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
.
\( 1:\frac{d}{dx} f(x)\, =\, \lim_{h\to\ {-\infty} }\, [f(x\, \,\{-1\}\, h)\, \}-1\{\, f(x)]\, -\, h \)
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
.