/biz/ - Business & Finance

[Anonymous No.8882053]

explain this seeming paradox to a brainlet
>liquidation of portolio in btc =x, usd = y
>(x(t1)-x(t0))/(y(t1)-y(t0)) > average btc price
>x(t1) < x(t0)
so I lost 'sats' but the difference is above btc price? me confused

[Anonymous No.8882263]

bump
>inb4 you're not going to make it

[Anonymous No.8882843]

help

[Anonymous No.8882876]

i don't know what the fuck this is

[Anonymous No.8882939]

>>8882876
that's not very helpful tbqh

[Anonymous No.8883301]

What the fuck is this delta? Liquidation are related to market. If its xbtusd or say dogeusd or dogebtc..

[Anonymous No.8883342]

>>8883301
it means value of portfolio in btc or usd though not margin liquidations. I don't use any leverage with this.
The two samples are at arbitrary times

[Anonymous No.8883486]

Dumb mong.

In this form, the equations correspond structurally to the condition that the Jacobian matrix is of the form. A matrix of this form is the matrix representation of a complex number. Geometrically, such a matrix is always the composition of a rotation with a scaling, and in particular preserves angles. The Jacobian of a function f(z) takes infinitesimal line segments at the intersection of two curves in z and rotates them to the corresponding segments in f(z). Consequently, a function satisfying the Cauchy–Riemann equations, with a nonzero derivative, preserves the angle between curves in the plane. That is, the Cauchy–Riemann equations are the conditions for a function to be conformal.

Moreover, because the composition of a conformal transformation with another conformal transformation is also conformal, the composition of a solution of the Cauchy–Riemann equations with a conformal map must itself solve the Cauchy–Riemann equations. Thus the Cauchy–Riemann equations are conformally invariant.

[Anonymous No.8883509]

Typically u and v are taken to be the real and imaginary parts respectively of a complex-valued function of a single complex variable z = x + iy, f(x + iy) = u(x,y) + iv(x,y). Suppose that u and v are real-differentiable at a point in an open subset of C (C is the set of complex numbers), which can be considered as functions from R2 to R. This implies that the partial derivatives of u and v exist (although they need not be continuous) and we can approximate small variations of f linearly. Then f = u + iv is complex-differentiable at that point if and only if the partial derivatives of u and v satisfy the Cauchy–Riemann equations (1a) and (1b) at that point. The sole existence of partial derivatives satisfying the Cauchy–Riemann equations is not enough to ensure complex differentiability at that point. It is necessary that u and v be real differentiable, which is a stronger condition than the existence of the partial derivatives, but in general, weaker than continuous differentiability.

[Anonymous No.8883763]

what

[Anonymous No.8883952]

>>8883763
Just sell now. You're not going to make it.

[Anonymous No.8884118]

>>8883952

[Anonymous No.8884133]

>>8883763
wat wat?