b\right)\big) =\lim_{b \to 1} \left(1+\frac{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}\right)=-\infty$

Identities for Nested Logarithms Similiar to Normal Logarithms:

Multiplication:

$latex \text{log}_{\, a}\left(x \ \times \ y\right)\ =\ \big(\text{nLog}_{\, a}\left(a^x\right)-1\big)\ +\ \big(\text{nLog}_{\, a}\left(a^y\right)-1\big)\ =\ \text{nLog}_{\, a}\left(a^x\right)\ +\ \text{nLog}_{\, a}\left(a^y\right)\ -\ 2$

$latex \text{log}_{\, a}\big( \text{log}_{\, a}\left(x\right) \ \times \ \text{log}_{\, a}\left(y\right)\big)\ =\ \big(\text{nLog}_{\, a}\left(x\right)-1\big)\ +\ \big(\text{nLog}_{\, a}\left(y\right)-1\big)\ =\ \text{nLog}_{\, a}\left(x\right)\ +\ \text{nLog}_{\, a}\left(y\right)\ -\ 2$

Division:

$latex \text{log}_{\, a}\left(\frac{x}{y}\right)\ =\ \big(\text{nLog}_{\, a}\left(a^x\right)-1\big)\ -\ \big(\text{nLog}_{\, a}\left(a^y\right)-1\big)\ =\ \text{nLog}_{\, a}\left(a^x\right)\ -\ \text{nLog}_{\, a}\left(a^y\right)$

$latex \text{log}_{\, a}\left( \frac{\text{log}_{\, a}\left(x\right)}{\text{log}_{\, a}\left(y\right)}\right)\ =\ \big(\text{nLog}_{\, a}\left(x\right)-1\big)-\big(\text{nLog}_{\, a}\left(y\right)-1\big)\ =\ \text{nLog}_{\, a}\left(x\right)\ -\ \text{nLog}_{\, a}\left(y\right)$

Exponents:

$latex \text{nLog}_{\, a}\left(x^{y}\right)=\text{log}_{\, a}\left(y\right)\ +\ \text{nLog}_{\, a}\left(x\right)$

Derivatives for Nested Logarithms (Derived using Mathematica):

$latex \frac{dy}{dx} \ \text{nLog}_{\, a}\left(x\right)=\frac{\text{nLog}_{\, a}\left(x\right)}{x\, \text{ln}\left(x\right) \left(\text{ln}\left(a\right) +\text{ln}\left(\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)\right)}=\frac{1}{x\, \text{ln}\left(x\right)\, \text{ln}\left(a\right)}$

$latex \frac{dy}{dx} \ \text{nLog}_{\, x}\left(a\right)=\frac{\text{ln}\left(x\right) \left(1-\text{nLog}_{\, x}\left(a\right)\right)-1}{x\, \text{ln}\left(x\right)^2}=\frac{\text{ln}\left(\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)-1}{x\, \text{ln}\left(x\right)^2}$

Integrals for Nested Logarithms (Derived using Mathematica):

The integrals for nested logarithms are based on the logarithmic integral or they are unkown.

 $latex \int \text{nLog}_{\, a}\left(x\right) \, dx$ $latex = \int 1\, +\, \frac{\text{ln}\left (\frac{\text{ln}\left(x\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)} \, dx$ $latex = x\ \text{nLog}_{\, a}\left(x\right)\, -\, \frac{\text{li}(x)}{\text{ln}(a)}\, +\, C$ $latex = x \left(1\, + \, \frac{\text{ln}\left(\frac{\text{ln}(x)}{\text{ln}(a)}\right)}{\text{ln}(a)}\right)\, -\, \frac{\text{li}(x)}{\text{ln}(a)}\, +\, C$ 

$latex \int {}_n\, \!\text{Log}_{\, x}\left(a\right) \, dx = \int 1+\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(x\right)}\right)}{\text{ln}\left(x\right)} \, dx$

The logarithmic integral:

$latex \text{li}(x)=\int \limits_{0}^{x} \frac{dt}{\text{ln}(t)}$

Interesting Relationships for Nested Exponentials and Nested Logarithms:

The following relationships and their proofs are just a few of the more interesting properties that can be derived by manipulating the equations. Even though most of these relationships may not have a usage in standard mathematics, some may prove to be useful when dealing with large numbers.

Relationship between nested exponentials and normal exponentials:

$latex a^b=a\ \frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}=a\ \text{log}_{\, a}\left(a^{\left \langle b \right \rangle}\right)$

Nested exponentiation by definition:

$latex a^{\left \langle b \right \rangle}=a^{a^{(b-1)}}$

Take the natural log of both sides:

$latex \text{ln}\left(a^{a^{(b-1)}}\right)=\text{ln}\left(a^{\left \langle b \right \rangle}\right)$

Simplify the result:

$latex a^{(b-1)}\, \text{ln}\left(a\right)=\text{ln}\left(a^{\left \langle b \right \rangle}\right)$

Divide both sides by ln(a):

$latex a^{(b-1)}=\frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}$

Multiply both sides by $latex a$:

$latex a^{b}=a\ \frac{\text{ln}\left(a^{\left \langle b \right \rangle}\right)}{\text{ln}\left(a\right)}$

Convert to a standard logarithm by “changing the base”:

$latex a^{b}=a\ \text{log}_{\, a}\left(a^{\left \langle b \right \rangle}\right)$

Relationship between logarithms with the base and input swapped:

$latex \text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right) + \text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right) = 0$

This is derived by subtracting both equations for nested logarithms:

$latex \left(1+\frac{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}\right)-\left(1-\frac{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)}\right)=0$

Simplify the left side:

$latex \frac{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right) + \text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}{\text{ln}\left(a\right)} = 0$

Multiply both sides by ln(a) to derive the relationship:

$latex \text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right) + \text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right) = 0$

We can prove this by raising $latex e$ to the above identity:

$latex e^{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right) + \text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}=e^0$

This is equal to:

$latex e^{\text{ln}\left (\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)} \times e^{\text{ln}\left (\frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)}\right)}=1$

Which simplifies to:

$latex \frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)} \times \frac{\text{ln}\left(a\right)}{\text{ln}\left(b\right)} = 1$

Relationship between natural logarithms and natural nested logarithms:

$latex \text{nLn}\left(b\right)-\text{nLn}\left(a\right)=\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)$

Natural nested logarithm by definition:

$latex \text{nLn}\left(x\right)=1+\text{ln}\left(\text{ln}\left(x\right)\right)$

Subtract the natural nested log of $latex a$ from the natural nested log of $latex b$:

$latex \text{nLn}\left(b\right)-\text{nLn}\left(a\right)=\left(1+\text{ln}\left(\text{ln}\left(b\right)\right)\right)-\left(1+\text{ln}\left(\text{ln}\left(a\right)\right)\right)$

Simplify the right hand side:

$latex \text{nLn}\left(b\right)-\text{nLn}\left(a\right)=\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)$

Simplify the division of the natural logarithms:

$latex \text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(\text{ln}\left(b\right)\right)-\text{ln}\left(\text{ln}\left(a\right)\right)$

We can see that they are both equal which brings us to the following relationship by raising $latex e$ to the power of the above identity:

$latex e^{\text{nLn}\left(b\right)-\text{nLn}\left(a\right)}=e^{\text{ln}\left(\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}\right)}=\frac{\text{ln}\left(b\right)}{\text{ln}\left(a\right)}=\text{log}_{\, a}\left(b\right)$

The nested logarithm of a double nested exponential with the same base:

$latex \text{nLog}_{\, a}\left(\left(a^{\left \langle b \right \rangle}\right)^{\left \langle c \right \rangle}\right)=\left(c-1\right)\, a^{(b-1)}+b$

Expand the double nested exponential using the general definition:

 $latex \left(a^{\left \langle b \right \rangle}\right)^{\left \langle c \right \rangle}$ $latex =\left(a^{\left \langle b \right \rangle}\right)^{\left(a^{\left \langle b \right \rangle}\right)^{(c-1)}}$ $latex =\left(a^{a^{(b-1)}}\right)^{\left(a^{a^{(b-1)}}\right)^{(c-1)}}$ $latex = \left(a^{a^{(b-1)}}\right)^{\left(a^{(c-1)\, a^{(b-1)}}\right)}$ $latex =a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}$ 

Take the natural log of the expanded version:

$latex \text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)=\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right) \text{ln}\left(a\right)$

Divide both sides by ln(a):

$latex \frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}=\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)$

Take the natural log of both sides:

$latex \text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)\right)$

Simplify the right hand side:

 $latex \text{ln}\left(\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)\right)$ $latex =\text{ln}\left(a^{(b-1)}\right) + \text{ln}\left(a^{(c-1)\, a^{(b-1)}}\right)$ $latex =(b-1) \, \text{ln}\left(a\right)\, + \, (c-1)\, a^{(b-1)} \, \text{ln}\left(a\right)$ $latex =\text{ln}\left(a\right)\, \left((b-1)\, +\, (c-1)\, a^{(b-1)}\right)$ 

Such that:

$latex \text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)=\text{ln}\left(a\right)\, \left((b-1)\, +\, (c-1)\, a^{(b-1)}\right)$

Divide both sides by ln(a):

$latex \frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=(b-1) \, + \, (c-1)\, a^{(b-1)}$

Complete the nested logarithm by adding one to both sides:

$latex 1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=1+(b-1) \, + \, (c-1)\, a^{(b-1)}$

Simplify the result:

$latex \text{nLog}_{\, a}\left(\left(a^{\left \langle b \right \rangle}\right)^{\left \langle c \right \rangle}\right)=1+\frac{\text{ln}\left(\frac{\text{ln}\left(a^{\left(a^{(b-1)}\right)\left(a^{(c-1)\, a^{(b-1)}}\right)}\right)}{\text{ln}\left(a\right)}\right)}{\text{ln}\left(a\right)}=(c-1)\, a^{(b-1)}\, +\, b$

Finding nested 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 nested roots or nested logarithms using this approach.

$latex x_{n+1}=x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}$

Nested Roots:

To find the nested root we must use the nested exponential function that has the constant as the nested exponent, $latex x^{\left \langle a \right \rangle}$. Using this function with Newton’s method yields the following algorithm for nested roots (Repeat the process until you have obtained the desired result):

$latex \sqrt[\left \langle a \right \rangle]{b}\ \ \ \ \text{approximated by} \ \ \ \ x_{n+1}=x_{n}-\frac{\left(x_{n}\right)^{\left(x_{n}\right)^{(a-1)}}-b}{\left(x_{n}\right)^{\left(x_{n}\right)^{(a-1)}}\left(\left(x_{n}\right)^{(a-2)}\left(1+\left(a-1\right)\, \text{ln}\left(x\right)\right)\right)}$

Nested Logarithms:

To find the nested logarithm we must use the nested exponential function that has the variable as thenested exponent, $latex a^{\left \langle x \right \rangle}$. Using this function with Newton’s method yields the following algorithm for nested logarithms (Repeat the process until you have obtained the desired result):

$latex \text{nLog}_{\, a}\left(b\right)\ \ \ \text{approximated by} \ \ \ x_{n+1}=x_{n}-\frac{a^{a^{\left(x_{n}\right)-1}}-b}{a^{a^{\left(x_{n}\right)-1}} a^{\left(x_{n}\right)-1} \text{ln}\left(a\right)^2}$

Graphs of nested exponentials and nested logarithms

$latex \text{The graph of } x^{\left \langle a \right \rangle}$:

This graph shows both the real part (blue) and the imaginary part (green).

$latex \text{The graph of } a^{\left \langle x \right \rangle}$:

This graph shows both the real part (blue) and the imaginary part (green).

$latex \text{The graph of } x^{\left \langle x \right \rangle}$:

This graph shows both the real part (blue) and the imaginary part (green).

$latex \text{The graph of } \text{nLog}_{\, a} \left(x\right):$

This graph shows both the real part (blue) and the imaginary part (green).

$latex \text{The graph of } \text{nLog}_{\, x} \left(a\right):$

This graph shows both the real part (blue) and the imaginary part (green).

$latex \text{The graph of } \text{nLog}_{\, x} \left(x\right):$

This graph shows both the real part (blue) and the imaginary part (green).

One Person Can Make A Difference

I wasn’t always the geeky math type lol. I spent most of my high school days playing guitar and chasing girls. And even though that hasn’t changed much, it was my ninth grade science teacher, Jon Myers, who would ultimately change my life forever. He was able to see something in me that I didn’t even know was there and he knew how to bring it out.

Mr. Myers was teaching us how to balance chemical formulas and after a couple of days of having us work on a few exercises, he kept me after class to see if he could challenge me. I’m not sure exactly why he did this, or how he knew that I might be capable of performing the task he was about to set forth. But, he went to his desk and pulled out his college organic chemistry book and proceeded to write out a chemical formula that spanned across two black boards. He told me to take my time and work the problem out. He then contacted the teacher of my next class so that I would not be counted late or absent.

I never really considered myself a mathematician. I didn’t even know I had a knack for solving mathematical problems. I was just a long haired kid who played heavy metal on the guitar. I spent most of my time with my girlfriend and thought very little about mathematics. But in that moment while staring at this enormous problem, I realized that I could somehow see the answer. Without hesitation, I walked up to the black board and wrote out the correct solution.

I was a little bewildered by Jon’s amazement when seeing that I had solved the problem. But, I did not use any paper or a calculator to find the solution. I managed to balance the formula within my head which only took me about seven seconds to solve. He explained to me that my ability was a gift and from that moment on I spent more time solving mathematical problems than playing music on my guitar. I probably should’ve substituted the time I spent with my girlfriend instead of the time I used to play music, but that is another story altogether.

I don’t claim to be a genius, prodigy, or anything of the sort. I just have a talent for mathematics which grew into a love for the subject. If it wasn’t for Mr. Myers, I would’ve never known about that hidden gift which enabled me to enjoy a wonderful career as a software engineer and discover many mathematical equations, some which might actually be new. I am currently pursuing a degree in Physics and I owe it all to him.