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I want to plot a complex momentum in complex plane where the y-axis shows imaginary values and x-axis shows the real values. I tried ReIm command but it plots Re and Im parts not Re vs Im.
Below is an example which plots Im and Re parts but I need Im vs Re. Any suggestions please?
Clear[m, M, s, t]
m = 0.34
M = 5;
t = m^2 + M^2 - s/2 + 1/2 Sqrt[s - 4 m^2] Sqrt[s - 4 M^2];
tplot = Plot[{Im@t, Re@t}, {s, 4 m^2, 4 M^2}, PlotRange -> All,
Frame -> True]
Definitions and some motivation:
Let $mathcal B$ be the set of bounded measurable functions from $(0, 1)$ to $mathbb R$. Denote by $mathcal N$ the set of measurable subsets of $(0, 1)$ with Lebesgue measure $0$.
Given a function $f in mathcal B$, define the function $mathcal Of$ by
$mathcal Of(x) := inf_{N in mathcal N} lim_{delta to 0} sup_{y, z in B_delta (x) setminus N} |f(y) – f(z)|$.
Thanks to Lusin’s theorem, we know that we can modify $f$ on an arbitrarily small set and get a continuous function, and so we force the oscillation to be $0$ everywhere. But can we force it to be whatever we want?
Question:
Does there exist, for any $f, g in mathcal B$ and $varepsilon > 0$, a function $f’ in mathcal B$ such that the following conditions are satisfied?
i) $f’ = f$ everywhere except for a set of measure at most $varepsilon$.
ii) $mathcal Of’ = mathcal Og$ everywhere.
Note: All functions are genuine functions and not equivalence classes modulo null sets of such.
I have a list (Table), mostly composed of real numbers. The list also contains some non-real elements. The list is similar to the following (but much much larger):
{{0.11011887269, 0.02087291466, 0.02082444233, 0.02083704779,
0.02084940959, 0.02086181292, 0.02087425501, 0.02088673503,
0.02089925331, 0.02091181001, 0.02092440529, 0.02093703933,
0.02094971232, 0.02096242441, test1(18, 15)}, {0.10941526373,
0.017604005802, 0.017549654594, 0.017558686829, test1(19, 5),
test1(19, 6), test1(19, 7), test1(19, 8), test1(19, 9),
test1(19, 10), test1(19, 11), test1(19, 12), test1(19, 13),
test1(19, 14), test1(19, 15)}, {0.10890647143, 0.014986243752,
0.014927392616, 0.014934004368, test1(20, 5), test1(20, 6),
test1(20, 7), test1(20, 8), test1(20, 9), test1(20, 10),
test1(20, 11), test1(20, 12), test1(20, 13), test1(20, 14),
test1(20, 15)}}
I would like to construct a single row or column list that preserves the last real number element of each row of the above list. For example, with the above table, the resulting list should contain only 3 elements as:
{0.02096242441, 0.017558686829, 0.014934004368}
How can this resulting list be automatically constructed from the original one?
Let $S$ be a real $n times n$ symmetric matrix, which is diagonalizable, of course, and let its Jordan normal form be $J=text{diag}(lambda_1,…,lambda_n)$, where each $lambda_i$ is an eigenvalue. Consider $M$ the change of basis, i.e. $M$ is an invertible matrix such that $S=M^{-1}JM$. We know it is possible to choose $M$ such that it is orthogonal, but can we guarantee that $M$ is always orthogonal?
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.
Let $H$ be a Hilbert space with an orthonormal basis $(e_n)$ and let $T:Hto H$ be the (only) bounded operator that satisfy:
$T(e_n)=e_{n+1}$ for all $nin N$.
Show that $T$ is isometric but not unitary.
Let $H_1,H_2$ Hilbert spaces and $T : H_1→ H_2$
T is called unitary if:
$T$ is surjective, and
$T$ preserves the inner product.
Let $X,Y$ be norm spaces.
$T:Xto Y$.
$T$ is called isometry if it preserves norm:
$||Tx||=||x|| forall xin X$.
T is Isometry
Since (e_n) is an orthonormal basis of H:
Let $xin H$ so we want to show that $||Tx||=||x||$.
From a theorem in functionl analysis, we have
$||x||^2=sum_{n} |<x,e_n>|^2$ for every $xin H$.So we can substitute $Txin H$ but i cannot see how to use then the given definition of the operator $T$.
From the same theorem, we get also:
$x=sum_{n} <x,e_n>e_n$ for all $xin H$ but then how to use this for showing that the operator is not unitary i.e. it does not preserve inner product, using the definition of $T$.
AFAIK they achieve this by
Parameterized Procedural Placement
They create a workflow where you create a handful assets in this case building floors and they procedurally tweak the parameters such as Floor count,size,tint and custom props such as air conditioning outside of the buildings.
I once a saw a video about ubisoft creating a procedural assasins creed city.They premade a bunch of houses and spread them randomly.
You should desing a workflow where you control how your building floors are stacked on each other and how each floor has its own specific unique props.
This is what i would do
with these parameters one can create wide range of different buildings considering you feed your procedural workflow with enough assets
AFAIK they achieve this by
Parameterized Procedural Placement
They create a workflow where you create a handful assets in this case building floors and they procedurally tweak the parameters such as Floor count,size,tint and custom props such as air conditioning outside of the buildings.
I once a saw a video about ubisoft creating a procedural assasins creed city.They premade a bunch of houses and spread them randomly.
You should desing a workflow where you control how your building floors are stacked on each other and how each floor has its own specific unique props.
This is what i would do
with these parameters one can create wide range of different buildings considering you feed your procedural workflow with enough assets
$$cos{(frac{2pi}{9})}$$
$$f(x) = cos{(2pi x)}$$
$$x = a + h$$
$$x = 0 + frac{1}{9}$$
$$f(a+h) quadtext{approximately is :} quad cos(0)-sin(0)*2pi*frac{1}{9}quad text{which is wrong…}$$
Why doesn’t it work this way? Where did I make a mistake?
Is $Isubseteqmathbb{R}$ an intervall and $f: Itomathbb{R}$ a differentiable function with bounded derivative $f’: Itomathbb{R}$, then $f$ is Lipschitz-continuous.
This is supposed to be an application of the mean-value theorem.
What gets me is the use of unspecified intervalls. So $I=(a,b), (a,b), (a,b), (a,b)$, as the mean-value theorem holds for differentiable functions defined on a compact intervall (a,b).
Every resource I looked it up proofs this result for compact intervalls, but I was unable to give a counterexample for say $I=(a,b)$, because of the bounded derivative.
But how does one relax the condition to $I=(a,b)$ to apply the mean-value theorem?
I thought that one might can proof that for $I=(a,b)$ you are able to continuously extend to $(a,b)$.
Thanks in advance.
