by size i mean a minimum volume for a computation of n bits, $R=(2ln(2)kTG/c^4)*n$ where n is the number of bits k is boltzman constant and T is temperature and R is radius and G is newton's gravitational contstant
based off of 1bit = ln(2)kT joules by szilards engine, e=mc^2 and schwarzild radius R=2mg/c^2
Is it true that if we're talking about the partial derivative of a function $f\colon R^2\to\mathbb R$ (with respect to $x$), we're actually talking about a family of functions? Because for each constant $y$, we actually have a different (specific) partial derivative.
@ShaVuklia You get a function with two inputs ($y$ as well as $x$). In other words, you get another function from $\Bbb R^2\to\Bbb R$.
But you can think of such functions as being infinitely many one-variable functions, one for each $y$. That's essentially the difference between writing $g(x,y)$ and $g_y(x)$.
when applying the chain rule to this function, we need to rearrange the intermediate results in column or row vectors depending on the needs when multiplying the resulting intermediate derivatives arbitrarily, or what?
in other words, if the derivative is something of the form df/dh * dh/dg * dg/dx, suppose the result of df/dh is a 3x1 vector and the result of dh/dg is a 3x1 vector, then, in theory, we can't multiply them together, so we need to transpose one of those
but when finding the Jacobian of another function of the form f(h(g(x)), I can use this gradient, but I need to transpose it, otherwise the dimensions do not match
it turns out that the gradient of this function with respect to either $j_1$ or $j_2$ can be calculated as follows $\frac{(g_i(t_k) \times j_) \times g_i(t_k)}{\lVert g_i(t_k) \times j_i \rVert_2}$
I think you need to consider as input either only $j_1$ or $j_2$ (but not both), so this should be a function of the form $\mathbb{R}^3 \mapsto \mathbb{R}$
and let $q$ be the functions vector $j_i$, which are actually vectors, but these vectors receive as input $x$, and produce their coordinates something of the form $\hat{j}_1={(\operatorname{cos}(\phi_1)\operatorname{cos}(\theta_1),\cos(\phi_1)\operatorname{sin}(\theta_1),\operatorname{sin}(\phi_1))}^T $
@PVAL-inactive Why wouldn't $g$ map from $\mathbb{R}^3$ to the same space? It takes as input vectors $j_i$ and $g_i$ and produces another vector
have you understood everything regarding $f$ and $e$ so far?
now, we can use the chain rule to find the Jacobian of $f$
how?
$J_h(g(q(x)) * J_g(q(x)) * J_q(x)$, right?
but now note that I have already $J_h(g(q(x)) * J_g(q(x))$, i.e. it's the gradient I found before, but it's transposed
I know it's not easy to follow
also because these vectors are not very friendly
Why is it the gradient transposed?
let me try to explain
$J_h(g(q(x))$ is actually just the derivative of $h$, so it's a map from $\mathbb{R}^3$ to $\mathbb{R}^k$, since $g$ is the cross product, which is the input, and it produces $k$ rows
A polynomial is recursively, multiplicatively defined if it is in the closure $K$ of the $k$ under the rules:
(1) $a ,b\in k \implies aX + b \in K$
(2) $f,g \in K \implies f\cdot g \in K$
(3) $f,g \in K \implies f\circ g \in K$
If $f(X) = aX + b$, then $X$ is solvable for, namely $X = (f(X) -...
And, finally: it seems incredibly fishy that you start with 3 axioms to define something, then at the very end you replace one axiom by another and add a fourth.
Let $(k, +, \cdot)$ be a field. Then associate this field to another field $K = (\star, \cdot)$, where $K$ is defined recursively (that's the easiest way seemingly):
$$
a,b,c \in k, \ f,g \in K \implies \\
$$
(1)$ \ aX + b \in K$
(2) $f/g \in K, g \neq 0 $
(3) $f\cdot g \in K$
(4) $f\star g ...
A polynomial is recursively, multiplicatively defined if it is in the closure $K$ of the $k$ under the rules:
(1) $a ,b\in k \implies aX + b \in K$
(2) $f,g \in K \implies f\cdot g \in K$
(3) $f,g \in K \implies f\circ g \in K$
If $f(X) = aX + b$, then $X$ is solvable for, namely $X = (f(X) -...
