text:
# Elasticsearch mapping文档相似性算法

Elasticsearch allows you to configure a scoring algorithm or similarity per field. The similaritysetting provides a simple way of choosing a similarity algorithm other than the default TF/IDF, such as BM25.

Similarities are mostly useful for text fields, but can also apply to other field types.

Custom similarities can be configured by tuning the parameters of the built-in similarities. For more details about this expert options, see the similarity module.

The only similarities which can be used out of the box, without any further configuration are:

BM25
The Okapi BM25 algorithm. The algorithm used by default in Elasticsearch and Lucene. See Pluggable Similarity Algorithms for more information.
classic
The TF/IDF algorithm which used to be the default in Elasticsearch and Lucene. See Lucene’s Practical Scoring Function for more information.

The similarity can be set on the field level when a field is first created, as follows:

PUT my_index
{
"mappings": {
"my_type": {
"properties": {
"default_field": {
"type": "text"
},
"classic_field": {
"type": "text",
"similarity": "classic"
}
}
}
}
}
 The default_field uses the BM25 similarity. The classic_field uses the classic similarity (ie TF/IDF).

+ 订阅

format: plain


text:
# Attributes for calculating a hit in a turn based game or Roguelikes

I'm trying to figure out a good formula for determining a hit when the player attacks an enemy in a Roguelike. I'd like to come up with something simplistic but, still true to the way it's done in D&D or popular Roguelikes like AngBand or NetHack.

What I have so far is adding the differences between player and enemy attributes, along with a random integer:

if ((player.dex - enemy.dex) + (player.speed - enemy.speed) +
(player.luck - enemy.luck) + random_integer_below_6()) >= 10) {
hit(enemy);
}

This makes for unnatural gameplay: the player never misses weak enemies and vice versa for strong ones. Does anyone know of a guide to attribute based combat used in turn based games etc? Should I (and can I) copy the D&D rules verbatim?

I'd figure out what you want the "guaranteed hit" and "guaranteed miss" chances to be (i.e. the chances for a kitten to hit a dragon and a dragon to miss a kitten). Store your random result, and before applying power differences, mark a hit if it's within those bounds. That will prevent the 100% hit/miss chance problem... although there's another option for always providing a chance of a hit, detailed below.

D&D is a "flat" system, where probability is evenly distributed across the possible roll results. Other systems use a more curved distribution, usually achieved by rolling multiple dice and adding them. It's easy enough to create a curve like this (rand() + rand())/2 comes to mind).

Another possibility is "exploding dice." In systems like Shadowrun and Savage Worlds, rolling the maximum possible die result lets you reroll that die and add the new result. In your game, you'd test to see if your random result was above a certain threshold, and if so, roll again. If you let dice explode indefinitely, then even the lowliest goblin has a chance to hit a god, and you don't need an explicit "guaranteed hit" mechanic.

So one possible approach with a curved roll, guaranteed chance of missing, and exploding dice would be something like:

function getRoll():Number
{
var result:Number = (rand() + rand()) * 3; //Curved result from 0 to 6
if(result > 5.75) result += getRoll(); //Explode over a certain threshold, recursively
return result;
}

function attackHits():Boolean
{
var roll:Number = getRoll();
if(roll < 0.25) return false; //Guaranteed chance of miss
else return ((player.dex - enemy.dex) + (player.speed - enemy.speed) +
(player.luck - enemy.luck) + roll) >= 10; //Your original test.
}

Of course, that's untested and unbalanced; you'd want to adjust all those magic numbers until they felt right.

As for copying rules verbatim, there are legal and ethical issues with that. You're fine pulling inspiration from D&D and other tabletop games, though; that's how they all got designed in the first place. You don't need to show the player exactly how the rules work, either, and you may find that using a smooth floating point dice curve leads to a better feel than restricting yourself to the integer constraints of physical dice.

• what language is that, with the funny type decls? Or is it just pseudocode? – tenpn Aug 11 '10 at 7:41
• I think it may be Actionscript or Javascript, but don't hold me to that. – The Communist Duck Aug 11 '10 at 9:44
• It's Actionscript 3. I'm a Flash developer. – <NAME> Aug 12 '10 at 13:29

I'm assuming this is a game balance question and not a coding question, and there are a lot of ways to handle this. From your existing algorithm, I suspect you are making it more complicated than it has to be (three attributes AND a random number make for an extremely muddy system from the player's perspective). My first instinct is to simplify!

Favor attributes that do multiple things, rather than results that are determined by multiple attributes. DEX works just fine as the primary determiner of to-hit chance; the addition of everything else just serves to make it harder to balance. Think of it this way: a player's expected damage on each hit is their to-hit % multiplied by their average damage when they do hit. How do you plan to calculate an "expected" or "typical" to-hit % with so many variables?

I see another issue, which is that your formula doesn't scale, and this is probably what you're experiencing. Assuming DEX, LUCK and SPEED tend to increase over the course of the game, differences between player and monster in those stats will also tend to increase. Just as an example, if a player's early-game DEX is in the 4 to 6 range and so are monsters, that's a 10-20% to-hit modifier. If the range is more like 10 to 20 in later stages of the game, that's up to a 100% modifier.

One way to fix that is to keep the stats fixed. Player and monster ranges are the same throughout the game, and getting something like +1 DEX is a huge deal, extremely rare, and treated as such (if it exists at all). Then your formula will work fine for the entire game, and you'll see a player with about the same to-hit chance throughout.

Another way is to change that hardcoded "10" to scale with level, so maybe it's "10 + monster level" or "10 + dungeon level" or something, which causes a slight increase in difficulty as the game progresses unless the player levels up their stats accordingly.

A third way, as others have said, is to set hard caps on the highest and lowest acceptable to-hit %, and simply say "if the calculated value is less than X, use X instead".

For the sake of completeness, I'll point out that there's no absolute law that you MUST include a to-hit roll in the first place. Taking an action in a game that's ineffective is not particularly fun, and doesn't really add to the experience except in rare cases. An alternative is to let players ALWAYS hit, and just mess with the damage range so that sometimes they'll do a lot less damage than others. Remember that complex systems are more fun for the designer than the player.

• This is a great answer. – seanicus Apr 19 '13 at 1:33

You might want to add critical hits (usually can't miss, bonus damage) and critical misses (flat chance to miss no matter what, possibly dazes the player).

A lot of games use a five percent change of a hit regardless of all factors in addition to the normal hit formula.


text:
# Rotational motion (confused about the variables)

we have many accelerations in rotational motion ,and I don't know the difference between them. first the linear acceleration which as I understand is the vector acceleration which has the same direction as the change of velocity and it affects the speed and direction then it's two components are centripetal acceleration which changes the angle and is towards the center and it doesnt change the magnitue of the velocity. Then there is the tangential acceleration which is perpendicular to the centripetal acceleration and it increases the magnitude of the velocity.
First is my understanding of the acceleration correct ? second I also see that we have the angular acceleration and it changes the the the angular velocity but this confuses me what is the difference between it and the centripetal acceleration? they both change the angle.

Related Introductory Physics Homework Help News on Phys.org
Doc Al
Mentor
Here's how I'd put it. Something moving in a circle has a linear (or translational) acceleration. That acceleration has two components: a centripetal component and a tangential component.

What is usually called the angular acceleration is the rate of change of angular speed. The angular speed relates to the tangential speed via: ω = r*vt. In turn, the angular acceleration relates to the tangential acceleration via: α = r*at.

Does that help?

but what I don't understand is whether the centripetal acceleration and the angular acceleration are measuring the same thing or not? I mean they all affect the change of the direction of the velocity right?

Doc Al
Mentor
but what I don't understand is whether the centripetal acceleration and the angular acceleration are measuring the same thing or not?
No. Centripetal acceleration is the rate of change in linear velocity (for constant speed). Angular acceleration is the rate of change of the angular speed. For something moving with constant speed, there will be a non-zero centripetal acceleration while the angular acceleration will be zero.

No. Centripetal acceleration is the rate of change in linear velocity (for constant speed). Angular acceleration is the rate of change of the angular speed. For something moving with constant speed, there will be a non-zero centripetal acceleration while the angular acceleration will be zero.
oh thanks I see it now but if angular acceleration is the rate of change of angular speed then what is the rate of change of angular velocity?

Doc Al
Mentor
oh thanks I see it now but if angular acceleration is the rate of change of angular speed then what is the rate of change of angular velocity?
As long as the axis of rotation doesn't change, as when you stick to two dimensions, angular speed and velocity are the same (for practical purposes). When you deal in three dimensions, and the axis of rotation can change, things get complicated fast. Angular velocity is a vector (pseudovector, really) whose magnitude is the angular speed and whose direction is along the axis of rotation. See: http://en.wikipedia.org/wiki/Angular_velocity" [Broken]

Last edited by a moderator:
but isnt the rotation negative if it is clockwise?

Doc Al
Mentor
but isnt the rotation negative if it is clockwise?
That's the convention, sure.


text:
2014
01-05

# Tower Parking

There is a new revolution in the parking lot business: the parking tower. The concept is simple: you drive your car into the elevator at the entrance of the tower, and the elevator and conveyor belts drag the car to an empty parking spot, where the car remains until you pick it up. When you return, the elevator and conveyor belts move your car back to the entrance and you’re done.

The layout of the tower is simple. There is one central elevator that transports the cars between the different floors. On each floor there is one giant circular conveyor belt on which the cars stand. This belt can move in clockwise and counterclockwise direction. When the elevator arrives on a floor, it becomes part of the belt so that cars can move through it.

At the end of the day the tower is usually packed with cars and a lot of people come to pick them up. Customers are processed in a first come first serve order: the elevator is moved to the floor of the first car, the conveyor belt moves the car on the elevator, the elevator is moved down again, and so on. We like to know how long it takes before the last customer gets his car. Moving the elevator one floor up- or downwards takes 10 seconds and moving a conveyor belt one car in either direction takes 5 seconds.

On the first line one positive number: the number of testcases, at most 100. After that per testcase:

One line with two integers h and l with 1 ≤ h ≤ 50 and 2 ≤ l ≤ 50: the height of the parking tower and the length of the conveyor belts.

h lines with l integers: the initial placement of the cars. The jth number on the ith line describes the jth position on the ith floor. This number is -1 if the position is empty, and r if the position is occupied by the rth car to pick up. The positive numbers form a consecutive sequence from 1 to the number of cars. The entrance is on the first floor and the elevator (which is initially empty) is in the first position. There is at least one car in the parking tower.

On the first line one positive number: the number of testcases, at most 100. After that per testcase:

One line with two integers h and l with 1 ≤ h ≤ 50 and 2 ≤ l ≤ 50: the height of the parking tower and the length of the conveyor belts.

h lines with l integers: the initial placement of the cars. The jth number on the ith line describes the jth position on the ith floor. This number is -1 if the position is empty, and r if the position is occupied by the rth car to pick up. The positive numbers form a consecutive sequence from 1 to the number of cars. The entrance is on the first floor and the elevator (which is initially empty) is in the first position. There is at least one car in the parking tower.

2
1 5
-1 2 1 -1 3
3 6
-1 5 6 -1 -1 3
-1 -1 7 -1 2 9
-1 10 4 1 8 -1

25
320

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
struct park
{
int x, y;
};
int main()
{
int i, j, n, m, T, k, t;
scanf("%d", &T);
while(T--)
{
scanf("%d%d", &n, &m);
park p[2505];
int turn[55];
memset(turn, 0, sizeof(turn));
int num=0;
for(i=1; i<=n; i++)
{
turn[i]=1;
for(j=1; j<=m; j++)
{
scanf("%d", &k);
if(k==-1)
continue;
num++;
p[k].x=i, p[k].y=j;
}
}
int total=0;
int mid;
for(i=1; i<=num; i++)
{
if(p[i].y>turn[p[i].x])
mid=p[i].y-turn[p[i].x];
else
mid=turn[p[i].x]-p[i].y;
total+=20*(p[i].x-1)+5*(min(mid, m-mid));
turn[p[i].x]=p[i].y;
}
printf("%d\n", total);
}
}

1. for(int i=1; i<=m; i++){
for(int j=1; j<=n; j++){
dp = dp [j-1] + 1;
if(s1.charAt(i-1) == s3.charAt(i+j-1))
dp = dp[i-1] + 1;
if(s2.charAt(j-1) == s3.charAt(i+j-1))
dp = Math.max(dp [j - 1] + 1, dp );
}
}
这里的代码似乎有点问题？ dp(i)(j) = dp(i)(j-1) + 1;这个例子System.out.println(ils.isInterleave("aa","dbbca", "aadbbcb"));返回的应该是false

2. 问题3是不是应该为1/4 .因为截取的三段，无论是否能组成三角形， x， y-x ，1-y,都应大于0，所以 x<y,基础应该是一个大三角形。小三角是大三角的 1/4.

3. 题本身没错，但是HDOJ放题目的时候，前面有个题目解释了什么是XXX定律。
这里直接放了这个题目，肯定没几个人明白是干啥

4. 博主您好，这是一个内容十分优秀的博客，而且界面也非常漂亮。但是为什么博客的响应速度这么慢，虽然博客的主机在国外，但是我开启VPN还是经常响应很久，再者打开某些页面经常会出现数据库连接出错的提示

5. 有两个重复的话结果是正确的，但解法不够严谨，后面重复的覆盖掉前面的，由于题目数据限制也比较严，所以能提交通过。已更新算法


text:
# Notation (or Name) of the function which is 0 at x=0 and 1 otherwise

Is there a traditional notation or name of this function:

$$\epsilon (x) = \begin{cases} 0 & \textrm{ if }x = 0 \\ 1 & \textrm{ if }x \neq 0 \end{cases}$$

I know one can use Indicator function $1_{A}(x)$ for $A = \mathbb{R} - \{0\}$, but I still want to know if there is a traditional, simpler, notation for it.

• Not that I know of. The simplest way of writing is either $1_{\mathbb R\setminus \{0\}}$ or $1 - 1_{\{0\}}$. Maybe even $1-\delta_0$, but that is already slightly ambiguous. – 5xum May 5 '15 at 7:28
• Well... $1_{\Re \backslash \{ 0 \} }$ is not a function. Moreover, $1 - \delta_0$ cannot be correct because the value of $\delta_0$ is infinite at $x=0$. This is a bizarre function, and I suspect the question poser isn't quite clear on what he or she needs. One can write the function as a limit, but again, that seems rather awkward. – <NAME> May 5 '15 at 7:31
• @David Yes, it is a function. And no, it does not take the value infinity. – <NAME>ildetoft May 5 '15 at 7:42
• @DavidG.Stork The first one is precisely defined like the function the OP is asking about. In general it is a common notation to use $1_A$ as the indicator function for the set $A$, meaning that $1_A(x)$ is $1$ if $x\in A$ and $0$ otherwise. $\delta_0$ is a common shorthand for the indicator function for the set $\{0\}$. – Tobias Kildetoft May 5 '15 at 8:01
• @DavidG.Stork why would you write it as a limit when there is a simple explicit definition as an indicator function? And why do you doubt indicator functions are functions? – <NAME> May 6 '15 at 4:39

The Kronecker delta function $\delta_{ij}$ is $1$ when $i=j$ and $0$ otherwise, so $\epsilon(x)=1-\delta_{0,x}$.

In the Iverson bracket notation you could write $$\epsilon(x)=[x\ne0].$$

You can write $\epsilon(x)=\mathrm{sgn}^2(x)$ where $\mathrm{sgn}$ is the signum function.

• It seems (to me) this is the best way. – user565739 May 5 '15 at 8:27
• Well, it's compact for sure, but I wouldn't recommend using it to communicate with other humans. :) – <NAME> May 5 '15 at 8:35
• $\epsilon(x)=\mathrm{sgn}|x|$ – charlie Mar 24 '19 at 8:49

The Heaviside step function is defined as

$$H (x) = \begin{cases} 0 & \textrm{ if }x < 0 \\ 1 & \textrm{ if }x => 0 \end{cases}$$

Clearly

$$H (-x) = \begin{cases} 1 & \textrm{ if }x <= 0 \\ 0 & \textrm{ if }x > 0 \end{cases}$$

Summing gives:

$$H(x)+ H (-x) = \begin{cases} 1 & \textrm{ if }x < 0 \\ 2 & \textrm{ if }x = 0 \\ 1 & \textrm{ if }x >0 \end{cases}$$

Subtract 1

$$H(x)+ H (-x)-1 = \begin{cases} 0 & \textrm{ if }x < 0 \\ 1 & \textrm{ if }x = 0 \\ 0 & \textrm{ if }x >0 \end{cases}$$

• You could also multiply two Heaviside functions: $\epsilon (x)=H(x)H(-x)$ – tomi May 5 '15 at 7:44
• All this H(x) + H(-x) reminds me of hyperbolic functions - is there any link? – tomi May 5 '15 at 7:45
• Yes, I do. Thanks - but I can't fix it... – tomi May 5 '15 at 7:59
• You can't edit a comment after 5 minutes, but you can copy, delete, and repost it. Anyway, I thought you wanted to fix your answer: by changing $H(x)+H(-x)-1$ to $2-H(x)-H(-x)$. – bof May 5 '15 at 8:07
• How is this better than writing out the definition? – <NAME> May 5 '15 at 8:13

This funtion is also known as the "discrete norm", which is to say an absolute value (in the sense of ring theory) whose metric induces the discrete topology. The discrete metric is $d(x,y)=1$ if $x\ne y$ and $d(x,x)=0$, while the discrete norm on any domain $X$ is $f(x)=1$ if $x\in X\setminus\{0\}$ and $f(0)=0$. For the case of $X=\Bbb R$, this reduces to your function.

I hope this answers the question posed by some of why this would be a natural or useful function, rather than a simple modification of some other function like $\delta_{x,0}$ which is useful in other contexts.

You could use the Heaviside step function $H(x)$. Your function is

$\epsilon(x)=|H(x)-H(-x)|$

• True, but see my post which avoids using absolute function. – tomi May 5 '15 at 7:43
• How is this better than writing out the definition? – <NAME> May 5 '15 at 8:13
• This isn't "traditional notation" (nor simpler) like the question asked, it seems more like obfuscation than anything.... unless this definition from the Heaviside function allows one to derive nontrivial information about the function, it seems pointless. – Thomas May 5 '15 at 8:23

Here's one possible definition, though extremely dependent on how one defines the value $0^0$.

If you happen to subscribe to the belief that $0^0=0$ (disclaimer, I personally define $0^0=1$), then $$\epsilon(x)=x^0$$

• Would the downvoter like explain why they did so? – Hayden May 6 '15 at 16:59


text:
# Make listing look normal (ttfamily and character spacing) [duplicate]

listings seems to like protrude characters. Every time I see that I have this strange urge of killing a kitten. How can I get rid of it?

\documentclass{article}

\parindent0pt

\usepackage{listings}
\lstset{basicstyle = \ttfamily}

\begin{document}

\texttt{test test test}

\begin{lstlisting}
test test test
\end{lstlisting}

\end{document}

• columns=fullflexible – egreg Mar 3 '15 at 18:08
• it works, but: WTF?! – yo' Mar 3 '15 at 18:09
• listings was written with normal type in mind, but spaced out so to get alignment in columns. This of course is automatic if a monospaced font is used, but columns=fullflexible is necessary nonetheless. – egreg Mar 3 '15 at 18:12


text:
User:<NAME> (usurped)

My name is <NAME>. I'm the editor of my own newspaper, published locally in Quesada, Costa Rica. I am a free software activist. Because of this, I use the GNU operating system. I have a minor degree on Software Engineering (specialized on Free Software) and I have advanced knowledge of ${\displaystyle \mathbf {L\!\!^{{}_{\scriptstyle A}}\!\!\!\!\!\;\;T\!_{\displaystyle E}\!X} }$.


text:
# What is limit of series an=1/(2n+1)+1/(2n+3)+-------+1/(4n-1)?

You may factor out `2n` at each denominator, such that:

`a_n = 1/(2n(1 + 1/(2n))) + 1/(2n(1 + 3/(2n))) + ... + 1/(2n(1 + (2n-1)/(2n))) `

You may factor out `1/(2n)` , such that:

`a_n = 1/(2n)*sum_(k=1)^n1/(1+(2k-1)/(2n))`

You need to notice that the summation that you have obtained represents a Riemann sum associated to the function `f:[0,1]->R` , f(x) = 1/(2(1+x)), having the partition` P = {[0,1/n),[1/n,2/n),[2/n,3/n),....,[(n-1)/n,1]}.`

Since the function is continuous, hence , you may evaluate the limit, such that:

`lim_(n->oo)a_n = int_0^1 f(x)dx`

Hence, you need to evaluate the definite integral of the function `f(x) = 1/(2(1+x))` , such that:

`int_0^1 f(x)dx = int_0^1 1/(2(1+x))dx`

`int_0^1 1/(2(1+x))dx = (1/2)*ln|1+x||_0^1`

Using the fundamental theorem of calculus, yields:

`int_0^1 1/(2(1+x))dx = (1/2)*(ln 2 - ln 1)`

`int_0^1 1/(2(1+x))dx = (1/2)*ln 2 = ln sqrt 2`

Hence, evaluating the limit of the given sequence `a_n` , using Riemann sum, yields `lim_(n->oo)a_n = ln sqrt 2.`


text:
# whitespace boxes

Status
Not open for further replies.

#### laroche73

##### New Member
here's a basic question. how do you enclose something in a whitespace box when writing a reply - for example, when quoting something from a previous posting? I think mechie mentioned enclosing things with "code" and "code/" markers, but I'm quite new to HTML, obviously :cry: . An example would help. Thank you
- CAL

#### mechie

##### New Member
Fancy formatting

Hi laroche73,
Here goes - I have used curly brackets {} instead of square ones [] to stop the commands working :?

Code:
When you post a reply the form has some buttons above the typing area
such as {b} to turn bold text on (cancelled with a {/b}) etc.

The list of options includes :-

{code}, very useful for tables of info as the font used is 'monospaced' -
every character is the same size so columns are easy to create

{quote}, anything after this will be inset in a box (similar to code) but
carries the title "Quote" above it.

Every command needs to be cancelled with the same command prefixed
with '/', all wrapped up in square brackets eg {/quote}

If you hover your mouse pointer over these options a quick explanation
of how to use them appears immediately above the typing area.
mechie said:
That's how I do it :wink:
... Note smartass use of {quote="mechie"} to add a name
I also do this 8)
... Plain old {quote} for comparison

Much better example of code ...
http://www.electro-tech-online.com/threads/pcb-transformer.2786/&postdays=0&postorder=asc&start=17 (bizzarely, the very post you were referring to, subsequently edited if you remember)

Ever wanted to indent text ? adding space characters won't work, they get deleted when you submit your post! The {list} option does the indenting for you :wink:

• (Sneaky use of {color=blue} command (note American spelling :roll This text is jibberish just to demonstrate the use of the {list} option, remember I am using curly brackets to replace square brackets in order to stop the command working. I hope this is enough text to cause a few line-wraps so the indenting effect can be seen.
[list:717438449f]
A second level of indent? a second {list} command.
This text is jibberish just to demonstrate the use of the {list} option, remember I am using curly brackets to replace square brackets in order to stop the command working. I hope this is enough text to cause a few line-wraps so the indenting effect can be seen.
[/list:u:717438449f]
Play, try things out; use the 'Preview' button instead of 'Submit' to see what happens. All the commands work similarly so once you get the idea of a few the rest seem easier !

#### laroche73

##### New Member
whitespace, etc.

Thanks, mechie! - CAL

Status
Not open for further replies.


text:
# Vieta's Formula

Consider a quadratic equation with complex coefficients and roots $r_1$, $r_2$ : $a_2(x-r_1)(x-r_2) = a_2x^2+a_1x+a_0.$ By comparing coefficients, we can see that $r_1+r_2=-\frac{a_1}{a_2}, r_1r_2=\frac{a_0}{a_2}.$

Generalizing this idea for a polynomial of degree $n$, we have

Vieta's Formula. Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$ be a polynomial with complex coefficients and degree $n$, having complex roots $r_n, r_{n-1}, \cdots, r_1$. Then for any integer $0 \leq k \leq n$:

$\sum\limits_{1\leq i_1 < i_2 < \cdots < i_k \leq n} r_{i_1}r_{i_2}\cdots r_{i_k} = (-1)^k \frac{a_{n-k}}{a_n}.$

This follows immediately by comparing coefficients on the polynomial $\prod ( x- r_i )$. We leave the proof to the reader.

## Worked Examples

### 1. Find all triples of complex numbers which satisfy the system of equations:

\begin{aligned} a + b + c & = 0,\\ ab + bc + ca & = 0, \\ abc & = 0. \\ \end{aligned}

Solution: Any 3 numbers $(a,b,c)$ can be considered as the roots of the monic cubic polynomial $P(x) = (x-a)(x-b)(x-c)$. Applying Vieta's Formula to the polynomial, we get $P(x)=x^3$. Hence, the only root of $P(x)$ is $x=0$ (with multiplicity 3), which implies that $(0,0,0)$ is the only triple of complex numbers that satisfies the system of equations.

### 2. Let $r_1, r_2$ and $r_3$ be the roots of the polynomial $5x^3 -11x^2+7x+3$. Evaluate $r_1(1+r_2+r_3) + r_2(1+r_3+r_1) + r_3(1+r_1+r_2) .$

Solution: The expression is equal to $r_1+r_2+r_3+2(r_1r_2+r_2r_3+r_3r_1)$. By Vieta's Formula, we know that

$r_1+r_2+r_3=-\frac{-11}{5}=\frac{11}{5}, r_1r_2+r_2r_3+r_3r_1=\frac{7}{5}, r_1r_2r_3 = - \frac{3}{5},$

so $r_1+r_2+r_3+2(r_1r_2+r_2r_3+r_3r_1) = \frac{11}{5}+2\times \frac{7}{5} = \frac{25}{5}=5$.

Note: A common approach would be to try and find each root of the polynomial, especially since we know that one of the roots must be real (why?). However, this is not necessarily a viable option, since it is hard for us to determine what the roots actually are.

### 3. Let $r_1, r_2$ and $r_3$ be the roots of the polynomial $5x^3 -11x^2+7x+3$. Evaluate $r_1^3+r_2^3+r_3^3.$

Solution: In this problem, the application of Vieta's Formulas is not immediately obvious, and the expression has to be transformed. From Factorization 4, we have that

$r_1^3+r_2^3+r_3^3 = 3r_1r_2r_3 + (r_1+r_2+r_3)\left[ r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1) \right].$

Now we only need to know how to calculate $r_1^2+r_2^2+r_3^2$. Again, from factorization, we have that

$r_1^2+r_2^2+r_3^2 = (r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1),$

so this allows us to conclude that

$\begin{array} { l l } r_1^3+r_2^3+r_3^3 & = 3r_1r_2r_3 + (r_1+r_2+r_3) \left[ r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1) \right] \\ & =3r_1r_2r_3 + (r_1+r_2+r_3 ) \left[ (r_1+r_2+r_3)^2-3(r_1r_2+r_2r_3+r_3r_1) \right] \\ & =3\times (-\frac{3}{5}) +(\frac{11}{5}) \left[ \left(\frac{11}{5} \right)^2-3\times \frac{7}{5} \right] \\ & = -\frac{49}{125}.\\ \end{array}$

### 4. [IMO shortlist] Determine all real values of the parameter $a$ for which the equation

$16x^4-ax^3+(2a+17)x^2-ax+16 = 0$ has exactly four distinct real roots that form a geometric progression.

Solution: Suppose that $a$ satisfies the requirements of the problem and that $x$, $qx$, $q^2x$, $q^3x$ are the roots of the given equation. Then $x \neq 0$ and we may assume that $|q|>1$, so that $|x|<|qx|<|q^2x|<|q^3x|$. Notice that the coefficients are symmetric, namely the first coefficient is the same as the fifth one, the second is the same as the fourth and the third is the same as the third. It guarantees us that if, $\alpha$ is a root, then its reciprocal (which is $\alpha ^{-1} = \frac{1}{\alpha}$) will also be a root. Hence, $\frac{1}{x} = q^3 x,$ so $q = x^{-\frac{2}{3} }$ and the roots are $x, x^{\frac{1}{3}}, x^{\frac{-1}{3}}, x^{-1}$.

Now, by Vieta's formula we have $x+x^{\frac{1}{3}}+x^{\frac{-1}{3}}+x^{-1}=\frac{a}{16}$ and $x^{\frac{4}{3}}+x^{\frac{2}{3}}+1 + 1+x^{\frac{-2}{3}}+x^{\frac{-4}{3}}=\frac{2a+17}{16}$. On setting $z =x^{\frac{1}{3}}+x^{\frac{-1}{3}}$ these equations become

$z^3-2z=\frac{a}{16}, (z^2-2)^2+z^2-2=\frac{2a+17}{16}.$

Substituting $a=16(z^3-z^2)$ in the second equation leads to $z^4-2z^3-3z^2+4z+\frac{15}{16} = 0$. We observe that this polynomial factors as $(z+\frac{3}{2})(z-\frac{5}{2})(z^2-z-\frac{1}{4})$. Since $|z|=|x^{\frac{1}{3}}+x^{\frac{-1}{3}}|\geq 2$, the only viable value is $z=\frac{5}{2}$. Consequently $a=170$

Rearranging the equation, we get $16 (x^2 + \frac{1}{x^2}) - 170(x + \frac{1}{x}) + 357 = 0$. To simplify it, we can call $y = x + \frac{1}{x}$ and therefore $y^2 = x^2 + 2 + \frac{1}{x^2}$, thus getting a new form $16 (y^2 - 2) - 170y + 357 = 0 \Rightarrow 16y^2 - 170y +325 = 0$, whose roots are $\frac{5}{2}$ and $\frac{65}{8}$. We have to plug the two back into $y = x + \frac{1}{x}$, leading us to two more quadratics, getting finally $\boxed{\dfrac{1}{8}, \dfrac{1}{2}, 2, 8}$.

## Test Yourself

1. If $\alpha, \beta$ are the roots to the equation $x^2 +2x+3 = 0$, what is the quadratic equation whose roots are $(\alpha - \frac{1}{\alpha} )^2$ and $( \beta - \frac{1}{\beta} ) ^ 2$?

2. ARML 2012, Team Problems, #6 The zeros of $f(x) = x^6 + 2x^5 + 3x^4 + 5x^3 + 8x^2 + 13x + 21$ are distinct complex numbers. Compute the average value of $A + BC + DEF$ over all possible permutations $(A, B, C, D, E, F)$ of these six numbers.

3. (Created by <NAME>. This is HARD) Show that $\sum_{j=1}^n \cot^2\Big(\tfrac{j \pi}{2n+1}\Big) \; = \; \tfrac13n(2n-1)$ for any integer $n \ge 1$. What is $\sum_{j=1}^n \cot^4\Big(\tfrac{j \pi}{2n+1}\Big) \qquad \mbox{?}$

Note by <NAME>
6 years, 6 months ago

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The solution to example two states r1r2r2=-3/5, not r1r2r3=-3/5.

- 5 years, 11 months ago

Could u please explain this to me in simple language as in just how do u apply it to quadratic equations and why is it even related to complex numbers? Thanks very much.

- 5 years, 11 months ago


text:
Leave one out cross validation for LSTM

I have a multivariate time series data set having 6 variables. I have to predict the sixth variable at the next time step given the expected values of other five at the next time step. I am using LSTM for prediction. The data set is small (170*6), have 170 observations. Splitting the data set into three i.e. train/test/split doesn't seem a probable option in this case.

As far as I know, input sequences (of fixed length) over which the LSTM learns are assumed independent. So my question was, can one sequence be left out for validation just like in the case of a cross sectional data problem.

Leave one out cross validation for LSTM, or any other time-series model, doesn't really make much sense because it would introduce missing values in the series and leaking information from the future.

Time series models learn from historical values, to predict future. In leave one out cross-validation, you remove observations from the series, including the past ones, what leads to missing data. What follows, sch cross-validation strategy would need the model to be able to deal somehow with missing data, or would need some external strategy of dealing with them and basically would test how well the model deals with missing data. Simple cross-validation can be used only if the samples can be assumed to be independent, this is never the case for time series (unless you assume it to be random noise).

With time series data, as described by Hyndman and Athanasopoulos (2018), we perform one step ahead cross validation, where we use $$1,\dots,k$$ samples, to predict $$k+1$$ value (or alternatively $$k+1,\dots, k+m$$ values), where we use different $$k$$ values, as illustrated on image below (also by Hyndman and Athanasopoulos, 2018).

Such strategy tests what you are trying to test, i.e. how well does your model predict future from the past. Moreover, it also tests how well does the model deal with changing data, a scenario that is very common in such cases, since you usually don't make single prediction using static training sample, but want a model that would be re-trained with new data and would adapt.

Commenting on LSTM networks

Input sequences (of fixed length) over which the LSTM learns are assumed independent. So my question was, can one sequence be left out for validation just like in the case of a cross sectional data problem.

let me comment on this. The above answer applies to LSTM network as well. LSTM network learns the time-dependence in your data by using the sliding LSTM window, so the sequences are not assumed as independent, as the idea of "long term memory" is the dependence assumption. Usually the sequences are generated as in this Keras code example (see also this tutorial for code with description):

sentences = []
next_chars = []
for i in range(0, len(text) - maxlen, step):
sentences.append(text[i: i + maxlen])
next_chars.append(text[i + maxlen])

As you can see, the loop is used to iterate through your data and create a sliding window, that makes jumps of step size, where step is smaller then maxlen. So the sequences are dependent. Even if you used step equal to maxlen, still you would be assuming time-dependence by choice of the model.

What follows, dropping one of the sequences leads to introducing missing data. Using the simple cross-validation would introduce bias, since you would be measuring how does your model work in missing data scenario, rather then how well does it make predictions for the future. For time-series, you should rather use the one step ahead cross-validation.

• thanks! I appreciate the technique you mentioned in the answer and I have edited the question to reflect my understanding of LSTM for time series. Please correct me if I am wrong. – naive Jun 17 '18 at 8:14


text:
# Polarization Insensitive, Wide-Angle, Ultra-wideband, Flexible, Resistively Loaded, Electromagnetic Metamaterial Absorber using Conventional Inkjet-Printing Technology

## Abstract

A novel, polarization insensitive, wide-angle and broadband electromagnetic metamaterial absorber, which can cover either a flat or a bent geometry, is presented in this work. The periodic geometry has a unit-cell, which consists of four split ring resonators, which are sequentially rotated around the unit-cell axis progressively by 90 deg. First, the metallic parts of the geometry consists of low resistivity copper traces. Next, in order to increase the frequency bandwidth, resistively loaded traces, printed by conventional inkjet printer are used to replace the copper ones. For normal incidence, simulated and measurement results shown that the proposed flat absorber exhibits absorption efficiency higher than 0.8 for 6.9–29.9 GHz (i.e., bandwidth of 125%) regardless of polarization, while the curved absorber for 6.6–29 GHz (i.e., bandwidth of 126%) or for 10.5–29.6 GHz (i.e., bandwidth of 95%), depending the polarization. For oblique incidence and for TE or TM polarized incident wave it presents bandwidth of 118% (7.7–29.9 GHz) or 100.5% (9.9–29.9 GHz), respectively, for an incident angular range of 0–45 deg. Finally, the proposed absorber has thickness of 3.89 mm, corresponding to λ/11.7 at its lowest operation frequency.

## Introduction

Metamaterials are inherently inhomogeneous structures that usually consist of periodically repeated unit cells. Their properties mainly arise from their geometric details rather than their constituting material properties. A particular type of metamaterial, the metamaterial perfect absorber, has attracted significant attention due to its ability to offer near unity absorption of electromagnetic waves. The first perfect metamaterial absorber was presented by N. <NAME> et al. in 20081. Since then, the scientific effort was focused on the design of polarization insensitive and wide-angle2,3,4,5,6,7,8, tunable9,10,11, flexible12, multi-band13,14,15,16,17,18,19,20,21, and wideband electromagnetic absorbers by using 3-dimensional (3D) structures22,23,24, copper traces and chip resistors25,26,27,28,29,30, or resistively loaded patters31,32,33,34,35,36,37 This work is focused on the latter group since usually this group of absorbers exhibit superior performance in terms of bandwidth: it will be shown that, the absorption mechanism in these structures is because of the losses in the resistive parts rather in the dielectrics and that, in turn, leads to wider absorption frequency bandwidth.

Specifically, in23 the presented electromagnetic absorber, which consists of a conductive plastic disk, was fabricated through 3D-printing technology and lies between two substrates (one of the later is grounded), presents simulated absorbance for normal incidence higher than 0.9 for 16.3–54.3 GHz, while measured results reveals operation from 23.3 GHz. Similarly, in24 the authors fabricated via 3D-printing technology and tested via measurements a 3D-resistive absorber, which is much like a honeycomb of resistive walls. The geometry presents absorbance bandwidth of 148.7% and is robust for oblique incidence of a plane wave with transverse magnetic (TM) polarization. However, both the above electromagnetic absorbers23,24 are relatively thick (around ~λmin/6.1 at their lowest operation frequency) and demanding in terms of fabrication process.

In28 the proposed geometry consists of chip resistors, soldered onto a copper pattern, which lies in between two substrates (one is grounded). For normal incidence the absorber exhibits a fractional bandwidth (FBW) of 100%, while for oblique incidence it presents FBW from 77.3% to 90.5% up to 60 deg, depending the polarization of the incident plane wave. The absorber in29 consists of two patterned layers (one is scaled to the other) and chip resistors, as well, but exhibits wider FBW (112%) than28 for normal incidence: in general, the use of multiple pattern layers leads to wider absorbance, since each of these, resonates at a different frequency, and thus, the final absorbance consists of multiple maxima, which produce a wide bandwidth38. A similar technique is used by the authors in30: now the scaled pattern and the original one lie on the same layer, resulting FBW = 126.8%.

In order to reduce the fabrication effort but to keep the FBW as wide as it is possible, the chip resistors, which need soldering during the fabrication process, are replaced by resistively loaded patterns: now the captured RF energy dissipates in the latter and not in the chip resistors, as it was previously. In35 a restively loaded electromagnetic absorber was presented: the structure is thin (thickness of only λmin/12.3) and for normal incidence has FBW = 102%. However, the absorber is weak in terms of absorbance bandwidth for oblique incidence since its FBW is reduced to 54% and 50%, for TE- and TM-polarization, up to 45 deg angle of incidence. In36, three scaled resistive patterns are used in a multilayer geometry, which remains relatively thin (λmin/11) and the FBW is increased to 132.9% for normal incidence, as expected.

The contribution of this work is the presentation of a systematic, versatile and low complexity method for the design of a type of polarization insensitive and wide-angle electromagnetic metamaterial absorber, wherein the basic idea is the form of super-cells with 90 rotational degree symmetry. In this work the proposed method was applied to a squared split ring resonator (SSRR) structure, but it could be applied to any resonator. The resulting absorber is shown to be an electrically thin (thickness of λmin/11.7), ultra-wideband, polarization insensitive having wide-angle absorption. The structure is easily fabricated and low-cost, since it was printed using a conventional inkjet printer deploying conductive ink, consisting of silver nano-particles diluted in aqua39. Additionally, the proposed absorber is flexible: it managed to reduce the radar cross section (RCS) of a metallic pole of radius about λmin/7 by an average of 10 dB for a frequency zone of about 30 GHz, depending the polarization of the incident field. Based on the literature, the proposed structure is the first flexible, ultra-wideband, polarization insensitive, wide-angle, easily fabricated through conventional inkjet printing techniques, electrically thin electromagnetic absorber (Table 1).

## Results and Discussion

This section presents performance in terms of absorptivity of two types metamaterial absorbers: first (from now on called copper-absorber), the metallic parts of the periodic geometry are made of copper, while second (from now on, called printed-absorber), are inkjet-printed resistively loaded traces. In the both cases, the unit-cell of the periodic geometry consists of four SSRRs, which are co-planar and are sequentially rotated by an angle of 90 deg. around the central, unit-cell axis.

Because of their rotational symmetry, both absorbers present polarization insensitivity, while because the unit-cell’s dimension, which is electrically small (λ/11.3 at resonance frequency and λmin/6.9 for the lower absorption frequency for the copper- and printed-absorber, respectively), they are wide-angle. The main difference between the two absorber types is the absorption frequency bandwidth: the copper-absorbers are narrow-band, with fractional bandwidth (FBW) of 80% absorbance on the order of 2%, while the resistively loaded absorbers present FBW of the order of 100%31,32,33,34,35,36,37 for 80% absorbance. The reason for this is that, in the copper-absorber the captured energy mainly dissipates in the substrate rather than in the copper traces, while, in the printed-absorber, energy mainly dissipates in the resistively loaded traces, which removing the resonant nature of the SSRRs by re-distributing their currents. Based on the latter, the absorption mechanism changes, resulting in wideband operation, as will be explained.

### Copper Absorber

The geometry of a typical SSRR is presented in Fig. 1a: the unit-cell has dimension of 3 mm with copper ring of conductivity 5.8 × 107 S/m, size, width and gap of 2.8, 0.32 and 0.42 mm, respectively, lying on an FR-4 grounded substrate with thickness of 0.8 mm, dielectric permittivity of εr = 4 and losses of tanδ = 0.02. A periodic geometry is formed by applying periodic boundary conditions around the unit-cell and the reflection coefficient for normal incidence and for both polarizations transverse electric (TE) and magnetic (TM) behaviour is estimated through electromagnetic simulation using the commercial solver COMSOL Multiphysics40: the electric field component is also depicted in the Fig. 1a. For an absorber periodic geometry, absorbance is given by,

$$A=1-{|{\rm{\Gamma }}|}^{2}-{|{\rm{T}}|}^{2},$$
(1)

where, Γ and T is the reflection and transmission coefficient, respectively, while since the geometry is grounded, thus T → 0, is

$$A=1-{|{\rm{\Gamma }}|}^{2}.$$
(2)

Figure 1b depicts A for both polarization cases at normal incidence of the single SSRR (Fig. 1a): it is evident that only for the TM case does the geometry act as an absorber, i.e., it presents A higher than 80%, at 9.4 GHz. Next a super-cell is formed by using four of the basic unit-cells, which are sequentially rotated by 90 deg. and the final geometry is depicted in Fig. 1c). A is simulated for normal incidence and for both TE, TM polarization, and the results are depicted in Fig. 1d: A is identical for the both TE-, TM-cases, as expected because of the unit-cell’s symmetry, presenting maximum of 98.5% at 8.84 GHz and has FBW = 2.1%. The super-cell’s size is λ/5.65 at the resonance frequency, and thus, it remains electrically compact.

In order to investigate the absorbance mechanism, the surface current distribution was simulated, for both the unit and the super-cell at the resonance frequency. Figure 1e,f depicts the results for TE and TM polarization, respectively, for normal incidence and for the unit-cell (Fig. 1a) at 9.4 GHz: it is observed that for the TE-case reverse currents (red lines) are induced, resulting in no absorption, while for the TM-case the current forms a loop, which runs along the whole SSRR periphery and results in absorption. Moreover, for the TE-case, the maximum current magnitude (≈450 A/m) is much smaller than the corresponding current at the TM-case (≈7 × 103 A/m). Similarly, Fig. 1g,h depict the surface current distribution, but now for the super-cell (Fig. 1c) at 8.84 GHz. For the TE-case, (Fig. 1g), the dominant SSRRs in terms of interaction with the incident wave is the upper-left and the bottom-right, as expected, while for the TM-case (Fig. 1h) the top-right and bottom-left are active. Hence, at each polarization (TE or TM), the super-cell absorbs energy since two of its four sub unit-cells resonate, and thus, the super-cell presents polarization insensitivity: the latter is evident in Fig. 2a, which depicts the simulated absorbance for normal incidence as the electric (or magnetic, as well) field component of the incident plane wave steers from 0 to 45 deg. (angles greater than 45 deg. were not tested since the super-cell presents 45 deg. symmetry). Thus, for normal incidence, and based on the surface current distribution, the absorption mechanism is due to the interaction of the electric field of the plane wave with the SSRRs. Specifically, the latter are coupled together and that causes strong resonance driven surface current loops, which in turns produces ohmic losses because of the finite conductivity of the copper and due to their interaction with the lossy substrate (it will be explained later how the energy dissipates). For this case, there is no normal magnetic field component on the SSRRs, hence, surface currents are not induced because of the latter, and thus, magnetic field does not contribute to the absorption mechanism.

Next, polarisation insensitivity was tested for oblique incidence. Figure 2b depicts the results for TE polarisation. As angle of incidence increases from 0 to 60 deg., absorbance remains higher than 96% at 8.84 GHz. However, a second maxima occurs at around 11.5 GHz. For this case, there is a magnetic field component normal to the SSRRs, and thus, surface loop currents are induced, which in turn, contribute to the absorption mechanism, as explained before. The second maxima observed in the absorbance is because of these currents: as the angle of incidence θ increases, the normal magnetic field component magnitude becomes higher, which in turn increases the induced surface loop currents, which lead to higher losses and thus, to higher absorbance. For θ > 45 deg., the latter phenomenon saturates, and there is no further increment for the absorbance’s second maxima. On the other hand, the first maxima at 8.84 GHz remains almost the same as θ increases, or equivalently as the normal magnetic field component increases, This is a result of the coupling between the tangential electric field of the plane wave and the SSRRs, which remains constant as θ increases.

Figure 2c again depicts the absorbance for oblique incidence but now for TM polarisation. It is clear that A always remains higher than 98.5%, as it was for the normal incidence case, but now there is a slight shift of the maxima to higher frequencies. The latter results from the tangential to the SSRRs electric field component of the incident plane wave is changing as θ increases: now, as the incident plane wave impinges on the SSRRs the apparent equivalent SSRR’s electrical size changes from the perspective of the incident signal: the inducted currents by the tangential electric field form loops with smaller equivalent electrical length, and thus, the resonance frequency increases.

Figure 2 also depicts the corresponding measurements for normal and oblique incidence, for both polarizations: here good agreement is observed. Also depicted is the fabricated (though chemical etching process) and measured geometry as inset. The measurement procedure and set-up will be explained in the Methods section. It is noted that time domain gating and smoothing techniques (i.e., moving average filter)41 were applied over the measurement’s data.

### Salisbury Screen Absorber

A typical resistively loaded absorber is the Salisbury screen (SC) geometry, which is used for reflection reduction of incident radio waves to a surface. Proposed by Salisbury42 in the early of 1950s it usually consists of three layers: a screen, which corresponds to a specific equivalent electric surface resistance ZS, a loss-less dielectric substrate of thickness λ/4, where λ is the wavelength of the absorbed radio wave, and a ground plane: the wave initially impinges on the surface and part of it is reflected, while the other part propagates into the substrate until it is reflected from the ground plane. At the ground plane, the first and the second reflected parts are out of phase (i.e., phase difference of 180 deg.), resulting in reflection suppression.

The Salisbury screen geometry and its equivalent circuit schematic are depicted in Fig. 3, where ηa, ka = 2π/λ and la is the characteristic impedance, wavenumber and thickness of the substrate, respectively. Based on transmission line theory (TLT) and since the transmission line is terminated in a short-circuit through a substrate of λ/4–thickness (i.e., la = λ/4) is

$${Z}_{{\rm{L}}}=j{\eta }_{{\rm{a}}}\,\tan ({k}_{{\rm{a}}}{l}_{{\rm{a}}})\to \infty ,$$
(3)

thus, the input impedance of the geometry is given by,

$${Z}_{{\rm{in}}}=\frac{{Z}_{{\rm{S}}}{Z}_{{\rm{L}}}}{{Z}_{{\rm{S}}}+{Z}_{{\rm{L}}}}\to {Z}_{{\rm{S}}}.$$
(4)

For perfect absorption, the structure should be impedance matched to the vacuum, and hence,

$${Z}_{{\rm{in}}}={\eta }_{0}^{\ast }={\eta }_{0}$$
(5)

where $${\eta }_{0}=\sqrt{{\mu }_{0}/{\varepsilon }_{0}}\approx 120\pi$$ $${\rm{\Omega }}$$ is the wave impedance of the vacuum and {*} denotes conjugated match. Based on (4), (5)

$${Z}_{{\rm{S}}}\to {\eta }_{0}.$$
(6)

Thus, in a typical Salisbury absorber the screen is represented by a real resistance, i.e., Zs = R, i.e. pure resistance.

Figure 3c shows the estimated frequency response of a Salisbury screen. The latter was designed to absorb at f0 frequency and operates (i.e., A ≥ 0.8) from 0.5f0 to 1.5f0, and thus, it has bandwidth of 100%. The above analysis has revealed that, by using resistively loaded surfaces it is possible to design a wideband absorber, which has thickness of at least λ/4. For thinner geometries with ever wider bandwidth, the above analysis leads Zs to be a complex impedance and its circuit equivalent is an RLC circuit43,44, as will be explained.

### Printed Absorber

In order to design a polarization insensitive, wide angle and wideband absorber, the rotated structure presented in the previous section, but now with resistively loaded traces has been used. The proposed geometry is depicted in Fig. 4. The super-cell (unit-cell) has now size of 12.59 mm (6.3 mm) with resistively loaded rings of 27 Ω/sq, size, width and gap of 5.84, 0.68 and 0.88 mm, respectively, while they lie on a substrate with a permittivity of εr = 2.5, losses of tanδ = 0.0009 and thickness of h1 = 0.14 mm. The latter is mounted on a grounded foam-substrate, with εr = 1.05, zero losses (tanδ = 0) and thickness of h2 = 3.75 mm. Thus, the absorber has thickness of h1 + h2 = 3.89 mm. The geometry was studied through full electromagnetic analysis is terms of absorbance. Specifically, periodic boundary conditions were applied again around the super-cell and the reflection coefficient of plane wave for normal incidence for both polarizations, TE and TM, was estimated using the COMSOL Multiphysics solver. Figure 4b depicts the results: absorber operates (A ≥ 0.8) for 6.9–29.9 GHz, i.e., has FBW = 125% for both polarizations. For the lower frequency, the super-cell has thickness of λmin/11.17, and hence, is much thinner than the Salisbury screen, while presenting wider bandwidth, as well. Additionally, two maxima of absorbance are observed, i.e., 0.985 and 0.999 at 9.7 and 24.7 GHz, respectively.

Next the absorber was fabricated and measured. The resistively loaded SSRRs were printed through the conventional inkjet printer Epson Stylus C88 + on a coated polyethylene terephthalate (PET) sheet (εr = 2.5, tanδ = 0.0009 of thickness 0.14 mm). Between the ground plane (i.e., metallic plate of copper) and the PET-sheet the foam-substrate (εr = 1.05) of thickness 3.75 mm was placed. In order to achieve resistively loaded SSRRs, the electrically conductive ink MetalonTM JS-B25P45 was used. The latter was diluted with water as the solvent, stored in an empty inkjet cartridge and used by the printer. The surface impedance in this work was fixed at Rs = 27 Ω/sq and it was adjusted by two ways: first by forming specific volume ratio of the conductive ink to water (i.e., 1:5) and second, by adjusting the grey scale (through the RGB-code) on the printer (i.e., RGB-code (29,29,29)), while the whole fabrication procedure will be explained in the Methods section and it is also well explained in39. The fabricated geometry is depicted in Fig. 4c, where the super-cell is presented, as well, enlarged. Figure 4d depicts the measured versus simulated results for the normal incidence: here good agreement is observed. It is noted that the printed absorber was tested for normal incidence only from 1 GHz to 26 GHz due to facilities limitations in our laboratory. Again, time domain gating and smoothing techniques (i.e., moving average filter)41 were applied over the measurement’s data.

It is known that the captured by the absorber power dissipates either on its dielectric or metallic parts. Specifically, for the dielectric parts the power losses can be obtained through46

$${P}_{d}=\pi \varepsilon f\,\tan \,\delta {{\iiint }_{{V}_{d}}|E|}^{2}{\rm{d}}V$$
(7)

where Vd, ε and tanδ is the substrate’s volume, permittivity and tangential losses, respectively, E the electric field and f the frequency of the incident plane wave, while for the metallic parts46,

$${P}_{m}=\frac{1}{2}\sqrt{\frac{\pi \mu f}{\sigma }}{{\iint }_{{S}_{m}}|H|}^{2}{\rm{d}}S$$
(8)

where, Sm, μ and σ is the conductor’s surface, permeability and conductivity, respectively and H the magnetic field of the incident plane wave.

Figure 5 presents in which medium (i.e., metallic (metal) or dielectric (diel)) the captured by the absorber energy dissipates. Specifically, it depicts the ratio to percentage of either the Pd or Pm to the total captured power for normal incidence for the copper (Fig. 5a) and the resistively loaded (Fig. 5b) absorber. Based on Fig. 5a it is evident that for the copper absorber the captured energy mainly dissipates in the dielectric parts (above 80%) rather than in the metallic parts (less than 20%). The situation is reversed for the resistively loaded absorber (Fig. 5b): now the losses in the metallic parts dominate (90% to 100%), while the losses in the dielectrics are limited to 0.06%.

The surface current distribution for the resistively loaded absorber for normal incidence and for both polarisations, TE or TE is depicted in Fig. 6. Again, the tangential electrical field of the incident plane wave is coupled with the SSRRs and forms surface currents: in the SSRRs, which are symmetrical to the electric component, reverse currents are observed, while in the other SSRRs the currents form loops. However, the mechanism of absorption has now been changed, compared to the copper absorber (Fig. 1g,h): the resistive traces now act as distributed resistors, which transform the captured energy to ohmic losses, and hence, absorbance occurs. All the SSRRs contribute to this phenomenon, but with different weights. Specifically, for the TE-case (Fig. 6a) the arms of the upper-left and down-right SSRR act as distributed resistors, where the captured energy is transformed to ohmic losses, while, the upper-right and the bottom-left SSRRs also contribute to the absorption mechanism as distributed resistors, but with lower impact, since the magnitude of the current in those loops is less. For the TM-case (Fig. 6b) the phenomenon is similar, as expected due to 90-deg. rotational symmetry. The maximum magnitude of the surface currents (≈40 A/m) is now much lower than the copper absorber (≈2.5 × 103 A/m, Fig. 1g,h). The latter occurs due to the fact that the SSRRs now act as distributed resistors and not as resonating elements (copper absorber): the captured energy now is transformed to ohmic losses in the resistively loaded traces and not because of vicinity of strong surface currents to the lossy substrate (copper absorber).

Next, the proposed geometry was modelled in terms of absorbance using a transmission line equivalent circuit model. The equivalent circuit schematic for the absorber is depicted in Fig. 7. Based again on the TLT and since the transmission line ends to a short-circuit (grounded absorber) is

$${Z}_{{\rm{L}}}=j{\eta }_{{\rm{b}}}\,\tan ({k}_{{\rm{b}}}{l}_{{\rm{b}}}),$$
(9)

where, ηb, kb and lb is the characteristic impedance, the wavenumber and the thickness of the foam-substrate. Then,

$${Z^{\prime} }_{{\rm{in}}}={\eta }_{{\rm{a}}}\frac{{Z}_{{\rm{L}}}+j{\eta }_{{\rm{a}}}\,\tan ({k}_{{\rm{a}}}{l}_{{\rm{a}}})}{{\eta }_{{\rm{a}}}+j{Z}_{{\rm{L}}}\,\tan ({k}_{{\rm{a}}}{l}_{{\rm{a}}})},$$
(10)

where similarly, ηa, ka and la is the characteristic impedance, the wavenumber and the thickness of the PET-substrate. Hence, the input impedance of the absorber is given by,

$${Z}_{{\rm{in}}}=\frac{{Z}_{{\rm{S}}}{Z^{\prime} }_{{\rm{in}}}}{{Z}_{{\rm{S}}}+{Z^{\prime} }_{{\rm{in}}}},$$
(11)

where ZS is the equivalent impedance of the resistively loaded pattern (i.e., SSRR loops), and thus, the reflection coefficient is given by,

$${\rm{\Gamma }}=\frac{{Z}_{{\rm{in}}}-{\eta }_{0}}{{Z}_{{\rm{in}}}+{\eta }_{0}},$$
(12)

and finally the absorbance is estimated through (2) as a function of ZS. It is noted that in an isotropic medium the characteristic impedance is given by,

$$\eta ={\eta }_{0}\sqrt{\frac{{\dot{\mu }}_{r}}{{\dot{\varepsilon }}_{r}}},$$
(13)

where, $${\dot{\varepsilon }}_{r}={\varepsilon }_{r}(1-j\,\tan \,{\delta }_{e})$$, $${\dot{\mu }}_{r}={\mu }_{r}(1-j\,\tan \,{\delta }_{m})$$ is the complex relative permittivity and permeability of a material with electrical ($$\tan \,{\delta }_{e}$$) and magnetic ($$\tan \,{\delta }_{m}$$) losses: e.g., in our case for the PET and the foam absorber is $${\dot{\varepsilon }}_{r}=2.5(1-j0.0009)$$ and $${\dot{\varepsilon }}_{r}=1.05$$, respectively, while for all cases $${\dot{\mu }}_{r}=1$$. Additionally, the wavenumber is given by,

$$k=\frac{2\pi }{{\lambda }_{0}}\sqrt{{\dot{\varepsilon }}_{r}{\dot{\mu }}_{r}},$$
(14)

where λ0 is the wavelength in the free space.

For the resistively loaded absorber it is assumed that ZS is a series RLC circuit, and not a resistance as in the Salisbury screen, and thus, is,

$${Z}_{{\rm{S}}}=R+j\omega L+\frac{1}{j\omega C},$$
(15)

where, R, L, C is the resistance, inductance and capacitance of the impedance and ω is the angular frequency. Based on curve fitting optimization analysis and on the absorbance, which has been estimated though full electromagnetic simulation (Fig. 4b) is R = 240.9315 Ω, L = 1.7637 nH and C = 66.9497 fF. Figure 7b depicts the resistively loaded absorbance versus frequency as estimated through full electromagnetic (EM) and TLT analysis: good agreement is observed, and thus the assumption that ZS could be considered as a series RLC circuit appears correct.

Figure 8a depicts the simulated resistively loaded structure’s absorbance for normal incidence as the electric (or magnetic, as well) field component of the incident plane wave steers from 0 to 45 deg. (again, angles greater than 45 deg. were not tested since the super-cell presents 45 deg. symmetry): it is evident that the absorber is polarisation insensitive for normal incidence. The latter occurs because of the 90 deg. super-cell symmetry.

Next, the polarization insensitivity was simulated, but now for oblique incidence. The TE-case is depicted in Fig. 8b. As angle of incidence θ varies from 0 to 45 deg., the absorbance remains higher than 0.8 from 7.7 to 29.9 GHz, resulting FBW = 118%, but is changing shape: a normal magnetic field component to the SSRRs arises as θ increases, which induces surface currents, which in turn contribute to the absorption mechanism constructively or destructively: the A decrement is because tangential electric field currents and the normal magnetic field currents operate destructively, and hence, the latter is most noticeable for higher θ. However, the major contribution to the absorbance is because of the tangential electric field, which remains constant as θ increasing. Thus, the A is changing shape but is not significantly shifting in frequency. Additionally, due to the wideband operation, there are no distinct resonance defined maxima, as there were for the copper-absorber.

Figure 8c presents the absorbance for oblique incidence and for TM polarization. As angle θ increases from 0 to 45 deg., A remains higher than 0.8 from 9.9 to 29.9 GHz, resulting FBW = 100.5%. Now, the tangential electric component is decreasing as the angle of incidence increases. Thus, this induces through the electric field surface currents form equivalently smaller electrical traces and loops, and hence, a shift of the bandwidth to higher frequencies is observed.

### Bent Printed Absorber

The resistively loaded absorber was curved and its absorbance was tested. In particular, a metallic circular cylinder with infinite height was covered by the absorber (Fig. 9a) and the monostatic radar cross section (RCS) versus frequency of the latter geometry (i.e., σa) was estimated through full electromagnetic analysis when the incident electrical $${{\bf{E}}}_{\parallel }^{i}$$ or magnetic $${{\bf{H}}}_{\parallel }^{i}$$ field is in parallel to the cylinder: 10 super-cells covered its periphery, and thus, the latter has radius of R = 5l/π, where l = 12.59 mm is the super-cell’s size, as mentioned before. The analytical and simulated (numerical) monostatic RCS of a metallic cylinder (i.e., σm) with exactly the same radius was estimated, also, for comparison. Based on the literature46, the two dimensional RCS (i.e., σ2D) of a metallic circular cylinder of radius R and of infinite height, when a plane wave of $${{\bf{E}}}_{\parallel }^{i}$$-polarisation normally impinges it, is given by46

$${\sigma }^{2{\rm{D}},{{\bf{E}}}_{\parallel }^{i}}=\mathop{\mathrm{lim}}\limits_{\rho \to \infty }[2\pi \rho \frac{{|{{\bf{E}}}_{\parallel }^{s}|}^{2}}{{|{{\bf{E}}}_{\parallel }^{i}|}^{2}}]=\frac{4}{k}{|\mathop{\sum }\limits_{n=-\infty }^{+\infty }\frac{{J}_{n}(kR)}{{H}_{n}^{(2)}(kR)}{e}^{jn\varphi }|}^{2},$$
(16)

where Jn is Bessel function of the first kind, $${H}_{n}^{(2)}$$ is Bessel function of third kind (Hankel function), k the wavenumber, R the radius of the cylinder, ϕ is the azimuthal angle and {i}, {s}, denotes the incident plane wave and the scattered by the cylinder electric field. In our case, where a ring of hight l is periodically repeated (Fig. 9a), the three dimensional RCS is given by46,

$${\sigma }^{3{\rm{D}},{{\bf{E}}}_{\parallel }^{{\rm{i}}}}\underline{\approx }\frac{4{l}^{2}}{\pi }{|\mathop{\sum }\limits_{n=-\infty }^{+\infty }\frac{{J}_{n}(kR)}{{H}_{n}^{(2)}(kR)}{e}^{jn\varphi }|}^{2}=\frac{k{l}^{2}}{\pi }{\sigma }^{2{\rm{D}},{{\bf{E}}}_{\parallel }^{i}}.$$
(17)

Similarly, for the $${{\bf{H}}}_{\parallel }^{i}$$-polarisation is46

$${\sigma }^{2{\rm{D}},{{\bf{H}}}_{\parallel }^{i}}=\mathop{\mathrm{lim}}\limits_{\rho \to \infty }[2\pi \rho \frac{{|{{\bf{E}}}_{\parallel }^{s}|}^{2}}{{|{{\bf{E}}}_{\parallel }^{i}|}^{2}}]=\frac{4}{k}{|\mathop{\sum }\limits_{n=-\infty }^{+\infty }\frac{{J^{\prime} }_{n}(kR)}{{H}_{n}^{(2)^{\prime} }(kR)}{e}^{jn\varphi }|}^{2}$$
(18)

and46

$${\sigma }^{3{\rm{D}},{{\bf{H}}}_{\parallel }^{i}}\underline{\approx }\frac{4{l}^{2}}{\pi }{|\mathop{\sum }\limits_{n=-\infty }^{+\infty }\frac{{J^{\prime} }_{n}(kR)}{{H}_{n}^{(2)}(kR)}{e}^{jn\varphi }|}^{2}=\frac{k{l}^{2}}{\pi }{\sigma }^{2{\rm{D}},{{\bf{H}}}_{\parallel }^{i}},$$
(19)

where {′} denotes the partial derivative with respect to the entire argument of the Bessel functions The results for the monostatic RCS are depicted in Fig. 9b,c. There are three distinct regions: first when σa = σm, where the absorber is outside the operating frequency range, the plane wave penetrates it and finally is reflected by the absorber’s ground (i.e., metallic circular cylinder of radius R − h1 − h2), second when σa < σm, where the absorber is inside the operating frequency range and captures the incidence RF power, and third when σa > σa, where the SSRRs become resonant and act as radiating elements in an array configuration which steer the scattered power back, as will explained. For $${{\bf{E}}}_{\parallel }^{i}$$-polarization, the cylinder absorbs power from 2.5 GHz to 28.7 GHz while for $${{\bf{H}}}_{\parallel }^{i}$$-polarization from 2.9 GHz to 34.1 GHz. In order to shed further light on the operation mechanism, the power flow (time average) of the scattered field (vectors and magnitude in dB) is also depicted for the mentioned three regions (at 2, 10 and 36 GHz) and for both polarizations ($${{\bf{E}}}_{\parallel }^{i}$$ and $${{\bf{H}}}_{\parallel }^{i}$$-polarization). At 2 GHz (Fig. 9d,g), the plane wave propagates inside the absorber until it is uniformly reflected by geometry’s ground. At 10 GHz (Fig. 9e,h), it is observed that the scattered energy mainly dissipates in the SSRRs, it is not reflected back, and thus, it is absorbed. At 36 GHz (Fig. 9f,i), it is observed that the incident wave is not scattered by the ground, as at 2 GHz, but mainly via specific SSRRs, which reflect the reflected energy to the exactly position of the incident wave. Figure 9j–l and Fig. 9m–o depicts the surface current distribution in the SSRRs for $${{\bf{E}}}_{\parallel }^{i}$$ and $${{\bf{H}}}_{\parallel }^{i}$$-polarization, respectively, at 2, 10 and 36 GHz. At 2 GHz, it is evident that the surface current distribution in the loops is very low, as expected. At 10 GHz, the SSRRs interact with the incident wave, and strong surface currents appear in all the loops, as has been explained previously, and the absorption mechanism is activated. At 36 GHz the incident wave forms again surface currents in the loops, mainly at the side of the cylinder, the mechanism changes and the geometry is not working as absorber. Specifically, the strong surface currents in the SSRRs are in distance of h1 + h2 = 3.89 mm from the ground, and thus, they resonate in distance ~0.47λ36 GHz from a metallic cylinder of radius R − h1 − h2, and hence, form an array which steers the energy back. The latter is the reason that the σa at that frequency is higher than the σm of a solid metallic cylinder of radius R.

Next, the captured power by the metallic and dielectric parts of the bent absorber was estimated through (8) and (7), respectively. The time averaged incident power to the curved absorber is found to be

$${P}_{in}=\mathop{\iint }\limits_{{S}_{m}}\,{{\bf{P}}}^{i}\cdot {\bf{dS}}=\mathop{\iint }\limits_{{S}_{m}}\,\frac{{|{{\bf{E}}}^{i}|}^{2}}{2{\eta }_{0}}\hat{{\bf{n}}}\cdot {\bf{dS}},$$
(20)

evaluated over the Sm surface of a cylinder with radius r, which encapsulates the bent absorber, where dS is the unit normal vector of latter surface, $$\hat{{\bf{n}}}={{\bf{k}}}^{i}/|{{\bf{k}}}^{i}|$$ is the unit vector which denotes the propagation direction and Pi is the time average of the power flow of the incident plane wave. Assuming that |Ei| = 1 V/m, the bent absorber lies on a cylinder with z-axis and the incident plane wave propagates to x-axis is found to be

$${P}_{in}=\frac{Rl}{{\eta }_{0}}.$$
(21)

It is noted that, only the sub-area where x < 0 was taken into account in our calculations, i.e., the part which is facing the incident plane wave.

Next, the ratio of the power dissipated in the metallic and dielectric parts of the bent absorber to the incident power (time average) it is defined,

$$A^{\prime} =\frac{{P}_{m}+{P}_{d}}{{P}_{in}},$$
(22)

which is a metric of the amount of the power which is captured by the absorber, i.e., equivalent metric for the absorbance but for the bend version: Fig. 10 depicts the A′ and A of the flat resistively loaded absorber for normal incident: for the $${{\bf{E}}}_{\parallel }^{i}$$ and $${{\bf{H}}}_{\parallel }^{i}$$-polarisation, A′ remains higher than 0.8 for 6.6–29 GHz and for 10.5–29.6 GHz, respectively, resulting FBW = 126% and FBW = 95%, respectively. Moreover, for higher frequencies, greater than 30 GHz, the equivalent absorbance for the bent geometry remains higher than 0.3, which is in alignment with the observation that at these frequencies strong surface currents appear in the SSRRs, making the latter act as resonators, which steer the scattered energy back.

Based on the above analysis, the proposed resistively loaded absorber presents high performance in terms of absorbance either when deployed as flat or curved, is wideband with FBW higher than 125% for normal incidence, polarisation insensitive and wide-angle, up to 45 deg.

## Methods

### Simulation setup

All the presented absorbers were simulated in terms of their absorbance via COMSOL Multiphysics40 with finite element method. For the flat periodic absorbers, periodic boundary conditions (PBC) were applied around the super-cell, which was illuminating by a plane wave (periodic port), which in tern, normally or by an angle impinges to the structure with a specific polarization. On the bottom of the supper-cell perfect electric condition (PEC) was applied, while in the top perfect matched layer (PML) condition. The absorbance A was estimated here through the reflection coefficient parameter. The printed, resistively loaded traces were simulated as ohmic sheets of zero thickness and surface resistance of 27 Ω/sq. For the bent printed absorber, only the ring presented in Fig. 9a was simulated and the RCS was estimated: PBC were applied on the top and bottom of the ring, PML was around the ring, while the structure was solved for the scattered field. The absorbance A′ was estimated here through (22), as has been explained.

### Fabrication Process

For the fabrication of the printed absorber the resistively loaded traces were printed on a 0.14 mm thick, polyethylene terephthalate (PET) sheet (εr = 2.5, losses of tanδ = 0.0009) by using the conventional, unmodified inkjet printer Epson Stylus C88+. The printer deposited ink consisting of electrically conductive ink MetalonTM JS-B25P45, consisting of silver nanoparticles (Ag content 25 wt%), diluted with water as the solvent. During the fabrication process, three factors determine the achieved surface resistance. First is the ratio of ink and solvent, the second is the RGB-code used during the printing process and finally the curing process after the printing (i.e. the post processing step) for the evaporation of the solvent. Thus, in order to achieve the desired surface resistance (i.e., 27 Ω/sq) first, a mixture of 1-part ink to 5-part solvent (i.e., 1:5 volume ink-solvent ratio) and second, RGB-code of (29,29,29) was utilised based on39 results. Any change on the mixture or RGB-code will modify the resulting surface resistance. In general, a curing process in an electric oven is often used evaporate any solution in the ink and bond the silver nanoparticles together, leading to a decrease in surface resistance, hence this technique is usually used for printing low resistively loaded structures. In our work, where the surface resistance is relativity high (i.e., 27 Ω/sq) for the specified ink-solvent mixture and RGB-code, we do not “bake” the printed surface, but instead we leave it to slowly evaporate the solvent at room temperature (48 hours at 20°C), thereby allowing fewer nano particles to bond and achieving a higher surface resistance. Based on39, where no oven was used after printing, the measured surface resistance for 1:5 volume ink-solvent ratio and for RGB-code (25,25,25), (27,27,27) and (30,30,30) is 5 Ω/sq, 10 Ω/sq and 43 Ω/sq, respectively. Thus, applying curve fitted in our work we have chosen RBG-code of (29,29,29) to achieve a surface resistance of 27 Ω/sq. Additionally, based on39, it is evident that the RGB-code controls the spatial density of the droplets dispensed by the printer: RGB-code (0,0,0) is black (maximum spatial density of the ink, hence minimum surface resistance) and RGB-code (255,255,255) is white (no ink, hence maximum surface resistance) with RGB values in between (i.e., grayscale) resulting in resistivity control between the available maximum and minimum values. Figure 11 depicts the resulting absorbance versus frequency for various levels of surface resistance: change on surface resistance from 5 Ω/sq to 60 Ω/sq modifies the resulting absorbance, however from 27 Ω/sq to 60 Ω/sq the change is marginal and affects mainly to the absorbance bandwidth. Thus, the proposed structure is robust in terms of surface resistance fluctuations during fabrication.

### Measurement setup

The measurement setup is depicted in Fig. 12. A two-ports vector network analyser was connected to the receiver (Rx) and transmitter (Tx) horn antennas, while at far-field distance, was placed the electromagnetic absorber. For normal incidence, Rx-Tx were collocated in front of the absorber, while for oblique incidence of angle θ, the horn antennas were moved in distance 2θ. For each combination (i.e., type of absorber, polarization, angle of incidence) the transmission coefficient S21 was measured for three cases: first for the absorber $${S}_{21}^{\,a}$$, second for a metallic plate of exactly the same dimensions to the absorber $${S}_{21}^{\,m}$$, and third, for the empty space (i.e., the absorber was removed) $${S}_{21}^{\,n}$$. The absorbance was estimated as $${S}_{21}^{\,m}-{S}_{21}^{\,a}-{S}_{21}^{\,n}$$, i.e., differential measurement. Time domain gating and smoothing techniques were applied over the measured data. All the above measurements took place inside anechoic chamber. In order to verify the exact thickness of the printed resistively loaded traces, measurement was performed through the stylus-based method and specifically by using the Tencor Stylus profilometer: the measured thickness was found to vary from 0.2 to 0.4 μm. Figure 4c depicts the measured, printed, resistively loaded traces through a microscope (10× zooming). It is noted that, since the resistively loaded traces have been simulated as ohmic sheets with surface resistance of 27 Ω/sq, the measured thickness was not taken into account in our calculations.

## Conclusion

This work presented the first flexible, polarization insensitive, wide-angle, ultra wideband, electrically thin electromagnetic absorber, which was easily fabricated by using a conventional inkjet printer by utilizing conductive ink, which diluted in water, stored in an empty inkjet cartridge and used by the printer. The design of the absorber is based on a versatile and systematic technique, which was presented and tested in this work: the basic idea is the form of super-cells with 90 degrees rotational symmetry. The presented absorber was tested in terms of absorbance through numerical analysis and measurements, where a very good agreement was observed between the simulated and measured results. This novel electromagnetic absorber is a perfect candidate for stealth technology applications, since it is able to cover not only flat surfaces, but also curved objects, operating for extended FBW, higher than 125%, regardless polarization, up to 45 deg. angle of incidence.

To the best of the authors knowledge and based on the literature, the proposed structure is the first flexible, ultra-wideband, polarization insensitive, wide-angle, easily fabricated through conventional inkjet printing techniques, electrically thin electromagnetic absorber.

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## Acknowledgements

The work of V.F. was supported by the UK Engineering and Physical Sciences Research Council (EPSRC) EP/P000673/1. The authors would like to thank Dr <NAME> and Dr <NAME>, members of the Queen’s Advanced Micro- Engineering Centre (QAMEC) for their help during the thickness measurement of the printed resistively loaded traces.

## Author information

S.D.A. conceived the idea. S.D.A. designed and simulated the electromagnetic absorbers, performed the measurements, interpreted the results and wrote the paper. V.F. supervised the research and contributed to the general concept and interpretation of the results. Both authors reviewed the manuscript.

Correspondence to <NAME>.

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# Intercepts of a Line 4

ACTMAT-JYNQBX

What are the coordinates of the x-intercept of the line passing through $(-4,-1)$ and $(6,-3)$?

A

$(-9,0)$

B

$(9,0)$

C

$(0, \cfrac{-9}{5})$

D

$(\cfrac{-1}{5},0)$

E

$(\cfrac{-9}{5},0)$


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# 33 true theorem 782 34 false consider f x 1 35 false

• 50

This preview shows page 38 - 41 out of 50 pages.

33. True, Theorem 7.8.2. 34. False; consider f ( x ) = 1. 35. False, neither 0 nor 3 lies in [1 , 2], so the integrand is continuous. 36. False, the integral is divergent. 37. Z + 0 e - x x dx = 2 Z + 0 e - u du = 2 lim + ( - e - u ) 0 = 2 lim + ( 1 - e - ) = 2. 38. Z + 12 dx x ( x + 4) = 2 Z + 2 3 du u 2 + 4 = 2 lim + 1 2 tan - 1 u 2 2 3 = lim + tan - 1 2 - tan - 1 3 = π 6 . 39. Z + 0 e - x 1 - e - x dx = Z 1 0 du u = lim 0 + 2 u 1 = lim 0 + 2(1 - ) = 2. 40. Z + 0 e - x 1 - e - 2 x dx = - Z 0 1 du 1 - u 2 = Z 1 0 du 1 - u 2 = lim 1 sin - 1 u 0 = lim 1 sin - 1 = π 2 . 41. lim + Z 0 e - x cos x dx = lim + 1 2 e - x (sin x - cos x ) 0 = 1 / 2. 42. A = Z + 0 xe - 3 x dx = lim + - 1 9 (3 x + 1) e - 3 x 0 = 1 / 9. 43. (a) 2 . 726585 (b) 2 . 804364 (c) 0 . 219384 (d) 0 . 504067 45. 1 + dy dx 2 = 1 + 4 - x 2 / 3 x 2 / 3 = 4 x 2 / 3 ; the arc length is Z 8 0 2 x 1 / 3 dx = 3 x 2 / 3 # 8 0 = 12.
Exercise Set 7.8 401 46. 1 + dy dx 2 = 1 + x 2 4 - x 2 = 4 4 - x 2 ; the arc length is Z 2 0 r 4 4 - x 2 dx = lim 2 - Z 0 2 4 - x 2 dx = lim 2 - 2 sin - 1 x 2 0 = 2 sin - 1 1 = π . 47. Z ln x dx = x ln x - x + C , Z 1 0 ln x dx = lim 0 + Z 1 ln x dx = lim 0 + ( x ln x - x ) 1 = lim 0 + ( - 1 - ln + ), but lim 0 + ln = lim 0 + ln 1 /‘ = lim 0 + ( - ) = 0, so Z 1 0 ln x dx = - 1. 48. Z ln x x 2 dx = - ln x x - 1 x + C , Z + 1 ln x x 2 dx = lim + Z 1 ln x x 2 dx = lim + - ln x x - 1 x 1 = lim + - ln - 1 + 1 , but lim + ln = lim + 1 = 0, so Z + 1 ln x x 2 = 1. 49. Z 0 e - 3 x dx = lim + Z 0 e - 3 x dx = lim + - 1 3 e - 3 x 0 = 1 3 . 50. Z 4 8 x 2 - 4 dx = lim + Z 4 8 x 2 - 4 dx = lim + 2 ln x - 2 x + 2 4 = 2 ln 3. 51. (a) V = π Z + 0 e - 2 x dx = - π 2 lim + e - 2 x 0 = π/ 2. (b) S = π + 2 π Z + 0 e - x p 1 + e - 2 x dx , let u = e - x to get S = π - 2 π Z 0 1 p 1 + u 2 du = π + 2 π u 2 p 1 + u 2 + 1 2 ln u + p 1 + u 2 1 0 = π + π h 2 + ln(1 + 2) i . 53. (a) For x 1 , x 2 x, e - x 2 e - x . (b) Z + 1 e - x dx = lim + Z 1 e - x dx = lim + - e - x i 1 = lim + ( e - 1 - e - ) = 1 /e . (c) By parts (a) and (b) and Exercise 52(b), Z + 1 e - x 2 dx is convergent and is 1 /e . 54. (a) If x 0 then e x 1 , 1 2 x + 1 e x 2 x + 1 . (b) lim + Z 0 dx 2 x + 1 = lim + 1 2 ln(2 x + 1) 0 = + . (c) By parts (a) and (b) and Exercise 52(a), Z + 0 e x 2 x + 1 dx is divergent. 55. V = lim + Z 1 ( π/x 2 ) dx = lim + - ( π/x ) i 1 = lim + ( π - π/‘ ) = π , A = π + lim + Z 1 2 π (1 /x ) p 1 + 1 /x 4 dx ; use Exercise 52(a) with f ( x ) = 2 π/x , g ( x ) = (2 π/x ) p 1 + 1 /x 4 and a = 1 to see that the area is infinite. 56. (a) 1 x 3 + 1 x for x 2, Z + 2 1 dx = + .
402 Chapter 7 (b) Z + 2 x x 5 + 1 dx Z + 2 dx x 4 = lim + - 1 3 x 3 2 = 1 / 24. (c) Z 0 xe x 2 x + 1 dx Z + 1 xe x 2 x + 1 Z + 1 dx 2 x + 1 = + . 57. The area under the curve y = 1 1 + x 2 , above the x -axis, and to the right of the y -axis is given by Z 0 1 1 + x 2 . Solving for x = r 1 - y y , the area is also given by the improper integral Z 1 0 r 1 - y y dy. 1 2 3 4 5 0.5 1 x y 58. (b) u = x, Z + 0 cos x x dx = 2 Z + 0 cos u du ; Z + 0 cos u du diverges by part (a). 59. Let x = r tan θ to get Z dx ( r 2 + x 2 ) 3 / 2 = 1 r 2 Z cos θ dθ = 1 r 2 sin θ + C = x r 2 r 2 + x 2 + C , so u = 2 πNIr k lim + x r 2 r 2 + x 2 a = 2 πNI kr lim + ( ‘/ p r 2 + 2 - a/ p r 2 + a 2 )] = 2 πNI kr (1 - a/ p r 2 + a 2 ). 60. Let a 2 = M 2 RT to get (a) ¯ v = 4 π M 2 RT 3 / 2 1 2 M 2 RT - 2 = 2 π r 2 RT M = r 8 RT πM . (b) v 2 rms = 4 π M 2 RT 3 / 2 3 π 8 M 2 RT - 5 / 2 = 3 RT M so v rms = r 3 RT M .


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# Finding pdf/cdf of continuous r.v.

I have a question. I'm given a continuous random variable X with the pdf:

$$f_X(x) = {1\over(1+\pi^2x^2)}\cdot I_{(-\infty, 0)\bigcup(0, \infty)}(x)$$

and asked to find the pdf and cdf of $Y = {1\over X}$. I'm given the hint that $F_Y(-\pi)=1/4$.

I'm having trouble solving for the cdf of the r.v. $Y$.

I get that $F_Y(-\pi)={3\over 4}$ (if $\arctan(-1) = -{\pi\over 4}$) or $-{1\over 4}$ (if $\arctan(-1) = {3\pi\over 4}$), however. This intuitively doesn't make much sense to me, even though I'm not seeing what I'm doing wrong with the integration.

Any ideas? Thanks!

• In your integral for $F_Y(y)$, why are your limits $\int_{1/y}^\infty$? Seems to me that they should be $\int_{-\infty}^{1/y}$. Mar 9 '14 at 21:37
• Because of the $P({1\over y} \le X)$ statement. But I think I have it. Per Andre's suggestion below, you (presumably) flip the inequality, which makes the limits what you suggest (and what works). Mar 9 '14 at 21:44

Hint: We will need to take some care when $y$ is negative, since division then reverses inequalities.

But let's take $y$ positive. What is the probability that $Y\le y$?
This event happens if either $X\lt 0$ or $X\ge 1/y$.

By symmetry $\Pr(X\lt 0)=\frac{1}{2}$. Your integral expression for $\Pr(X\ge 1/y)$ is correct.

Added: For negative $y$, I would suggest not doing a separate calculation, negative numbers are treacherous. If $y$ is negative, then by symmetry the probability that $Y$ is $\le y$ is the same as the probability that $Y\ge |y|$.

• Thanks! That's what I figure (re: your hint), but not sure where to run with that. Definitely get the symmetry part -- I figure that's why the integral produces the ${1\over 2}$ term. Just not sure about the rest. (If the arctan() term was positive instead of negative, it'd work out. But that's voodoo math to me unless I can actually compute that.) Mar 9 '14 at 21:39
• Are you hinting that I should add ${1\over 2}$ to my current expression? That'd work for $arctan(-1)= {3\pi\over 4}$. Mar 9 '14 at 21:49
• For positive $y$, we have $F_Y(y)=\frac{1}{2}+\int_{1/y}^\infty \frac{dx}{1+\pi^2x^2}$. You integrated correctly, we get $F_Y(y)=1-\frac{1}{\pi}\arctan(\pi/y)$. This makes sense, it approaches $1$ as $y$ goes to infinity. Mar 9 '14 at 21:51
• It is not quite a hint, I think it says explicitly you should add $\frac{1}{2}$. Mar 9 '14 at 21:52
• Thank you! I get your explanation conceptually, just didn't know how I could reconcile that with the math. But this all makes sense. Mar 9 '14 at 21:57


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Cauchy sequence in vector topological and metric space

For a vector topological space, a sequence $(x_n)$ is Cauchy when $\forall$ neighbourhood $U$ of $0$, there exists $N$ such that $m,n>N$ implies $x_m-x_n\in U$.

For a metric topological space $(E,d)$, a sequence $(x_n)$ is Cauchy when $\forall \epsilon>0$, there exists $N$ such that $m,n>N$ implies $d(x_m,x_n)<\epsilon$.

QUESTION : are these two definitions equivalent in the case of a metric topological vector space ?

If the distance is invariant by translation, I can prove it. But in general no.

QUESTION : if the previous question has answer "no", what is the "good" definition in the intersection ?

The answer is : yes if the vector topological space is metrizable, that is IFF it has a countable basis of topology on every point.

See theorem 3.21 here : http://ma.huji.ac.il/~razk/iWeb/My_Site/Teaching_files/TVS.pdf (difficult)

And also the fact that a metrizable topological vector space has always a countable local basis of topology. This one is easy : if $A$ is closed in $V$, then consider the open sets $V_n=\{x\in X\,st\, d(A,x)<1/n\}$. We have $A=\bigcap_{n\geq 1}V_n$.

If you read French, I'll write the answer with some details here : http://laurent.claessens-donadello.eu/pdf/mazhe.pdf in the chapter "Topologie générale"


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# Finding simple sequence from reads with significant overlap

I wrote a "script" to pull out reads from a huge fastq file in an iterative manner, by finding homology to the previous sequence.

It should be relatively easy to overlap them and assemble the gene, however traditional assemblers like SPAdes fail due to obviously low coverage.

I am hesitant to just feed it more faked reads, when I know I have all the data necessary to just overlap them into a single sequence.

The reads are set up like this:

GATCAAACATC
AACATCAGTTAG
TCAGTTAG
TAGAGGATAGC

What kind of assembler is out there where I can feed it this kind of terribly-formatted data?

By default it looks for overlap regions that have a coverage of 3 reads when joining (option -c), but I assume that could be reduced down to 2 reads, as in your example (which has a minimum overlapping coverage of 2).


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# The 6th Conference of the Polish Society on Relativity

Sep 23 – 26, 2019
Other Institutes
Europe/Warsaw timezone

## Cosmography as a tool to discriminate between modified gravity and dark energy

Sep 25, 2019, 9:00 AM
45m
Other Institutes

#### Other Institutes

Maritime Academy, <NAME> 1-2, Szczecin, Poland
Talk/Seminar Plenary Sessions

### Speaker

Prof. <NAME> (INFN - National Institute for Nuclear Physics)

### Description

Cosmography can be considered as a sort of a
model-independent approach to tackle the dark energy/modified gravity
problem. In this talk, the success and the shortcomings of the
$\Lambda$CDM model, based on General Relativity and standard model of
particles, are discussed in view of the most recent observational
constraints. The motivations for considering extensions and modifications
of General Relativity are taken into account, with particular attention to
$f(R)$ and $f(T)$ theories of gravity where dynamics is represented by
curvature or torsion field respectively. The features of $f(R)$ models
are explored in metric and Palatini formalisms.
Cosmological dynamics of $f(R)$ models is investigated through the
corresponding viability criteria. Afterwards, the equivalent formulation
of General Relativity (Teleparallel Equivalent General Relativity) in
terms of torsion and its extension to $f(T)$ gravity is considered.
Finally, the cosmographic method is adopted to break the degeneracy among
dark energy models. A novel approach, built upon rational Pad\'e and
Chebyshev polynomials, is proposed to overcome limits of standard
cosmography based on Taylor expansion. The approach provides accurate
model-independent approximations of the Hubble flow. Numerical analyses
are presented to bound coefficients of the cosmographic series. These
techniques are thus applied to reconstruct $f(R)$ and $f(T)$ functions
and to frame the late-time expansion history of the universe with no
\emph{a priori} assumptions on its equation of state. A comparison
between the $\Lambda$CDM cosmological model with $f(R)$ and $f(T)$ models
is reported.

### Primary author

Prof. <NAME> (INFN - National Institute for Nuclear Physics)

### Presentation materials

There are no materials yet.