Expression involving square roots not simplifying

I have a relatively simple expression here that is not simplifying:

$$
frac{2 s_0 left(sqrt{gamma ^5 s_0}+sqrt{gamma ^9 s_0}right)+sqrt{gamma ^3
s_0}+2 sqrt{gamma ^7 s_0}+sqrt{gamma ^{11} s_0}+sqrt{gamma ^7 s_0^5}}{gamma
left(gamma ^2+gamma s_0+1right){}^2}
$$

$Assumptions = {(s0 | (Gamma)) (Element) Reals, (Gamma) > 0, 
   s0 > 0};
(Sqrt(s0 (Gamma)^3) + 2 Sqrt(s0 (Gamma)^7) + Sqrt(s0^5 (Gamma)^7) +
   Sqrt(s0 (Gamma)^11) + 
  2 s0 (Sqrt(s0 (Gamma)^5) + Sqrt(s0 (Gamma)^9)))/((Gamma) (1 + 
    s0 (Gamma) + (Gamma)^2)^2) // Simplify
(Sqrt(s0 (Gamma)^3) + 2 Sqrt(s0 (Gamma)^7) + Sqrt(s0^5 (Gamma)^7) +
    Sqrt(s0 (Gamma)^11) + 
   2 s0 (Sqrt(s0 (Gamma)^5) + Sqrt(s0 (Gamma)^9)))/((Gamma) (1 + 
     s0 (Gamma) + (Gamma)^2)^2) == Sqrt(s0 (Gamma)) // Simplify

The output is:

(Sqrt(s0 (Gamma)^3) + 2 Sqrt(s0 (Gamma)^7) + Sqrt(
 s0^5 (Gamma)^7) + Sqrt(s0 (Gamma)^11) + 
 2 s0 (Sqrt(s0 (Gamma)^5) + Sqrt(s0 (Gamma)^9)))/((Gamma) (1 + 
   s0 (Gamma) + (Gamma)^2)^2)
True

Why is Mathematica not simplifying to this much simpler form $sqrt{s_0 gamma}$, I think my assumptions should be enough. I can do the simplification by hand

parametric – Least square Estimation in nonlinear Regression

Consider the following Model:

$Y_i = f(theta, x_i) + e_i$

Where $theta$ is the unknown d-dimesional parameter, and $e_i$ are some nice stochastic errors so maybe identical and normal distributed. $x_i$ are deterministic designpoints in $(0,1)$

Does someone know what optimal rate can be achieved for some parametric Estimator? So for example
let $theta_n$ denote the LSE, Then i am intrestet in

$E||theta_n-theta||^2lesssim $ ???

with $||theta_n-theta||^2:=sum_{k=1}^d |theta_{n,i}-theta_i|^2$

algebra precalculus – Prove that the square root of distinct prime numbers is irrational

Prove that if $p_1,…,p_k$ are distinct prime numbers, then $sqrt{p_1p_2…p_k}$ is irrational.

I do not usually prove theorems, so any hint is appreciated. I have taken a look at this and tried to repeat that argument over and over, but I messed up. Perhaps there is an easier to do it. Thanks in advance.

ag.algebraic geometry – Commutative square of module of differential is cartesian?

Let $R$ be a regular local $mathbb{Q}$-algebra and $f$ be a normal crossing divisor(i.e. $f = x_{1}x_{2}…x_{r}$ such that $R/(x_{i})$ is regular for each $i$). Then we have commutative diagram

$require{AMScd}$
begin{CD}
R/fR @>{d}>> Omega^{1}_{R/fR} \
@VVV @VVV\
widehat{R}/fwidehat{R} @>{ widehat{d}}>> Omega^{1}_{widehat{R}/fwidehat{R}}
end{CD}

where $widehat{R}$ denotes the completion of local ring and maps are induced by completion map and d denote the differential map. Is the above square is cartesian?

I am thinking that this should be a cartesian square since there is similar square for $R$ and $widehat{R}$ namely $require{AMScd}$
begin{CD}
R @>{d}>> Omega^{1}_{R} \
@VVV @VVV\
widehat{R} @>{ widehat{d}}>> Omega^{1}_{widehat{R}}
end{CD}
above this and that seems to be cartesian but I am not sure. Even if second square is cartesian it does not mean that first square is cartesian but it is definitely necessary condition. Any help would be great.

complex – Symplifying expressions with exponentials inside square root

I have an expression
$$
exp (i k x) sqrt{y^2 exp (-2 i k x)}
$$

When I put this in Mathematica and do FullSimplift, it gives

FullSimplify(Exp(I k x) Sqrt(Exp(-2 I k x) y^2))

The output is
$$
e^{i k x} sqrt{y^2 e^{-2 i k x}}
$$

Even if I give all proper assumptions ${x,y, k} in mathbb R$ and $ -pi < k leq pi$ like this

FullSimplify(Exp(I k x) Sqrt(Exp(-2 I k x) y^2), {x, y, k} (Element) Reals && -(Pi) < k <= (Pi))

The output comes as
$$
left| yright| e^{i k x} sqrt{e^{-2 i k x}}
$$

But the exponentials should not be there anymore, the result should be only $left| yright|$.

What simplification or assumptions to make, to get the desired result?

plotting – ListContourPlot returns blank square

I have a set of values of a function of X and Y defined as sample={{x1, y1, f(x1,y1)},{x2, y2, f(x2,y2)}{x3, y3, f(x3,y3)}...} I want to do a list contour plot of the data in this vector as ListContourPlot(sample, PlotLegends -> Automatic). The output of this is a blank square, however both sides of the square have the dimensions of X and Y and the legend also has the right values of f(X, Y). Can anybody please tell me what is wrong with this code?

Data and code:

sample = {{0., 0., 2.25}, {0.00202264, 0.00404527, 2.26953}, {0.008085, 
  0.01617, 2.32934}, {0.0181705, 0.036341, 2.42506}, {0.0322514, 
  0.0645029, 2.55508}, {0.0502892, 0.100578, 2.7123}, {0.0722345, 
  0.144469, 2.89297}, {0.098027, 0.196054, 3.08826}, {0.127596, 
  0.255192, 3.29345}, {0.160861, 0.321722, 3.49927}, {0.19773, 
  0.39546, 3.70137}, {0.238102, 0.476204, 3.89137}, {0.281867, 
  0.563734, 4.06654}, {0.328905, 0.65781, 4.22049}, {0.379086, 
  0.758173, 4.35298}, {0.432274, 0.864548, 4.4602}, {0.488322, 
  0.976644, 4.54472}, {0.547077, 1.09415, 4.60527}, {0.608378, 
  1.21676, 4.64685}, {0.672056, 1.34411, 4.67003}, {0.737938, 1.47588,
   4.68126}, {0.805842, 1.61168, 4.6817}, {0.875583, 1.75117, 
  4.67777}, {0.946969, 1.89394, 4.66967}, {1.01981, 2.03961, 
  4.66215}, {1.09389, 2.18778, 4.65288}, {1.16902, 2.33805, 
  4.64344}, {1.245, 2.49, 4.62778}, {1.3216, 2.64321, 
  4.6033}, {1.39863, 2.79727, 4.55967}, {1.47588, 2.95175, 
  4.48985}, {1.55312, 3.10623, 4.37954}, {1.63015, 3.26029, 
  4.21782}, {1.70675, 3.41351, 3.98769}, {1.78273, 3.56546, 
  3.67539}, {1.85786, 3.71572, 3.26354}, {1.93195, 3.86389, 
  2.73331}, {2.00478, 4.00956, 2.10516}, {2.07617, 4.15234, 
  1.55239}, {2.14591, 4.29182, 1.13432}, {2.21381, 4.42763, 
  0.815649}, {2.2797, 4.55939, 0.580065}, {2.34337, 4.68675, 
  0.407037}, {2.40467, 4.80935, 0.283192}, {2.46343, 4.92686, 
  0.1958}, {2.51948, 5.03896, 0.135243}, {2.57267, 5.14533, 
  0.0941795}, {2.62285, 5.24569, 0.0662465}, {2.66988, 5.33977, 
  0.0478592}, {2.71365, 5.4273, 0.0350222}, {2.75402, 5.50804, 
  0.02657}, {2.79089, 5.58178, 0.0200009}, {2.82416, 5.64831, 
  0.0155287}, {2.85372, 5.70745, 0.0114623}, {2.87952, 5.75903, 
  0.00866999}, {2.90146, 5.80292, 0.00584001}, {2.9195, 5.839, 
  0.00405121}, {2.93358, 5.86716, 0.00217192}, {2.94367, 5.88733, 
  0.00124447}, {2.94973, 5.89946, 0.000276389}}

ListContourPlot(sample, PlotLegends -> Automatic)

simplifying expressions – How to find all the complete square results

I find the square result of the bivariate polynomial by the following method:

   5 x^2 + 2 x y - 14 x + 2 y^2 - 10 y + 
  17 //. (a : _ : 1)*s_Symbol^2 + (b : _ : 1)*s_ + rest__ :> 
  a (s + b/(2 a))^2 - b^2/(4 a) + rest

But I found that the following results also meet the requirements:

(x - y + 1)^2 + (2 x + y - 4)^2 // Expand
1/2 (x + 2 y - 5)^2 + 9/2 (x - 1)^2 // Expand
9/17 (y - 2 x)^2 + 1/17 (7 x + 5 y + 17)^2 // Expand
1/5 (5 x + y - 7)^2 + 9/5 (y - 2)^2 // Expand

I want to know how to find out all the possible square results.

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algorithms – How to find the square with the highest total sum

I have an integer matrix of size 4n x 4n. I need to select a part of the matrix of size n^2 from which adds up to the most.

For example if n = 3. I have a matrix of size 12 x 12. In the below picture, you can clearly see that n^2 square (outlined in red) adds up to the most. For simplicities sake, I made all the numbers 5, expect for 9 of the boxes which are 9999 so its clear that that is the n^2 squares that add up to the most.

enter image description here

My approach to solving this problem was to essentially create a n^2 square and brute force the entire 4n x 4n matrix. However, that runs in O(n^4) time complexity. How can I do it in O(n^2)?

Wouldn’t it be better to just take the square root of the numerator and divide it by N instead in the expression for standard deviation?

The equation of the standard deviation of a dataset is given by $sumsqrt{frac{x_{i} – bar{x}}{N}}$. Why is that the case and why can’t we use $sumfrac{sqrt{x_{i} – bar{x}}}{N}$ instead? The units line up and we don’t have to worry about negatives in this case too. Thanks a million in advance!