## 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.

## Centrium Square (88)

Centrium Square was the former Serangoon Plaza en-bloc redevelopment project produced by renowned developer Feature Development an associate company of Tong Eng Group. A Developer having an excellent reputation for its’ uncompromising high quality and distinction

It is strategically situated at 320 Serangoon Road, close to Mustafa Shopping Centre and City Square Mall. With access to Farrer Park MRT Station.

The 115 metre long podium façade makes use of a flat 2 dimensional geometric tessellation patterned screen, setting up a visual effect of repeated three-dimensional cubes.

Perfectly situated in the heart of Serangoon Road, at the fringe of the city, Centrium Square gives unparalleled access to the Central Business District, Marina Bay and Orchard Road.

Within walking distance to Farrer Park MRT and easy access to major expressways such as the CTE and PIE shows that getting to and from work has never been so easy. In the middle of a wealth of amenities together with the charms of Little India, Centrium Square truly gives the best of both worlds.

Register to view Centrium Square Showflat
Centrium Square Price will be released during the preview
Request for Centrium Square Floor Plan
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.
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)`?
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!