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 The "little" differential operator and applications to tetration JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/19/2011, 08:19 PM (This post was last modified: 04/27/2011, 04:48 PM by JmsNxn.) The "little" differential operator is defined by the following: $1:\frac{d}{dx} f(x)\, =\, \lim_{h\to\ {-\infty} }\, [f(x\, \,\{-1\}\, h)\, \}-1\{\, f(x)]\, -\, h$ where $x\, \{-1\}\, y = ln(e^x + e^y)$ and $x\, \}-1\{\, y = ln(e^x - e^y)$ The lowered differential operator is a special case of the logarithmic semi operator differential operator $p \epsilon \R^+$, and S(p) is the identity function. $p:\frac{d}{dx} f(x)\,=\, \lim_{h\to\ {S(-p)} }\,\,\, [f(x\, \{-p\}\, h)\, \}-p\{\, f(x)]\, \}1-p\{ h$ we have very interesting results for the little differential operator. We have a product rule that is spread across addition, for convenience sake f' is taken to mean the little derivative of f: $1:\frac{d}{dx} (f(x) + g(x)) = [f'(x) + g(x)]\, \{-1\}\, [f(x) + g'(x)]$ we have a chain rule that is neat and elegant. $1:\frac{d}{dx} f(g(x)) = f'(g(x)) + g'(x)$ It's spread across {-1} $1:\frac{d}{dx} (f(x)\,\{-1\}\,g(x)) = f'(x)\,\{-1\}\,g'(x)$ At first, I thought the little derivative wouldn't work because the identity of {-1} is $-\infty$ and so therefore: $1:\frac{d}{dx} C = -\infty$ but otherwise everything is very consistent. Here are some derivatives: $1:\frac{d}{dx} e^x = e^x$ $1:\frac{d}{dx} ln(x) = -x$ this one was a bitch to work out $1:\frac{d}{dx} sin(x) = ln(cos(x)) + sin(x) - x$ and its cos counterpart: $1:\frac{d}{dx} cos(x) = ln(-sin(x)) + cos(x) - x$ The little polynomial has a slightly altered power rule: $1:\frac{d}{dx} xn = x(n-1) + ln(n)$ Its laws for exponentiation: $1:\frac{d}{dx} b^x = b^x + (lnb-1)x + ln^{[2]}(b)$ $1:\frac{d}{dx} x^n = x^n + (n-1)ln(x) + ln(n) - x$ And now here's where I wonder about tetration. Since the little derivative works across {p-1} operators the same way the normal derivative works across {p} operators. (And we know this to be true if only by heuristically looking at the derivatives I just showed you.) Is there anyway we can use the knowledge we know about the little derivative and how it works across exponentiation and apply that to how the normal derivative works across tetration. I can see where there maybe problems with that, but I'm thinking optimistically. Hmm, I can see where my reasoning might be considered aesthetic, but I'll post this now and come back to it in a bit. Maybe somebody here has some info they can give me. Edit: if you couldn't figure it out for yourself, the little derivative taylor series is given by the following: $f(x) = \{-1\} \sum_{n=0}^{\infty} f^{(n)}(a) + n(x \{-1\} a) - ln(n!)$ where $f^{(n)}(x)$ is the n'th little derivative of f and $\{p\} \sum_{n=0}^{R} f(n) = f(0) \{p\} f(1) \{p\} ... f( R )$ edit again: The little derivative is related to the normal derivative by the following equation: $\frac{d}{dx} f(x) = e^{[1:\frac{d}{dx} f(x)] + x - f(x)}$ which should hold for any analytic function. And if anyone would like to see the full proofs of these I'd be happy to give them, but they tend to be very cumbersome and long, especially typing them out in Latex . Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 04/20/2011, 06:31 AM Nice! One reason why I liked the fiddling with the "tetration" was always its openness for creativity. I've currently no space to read into your idea - I'll look at it next week (perhaps) Gottfried Gottfried Helms, Kassel JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/20/2011, 08:42 PM (This post was last modified: 04/20/2011, 08:48 PM by JmsNxn.) I've had another little thought bubble, this time it involves the geometric derivative. if $-1:\frac{d}{dx}f(x) = \lim_{h\to\0}(\frac{f(x+h)}{f(x)})^{\frac{1}{h}}$ is the geometric derivative, abbreviated as: $-1:\frac{d}{dx}f(x) = e^{\frac{f'(x)}{f(x)}}$ then: $-1:\frac{d}{dx} sexp_e(x) = e^{\frac{d}{dx}sexp_e(x-1)}$ Now what I am looking for are two functions that behave the same for the little derivative and the little geometric derivative. I believe that one of these functions will have a little derivative which will essentially "look the same" as the tetration derivative, we'll only have to raise the operators by one and remove lns where applicable. I'm not sure how rigorous this is all sounding now, but I'll try to be as rigorous as possible when writing it out mathematically. $1:\frac{d*}{dx} f(x) = \lim_{h\to\0}\,\frac{f(x\, \{-1\}\, ln(h)) - f(x)}{h}$ is the little geometric derivative. closest abbreviation is $1:\frac{d*}{dx} f(x) = \lim_{h\to\0}\,\frac{f(ln(e^x+h)) - f(x)}{h}$ $1:\frac{d*}{dx} e^x = 1$ but $1:\frac{d*}{dx} e^{e^x} = e^{e^x}$ just like the geometric derivative. The fact that they both coincide there leaves me optimistic. This is leading me somewhere but I have to work out some more on paper first. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/21/2011, 11:12 PM (04/20/2011, 08:42 PM)JmsNxn Wrote: I've had another little thought bubble, this time it involves the geometric derivative. if $-1:\frac{d}{dx}f(x) = \lim_{h\to\0}(\frac{f(x+h)}{f(x)})^{\frac{1}{h}}$ is the geometric derivative, abbreviated as: $-1:\frac{d}{dx}f(x) = e^{\frac{f'(x)}{f(x)}}$ then: $-1:\frac{d}{dx} sexp_e(x) = e^{\frac{d}{dx}sexp_e(x-1)}$ Now what I am looking for are two functions that behave the same for the little derivative and the little geometric derivative. I believe that one of these functions will have a little derivative which will essentially "look the same" as the tetration derivative, we'll only have to raise the operators by one and remove lns where applicable. I'm not sure how rigorous this is all sounding now, but I'll try to be as rigorous as possible when writing it out mathematically. $1:\frac{d*}{dx} f(x) = \lim_{h\to\0}\,\frac{f(x\, \{-1\}\, ln(h)) - f(x)}{h}$ is the little geometric derivative. closest abbreviation is $1:\frac{d*}{dx} f(x) = \lim_{h\to\0}\,\frac{f(ln(e^x+h)) - f(x)}{h}$ $1:\frac{d*}{dx} e^x = 1$ but $1:\frac{d*}{dx} e^{e^x} = e^{e^x}$ just like the geometric derivative. The fact that they both coincide there leaves me optimistic. This is leading me somewhere but I have to work out some more on paper first. returning to standard notation and at first glance , it seems the function you search for is any one ( coo ) that satisfies f(x) = exp(f(x)) that function has a " history " here. ... at first glance ... im in a hurry ... JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/21/2011, 11:34 PM (This post was last modified: 04/21/2011, 11:42 PM by JmsNxn.) (04/21/2011, 11:12 PM)tommy1729 Wrote: returning to standard notation and at first glance , it seems the function you search for is any one ( coo ) that satisfies f(x) = exp(f(x)) that function has a " history " here. ... at first glance ... im in a hurry ... I'm not really sure how you got there? Basically, I'm looking for a function that will have a little derivative that is translated from the tetration derivative. By "translated" I mean: $1:\frac{d}{dx} xn = x(n-1) + ln(n)$ is a translation of $\frac{d}{dx} x^n = nx^{n-1}$ and $1:\frac{d}{dx} x + y = y$ is a translation of $\frac{d}{dx} xy = y$. So I'm looking for a function that will have a translated little derivative of sexp, so that we can derive the derivative of sexp. So basically we first have to design a law as to how translated derivatives are related, and then try to derive one for tetration. My first instinct was that it would be exponentiation, but I doubt that now. I'm still a little sketchy now but it's coming to me slowly, I'm sure you know that feeling. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/22/2011, 12:29 PM (04/21/2011, 11:34 PM)JmsNxn Wrote: (04/21/2011, 11:12 PM)tommy1729 Wrote: returning to standard notation and at first glance , it seems the function you search for is any one ( coo ) that satisfies f(x) = exp(f(x)) that function has a " history " here. ... at first glance ... im in a hurry ... I'm not really sure how you got there? Basically, I'm looking for a function that will have a little derivative that is translated from the tetration derivative. By "translated" I mean: $1:\frac{d}{dx} xn = x(n-1) + ln(n)$ is a translation of $\frac{d}{dx} x^n = nx^{n-1}$ and $1:\frac{d}{dx} x + y = y$ is a translation of $\frac{d}{dx} xy = y$. So I'm looking for a function that will have a translated little derivative of sexp, so that we can derive the derivative of sexp. So basically we first have to design a law as to how translated derivatives are related, and then try to derive one for tetration. My first instinct was that it would be exponentiation, but I doubt that now. I'm still a little sketchy now but it's coming to me slowly, I'm sure you know that feeling. you asked for a function f such that little derivative = little geo derivative ? i was talking about that. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 04/23/2011, 06:25 PM (04/22/2011, 12:29 PM)tommy1729 Wrote: you asked for a function f such that little derivative = little geo derivative ? i was talking about that. oh no no, I was asking for a function f that satisfied this identity: $1:\frac{d*}{dx}f(x) = e^{1:\frac{d}{dx} f(x-1)}$ where the left side is the little geometric derivative and the right side is e to the power of the little derivative of f(x-1). I just have a curiosity that this function is related to tetration; because tetration satisfies: $\frac{d*}{dx}sexp(x) = e^{\frac{d}{dx}sexp(x-1)}$ but it's just a hunch. « Next Oldest | Next Newest »

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