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## 数学代写|交换代数代写commutative algebra代考|MAST90025

2022年9月26日

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## 数学代写|交换代数代写commutative algebra代考|Dedekind-Mertens Lemma

D First of all, notice that the products $f_i g_j$ are the coefficients of the polynomial $f(Y) g(X)$. Similarly, for some indeterminates $Y_0, \ldots, Y_m$, the content of the polynomial $f\left(Y_0\right) \cdots f\left(Y_m\right) g(X)$ is equal to $c(f)^{m+1} \mathrm{c}(g)$.

Let $h=f g$. Imagine that in the ring $\mathbf{B}=\mathbf{A}\left[X, Y_0, \ldots, Y_m\right]$ we are able to show the membership of the polynomial $f\left(Y_0\right) \cdots f\left(Y_m\right) g(X)$ in the ideal
$$\sum_{j=0}^m\left(h\left(Y_j\right) \prod_{k, k \neq j}\left(f\left(Y_k\right)\right\rangle\right) .$$
We would immediately deduce that $\mathrm{c}(f)^{m+1} \mathrm{c}(g) \subseteq \mathrm{c}(f)^m \mathrm{c}(h)$.
This is more or less what is going to happen. We get rid of the denominators in Lagrange’s interpolation formula (we need at least $1+\operatorname{deg} g$ interpolation points):
In the ring $\mathbf{B}$, by letting $\Delta=\prod_{j \neq k}\left(Y_j-Y_k\right)$, we get:
$$\Delta \cdot g(X) \in \sum_{j=0}^m\left\langle g\left(Y_j\right)\right\rangle .$$
Thus by multiplying by $f\left(Y_0\right) \cdots f\left(Y_m\right)$ :
$$\Delta \cdot f\left(Y_0\right) \cdots f\left(Y_m\right) \cdot g(X) \in \sum_{j=0}^m h\left(Y_j\right) \prod_{k, k \neq j}\left\langle f\left(Y_k\right)\right\rangle .$$
If we show that for any $Q \in \mathbf{B}$ we have $\mathrm{c}(Q)=\mathrm{c}(\Delta \cdot Q)$, the previous membership gives $\mathrm{c}(f)^{m+1} \mathrm{c}(g) \subseteq \mathrm{c}(f)^m \mathrm{c}(h)$.
Note that $\mathrm{c}\left(Y_i Q\right)=\mathrm{c}(Q)$ and especially that
$$\mathrm{c}\left(Q\left(Y_0 \pm Y_1, Y_1, \ldots, Y_m\right)\right) \subseteq \mathrm{c}\left(Q\left(Y_0, Y_1, \ldots, Y_m\right)\right) .$$
Therefore, by putting $Y_0=\left(Y_0 \pm Y_1\right) \mp Y_1, \mathrm{c}\left(Q\left(Y_0 \pm Y_1, Y_1, \ldots, Y_m\right)\right)=\mathrm{c}(Q)$. The following polynomials thus all have the same content:
$$\begin{gathered} Q, Q\left(Y_0+Y_1, Y_1, \ldots, Y_m\right), Y_0 Q\left(Y_0+Y_1, Y_1, \ldots, Y_m\right), \ \left(Y_0-Y_1\right) Q\left(Y_0, Y_1, \ldots, Y_m\right) . \end{gathered}$$

## 数学代写|交换代数代写commutative algebra代考|The Universal Splitting Algebra for a Monic Polynomial

Disclaimer In a context where we manipulate algebras, it is sometimes preferable to keep to the intuition that we want to have a field as the base ring, even if it is only a commutative ring. In which case we choose to give a name such as $\mathbf{k}$ to the base ring. This is what we are going to do in this section dedicated to the universal splitting algebra.
When we are truly dealing with a discrete field, we will use $\mathbf{K}$ instead.
We now proceed to the inverse operation to that which passes from the polynomial ring to the subring of symmetric polynomials.

In the presence of a monic polynomial $f=T^n+\sum_{k=1}^n(-1)^k s_k T^{n-k} \in \mathbf{k}[T]$ over a ring $\mathbf{k}$, we want to have at our disposal an extension of $\mathbf{k}$ where the polynomial is decomposed into linear factors. Such an extension can be constructed in a purely formal way. The result is called the universal splitting algebra.
4.1 Definition and notation Let $f=T^n+\sum_{k=1}^n(-1)^k s_k T^{n-k} \in \mathbf{k}[T]$ be a monic polynomial of degree $n$. We denote by Adu $\mathbf{k}{\mathbf{k}, f}$ the universal splitting algebra of $f$ over k defined as follows $$\mathrm{Adu}{\mathbf{k}, f}=\mathbf{k}\left[X_1, \ldots, X_n\right] / \mathcal{J}(f)=\mathbf{k}\left[x_1, \ldots, x_n\right]$$
where $\mathcal{J}(f)$ is the ideal of symmetric relators necessary to identify $\prod_{i=1}^n\left(T-x_i\right)$ with $f(T)$ in the quotient. Precisely, if $S_1, S_2, \ldots, S_n$ are the elementary symmetric functions of the $X_i$ ‘s, the ideal $\mathcal{J}(f)$ is given by
$$\mathcal{J}(f)=\left\langle S_1-s_1, S_2-s_2, \ldots, S_n-s_n\right\rangle$$
The universal splitting algebra $\mathbf{A}=\mathrm{Adu}_{\mathbf{k}, f}$ can be characterized by the following property.
4.2 Fact (Universal decomposition algebra, characteristic property)

1. Let $\mathbf{C}$ be a $\mathbf{k}$-algebra such that $f(T)$ is decomposed into a product of factors $T-z_i$. Then, there exists a unique homomorphism of $\mathbf{k}$-algebras of $\mathbf{A}$ to $\mathbf{C}$ which sends the $x_i$ ‘s to the $z_i$ ‘s.
2. This characterizes the universal splitting algebra $\mathbf{A}=\mathrm{Adu}_{\mathbf{k}, f}$, up to unique isomorphism.
3. Moreover, if $\mathbf{C}$ is generated (as a $\mathbf{k}$-algebra) by the $z_i$ ‘s, the universal splitting algebra is isomorphic to a quotient of $\mathbf{A}$.

# 交换代数代考

## 数学代写|交换代数代写交换代数代考|Dedekind-Mertens引理

$$\sum_{j=0}^m\left(h\left(Y_j\right) \prod_{k, k \neq j}\left(f\left(Y_k\right)\right\rangle\right) .$$我们马上就能推断出来 $\mathrm{c}(f)^{m+1} \mathrm{c}(g) \subseteq \mathrm{c}(f)^m \mathrm{c}(h)$这或多或少就是将要发生的事情。我们去掉了拉格朗日插值公式中的分母(我们至少需要 $1+\operatorname{deg} g$ 插补点):

$$\Delta \cdot g(X) \in \sum_{j=0}^m\left\langle g\left(Y_j\right)\right\rangle .$$

$$\Delta \cdot f\left(Y_0\right) \cdots f\left(Y_m\right) \cdot g(X) \in \sum_{j=0}^m h\left(Y_j\right) \prod_{k, k \neq j}\left\langle f\left(Y_k\right)\right\rangle .$$
$Q \in \mathbf{B}$ 我们有 $\mathrm{c}(Q)=\mathrm{c}(\Delta \cdot Q)$，以前的会员给予 $\mathrm{c}(f)^{m+1} \mathrm{c}(g) \subseteq \mathrm{c}(f)^m \mathrm{c}(h)$.

$$\mathrm{c}\left(Q\left(Y_0 \pm Y_1, Y_1, \ldots, Y_m\right)\right) \subseteq \mathrm{c}\left(Q\left(Y_0, Y_1, \ldots, Y_m\right)\right) .$$

$$\begin{gathered} Q, Q\left(Y_0+Y_1, Y_1, \ldots, Y_m\right), Y_0 Q\left(Y_0+Y_1, Y_1, \ldots, Y_m\right), \ \left(Y_0-Y_1\right) Q\left(Y_0, Y_1, \ldots, Y_m\right) . \end{gathered}$$

## 数学代写|交换代数代写交换代数代考|一个Monic多项式的通用分裂代数

4.1定义和符号Let $f=T^n+\sum_{k=1}^n(-1)^k s_k T^{n-k} \in \mathbf{k}[T]$ 是次的一多项式 $n$。我们用阿杜表示 $\mathbf{k}{\mathbf{k}, f}$ 的普适分裂代数 $f$ 除以k，定义如下 $$\mathrm{Adu}{\mathbf{k}, f}=\mathbf{k}\left[X_1, \ldots, X_n\right] / \mathcal{J}(f)=\mathbf{k}\left[x_1, \ldots, x_n\right]$$
where $\mathcal{J}(f)$ 是否有必要确定理想的对称关系式 $\prod_{i=1}^n\left(T-x_i\right)$ 用 $f(T)$ 在商中。没错，如果 $S_1, S_2, \ldots, S_n$ 初等对称函数 $X_i$ ’s，理想 $\mathcal{J}(f)$
$$\mathcal{J}(f)=\left\langle S_1-s_1, S_2-s_2, \ldots, S_n-s_n\right\rangle$$

1. 设$\mathbf{C}$为$\mathbf{k}$ -代数，使$f(T)$分解为因子$T-z_i$的乘积。然后，$\mathbf{A}$到$\mathbf{C}$的$\mathbf{k}$ -代数存在一个唯一同态，它将$x_i$ ‘s发送到$z_i$ ‘s。
2. 这表征了普遍分裂代数$\mathbf{A}=\mathrm{Adu}_{\mathbf{k}, f}$，直到唯一同态。

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。


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# What's inside the slit in double slit experiment?

If double slit experiment is done in a environment with air, then slits could also contain air made up of (approx. 80% nitrogen, 20% oxygen), then there is not empty space inside the slit. How can it affect the experiment? Has it been taken in account? What are specific requirement to define the slit?

If it were not that way. Why it doesn't affect the measurement?

Air is everywhere, not only in slit, but also in the optical path before and after the slit.

However, interaction with transparent media (e.g. air, water, glass) can be easily included: by use of refractive index $n$. Then you know that wavelength in medium is $$\lambda = \frac{\lambda_0}{n},$$ where $\lambda_0$ is the wavelength in the vacuum.

If in the slit there were glass - still it is easy to calculate the outcome. (Just you need to include the phase shift caused by it.)

• +1, Thanks, I am trying to understand the required properties of the wall material, then understanding the difference with the envieroment helps a lot – HDE Mar 4 '11 at 13:22
• @HDE You can google meta materials and then things get very interesting! – PhysicsDave Aug 20 '19 at 12:45

The air will simply introduce a small amount of additional noise into the experiment due to scattering. The majority of the photons will not scatter off the air molecules.

The optical path lengths will also be changed, but not by much. For example, the troughs in the interference pattern will occur when the optical path lengths of the paths to that point are out of phase. Since the optical path lengths in vacuum and air differ by a factor of $n_\mathrm{Air} = 1.00027$ for an arbitrarily chosen visible wavelength of 500 nm, the air has hardly any effect.

The properties of the wall do have a moderate efect on the interference pattern. However for most opaque walls the simple physical optics model sort of ignoring the wall seems to work reasonably well. Precise mathematics based on Maxwell equations for light are complex and details change significantly with say horizontal or vertical polarisation for a metallic grate. See Sommerfeld edge diffraction or JB Keller geometric theory of diffraction which have good math development in terms of series expansionin wavelength but one needs knowledge of wall properties and seems rather at odds with PO in many ways. However PO gives us useful answers even when we dont know wall properties and where GTD would be rather at a loss.

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# Algebra - Infinite Dihedral Group

Let $G$ be the set of bijections $\mathbb{R} \to \mathbb{R}$ which preserve the distance between pairs of points, and send integers to integers. Then $G$ is a group under composition of functions. The following two elements are obviously in $G$: the function $t$ (translation) where $t(x)=x+1$ for each $x \in \mathbb{R}$ and the function $r$ (reflection) where $r(x)=-x$ for each $x \in \mathbb{R}$. The subgroup of $G$ generated by $r$ and $t$ is called the infinite dihedral group and denoted by $D_{\infty}$. Note that this information describes an action of $D_{\infty}$ on $\Re$.

1. Show that every element of $D_{\infty}$ can be written uniquely in the form $r^it^j$ for suitably restricted values of $i$ and $j$. Explain how to multiply two such elements.

2. Describe geometrically the possible sorts of actions of elements of $D_{\infty}$ on the real line $\mathbb{R}$.

I have shown that every element is of the form $r^it^j$ but I am unsure how two elements would be multiplied together and how the actions should be described geometrically.

Can anybody help?

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Homework should be tagged as homework – hmmmm Oct 17 '12 at 19:53
I've changed algebra tag to abstract-algebra, since we don't use algebra tag anymore, see meta for details. – Martin Sleziak Oct 19 '12 at 13:54

I assume, you have also noted that $i\in\{0,1\}$ when writing an element as $r^it^j$. Essentially, there are just four cases to consider:

• $t^j \cdot t^k$: Clearly, this is $t^{j+k}$.
• $rt^j \cdot t^k$: Clearly, this is $rt^{j+k}$.
• $t^j \cdot rt^k$: Note that $(t^j rt^k)(x) = (t^j r)(x+k)=t^j(-x-k)=-x-k+j$, hence $t^j \cdot rt^k = r t^{k-j}$.
• $rt^j \cdot rt^k$: Using the previous, this is $r \cdot r t^{k-j}= t^{k-j}$.

The elements of $D_\infty$ are $t^j\colon x\mapsto x+j$ with $j\in \mathbb Z$, i.e. translations by an integer amount, and $rt^j\colon x\mapsto -x-j$, i.e. reflection around integers or half-integers (note that $-\frac j2$ is the fix pint of $x\mapsto -x-j$).

Bonus Question: Can you find any element of $G$ that is not in $D_\infty$?

Remark: How could you show that all elements have the form $r^it^j$ without observing how elemnts of $D_\infty$ are multiplied?

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Would $t^j.rt^k$ not be equal to $rt^{k-j}$ rather than $rt^{j-k}$? – HHHHHH Oct 17 '12 at 19:18
Yes as , $rt^k=t^{-k}r$ and so $t^j rt^k=t^jt^{-k}r=t^{j-k}r=rt^{k-j}$ – hmmmm Oct 17 '12 at 20:11
Thanks for the hint – Hagen von Eitzen Oct 23 '12 at 10:56

This is not an answer but a different construction of this group (which shows that it is a Coxeter group).

Let $M=\{1, -1 \}$. Then $M$ acts on $\mathbb{Z}$ by $(-1) x=-x$.

$D_{\infty}$ is the semidirect product of $M$ with $\mathbb{Z}$. The group law is given by

$$(\epsilon, x ) (\epsilon^{\prime}, x^{\prime} )=(\epsilon \epsilon^{\prime}, \epsilon^{\prime} x+ x^{\prime} ) \ .$$

If you set $\rho=(-1,0)$, $\mu=(-1,1)$, $\pi=(1,1)$ then $$\rho^2=\mu^2=1, \ \ \ \pi=\rho \mu$$

The group is generated by $<\rho, \mu>$.

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# Definite Integral Work

chukie
A bucket that weighs 3 lb and a rope of negligible weight are used to draw water from a
well that's 60 feet deep. Suppose the bucket starts with 37 lb of water and is pulled up by a rope at 2 ft/sec, while water leaks out of the bucket at a rate of 1/4 lb/sec.

A. How long does it take for the bucket to get to the top of the well? Write an equation that
expresses the total weight of the bucket as a function of time, as the time varies from 0 until the time the buckets gets to the top.

B. Recall that work equals force times distance. Calculate the work done in lifting the bucket
to the top of the well, keeping in mind that here force is equal to weight.

For A:I did 60ft/2ft/s=30 sec so it takes 30 secs for the bucket ot get to top of well
equation: Weight=40+ integral sign from 0 to 30 (-1/4)dx =32.5 lb

For B: i did total work= integral sign from 0 to 60 (32.5x)dx

i dunno if im doing this right. ive searched the internet and found a question similar to mine (question #6):
http://count.ucsc.edu/~bauerle/images/19BW07HWK/Homework 6.4 Math 19B Winter 2007, Bauerle.pdf"

and ive used the same steps that i did for this question but i got the wrong answer.

Last edited by a moderator:

## Answers and Replies

Homework Helper
For A:I did 60ft/2ft/s=30 sec so it takes 30 secs for the bucket ot get to top of well
equation: Weight=40+ integral sign from 0 to 30 (-1/4)dx =32.5 lb
There isn't any integral involved here yet. You have to express your answer in terms of t, time. The weight would be wt of bucket + water remaining. Find an expression for the amount of water remaining at some given time t. No integration is required.

For B: i did total work= integral sign from 0 to 60 (32.5x)dx
One way you could do this would be to change the variables of your integration. Work done is given by $$\int^b_a \vec{F} \cdot d \vec{r} = \int^{b(t)}_{a(t)} \vec{F}(t) \cdot \frac{d\vec{r}}{dt} dt$$.

In this case since the problem is strictly one dimensional you can treat the vector functions as scalar functions. You have to do a change of variables from r to t. Firstly express r in terms of t. r is the distance from the bottom of the well to the top. Then you can continue from there.

chukie
There isn't any integral involved here yet. You have to express your answer in terms of t, time. The weight would be wt of bucket + water remaining. Find an expression for the amount of water remaining at some given time t. No integration is required.

One way you could do this would be to change the variables of your integration. Work done is given by $$\int^b_a \vec{F} \cdot d \vec{r} = \int^{b(t)}_{a(t)} \vec{F}(t) \cdot \frac{d\vec{r}}{dt} dt$$.

In this case since the problem is strictly one dimensional you can treat the vector functions as scalar functions. You have to do a change of variables from r to t. Firstly express r in terms of t. r is the distance from the bottom of the well to the top. Then you can continue from there.

so for A:the equation would be W(t)=40-(1/4)t

for B: i dun quite understand wut u mean.do u mean i do the integral from 0 to 30? if that's the case i dunno wut the equation that im trying to integrate is


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# ConTeXt: Extend figure into margin with \placefigure

I would like to extend a figure into the margin pretty much as done here in LaTeX: Place figures side by side, spill into outer margin

My idea was to place the figure right and make it wider than \textwidth.

\placefigure[right]{My caption}{\externalfigure[foo.png][width=1.2\textwidth]}

But that's not working since the figure is then extending into the right margin and not into the left :(

• Have you looked in the details document? I think it is covered there. (I'm not able to check it right now.) – mickep Oct 29 '18 at 18:07

Apparently, setting the location parameter in \setupfloat does the trick.
\setupfloat[figure][default=here,location=right] \placefigure[]{My caption}{\externalfigure[foo.png][width=1.2\textwidth]}


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Letter

# Detection of nanoflare-heated plasma in the solar corona by the FOXSI-2 sounding rocket

Accepted:
Published online:

## Abstract

The processes that heat the solar and stellar coronae to several million kelvins, compared with the much cooler photosphere (5,800 K for the Sun), are still not well known1. One proposed mechanism is heating via a large number of small, unresolved, impulsive heating events called nanoflares2. Each event would heat and cool quickly, and the average effect would be a broad range of temperatures including a small amount of extremely hot plasma. However, detecting these faint, hot traces in the presence of brighter, cooler emission is observationally challenging. Here we present hard X-ray data from the second flight of the Focusing Optics X-ray Solar Imager (FOXSI-2), which detected emission above 7 keV from an active region of the Sun with no obvious individual X-ray flare emission. Through differential emission measure computations, we ascribe this emission to plasma heated above 10 MK, providing evidence for the existence of solar nanoflares. The quantitative evaluation of the hot plasma strongly constrains the coronal heating models.

During an observation on 11 December 2014, the second flight of the Focusing Optics X-ray Solar Imager (FOXSI-2) sounding (research) rocket detected hard X-rays (HXRs, above ~ 3 keV in this article) from several active regions3. FOXSI is the first instrument to directly focus solar HXRs and achieve high-sensitivity imaging and spectroscopic observations3,4,5. Figure 1a shows a FOXSI-2 HXR image (cyan contours) of two such regions near disk centre overlaid on background soft X-ray (SXR, below ~ 3 keV) images taken by the X-ray Telescope (XRT) on-board the Hinode spacecraft. The XRT filter shown is sensitive to >2 MK plasma. Both northern and southern active regions (National Oceanic and Atmospheric Administration designations 12234 and 12230, respectively) showed HXR intensity changes at some point during the flight, but we were able to isolate a time with no significant HXR changes for active region 12234 (yellow box in Fig. 1a and blue box in Fig. 1b). We checked the SXR emissions detected by the Geostationary Operational Environmental Satellite (GOES) X-ray Sensor (XRS), which is typically used to identify flares. As can be seen in Fig. 1b, GOES/XRS observed no flare during the time interval shown in the blue box. SXRs and HXRs from active region 12234 (Hinode/XRT and FOXSI-2 lightcurves in Fig. 1b) in this time interval were constant within 1σ standard deviation uncertainties (indicated by the error bars in Fig. 1b). We thus conclude that active region 12234 was quiescent during this time interval. We define such a region a ‘quiescent active region’. Despite the lack of distinct, resolved flares, FOXSI-2 successfully detected HXR emission up to 9 keV from the region.

The temperature structure of solar plasma can be evaluated by estimating the differential emission measure (DEM), which is the temperature-dependent brightness. Figure 2a shows the DEM of active region 12234, produced by combining HXR and SXR data for the time interval indicated in the blue box in Fig. 1b. (See Methods for details on the DEM computation.) Magenta curves in Fig. 2a are loci curves corresponding to observed values with individual XRT filter configurations using the latest calibration6. Blue loci curves show the detected HXR emissions by FOXSI-2 in three energy bands: 5–6 keV, 6–7 keV and 7–9 keV. Each curve indicates the amount of emission measure that would be needed to produce the observed intensity if the emitting plasma were isothermal at the temperature given in the abscissa. The true DEM distribution must lie below these loci curves. However, it must not lie too far below, because the total emission from the entire temperature distribution must produce the observed counts in each energy band. A precise DEM over a wide range of coronal temperatures was successfully obtained thanks to the combination of high-sensitivity SXR and HXR observations; the SXR emission determines the lower-temperature plasma, while HXRs indicate the presence of a faint, high-temperature component. The DEM value of this hot component at 10 MK is more than five orders of magnitude smaller than that at the peak of ~ 2.5 MK. This challenging dynamic range is successfully achieved by the combination of the different observational wavelength bands with a range of temperature sensitivities. Between the temperatures, T, of 4 MK and 10 MK, the DEM has approximately a steep power-law relation, $∝ T - 12$.The slope $β=12$ is similar to the highest values for the upper power-law index of the distribution DEM $∝ T - β$ in previous studies using extreme ultraviolet (EUV) line spectroscopy ($β=$ 5–13 in ref. 7).Figure 2b shows the HXR spectrum observed by FOXSI-2, overplotted with the spectrum predicted by the estimated DEM.

This hot plasma above 10 MK, although faint, is significant. The detected HXR counts above 7 keV (locus curve shown by the thick blue line in Fig. 2a) require the existence of hot (≥10 MK) plasma. The <10 MK plasma in this DEM distribution can account for only 7% or less of the counts detected in the 7–9 keV band. Even if we extend the power law at 4–10 MK to high temperatures, a total of only 10% of the detected 7–9 keV counts would be accounted for. We confirmed that the instrumental background is notably smaller than the detected intensity and that no detector pileup is expected (see Methods for details). This means that more than 90% of the observed >7 keV HXR emissions must come from the ≥10 MK plasma, revealing a significant excess above the power-law DEM. The emission measure of the >10 MK plasma is in the range 1.8 × 1022–1.5 × 1023 cm−5 in the solutions with chi-squared values within the 95% occurrence probability in trial runs (see Methods for details of the analysis). The >10 MK emission measure is >1022 cm−5 for 99.8% of the Monte Carlo solutions, and is >1021 cm−5 for all the solutions. Therefore, we conclude that significant HXRs from the >10 MK plasma component were detected.

Simulations show that nanoflare models can have temperatures beyond 10 MK (refs 8,9,10). An alternative set of theories explains coronal heating via waves propagating from the photosphere to the corona, including shearing Alfvén waves11. However, it is difficult to produce plasma of 10 MK and above using wave heating models12, and thus we interpret such a hot component to be an observational signature of frequent energy input by nanoflares13. Our observational result supplies quantified parameters (DEM power-law slope, quantity of >10 MK plasma) that will constrain future modelling of nanoflare distributions. Recently, a two-fluid simulation suggested that detectable >10 MK plasma could be generated only by short nanoflares with non-ionization equilibrium and an ignorable difference in ion and electron heating14.

In past observations, the Reuven Ramaty High Energy Solar Spectroscopic Imager (RHESSI) detected HXRs from the Sun during periods with no flares by examining the day–night transitions of emission from the entire solar disk15. Those HXRs were from hot plasma with temperatures of 5–10 MK, and no >11 MK plasma was found. If we fit the FOXSI-2 spectrum with an isothermal model, the temperature of active region 12234 during the time period of this study is ~10 MK, towards the higher end of the range found for the Sun15. The emissions measure for 12234 is also more than three orders of magnitude smaller than the lowest emission measure detected by RHESSI from the non-flaring Sun. The fact that the temperature is within the range of the statistical study suggests that our target active region is not an exceptionally hot region.

This observation provides the most direct evidence to date for the presence of hot plasma in a quiescent solar active region. With the advent of directly focusing HXR instruments, we expect future observations to enable statistical studies and investigations of nanoflare-heated plasma and energy input to the corona. More high-sensitivity X-ray imaging and spectroscopic observations are expected with current and future instruments, including the next launch of FOXSI (FOXSI-3), the Nuclear Spectroscopic Telescope Array (NuSTAR) spacecraft16 and a potential future solar-dedicated mission that is in the early stages of development.

## Methods

Coronal DEMs are usually evaluated by EUV and SXR observations7,17,18. Because intensities of the EUV lines sensitive to high temperatures (~10 MK and above; for example, Fe xxiii and Fe xxiv lines with sensitivity peaks at 10–20 MK) are weak in the absence of large flares, only a few detections of those lines in quiescent active regions have been reported. The few positive reports19 are based on multi-filter SXR observations by the XRT, especially using the thick filters. However, the XRT uncertainties are large in the temperature range of flare-heated plasma20, and so the high-temperature components are poorly constrained without the additional use of bremsstrahlung HXRs. The primary solar HXR instrument is the RHESSI spacecraft, which has never detected significant emission localized in individual quiescent active regions due to limited sensitivity. In the case of ref. 19, follow-on work using RHESSI’s non-detection was used to significantly constrain the hot plasma present12. In this case, RHESSI supplies only an upper limit, and the hot plasma could be constrained much more, or perhaps directly measured, with a lower-background, higher-sensitivity HXR instrument. HXR observations provide a more reliable method to investigate hot plasma than SXRs or EUV because the contributions of hot plasma and high-energy (non-thermal) electrons dominate this range. To investigate hot plasma in quiescent active regions using HXRs therefore requires the higher sensitivity that can be provided by newly available focusing optics that directly image HXRs.

FOXSI is a sounding rocket experiment to investigate HXR emissions in the 4–20 keV range from the Sun with superior sensitivity over previous, indirectly imaging, instruments3,5. The high sensitivity is achieved by a combination of directly focusing optics and semiconductor detectors for imaging and spectroscopy. Using a grazing incidence telescope with precision mirrors, we can increase effective area with a much smaller detector volume, drastically increasing the signal to background ratio and therefore the sensitivity. The first flight of FOXSI on 2 November 2012 successfully demonstrated the concept and effectivity of HXR focusing optics for solar observation by the observation of a microflare. FOXSI clearly showed greater sensitivity and imaging dynamic range than a simultaneous observation of the microflare by RHESSI4. During the same flight, FOXSI targeted a quiescent active region, but did not detect emission. This observation was used to constrain the hot plasma component of that active region21. On 11 December 2014, FOXSI-2 was flown for a 6.5 min observation at the White Sands Missile Range in New Mexico, United States, during which it successfully performed imaging and spectroscopy of multiple regions.

The FOXSI-2 instrument consists of seven optics modules, each paired with a dedicated semiconductor strip detector. The FOXSI optics are Wolter type-I grazing incidence telescopes developed at the NASA/Marshall Space Flight Center, with heritage from previous designs for non-solar astrophysics22. Multiple mirror shells (10 shells for two modules, and 7 shells for the other five) are nested together to increase collecting area. The optics are the highest-resolution HXR focusing optics ever used for astrophysics, with a measured spatial resolution of 5 arcsec (full-width at half-maximum) and a 27 arcsec half-power diameter5.

The FOXSI-2 detectors are low-noise, fine-pitch, double-sided silicon (Si) and cadmium telluride (CdTe) detectors developed at the JAXA/Institute of Space and Astronautical Science by a team including development members of the Hard X-ray Imager and Soft Gamma-ray Detector on-board the Hitomi satellite. Spatial samplings are 75 μm and 60 μm, and measured energy resolutions are 0.5 keV and 1.3 keV (full-width at half-maximum) for the Si and CdTe detectors, respectively23,24. The sensitive areas are 9.6 mm × 9.6 mm and 7.68 mm × 7.68 mm for the Si and CdTe detectors, respectively, corresponding to respective fields of view of 16.5 arcmin × 16.5 arcmin and 13.2 arcmin × 13.2 arcmin. We newly developed the CdTe detectors for FOXSI-2 for higher efficiency for energies above 10 keV, and almost 100% efficiency was achieved for the FOXSI-2 energy range of 4–20 keV. We used five Si and two CdTe detectors for the FOXSI-2 flight. The FOXSI data consist of a list of detected photons with energies, incident locations and times. Images, lightcurves and energy spectra can be derived from this event list.

For FOXSI-2, we coordinated an observation of Hinode (Hinode Observation Plan 221). Our request to XRT included multi-filter images with a large field of view covering an area of 2,107 arcsec × 2,107 arcsec (more than the full solar disk size). XRT took images with 11 filter configurations, and it took ~6 min to take a full set. The region of interest in this study was outside the fields of view of the other Hinode instruments. In addition to the multi-filter observation, XRT took images with an identical filter configuration to check for flare occurrence, called the ‘flare patrol’ images. (The flare patrol can trigger a flare mode observation if a brightening is found, although none occurred during the FOXSI flight.) The XRT data in Fig. 1b are from flare patrol images.

We calculated the DEM by combining FOXSI-2 data integrated from 19:13:47 to 19:14:25 Universal Time (ut) and a single set of the XRT observations at 19:08:56–19:14:57 ut. We checked the flare patrol images and FOXSI lightcurves, and confirmed that the X-ray flux from this region was constant during this set of observations. The DEM calculation was performed in an area of 100 arcsec × 100 arcsec, centred at (−90 arcsec, 60 arcsec), shown in the yellow box in Fig. 1a, to cover the entire HXR emission detected by FOXSI-2. The DEM was estimated by a forward-fitting method to find the DEM solution to reproduce FOXSI-2 and XRT observations by assuming a smooth distribution represented by a spline function for the DEM curve. The procedure used in this work is xrt_dem_iterative2.pro, which is a standard tool set for solar data analysis in the Solar Software package. The details of this method are described in our previous paper21. We had 10,000 trial runs for the DEM estimation by varying each observational value randomly according to a Poisson distribution for the statistical error, and a range of solutions with chi-squared values within the 95% occurrence probability is shown in Fig. 2.

Although it is not statistically significant, a slight intensity increase in the FOXSI-2 lightcurve is seen at approximately 19:14:00 ut. We confirmed that the >7 keV HXRs were not concentrated at the time of the intensity increase, but instead were spread over the time range for the analysis. These photons are also not spatially concentrated at any preferred location within the active region. Therefore, we confirmed that the >7 keV emissions do not come predominantly from a single flaring event.

FOXSI-2 detected 15 photons within the 7–9 keV band from the region of interest during the time interval of this work, and no signal above 9 keV was detected. We examined, and ruled out, several other possibilities for the origin of these photons. The estimated background over the surrounding detector area is 0.9 ± 0.1 counts. This background includes non-solar charged particles, scattered photons from other solar sources (active region 12230 in Fig. 1) and any quiet Sun emission, if present. The background-subtracted measurement is $14. 1 - 4 + 5$ counts, and this value was used for the DEM estimation. The expected 7–9 keV counts without the presence of the >10 MK plasma is 0.54 counts for the least chi-squared solution, meaning that the high-energy emission is not produced by the lower-temperature part of the DEM. We used the FOXSI-2 observation >5 keV because the uncertainty in the detection efficiency is greater at the lower detection limit energy of ~4 keV.

High energy counts could, in principle, be falsely recorded due to pileup of multiple incident photons. Pileup happens if two photons hit a single detector strip within a timescale of ~10 μs. The detector strip with the highest count rate from the target region has 1.0 counts per second in the 4–15 keV range during this pointing. Even if we assume the rate of the incident photons down to 3 keV is ten times more, the expected pileup count in this strip is $2.0×1 0 - 3$ counts per second. By calculating the expected pileup counts for all the pixels assuming that there are ten times more incident ≥3 keV photons than the detected 4–15 keV counts for all the strips, the total expected pileup count by multiple ≥3 keV photons by all the detectors in this analysis is 0.33 counts. Therefore, the effect of pileup is negligible for the 7–9 keV counts, because the expected pileup count is well below the statistical error. After ruling out these alternative explanations for the high-energy emission, we can state with confidence that our detected photons are primarily due to >10 MK plasma in active region 12234.

### Data availability

The data that support the plots within this paper and other findings of this study are available from the corresponding author upon reasonable request.

Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## References

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Winebarger, A. R., Schmelz, J. T., Warren, H. P., Saar, S. H. & Kashyap, V. L. Using a differential emission measure and density measurements in an active region core to test a steady heating model. Astrophys. J. 740, 2 (2011).

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## Acknowledgements

We acknowledge the FOXSI instrument team for the development of the experiment. The X-ray focusing optics were provided by a team at NASA/Marshall Space Flight Center led by B. Ramsey. The focal plane detectors were developed by a team at ISAS/JAXA led by T. Takahashi and S. Watanabe. This work was supported through KAKENHI grants 24244021 and 20244017 from the Japan Society for the Promotion of Science. FOXSI was funded by NASA’s Low-Cost Access to Space program, grant NNX11AB75G. L.G. is supported by an NSF grant (AGS-1429512). Hinode is a Japanese mission developed and launched by ISAS/JAXA, with NAOJ as domestic partner and NASA and STFC (UK) as international partners. It is operated by these agencies in cooperation with ESA and NSC (Norway). The FOXSI team thanks NASA’s Sounding Rockets Program Office and the NSROC teams who helped to attain a successful flight.

## Author information

### Affiliations

1. #### Institute of Space and Astronautical Science, Japan Aerospace Exploration Agency, 3-1-1 Yoshinodai, Chuo-ku, Sagamihara, Kanagawa, 252-5210, Japan

• Shin-nosuke Ishikawa
2. #### School of Physics and Astronomy, University of Minnesota, Minneapolis, MN, 55455, USA

• Lindsay Glesener
•  & Juliana Vievering
3. #### Space Sciences Laboratory, University of California, Berkeley, Berkeley, CA, 94720, USA

• Säm Krucker
•  & Juan Camilo Buitrago-Casas
4. #### Institute of 4D Technologies, School of Engineering, University of Applied Sciences and Arts Northwestern Switzerland, 5210, Windisch, Switzerland

• Säm Krucker
5. #### NASA Goddard Space Flight Center, Greenbelt, MD, 20771-0001, USA

• Steven Christe
6. #### National Astronomical Observatory of Japan, 2-21-1 Osawa, Mitaka, Tokyo, 181-8588, Japan

• Noriyuki Narukage

### Contributions

S.I., L.G., S.K., S.C. and J.B. contributed to the instrument development, analysis of the observational data, scientific discussions and text writing. N.N. contributed to the analysis of the data observed by the Hinode satellite and scientific discussions. J.V. contributed to software developments for the FOXSI data analysis.

### Competing interests

The authors declare no competing financial interests.

### Corresponding author

Correspondence to Shin-nosuke Ishikawa.


text:
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text:
# How do you solve -\frac { 1} { 2} + y = - \frac { 3} { 5}?

$y = - \frac{1}{10}$
$- \frac{1}{2} + y + \frac{1}{2} = - \left(\frac{3}{5}\right) + \frac{1}{2}$
$y = - \frac{6}{10} + \frac{5}{10}$
$y = - \frac{1}{10}$


text:
# Undo the square roots

Your job is to convert decimals back into the sum of the square roots of integers. The result has to have an accuracy of at least 6 significant decimal digits.

Input:

A number indicating the number of square roots and a decimal indicating the number to approximate.

Example input:

2 3.414213562373095

Output: Integers separated by spaces that, when square rooted and added, are approximately the original decimal accurate to at least 6 significant decimal digits.

Zeros are not allowed in the solution.

If there are multiple solutions, you only have to print one.

Example output (in any order):

4 2

This works because Math.sqrt(4) + Math.sqrt(2) == 3.414213562373095.

This is code golf. Shortest code (with optional bonus) wins!

There is always going to be a solution but -10 if your program prints "No" when there is no solution with integers. In addition, -10 if your program prints all solutions (separated by newlines or semicolons or whatever) instead of just one.

Test cases:

3 7.923668178593959 --> 6 7 8
2 2.8284271247461903 --> 2 2
5 5.0 --> 1 1 1 1 1
5 13.0 --> 4 4 9 9 9 --> 81 1 1 1 1 --> 36 9 4 1 1 etc. [print any, but print all for the "all solutions bonus"]

And yes, your program has to finish in finite time using finite memory on any reasonable machine. It can't just work "in theory," you have to be able to actually test it.

• If there are multiple solutions, does it matter which solution we print?. E.g. for your last test case (5 13.0), this is also a valid solution: 81 1 1 1 1 Nov 26 '14 at 11:11
• And are zeros allowed in the solution? Nov 26 '14 at 11:20
• Is the input always space-separated? Nov 26 '14 at 16:43
• And is entering input via function call allowed? Nov 26 '14 at 16:46
• Also, what about duplicate solutions? For the first example, is our code allowed to print all six permutations of 6 7 8 for the second bonus? Nov 26 '14 at 16:49

# Python 3, 90 - 10 = 80

def S(N,x,n=[],i=1):
if x*x<1e-12>N==0:print(*n)
while.1+x*x>i:S(N-1,x-i**.5,n+[i]);i+=1

(Mega thanks to @xnor for tips, especially the restructuring of the for loop into a while)

A simple recursive attempt. It starts off with the target number and continuously subtracts square roots until it hits 0 or lower. The function S can be called like S(2,3.414213562373095) (the second argument is assumed to be positive).

The program doesn't just print out all solutions, it prints out all permutations of solutions (a little extraneous, I know). Here's the output for the last case: Pastebin.

A slight tweak gives a 98 - 10 = 88 solution which doesn't print permutations, making it more efficient:

def S(N,x,n=[]):
*_,i=[1]+n
if x*x<1e-12>N==0:print(*n)
while.1+x*x>i:S(N-1,x-i**.5,n+[i]);i+=1

And just for fun, this 99 - 10 = 89 one is about as efficient as it gets (unlike the others, it doesn't blow the stack on S(1,1000):

def S(N,x,n=[]):
*_,i=[1]+n
if x*x<1e-12>N:print(*n)
while(.1+x*x>i)*N:S(N-1,x-i**.5,n+[i]);i+=1

Note that, although we have a mutable default argument, this never causes a problem if we rerun the function since n+[i] creates a new list.

## Proof of correctness

In order to end up in an infinite loop, we must hit some point where x < 0 and 0.1 + x2 > 1. This is satisfied by x < -0.948....

But note that we start from positive x and x is always decreasing, so in order to hit x < -0.948... we must have had x' - i0.5 < -0.948... for some x' > -0.948... before x and positive integer i. For the while loop to run, we must also have had 0.1 + x'2 > i.

Rearranging we get x'2 + 1.897x' + 0.948 < i < 0.1 + x'2, the outer parts implying that x' < -0.447. But if -0.948 < x' < -0.447, then no positive integer i can fit the gap in the above inequality.

Hence we'll never end up in an infinite loop.

• You can avoid abs with x*x<1e-12.
– xnor
Nov 26 '14 at 18:45
• I think this while loop works to replace the for: while.1+x*x>i:S(x-i**.5,n+[i]);i+=1, having initialized i=1 in the function parameters. The idea is to avoid needing to convert to ints. The .1 is to handle float inaccuracies; I think it's safe against infinite loops.
– xnor
Nov 26 '14 at 19:06
• @xnor I've implemented the first tip for now. I'm still checking the correctness of the second one, but if it's good then that's a lot of bytes saved! (Also I actually expected you to post a solution :P) Nov 27 '14 at 4:59
• And with N now a function argument, it's shorter to recurse down with N-1 and check when N==0 rather than len(n)==N.
– xnor
Nov 29 '14 at 10:48
• @Sp3000 I'm convinced now that the .1 is safe; I can chat you an argument if you'd like.
– xnor
Dec 2 '14 at 7:15

## ECLiPSe Prolog - 118 (138-20)

I used the following implementation of Prolog: http://eclipseclp.org/

:-lib(util).
t(0,S,[]):-!,S<0.00001,S> -0.00001.
t(N,S,[X|Y]):-A is integer(ceiling(S*S)),between(1,A,X),M is N-1,T is S-sqrt(X),t(M,T,Y).

This is a very naive, exponential approach. Listing all possible solutions takes time to cover all combinations (edit: the range of visited integers now decreases at each step, which removes a lot of useless combinations).

Here follows a transcript of a test session. By default, the environment will try to find all possible solutions (-10) and prints "No" when it fails to do so (-10).

As Sp3000 properly noted in the comment, it also prints "Yes" when it succeeds. That surely means I can remove 10 more points ;-)

[eclipse 19]: t(1,0.5,R).

No (0.00s cpu)
[eclipse 20]: t(2,3.414213562373095,R).

R = [2, 4]
Yes (0.00s cpu, solution 1, maybe more) ? ;

R = [4, 2]
Yes (0.00s cpu, solution 2, maybe more) ? ;

No (0.01s cpu)
[eclipse 21]: t(3,7.923668178593959,R).

R = [6, 7, 8]
Yes (0.02s cpu, solution 1, maybe more) ? ;

R = [6, 8, 7]
Yes (0.02s cpu, solution 2, maybe more) ? ;

R = [7, 6, 8]
Yes (0.02s cpu, solution 3, maybe more) ?
[eclipse 22]: t(5,5.0,R).

R = [1, 1, 1, 1, 1]
Yes (0.00s cpu, solution 1, maybe more) ? ;
^C

interruption: type a, b, c, e, or h for help : ? abort
Aborting execution ...
Abort
[eclipse 23]: t(5,13.0,R).

R = [1, 1, 1, 1, 81]
Yes (0.00s cpu, solution 1, maybe more) ? ;

R = [1, 1, 1, 4, 64]
Yes (0.00s cpu, solution 2, maybe more) ? ;

R = [1, 1, 1, 9, 49]
Yes (0.00s cpu, solution 3, maybe more) ?
[eclipse 24]:

(Edit) Regarding performance, it is quite good, at least compared to other ones (see for example this comment from FryAmTheEggman). First, if you want to print all results, add the following predicate:

    p(N,S):-t(N,S,L),write(L),fail.
p(_,_).

See http://pastebin.com/ugjfEHpw for the (5,13.0) case, which completes in 0.24 seconds and find 495 solutions (but maybe I am missing some solutions, I don't know).

• It also prints "Yes" when it succeeds! Oh Prolog. Nov 26 '14 at 16:20

# Erlang, 305-10 302-10

f(M,D)->E=round(D*D),t(p(E,M,1),{M,E,D}).
p(_,0,A)->A;p(E,N,A)->p(E,N-1,A*E).
t(-1,_)->"No";t(I,{N,E,D}=T)->L=q(I,N,E,[]),V=lists:sum([math:sqrt(M)||M<-L])-D,if V*V<0.1e-9->lists:flatten([integer_to_list(J)++" "||J<-L]);true->t(I-1,T)end.
q(I,1,_,A)->[I+1|A];q(I,N,E,A)->q(I div E,N-1,E,[I rem E+1|A]).

This function returns string "No" or a string with values separated by spaces. It (inefficiently) processes every possible values encoding them into a large integer, and starting with higher values. 0 are not allowed in the solution, and encoded 0 represents all ones. Error is squared.

Example:

f(1,0.5).               % returns "No"
f(2,3.414213562373095). % returns "4 2 "
f(3,7.923668178593959). % returns "8 7 6 "
f(5,5.0).               % returns "1 1 1 1 1 "
f(5,13.0).              % returns "81 1 1 1 1 "

Please be patient with f(5,13.0) as function search space is 13^10. It can be made faster with 2 additional bytes.

### Python 3 2: 173 159 - 10 = 149

Explanation: Each solution is of the form x_1 x_2 ... x_n with 1 <= x_1 <= x^2 where x is the target sum. Therefore we can encode each solution as an integer in base x^2. The while loop iterates over all (x^2)^n possibilities. Then I convert the integer back and test the sum. Pretty straight forward.

i=input;n=int(i());x=float(i());m=int(x*x);a=m**n
while a:
s=[a/m**b%m+1for b in range(n)];a-=1
if abs(x-sum(b**.5for b in s))<1e-5:print' '.join(map(str,s))

It finds all solutions, but the last test case takes way too long.

# JavaScript (ES6) 162 (172 - 10) 173

Edit A little shorter, a little slower.

As a function with 2 parameters, output to javascript console. This prints all solutions with no repetitions (solutions tuples are generated already sorted).
I cared more about timing than about char count, so that it's easily tested in a browser console within the standard javascript time limit.

(Feb 2016 update) Current time for last test case: about 1 150 secs. Memory requirements: negligible.

F=(k,t,z=t- --k,r=[])=>{
for(r[k]=z=z*z|0;r[k];)
{
for(;k;)r[--k]=z;
for(w=t,j=0;r[j];)w-=Math.sqrt(r[j++]);
w*w<1e-12&&console.log(r.join(' '));
for(--r[k];r[k]<1;)z=--r[++k];
}
}

ES 5 Version Any browser

function F(k,t)
{
var z=t- --k,r=[];
for(r[k]=z=z*z|0;r[k];)
{
for(;k;)r[--k]=z;
for(w=t,j=0;r[j];)w-=Math.sqrt(r[j++]);
w*w<1e-12&&console.log(r.join(' '));
for(--r[k];r[k]<1;)z=--r[++k];
}
}

Test Snippet it should run on any recent browser

F=(k,t)=>
{
z=t- --k,r=[];
for(r[k]=z=z*z|0;r[k];)
{
for(;k;)r[--k]=z;
for(w=t,j=0;r[j];)w-=Math.sqrt(r[j++]);
w*w<1e-12&&console.log(r.join(' '));
for(--r[k];r[k]<1;)z=--r[++k];
}
}

console.log=x=>O.textContent+=x+'\n'

t=~new Date
console.log('\n2, 3.414213562373095')
F(2, 3.414213562373095)
console.log('\n5, 5')
F(5, 5)
console.log('\n3, 7.923668178593959')
F(3, 7.923668178593959)
console.log('\n5, 13')
F(5, 13)

t-=~new Date
O.textContent = 'Total time (ms) '+t+ '\n'+O.textContent

(Edit) Below are the result on my PC when I posted this answer 15 months ago. I tried today and it is 100 times faster on the same PC, just with a 64 bit alpha version of Firefox (and Chrome lags well behind)! - current time with Firefox 40 Alpha 64 bit: ~2 sec, Chrome 48: ~29 sec

Output (on my PC - last number is runtime in milliseconds)

2 4
1 1 1 1 1
6 7 8
1 1 1 1 81
1 1 1 4 64
1 1 1 9 49
1 1 4 4 49
1 1 1 16 36
1 1 4 9 36
1 4 4 4 36
1 1 1 25 25
1 1 4 16 25
1 1 9 9 25
1 4 4 9 25
4 4 4 4 25
1 1 9 16 16
1 4 4 16 16
1 4 9 9 16
4 4 4 9 16
1 9 9 9 9
4 4 9 9 9
281889

# Mathematica - 76 - 20 = 56

f[n_,x_]:=Select[Union[Sort/@Range[x^2]~Tuples~{n}],Abs[Plus@@√#-x]<10^-12&]

Examples

f[2, 3.414213562373095]
> {{2, 4}}
f[3, 7.923668178593959]
> {{6, 7, 8}}
f[3, 12]
> {{1, 1, 100}, {1, 4, 81}, {1, 9, 64}, {1, 16, 49}, {1, 25, 36}, {4, 4, 64}, {4, 9, 49}, {4, 16, 36}, {4, 25, 25}, {9, 9, 36}, {9, 16, 25}, {16, 16, 16}}

• How does this print No? Also, the output isn't space separated. Also, can't you use Tr@ instead of Plus@@? And you might be able to save some characters by changing Select to Cases, the function at the end to a pattern, and making f an unnamed pure function. Nov 28 '14 at 12:37

# Haskell, 87 80 - 10 = 70

This is a recursive algorithm similar to the Python 3 program of @Sp3000. It consists of an infix function # that returns a list of all permutations of all solutions.

0#n=[[]|n^2<0.1^12]
m#n=[k:v|k<-[1..round$n^2],v<-(m-1)#(n-fromInteger k**0.5)]  With a score of 102 99 92 - 10 = 82 we can print each solution only once, sorted: 0#n=[[]|n^2<0.1^12] m#n=[k:v|k<-[1..round$n^2],v<-(m-1)#(n-fromInteger k**0.5),m<2||[k]<=v]

## Pyth55 54 47-20 = 27

DgGHKf<^-Hsm^d.5T2^10_12^r1hh*HHGR?jbmjdkKK"No

Try it online.

Shamelessly borrows from xnor's comment ;)

This will run out of memory on any sane computer even for a value like 5,5.0. Defines a function, g, that can be called like g 3 7.923668178593959.

This python 3 program uses essentially the same algorithm (just doesn't do the "No" printing at the end, which could be done by assigning a variable to all the results, then writing print(K if K else "No")), but uses a generator, so it doesn't get a memory error (it will still take extremely long, but I've made it print as it finds the values):

This gave the exact same results that @Sp3000 got. Also, this took several days to finish (I didn't time it, but around 72 hours).

from itertools import*
def g(G,H):
for x in product(range(1,int(H*H+2)),repeat=G):
if (H-sum(map(lambda n:n**.5,x)))**2<1e-12:print(*x)

# Python 3 - 157174 169 - 10 = 159

Edit1: Changed the output format to space-separated integers instead of comma-separated. Thanks for the tip of removing the braces around (n,x).

Edit2: Thanks for the golfing tips! I can trim off another 9 chars if I just use an == test instead of testing for approximate equality to within 1e-6, but that would invalidate approximate solutions, if any such exist.

Uses itertools to generate all possible integer combinations, hopefully efficiently :)

I haven't found a way to add printing "No" efficiently, it always seems to take more than 10 extra characters.

from itertools import*
n,x=eval(input())
for c in combinations_with_replacement(range(1,int(x*x)),n):
if abs(sum(z**.5for z in c)-x)<1e-6:print(' '.join(map(str,c)))

• Your program has the wrong output format (commas instead of spaces). Also, you can shave off 2 bytes by removing the braces around n,x. Nov 26 '14 at 15:51
• I seem to be getting SyntaxErrors when I try the eval line... Nov 26 '14 at 15:56
• @Sp3000: try enter 3,7.923668178593959. You need the ',' Nov 26 '14 at 16:35
• 4 little improvements: from itertools import* saves 1, removing the space z**.5for saves 1, and removing the [] in sum(z**.5for z in c) saves 2 and removing the () in if(...) saves 1. Nov 26 '14 at 16:40
• Would a change to Python 2 and using n,x=input() be more compact? Nov 26 '14 at 23:22

## Scala (397 bytes - 10)

import java.util.Scanner
object Z extends App{type S=Map[Int,Int]
def a(m:S,i:Int)=m updated(i,1+m.getOrElse(i,0))
def f(n:Int,x:Double):Set[S]={if(n==0){if(x.abs<1e-6)Set(Map())else Set()}
else((1 to(x*x+1).toInt)flatMap{(i:Int)=>f(n-1,x-Math.sqrt(i))map{(m:S)=>a(m,i)}}).toSet}
val s=new Scanner(System.in)
f(s.nextInt,s.nextDouble)foreach{(m:S)=>m foreach{case(k,v)=>print(s"\$k "*v)};println}}

If there are no permutations, then this program prints nothing.


text:
Last edited by Yobar
Sunday, May 10, 2020 | History

2 edition of Numerical methods for integrating oscillatory functions. found in the catalog.

Numerical methods for integrating oscillatory functions.

M. Newby

# Numerical methods for integrating oscillatory functions.

## by M. Newby

Written in English

Edition Notes

M.Sc. dissertation. Typescript.

The Physical Object ID Numbers Series Dissertations Pagination 121p. Number of Pages 121 Open Library OL13693946M

NIntegrate symbolically analyzes its input to transform oscillatory and other integrands, subdivide piecewise functions, and select optimal algorithms. The method suboption "SymbolicProcessing" specifies the maximum number of seconds for which to attempt performing symbolic analysis of . Numerical experiments are carried out for the algorithms discussed in Section 3 and Section 4 for linear hyperbolic equations with oscillatory coefficients and oscillatory initial values. In the numerical experiments, we take the time step size At equal to the space grid size h (the CFL condition is satisfied) and always com-Cited by: 6.

An introduction to highly oscillatory problemsThe wonderful world of asymptotic expansionsOscillatory integrals Design goal of hybrid numerical-asymptotic methods Say f (n) is the solution to Lf = 0. Say Q[f ] is the numerical solution to the approximate equation L hf = 0. Then if f (n) ˘ X1 k=1 a kn k; n ˛1 and g(n) ˘ X1 k=1 b kn k; n ˛1. formula (DIFSUB) of Gear [13], [14] and the blended linear multistep methods of Skeel and Kong [24], and the symmetric multistep methods of Lambert and Watson [17]. 1. Introduction. The development of numerical integration formulas for stiff as well as highly oscillatory systems of differential equations has attracted considerable.

How to integrate a highly oscillatory function. Follow 16 views (last 30 days) Mila Z on 22 Nov Vote. 0 ⋮ You can't prove divergence of this integral using numerical methods. Try analytical methods instead (e.g. show that the function behaves like constant*cos(2*y) for large y). DRAFT: Integration of oscillatory integrals, a computer-algebra approach Richard Fateman Computer Science University of California Berkeley, CA, USA Novem Abstract The numerical integration of oscillatory integrals is an important and well-studied area of mathemat-ical inquiry.

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### Numerical methods for integrating oscillatory functions by M. Newby Download PDF EPUB FB2

Functions Gradimir V. Milovanovic and Marija P. Stani´ ´c Abstract 1 Introduction By a highly–oscillating function we mean one with large number of local maxima and minima over some interval. The computation of integrals of highly–oscillating functions is one of the most important issues in numerical analysis since such inte-File Size: KB.

Numerical methods for integrating oscillatory functions. book specific nonstandard methods for numerical integration of highly oscillating functions, mainly based on some contour integration methods and applications of some kinds of Gaussian quadratures.

NIntegrate of a highly oscillatory integral double exponential oscillatory. Ask Question Asked 3 years, 8 I see the problem from another point of view. Basically, the integral is a Gaussian multiplying a highly oscillatory function (two functions indeed, a Sin and a Cos). Thanks for contributing an answer to Mathematica Stack Exchange.

Abstract. Some specific nonstandard methods for numerical integration of highly oscillating functions, mainly based on some contour integration methods and applications of some kinds of Gaussian quadratures, including complex oscillatory weights, are presented in this by: 5.

Many effective methods have been proposed for the oscillatory integrals in order to overcome the difficulty caused by the high oscillation such as Filon-type methods [10,11,27], Levin methods [15 Author: Sheehan Olver.

4. Buyst and L. Schotsmans,A method of Gaussian type for the numerical integration of oscillating functions, ICC Bull. 3 (), – Google ScholarCited by: A Method for the Numerical Evaluation of Finite Integrals of Oscillatory Functions By I.

Longman 1. Introduction. In two previous publications [1, 2] the author has demon-strated a method, based on Euler's transformation of slowly convergent alternat. The integration of systems containing highly oscillatory functions is a central point in many practical problems in physics, chemistry and engineering.

Highly oscillatory integrals are allegedly difficult to calculate by the standard classic integration formulae when the frequency is significantly larger than the number of quadrature by: Method for numerical integration of difficult oscillatory integral.

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Highly oscillatory functions appear to be at the forefront of the research in numerical analysis. In this work we developed efficient numerical algorithms for solving highly oscillatory differential equations.

The main important achievements are: to the contrary of classical Author: Marianna Khanamiryan. on haar wavelets and hybrid functions, Comput Math Applpp.

– [9] Siraj-ul-Islam,Aziz, I, KhanW, Numerical integration of multi-dimensional highly oscillatory, gentle oscillatory and non-oscillatory integrands based on wavelets and radial basis functions, Engineering Analysis with Boundary Elements,pp.

–pose of the research is to nd appropriate integration methods for the oscillatory integrals, we made use of di erent in-built Mathematica functions. We calculated the oscillatory integrals by the use of those various functions.

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text:
After learning the definitions and concepts of proportional and inversely proportional in mathematics, the next step to study is linear functions. Similar to the graph of direct proportionality, it is linear functions that if the value of $x$ increases, the value of $y$ increases in the same ratio.

In linear functions, there are two concepts: slope and intercept. The shape of the graph will change depending on what number of slopes and intercepts are used. Linear functions also require us to write graphs and create linear equations from them.

There are also many problems to find the intersection, in which case we can use simultaneous equations to find the intersection point.

Linear functions are almost identical to the concept of direct proportionality. Therefore, we frequently use linear functions in our daily life. We will explain how to solve problems with linear functions.

## Linear functions Are Special Proportional

There is often a relationship between direct proportionality. Therefore, we frequently use proportional in everyday life. Another concept that is also essential in everyday life is linear functions. Proportional and linear functions are similar, and all people use linear functions in everyday life.

For example, we use linear functions when we think about our daily living expenses. It is the calculation of a linear function to figure out how much you can spend each day based on your current savings.

The proportional formula is $y=ax$. For example, if you get an allowance of \$20 per month and you don’t spend any money at all, your savings after$x$months will be expressed by the following equation. •$y=20x$If you want to save money from zero savings, it becomes a linear equation like this On the other hand, if you already have some savings, you have to take into account the amount of savings. For example, if you already have \$50 in savings, the total amount of savings after $x$ months will be following.

• $y=20x+50$

In real life, it is common to start from a specific point, rather than starting from zero. In linear functions, we create an equation that takes this into account.

Think of linear functions as a special case of proportional. Linear functions are proportional that start from a specific point, rather than starting from zero.

### Linear Functions Have a Slope and Intercept, Resulting in the Formula $y=ax+b$

The formula for direct proportionality is $y=ax$. The formula for linear functions, on the other hand, is $y=ax+b$. Because it starts at a specific point, we add $+b$ to the formula of proportional.

## Create and Solve an Equation for Linear Functions

A very similar field to direct proportionality is linear functions. Think of linear functions in mathematics as a special case of a proportional.

There are so many situations where we have to use linear functions to make calculations. For example, when you calculate your future savings, you use linear functions. When calculating, we must find the slope and intercept to create an equation.

In linear functions, there are also many graph-based problems. So you need to be able to calculate the slope and intercept from the graph. By using coordinates, we can get an expression for a linear function. Also, by using simultaneous equations, we can find the intersection of two lines.

Once we understand the properties of these linear functions, we will be able to solve word problems. If you can solve word problems, you will be able to use mathematics in all aspects of your daily life.


text:
# How do you find the limit of $\frac{4x^4 + 5y^4}{x^2 + y^2}$?

Find the limit of $\frac{4x^4 + 5y^4}{x^2 + y^2}$ as $(x,y)\to (0,0)$.

Which method do I use to find the limit of that? I tried paths but the limits all came out to be $0$... (as a side question, when do you stop trying paths? I mean there are so many ways to try out when $x$ approaches $0$. You can try $y=0$, $y=x$, $y=x^2$, $y=mx$, and so many more ways. After you get like $0$ for 4 limits, do you just stop there and assume to try another method?) (Also, when I try different ways for paths, will the limits always be either $0$ or a finite number and never DNE?)

Thank you

-
And what is wrong with 0? It is a respectable member of $\mathbb{R}$ too, you know... –  vonbrand Apr 13 '13 at 20:23

The euclidean norm is $$||(x,y)||=\sqrt{x^2+y^2}$$ and we have $$x^4\leq||(x,y)||^4\quad;\quad y^4\leq||(x,y)||^4$$ so we have $$0\leq\frac{4x^4 + 5y^4}{x^2 + y^2}\leq9||(x,y)||^2\to_{(x,y)\to(0,0)}0$$

-
@PeterTamaroff: No, there's a norm squared in the denominator. –  Javier Apr 14 '13 at 2:25
@JavierBadia Ooops. –  Pedro Tamaroff Apr 14 '13 at 2:25

There is a nearly universal strategy when the denominator is $x^2+y^2$ or a close relative. Let $x=r\cos \theta$, $y=r\sin\theta$.

Here the bottom becomes $r^2$, and the top is $4r^4\cos^4\theta+5r^5\sin^5\theta$. Divide. We get $4r^2\cos^4\theta+5r^3\sin^5\theta$. The trigonometric functions are bounded, so the limit is $0$.

-

In principle, you can never stop looking for new paths and may have to be creative. However, you are right to suspect that the limit is indeed $0$. To show this, you better prove that for each $\epsilon>0$ there exists $\delta>0$ such that $\sqrt{x^2+y^2}<\delta$ implies $|f(x,y)|<\epsilon$.

For your second question: Virtually anything can happen along different paths - different limits, divergence to infinity, and and of course proper divergence (think of zigzagging between two paths with different limits - the zigzag path will not have a limit).

-
good advice ..+1 –  Halil Duru Apr 13 '13 at 20:20

Is this what you mean?

$$\underset{x,y\rightarrow0}{\lim}\frac{4x^4+5y^4}{x^2+y^2}=$$

$$=\underset{x,y\rightarrow0}{\lim}\frac{4x^4}{x^2+y^2}+\underset{x,y\rightarrow0}{\lim}\frac{5y^4}{x^2+y^2}=$$ $$=\underset{x\rightarrow0}{\lim}\frac{4x^4}{2x^2}+\underset{y\rightarrow0}{\lim}\frac{5y^4}{2y^2}=$$ $$=\underset{x\rightarrow0}{\lim}2x^2+\underset{y\rightarrow0}{\lim}\frac{5}{2}y^2=0+0=0$$

-
You need to justify some passages to limit their operations here. –  MathOverview Apr 13 '13 at 19:24
Well this is the intuitive approach. –  Ambesh Apr 13 '13 at 19:27
But its construction is very good. The justifications just saw that can be fitted perfectly in the middle of his solution. –  MathOverview Apr 13 '13 at 19:33

The standard procedure is indeed changing the coordinates from Cartesian to Polar. Your limit:

$$\lim_{(x, y)\to(0, 0)} \frac{4x^4 + 5y^4}{x^2 + y^2}$$

Will become (with $x=r\cos(\theta)$ and $y=r\sin(\theta)$ and using the fact that $r\to0$ when $(x, y)\to(0, 0)$): $$\lim_{r\to0} \frac{4r^4\cos(\theta)^4 + 5r^4\sin(0)^4}{r^2\cos(\theta)^2 + r^2\sin(\theta)^2}$$

which ONLY exists if it is the same for all values of $\theta$ (the limit does not vary with $\theta$).

Using: $\sin^2(\theta)+\cos^2(\theta)=1$ your limit will further reduce to:

$$\lim_{r\to0} 4r^2 \to 0 \text{ for all } \theta \text{; therefore the limit exists.}$$

-

Well , let $x^2+y^2=r^2$ then $$\frac{4x^4 + 5y^4}{x^2 + y^2}\leq\frac {5 r^4}{r^2} =5r^2 \to 0$$ as $r \to 0$ $\hspace{99mm} \blacksquare$

-
$\frac{5r^4}{r^2}$. –  Eric Jablow Apr 14 '13 at 1:06
this version better i guess --thank you... –  Halil Duru Apr 14 '13 at 5:15