12:55 PM
@robjohn So last night, had an idea. Did some experimental research with power of two decomposition of numbers that is... what's the formal term? Suboptimally? Non-minimally? Something like that. Anyways...
You find the smallest power of two greater than the divisor for which we already know fits in one time. We can then use the modulus of the power of two on the dividend to get a remainder. We then end up with an identity of $\frac{x}{y}$ which I'll describe in the next comment.
We find the smallest power of two greater than $y$. We'll call it $2^n$. $y$ is then decomposed into $2^n + (y\bmod 2^n)$.
For $2^n$, we further decompose this into $(y + (2^n - y))$
You can probably see where I'm going with this.
We compute $a = \lfloor\frac{x}{2^n}\rfloor$.
We then multiply this by our decomposition of $2^n$ as $a(y + (2^n - y))$. We then check if $2^n - y \geq y$, and if so, repeat the process of decomposition once again and determine once more the coefficient, and that's about as far as I've gotten. The modular part is pretty straightforward as we just add 0 or 1 depending on whether or not the modulus is greater than or equal to y.
I wouldn't be surprised, however, if this was just long division in disguise, and not actually a different algorithm.
Or maybe it's a form of long division that takes fewer iterations in base two, not sure.
The whole point is, this method can be used to once again reduce the size of $x$ for our modulus computation.
(or just compute floored quotients directly).
It took me long enough to realize, but I finally did figure out that you can in fact divide multiple sums in the denominator just not in the conventional manner, and certainly not by just expanding the sum to same number with each operand as the denominator.
If we consider the quotient of $\frac{27}{3}$, for instance, this method of decomposition gives us $\frac{27}{6\cdot (3 + 1) + 3}$. Distributing the coefficient, we obtain $\frac{27}{(18 + 6) + 3)}$. We see that the right hand term in the left-most parenthetical expression is greater than three. At a glance, we can already see that the sum of coefficients as-is gives us the correct quotient as 6 + 2 + 1 = 9, but of course, we are interested in determining this algorithmically.
Yet all we did was compute the floored quotient of one or more powers of two into the dividend.
But basically the recursive sequence here is a matter of doing the above, and then for that right operand in the power of two's decomposition of the divisor, we then compute that quotient. Namely, if we have a number in the form of $a(y + (2^n - y)) + (y\bmod 2^n)$, after computing the coefficient $a$ and the quotient $\frac{y\bmod 2^n}{y}$, we compute $\frac{2^n - y}{y}$.
Since we can once again decompose $2^n - y$ and use the same method to extract coefficients from it, we can do this ad infinitum to compute the quotient, and you could probably extend it to get the fractional portion, but I just think it's easier to just compute $\frac{2^N - 1}{y}$ for $N$ bits to compute the reciprocal of $y$ and multiply it by $x$.
Given $f(z)=|z|^{1/2}z$, it is obviously complex differentiable at $z=0$ because $$f'(0)=\lim_{z\rightarrow 0}\frac{|z|^{1/2}z}{z}=0$$ I have done similar examples with $|z|^2$, but the $|z|^{1/2}$ is throwing me off because you cannot simply expand it like you would with $|z|^2$. How can I show...
You are developing the right idea, but your Eq. 1 is a bit awkward. To understand why, let me say a bit about functionals and functional derivatives.
A functional acts on a function in its entirety, for example we need to know the full input function $n(r)$ in order to obtain the corresponding va...
