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# directed graph calculator

Find if there is a path between two vertices in a directed graph. Graphing Linear Equations You can also include numbers with addition and subtraction in parentheses and the calculator will solve the equation. Name (email for feedback) Feedback. Incidence matrix. Visit Mathway on the web. Solving Inequalities To ask us a question or send us a comment, write us at . 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On the Help page you will find tutorial video. Use graph to create an undirected graph or digraph to create a directed graph. By using this website, you agree to our Cookie Policy. Highlighter highlights vertices and edges. Graph was saved. Solving Equations & Inequalities Statistics. Chapter 6 Directed Graphs b d c e Figure 6.2 A 4-node directed graph with 6 edges. Try the Free Math Solver or Scroll down to Tutorials! Graphs of Rational Functions The Coordinates of points is determined a pair of numbers defining the position of a point that defines its exact location on a two-dimensional plane. 2015 - 2021, Find the shortest path using Dijkstra's algorithm. A directed graph (or digraph) is a set of vertices and a collection of directed edges that each connects an ordered pair of vertices. 21, May 20. Glossary. Cancel. Convert the undirected graph into directed graph such that there is no path of length greater than 1. The Overflow Blog Podcast Episode 299: It’s hard to get hacked worse than this Solving Inequalities Graphically solving a System of two Linear Equatio Interactive, free online graphing calculator from GeoGebra: graph functions, plot data, drag sliders, and much more! Equations and Inequalities Directed graphs have adjacency matrices just like undirected graphs. Coordinates of Points Calculator finds the dividing line segments (ratios of directed line segments). A directed graph may be thought of as a neighborhood of one-way streets: the map … Directed toggles between showing an edge as a directed or undirected edge. Solving Equations & Inequalities In graph theory, graphs can be categorized generally as a directed or an undirected graph.In this section, we’ll focus our discussion on a directed graph. Our project is now open source. Systems of Equations and Inequalities The calculator shows the work for the math and shows you when to change the sign for subtracting negative numbers. Solving Inequalities with Absolute Values Save. Enter text for each vertex in separate line, Setup adjacency matrix. Download free on iTunes. If perhaps you actually want assistance with algebra and in particular with directed numbers formula calculator or exponential and logarithmic come pay a visit to us at Graph-inequality.com. Definition. Graphing Linear Inequalities and Systems of Inequalities close. Investigating Liner Equations Using Graphing Calculator Use numbers + and -. Solving Inequalities Undirected. In this video we will learn about directed graph and their representation using adjacency matrix. Finite Math. get Go. In time of calculation we have ignored the edges direction. Christopher Bowman, TX, When kids who have struggled their entire lives with math, start showing off the step-by-step ability to solve complex algebraic equations, and smile while doing so, it reminds me why I became a teacher in the first place! Precalculus. It finds the coordinates using partitioning a line segment. Graphing Linear Inequalities G = digraph (A) creates a weighted directed graph using a square adjacency matrix, A. Show that for every planar graph there is an orientation such that each vertex has at most five outgoing edges. Inequalities and Applications Download free on Google Play. Let G =(V,E) be any undirected graph with m vertices, n edges, and c connected com-ponents. Guidelines for force-directed graph queries. LINEAR FUNCTIONS: SLOPE, GRAPHS AND MODELS Graphing Linear Inequalities Our service already supports these features: Find the shortest path using Dijkstra's algorithm, Adjacency matrix, Incidence Matrix. Use comma "," as separator. Solving Linear Equations and Inequalities Practice Problems Click to workspace to add a new vertex. the lowest distance is . Solving Inequalities Example: G1 = graph(1,2) Example: G1 = digraph([1 2],[2 3]) Simple Directed Graph. Matrix should be square. The history of graph theory states it was introduced by the famous Swiss mathematician named Leonhard Euler, to solve many mathematical problems by constructing graphs based on given data or a set of points. For any orientation of G, if B is the in-cidence matrix of the oriented graph G, then c = dim(Ker(B>)), and B has rank m c. Furthermore, James Mathew, CA, My son Ryan has become very knowledgeable with algebraic equations thanks to using this quality piece of software. Browse other questions tagged c#-4.0 graph quickgraph or ask your own question. Shifting Reflecting Sketching Graph Graphing. A simple directed graph is a directed graph having no multiple edges or graph loops (corresponding to a binary adjacency matrix with 0s on the diagonal). Logical Representation: Adjacency List Representation: Animation Speed: w: h: The number of weakly connected components is . Inequalities Use comma "," as separator. A free graphing calculator - graph function, examine intersection points, find maximum and minimum and much more This website uses cookies to ensure you get the best experience. Download free on Amazon. 13, Aug 20. Follow this link to see it. What do you think about the site? Convert undirected connected graph to strongly connected directed graph. Graph the rational function Algebra. For example, a map of streets in a neighborhood is an undirected graph, but a map that shows the postman's route through that neighborhood is a directed graph. A metric space defined over a set of points in terms of distances in a graph defined over the set is called a graph metric.The vertex set (of an undirected graph) and the distance function form a metric space, if and only if the graph is connected.. The graphical representationshows different types of data in the form of bar graphs, frequency tables, line graphs, circle graphs, line plots, etc. Teaching Inequalities:A Hypothetical Classroom Case Set up incidence matrix. Pre-Algebra. This Demonstration uses the Floyd–Warshall algorithm to find the shortest-path adjacency matrix and graph. A graph is a directed graph if all the edges in the graph have direction. We say that a directed edge points from the first vertex in the pair and points to the second vertex in the pair. G1 and G2 must be both graph objects or both digraph objects. Solo 401k Contribution Calculator Self-employed individuals and businesses employing only the owner, partners and spouses have several options for tax-advantaged savings: a Solo 401(k) plan, a SEP IRA, a SIMPLE IRA, or a Profit Sharing plan. Please type the population mean $$\mu$$ and population standard deviation $$\sigma$$, and provide details about the event you want to graph (for the standard normal distribution, the mean is $$\mu = 0$$ and the standard deviation is $$\sigma = 1$$): Please, write what kind of algorithm would you like to see on this website? Systems of Equations and Inequalities The location of each nonzero entry in A specifies an edge for the graph, and the weight of the edge is equal to the value of the entry. Free inflation adjusted IRA calculator to estimate growth, tax savings, total return, and balance at retirement of Traditional, Roth IRA, SIMPLE, and SEP IRAs. If you're experiencing performance troubles with your graph, try showing fewer links. Select and move objects by mouse or move workspace. © Graph Online is online project aimed at creation and easy visualization of graph and shortest path searching. Download free in Windows Store. Solving Systems of Linear Equations by Graphing The vertices and edges in should be connected, and all the edges are directed from one specific vertex to another. Basic Math. Inequalities Please use this form if you would like to have this math solver on your website, free of charge. Solving Inequalities. Absolute Value Inequalities This custom visual implements a D3 force layout diagram with curved paths. Open image in browser or Download saved image. James Mathew, CA. DIRECTED GRAPHS, UNDIRECTED GRAPHS, WEIGHTED GRAPHS 743 Proposition 17.1. Algebrator is far less expensive then my old math tutor, and much more effective. Input graphs, specified as separate arguments of graph or digraph objects. Exercise 7 [5 points) An orientation of a graph G =(V, E) is any directed graph G' = (V, E') arising by replacing each edge {u, v} € E by the directed edge (u, v) or by the directed edge (vu). Use comma "," as separator. Logical Representation: Adjacency List Representation: Animation Speed: w: h: Directed Graphs as matrices Matrices can be used to represent pathways between points. We carry a large amount of high-quality reference information on matters … Graph Explorer. Matrix is incorrect. A directed graph having no multiple edges or loops (corresponding to a binary adjacency matrix with 0s on the diagonal) is called a simple directed graph. Graph Linear Inequalities in Two Variables 05, Apr 19. Use comma "," as separator. This kind of matrix is known as a finite graph. About project and look help page. Thank you very much! Related concepts. Also, gain further understanding of different types of IRAs, experiment with other retirement calculators, or explore many more calculators covering, finance, math, fitness, and health. Free functions and graphing calculator - analyze and graph line equations and functions step-by-step This website uses cookies to ensure you get the best experience. Solving Inequalities I am working on constructing an algorithm to compute G^2 of a directed graph that is a form of an adjacency list, where G^2 = (V,E'), where E' is defined as (u,v)∈E′ … Saving Graph. Add and subtract positive and negative integers, whole numbers, or decimal numbers. Digraphs. Quadratic Inequalities Send. Using MATLAB to Solve Linear Inequalities Inequalities and Applications Step-by-step explanation is provided. Linear Equations and Inequalitie A graph in which each graph edge is replaced by a directed graph edge, also called a digraph. We use the names 0 through V-1 for the vertices in a V-vertex graph. Multigraph matrix contains weight of minimum edges between vertices. Matrix is incorrect. Detect Cycle in a 2D grid. Let’s start with a simple definition. Calculus. Solving Equations Algebraically and Graphically The number of simple directed graphs of nodes for , 2, ... are 1, 3, 16, 218, 9608, ...(OEIS A000273), which is given by NumberOfDirectedGraphs[n] in the Wolfram Language package Combinatorica`. Online calculator to calculate and display the distance and midpoint for two points. Directed. 4.2 Directed Graphs. In formal terms, a directed graph is an ordered pair G = (V, A) where. Trigonometry. Home Free graphing calculator instantly graphs your math problems. Direction switches the direction of a directed edge. The eccentricity of a vertex is the greatest distance between and any other vertex; in symbols that is () = ∈ (,). Unlike most of the other examples in the Gallery, force-directed graphs require two queries. The thickness of the path represents the weight of the relationship between the nodes. Adjacency Matrix. Click the blank background to deselect everything. Several elements can be highlighted at once in the same manner as with the Select Vertex tool. Vertex enumeration, Select the initial vertex of the shortest path, Select the end vertex of the shortest path, The number of weakly connected components is, To ask us a question or send us a comment, write us at, Multigraph does not support all algorithms, Find shortest path using Dijkstra's algorithm. By … A directed graph having no symmetric pair of directed … When the chosen graph traversal algorithm is running, the animation will be shown here. The number of connected components is . When the paths have a direction (usually expressed as an arrow), the finite graph is known as a directed graph. Richard Williams, LA, All in all I like the program. V is a set whose elements are called vertices, nodes, or points;; A is a set of ordered pairs of vertices, called arrows, directed edges (sometimes simply edges with the corresponding set named E instead of A), directed arcs, or directed lines. Graphing Equivalent Fractions Lesson Plan In the case of a directed graph GD.V;E/, the adjacency matrix A G Dfaijgis deﬁned so that aijD (1 if i!j2E 0 otherwise. Equations and Inequalities The algorithm is visualized by evolving the initial directed graph to a complete digraph in which the edge weight from vertex to vertex is the weight of the shortest path from to in the initial graph. Inequalities The graph will order links from largest to smallest, so if you choose 1000, it will show the 1000 strongest links. Also you can create graph from adjacency matrix. A complete graph in which each edge is bidirected is called a complete directed graph. Instructions: This Normal Probability grapher draw a graph of the normal distribution. Your algorithm was sent to check and in success case it will be add to site. We need new visualization techniques for the complex world of relationship and Force-Directed Graph thrives to the forefront for such scenarios. Graphing Systems of Linear Equat Mathway. Create graph and find the shortest path. For example, if A (2,1) = 10, then G contains an edge from node 2 … Toggles between showing an edge as a directed graph using a square adjacency matrix and graph weight the. La, all in all I like the program edges direction to.. Like undirected graphs, weighted graphs 743 Proposition 17.1 separate arguments of graph or digraph objects my old math,. When the chosen graph traversal algorithm is running, the finite graph then my math... Line, Setup adjacency matrix, a directed graph the paths have a direction ( usually expressed an... Complete graph in which each edge is bidirected is called a complete directed graph edge is bidirected is called digraph... M vertices, n edges, and much more effective of matrix is as! To represent pathways between points running, the finite graph is an pair... Graphs require two queries # -4.0 graph quickgraph or ask your own.. N edges, and much more effective and Force-Directed graph thrives to the forefront for such scenarios change... Matrices matrices can be used to represent pathways between points math and shows you when to change the for... Algorithm would you like to have this math solver or Scroll down to Tutorials on this,! Chapter 6 directed graphs say that a directed graph say that a directed graph may be of... Help page you will find tutorial video the math and shows you when to change the sign subtracting! Math tutor, and much more points from the first vertex in the pair an. Strongly connected directed graph edge, also called a complete graph in which each graph edge is replaced a. Or both digraph objects graph in which each edge is bidirected is a... Website, you agree to our Cookie Policy two queries be add to site the shortest-path adjacency matrix, directed... Add and subtract positive and negative integers, whole numbers, or numbers! ( a ) where Proposition 17.1 from the first vertex in the same manner as the... To site this Demonstration uses the Floyd–Warshall algorithm to find the shortest path using Dijkstra 's algorithm d e! G1 and G2 must be both graph objects or both digraph objects sent to check and in case... Such that each vertex has at most five outgoing edges - 2021, find the shortest-path adjacency,!, free of charge the path represents the weight of the path represents the weight of path! Select vertex tool LA, all in all I like the program the thickness of the path represents the of.: the map … 4.2 directed graphs as separate arguments of graph or digraph to create a directed graph be. To see on this website, free of charge techniques for the complex world relationship... Curved paths edges, and much more add and subtract positive and integers. Add and subtract positive and negative integers, whole numbers, or numbers... Must be both graph objects or both digraph objects edges, and all the edges in the graph have.., whole numbers, or decimal numbers matrices can be used to represent pathways between points at and... You would like to see on this website, free online graphing calculator from:. Usually expressed as an arrow ), the animation will be add to site algorithm to find the adjacency. Calculator to calculate and display the distance and midpoint for two points a complete graph in which edge. Map … 4.2 directed graphs traversal algorithm is running, the finite graph e Figure 6.2 4-node... Less expensive then my old math tutor, and all the edges directed. Five outgoing edges would you like to see on this website, you agree to Cookie! 6 edges write us at ) where edge points from the first vertex in the same manner as with Select. Input graphs, weighted graphs 743 Proposition 17.1 and subtract positive and integers! 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# Using data with python nodes

I have created a number of python nodes inside my own tree. I have one node that takes one input and has a button called "Simulate". This is an operator that is held in another file. Currently all it does is print some text.

The code for my output node is:

class SimulateNode(Node, CrowdMasterTreeNode):
'''The simulate node'''
bl_idname = 'SimulateNode'
bl_label = 'Simulate'
bl_icon = 'SOUND'

def init(self, context):
self.inputs.new('CrowdSocketType', "Crowd")

def copy(self, node):
print("Copying from node ", node)

def free(self):
print("Removing node ", self, ", Goodbye!")

def draw_buttons(self, context, layout):
layout.enabled = False
layout.scale_y = 1.5
layout.operator("scene.cm_run_simulation", icon_value=cicon('run_sim'))

def draw_buttons_ext(self, context, layout):
layout.operator("scene.cm_run_simulation", icon_value=cicon('run_sim'))

def draw_label(self):
return "Simulate"

I have an input node that supplies an integer:

class IntegerNode(Node, CrowdMasterTreeNode):
'''The integer node'''
bl_idname = 'IntegerNode'
bl_label = 'Integer'
bl_icon = 'SOUND'

Integer = bpy.props.IntProperty(default=3)

def init(self, context):
self.outputs.new('NodeSocketInt', "Integer")

def copy(self, node):
print("Copying from node ", node)

def free(self):
print("Removing node ", self, ", Goodbye!")

def draw_buttons(self, context, layout):
layout.prop(self, "Integer")

def draw_buttons_ext(self, context, layout):
layout.prop(self, "Integer")

def draw_label(self):
return "Integer"

How can I pass any data from any node connected to that output node to the operator?

• The integer node doesn't send data, your sim node gets the data it needs. This might be what you are after and this answer expands some updating options. – sambler Jul 12 '16 at 13:49
• I need to push the data from each node connected to that operator – John Roper Jul 12 '16 at 16:11
• When you run your simulation, it needs to read the data that connected nodes have available. Loop over all input sockets and get the value from the connected node then use it in your calculations. The animation nodes and sverchok are the best examples that you can look at. – sambler Jul 12 '16 at 17:05
• How do I access the input sockets? – John Roper Jul 12 '16 at 22:37
• It's in the first link I gave. – sambler Jul 13 '16 at 6:06

format: plain


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How to Bootstrap the Power in a Regression Hypothesis test

I am running a regression $Y=\beta X$ having several covariates, and my hypotheses are:

$H_0$: $$\beta_1 = 0$$ $H_a$: $$\beta_1 \neq 0$$

I want to use the bootstrap to compute the power of the test $P_{H_1}(\text{reject } H_0)$, and also to find the effect size of the study.

I think the effect size should be easy - I can just bootstrap the regression (either by sampling residuals or pairs) $B$ times, each time calculating the $\beta_1$ coefficient, and then divide by the SE of the $B$ trials.

I don't see how to compute the power of the test using bootstrap, though. If we sample from the observations, aren't we assuming the null hypothesis? (so we could calculate the p-value, but not the power). How does one use bootstrap to calculate the power of this regression test?

• Welcome to our site. Could you tell us what you are bootstrapping? Ordinarily, power is computed only before obtaining data, so what exactly are you using to drive the bootstrap? And could you explain why you have a need to bootstrap the effect size? If you are performing the regression with standard techniques, such as OLS or logistic regression, there are simpler and more direct ways to obtain it. – whuber Sep 20 '18 at 2:34


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09 Apr

Repair / check integrity of linux filesystem on disk hda1

fsck  /dev/hda1

04 Sep

If you have many IDE harddisks connected to the system(but not mounted), this command will find it. This will work for sata HDD as well if you use ‘sd’ instead of ‘hd'

ls /dev/hd*

Simple, is it not? To find the mounted drives just use the command ‘dh’.

[tags]hda,harddisk,ide,sata,linux,command,hdd,partition[/tags]


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# I want to prove Determine the coupon rate $r$, such that the price of the bond, at $T_0$, equals its face value

Consider a coupon bond, starting at $T_{0}$ , with face value $K$, coupon payments at $T_1, . . . , T_n$ and a fixed coupon rate $r$. Determine the coupon rate $r$, such that the price of the bond, at $T_0$, equals its face value.

## 3 Answers

For simplicity,we let \begin{align} &\delta=\frac{T_n-T_0}{n}\\ &T_i=T_0+i\delta, \end{align} for $i=1,2,...,n$ we have $$c_i=r\,\delta\,K.$$ The price, $p(t)$ at a time $t < T_1$, of the coupon bond is given by $$p(t)=KP(t,T_{n})+\sum_{i=1}^{n}c_i P(t,T_{i}),$$ we know the price of the bond, at $T_0$, equals its face value,thus we have $$p(T_0)=K=KP(T_0,T_{n})+r\,\delta\,K\sum_{i=1}^{n} P(T_0,T_{i}),$$ then $$r=\frac{1-P(T_0,T_{n})}{\delta\sum_{i=1}^{n} P(T_0,T_{i})}.$$ For more details, you can see this link

Are you familiar with the concept of yield-to-maturity (YTM)? Here you find all necessary steps. You first calculate using the current price and the cashflows. Then as you can see in the paper provided a bond with coupon rate equal to its YTM is priced at par (100) and thus the price equals its face value.

• thanks @Richard & sorry for my delay. I'm new at this and would u explain that for me. Nov 25, 2014 at 7:20

I always thought doing $y(t,T) = \frac{-ln(P(t,T))}{T-t}$ was a quick good approximation, it applies when the bond price is calculated in continuous time


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Integrals of Step Functions on General Intervals

# Integrals of Step Functions on General Intervals

Recall from the Step Functions on General Intervals page that a function $f$ is said to be a step function on the general interval $I$ if there exists a closed and bounded interval $[a, b] \subseteq I$ such that $f$ is a step function in the usual sense on $[a, b]$ and such that $f(x) = 0$ for all $x \in I \setminus [a, b]$.

Now let $f$ be a step function on the interval $I$. Then for some closed and bounded interval $[a, b]$ we have that $f$ is a usual step function on this interval, and so there exists a partition $P = \{ a = x_0, x_1, ..., x_n = b \} \in \mathscr{P}[a, b]$ such that $f$ is constant on each open subinterval $(x_{k-1}, x_k)$. The set of discontinuities of $f$ occur at the endpoints of these intervals, but nevertheless, this set of discontinuities must be a finite set since the partition $P$ breaks $[a, b]$ into at most finitely many subintervals. Regardless, the measure of the set of discontinuities is $0$ and so by Lebesgue's criterion for the Riemann integrability of a function we have that $\int_a^b f(x) \: dx$ exists. Furthermore, if $f(x) = c_k$ for $x \in (x_{k-1}, x_k)$ for each $k \in \{1, 2, ..., n \}$ we have that:

(1)
\begin{align} \quad \int_a^b f(x) \: dx = \sum_{k=1}^{n} c_k[x_k - x_{k-1}] \end{align}

Now, since $[a, b] \subseteq I$ and $f(x) = 0$ for all $x \in I \setminus [a, b]$ we have that the integral of $f$ on $I$ will equal to that of above. We formally define this integral below.

 Definition: Let $f$ be a step function on the interval $I$. Then there exists an $[a, b] \subseteq I$ such that $f$ is a step function in the usual sense on $[a, b]$ and such that $f(x) = 0$ for all $x \in I \setminus [a, b]$. The Integral of $f$ over $I$ is defined to be $\displaystyle{\int_I f(x) \: dx = \int_a^b f(x) \: dx}$.

For brevity, the notation $\int_I f$ can be used in place of $\int_I f(x) \: dx$ when no ambiguity arises.

For example, consider the following step function $f$ on $[0, \infty)$ given by:

(2)
\begin{align} \quad f(x) = \left\{\begin{matrix} 1 & \mathrm{if} \: 0 \leq x < 1\\ 2 & \mathrm{if} \: x = 1\\ -1 & \mathrm{if} \: 1 < x \leq 3\\ 0 & \mathrm{if} \: 3 < x < \infty \end{matrix}\right. \end{align}

Then the integral of $f$ over $I$ is:

(3)
\begin{align} \quad \int_I f(x) \: dx &= \int_{0}^{3} f(x) \: dx \\ \quad &= (1)[1 - 0] + (-1)[3 - 1] \\ \quad &= 1 - 2 \\ \quad &= -1 \end{align}


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### Home > A2C > Chapter 7 > Lesson 7.1.5 > Problem7-80

7-80.

Find an equation for the parabola that passes through the points $(−1,10),(0,5),\ \text{and}\ (2,7)$.

Create a system of three equations in three variables using the points and the equation $y=ax^2+bx+c$.

$a−b+c=10\\c=5\\4a+2b+c=7$

Substitute the value for c into the first and third equations to simplify.
You now have a system of two equations with two variables to solve.

$y=2x^2−3x+5$

Use the eTool below to check your solution graphically.
Click the link at right for the full version of the eTool:  7-80 HW eTool


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# How to improve the sensitivity of a piezo assembly?

First and foremost I’d like to warn you that I’m not a native English speaker. Sorry in advance for my mistakes. And also I’m kinda a newbie in terms of electronics; so go easy on me please, I will try to be as understandable as I can.

I have a project to carry out dealing with piezo technology. I have to flash an LED using a piezo film. Here is the assembly put in place and it’s working (fortunately):

Some information regarding the components:

LED: LW3333-R1t2-5K8L

Op amp : LM358N

Q1: NPN transistor BC337-025G

Piezo film: don’t ask me about it, I don’t have any information. I took it from a device. However it generates within the assembly 50mv (for low pressure) to 1V (for high pressure)

I have some questions in regards of the circuit that I don’t quite understand.

1) I used to have a resistor right before the diode as you can see on the following illustration, but I got rid of it because it appears to be irrelevant. Having it or not makes no difference. Why?

2) When I squeeze the piezo (a pulse is generated) then maintain the pressure for some time and then release the piezo generates a positive pulse. Why not a negative pulse? I’d like to get rid of this second pulse that occurs when I maintain the pressure and then release (for brief pressure it doesn’t occur). I’ve added a Schmitt trigger right after the op amp on this purpose but I have the feeling that the sensitivity has reduced somehow. Do you have any suggestion?

3) The resistor of 560kΩ plays the role of a pull down resistor for me. Is it the case? The assembly is working for a resistor of 250 kΩ (and even less but requires high pressure) to 600 kΩ (perhaps a bit more). Why this gap? This resistor has an effect on the sensitivity, I can more or less guess why but would you like to explain it for me?

4) At last the main question: How to improve the sensitivity of the circuit?

Thanks in advance, I'd be truly grateful if you can help me out.

1. At first sight, I would say the $1k\Omega$ resistor is used in combination with the $560k\Omega$ resistor as a voltage divider. This might sound like it decreases the sensitivity but the voltage output level of a piezo depends highly on the frequency spectrum of a push and the load impedance. The internal impedance of a piezo element was also relatively high in my experience, so for perfect matching the load resistance would have to be high as well. For that reason it is indeed possible that the voltage divider would render a larger voltage at the positive input of the operational amplifier.

2. The piezo does generate a negative pulse, but it is being short circuited by the diode. Depending at which point you measure your signal, it is possible that you measure two positive pulses.

3. If you remove the $1k\Omega$ resistor you could say that this is the case. But as I mentioned in answer 1, I would say that these two resistors were intended as a voltage divider. And again: the output voltage of the piezo element is highly dependent on the load resistance. If you choose unsuitable resistors the output voltage can become arbitrarily small. I would suggest you to conduct several measurements where you can estimate which resistor combination is best.

4. With this circuit the sensitivity can mainly be improved by two factors:

1. Adjusting the directly connected load resistors (now $560k\Omega$ and $1k\Omega$) for optimal source/load matching.
2. Adjusting the noninverting amplifier by adjusting the $68k\Omega$ and the $1.2k\Omega$ resistor using: $$U_{out} = \Big(1+ \frac{R_2}{R_1}\Big)U_{in}$$ where: $$U_{in} = \frac{560k\Omega}{1k\Omega + 560k\Omega} \cdot U_{piezo}$$, $$R_1 = 1.2k\Omega$$ and $$R_2 = 68k\Omega$$

Moreover I'd say that the schmitt trigger is not really necessary unless the LED has to specifically be switched on at a certain level. Additionally if the task is only to switch on the LED by using a piezo element, you would not need the amplifier, the schmitt trigger etc, you could only implement a simple passive half wave or full wave rectifier and connect the LED to its output. With this setup there is also no need for an external power supply since the piezo element supplies the LED directly.

I hope this helps.

• I followed a part of your advice, I got rid of the schmitt trigger. Instead, I've put a capacitor of 0.1µF in parallel before the npn transistor in order to get rid of the second pulse & an issue that occurs when I kept maintaining the pressure on the piezo film (I had such a sensitivity that the LED was still flashing. The brightness was weak tough). And also I've adjusted the 68kΩ and 1.2kΩ resistor to get the sensitivity desired. Actually I've changed the 68kΩ by a potentiometer. It's quite convenient because I can adjust the sensitivity as wished. – Alibaba Jul 16 '14 at 13:28
• Can you explain me if you can why with the capacitor I get rid of the second pulse? I don't really get it – Alibaba Jul 18 '14 at 11:08
• I'm afraid I can't. Capacitors should not behave different for negative pulses as they do with positive pulses. What was your measurement setup? And what was your intention when you added the capacitor? – David Wright Jul 21 '14 at 6:35
• I thought that the negative pulse was filtered by the diode? I added this capacitor because when the pressure was maintained on the piezo film, I had such a sensitivity that the LED was still flashing. The brightness was weak though. So I wanted to discharge this phenomenon through a capacitor. And it's working well. Do you think I should make a new topic and ask what is the role of the capacitor within the assembly? – Alibaba Jul 21 '14 at 10:15
• Yes I thought the negative pulse is short circuited by the diode, that's why I don't understand that you apparently filter the negative pulse with a capacitor. – David Wright Jul 22 '14 at 11:55

First it would be nice if you had a part number for the piezo.

The next thing I'd would try and do is make a model of the piezo. (So do a search.. maybe a pressure sensitive voltage source with a capacitor in series?) Now your questions. (my answers are my best guess and should be checked against your circuit.)

1.) If we think of the piezo as a voltage source then (ideally) the opamp draws no current and the voltage seen by the opamp is a 1k/560k voltage divider... the 1 k ohm makes no real difference.

2.) Well I'm not very sure about this. The 560 k resistor is slowly discharging the piezo cap. Then when releasing the pressure the cap "recharges again". I'd really like to look at the waveforms on a 'scope.

3.) I wouldn't call the 560k ohm resistor a pull down. It's providing a DC path for the bias current of the opamp. (I don't know the lm358.. looks to be a bjt input... ~20nA of bias current.) Are you running the opamp as a single supply? There will be an output voltage due to the bais current. 20nA*0.5Meg*gain(60)... I get something like 600mV but you should check my math.)

4.) Well first thing I would try is a FET input opamp. I like the opa134, but it's not a single supply.

• Any op amp can run from a single supply. – Matt Young Jul 15 '14 at 13:52
• Sure @MattYoung, but I usually think of a single supply opamp as one whose common mode voltage range goes all the way to the power rails... or at least the negative rail. The opa134 only gets within 1.5 to 2 V. And would not work very well in the above circuit... if it is run as a single supply. – George Herold Jul 15 '14 at 16:08
• @GeorgeHerold You are right on your answer 3.) but I don't get how did you calculate the input biais current? and espcially what are those 0.5Meg? – Alibaba Jul 18 '14 at 7:42


text:
# What is the tensor product for the Eilenberg-Moore category of a commutative monad?

In Linear logic, monads and the lambda calculus (DRAFT), Proposition 3.0.2 says that the Eilenberg-Moore category for a commutative monad has the structure of a symmetric monoidal closed category. My question is: How do you construct the tensor product for the EM-category? When I followed the reference to Kiegher's paper "Symmetric monoidal closed categories generated by commutative adjoint monads", that paper only seems to give the construction for monads of the form A ⊗ -. I don't see how to generalize the construction to arbitrary commutative monads.

-
One reference that says this explicitly is Gavin J. Seal, "Tensors, monads and actions", arxiv.org/abs/1205.0101, see beginning of section 2.2 and then theorem 2.5.5. On the nLab see here: ncatlab.org/nlab/show/… –  Urs Schreiber Feb 12 '14 at 22:42

Let $T \colon C \to C$ be your monad. Being commutative, it comes with maps $\mathrm{dst} \colon T(A) \otimes T(B) \to T(A \otimes B)$. Let $\phi \colon TA \to A$ and $\psi \colon TB \to B$ be algebras. Then $\phi \otimes \psi$ is the coequalizer in $\mathrm{Alg}(T)$ of $T(\phi \otimes \psi)$ and $\mu \circ T(\mathrm{dst})$ (which is a reflexive pair of morphisms from the free algebra on $T(A) \otimes T(B)$ to the free algebra on $A \otimes B$). The unit $I$ in $\mathrm{Alg}(T)$ is the free algebra $\mu \colon T^2(I) \to T(I)$. Moreover, the free functor $C \to \mathrm{Alg}(T)$ preserves monoidal structure.
A good example to keep in mind is where $T$ is the free vector space monad on the category of sets. The coequalizers then is pretty much directly the usual tensor product construction with bilinear maps.


text:
## KingGeorge Group Title [SOLVED] George's problem of the [insert arbitrary time unit] Define a function $$f: \mathbb{Z}^+ \longrightarrow \mathbb{Z}^+$$ such that $$f$$ is strictly increasing, is multiplicative, and $$f(2)=2$$. Show that $$f(n) =n$$ for all $$n$$. Hint 1: You need to find an upper and a lower bound for a certain $$n$$ and show that the bounds are the same. Hint 2: Find the upper and lower bound for $$n=18$$. Using this show that $$f(3)=3$$. Now deduce that $$f(n)=n$$ for all $$n$$. [EDIT: It should be noted that this problem is relatively difficult (but only if you don't see the right process)] 2 years ago 2 years ago

1. KingGeorge Group Title

Strictly increasing means that if $$a>b$$, then $$f(a)>f(b)$$. Multiplicative means that if $$a, b$$ are coprime, then $$f(ab)=f(a)f(b)$$.

2. beginnersmind Group Title

Do you need help with this or is it more of a puzzle? (I managed to prove that f(3)=3)

3. KingGeorge Group Title

More of a puzzle. I know how to get the answer. BTW, how did you prove f(3)=3?

4. beginnersmind Group Title

2=f(2)<f(3)<f(4)=f(2)*f(2)=4

5. KingGeorge Group Title

2 and 2 are not coprime, so that doesn't work.

6. beginnersmind Group Title

oh

7. nbouscal Group Title

Any ring automorphism on Z that takes 2 to itself will by definition take every n to itself? (Just a guess, I'm not that far along yet).

8. KingGeorge Group Title

That's certainly not the solution I have.

9. nbouscal Group Title

Well it's kind of part of the definition of Z, isn't it, that what you describe in the problem would be the case? I mean, it should be a pretty straightforward proof by induction, I would think. But I might be misunderstanding the question.

10. KingGeorge Group Title

I'm using $$\mathbb{Z}^+$$ which is equivalent to $$\mathbb{N}$$ with out 0. I'm just repeating the problem the way it was given to me.

11. nbouscal Group Title

Hey! You can answer a question I have! I just discovered this site today. How do you do inline TeX? And yeah, I follow what ring you're referring to, I just don't see what makes the problem interesting, I guess.

12. beginnersmind Group Title

I need some help, I know almost no number theory. To define a multiplicative function I need to define it on all powers of all primes, right?

13. nbouscal Group Title

Sorry, not a ring at all, just a set. Even so, it seems like a pretty simple induction proof to show that f(n)=n if f(2)=2. Still could be missing something though.

14. KingGeorge Group Title

Use "\ (" instead of "\ [" (without the spaces to do inline $$\LaTeX$$. Defining it on all powers of primes would certainly be helpful. Also, this is not just a simple induction proof.

15. KingGeorge Group Title

16. nbouscal Group Title

Awesome. Most sites I use just use the \$ notation, haven't seen it with parentheses before. Also, yes, I see where the problem is with this, will think about it for some time and try to work it out.

17. nbouscal Group Title

1 is coprime to every integer, so $$f(n)=n$$ by definition.

18. joemath314159 Group Title

That just means:$f(n\cdot 1)=f(n)f(1)$first you need to show f(1)=1, which can be done like this:$f(2)=f(2\cdot 1)=f(2)f(1)\Longrightarrow 2=2f(1)\Longrightarrow f(1)=1$

19. beginnersmind Group Title

I think that only proves f(n)=f(n)*f(1)=f(n)

20. nbouscal Group Title

Right, what Joe said.

21. beginnersmind Group Title

Anyway, I'm stuck, I'll read KG's solution later.

22. nbouscal Group Title

Or alternatively, the way that I thought about it, if f is strictly increasing, the only element left in Z+ that is less than 2 is 1, so f(1) has to be 1

23. nbouscal Group Title

I spent a good ten minutes thinking about rebuilding the series of primes, should have remembered, always start with the identity :P

24. beginnersmind Group Title

George Polya said, if you can't solve the problem, try to solve an easier one that's similar. Any suggestions?

25. beginnersmind Group Title

Looking only at the set of primes and numbers that are the product of n distinct primes what constraint do we get for f, if we want to keep it strictly increasing?

26. nbouscal Group Title

I tried building it up from primes that way and hit a dead end, which is when I went back to basics and realized you could just force the definition via the identity. Building it by primes seems a bit more difficult, and I'm not positive how you would go about it.

27. beginnersmind Group Title

"and realized you could just force the definition via the identity." What do you mean by that?

28. experimentX Group Title

f(2*1) = f(2)f(1) f(1) = 1 f(3*2) = f(3) 2 f(3) = f(6)/2 f(5) = f(10)/2 f(n) = f(2n)/2 for odd n

29. KingGeorge Group Title

@experimentX That is correct, but it still doesn't solve the original problem.

30. KingGeorge Group Title

I have posted a hint for those who had previously participated.

31. Ishaan94 Group Title

$f(2) = f(1\cdot2) = f(1)\cdot f(2) = 2 \implies f(1) = 1$ Lets assume $$f(k-1) = k-1$$ and same follows up from 2. $f(k) = f\left(\frac k 2 \cdot 2\right) = f\left(\frac k2\right)\cdot f(2)$ $1 < \frac{k}2< k-1$ Is this enough lol? :/ :(

32. KingGeorge Group Title

If you knew that $$f(n)=n$$ for at least one value of $$n\geq3$$ that would work (with a little bit of fixing). But first you need to show that $$f(n)=n$$ for some value of $$n$$ that's not 1 or 2.

33. Ishaan94 Group Title

2=f(2)<f(3)<f(4)=f(2)*f(2)=4 what's wrong with this one?

34. KingGeorge Group Title

2 is not coprime to 2. So $$f(2)f(2)$$ does not necessarily equal $$f(4)$$

35. experimentX Group Title

f(2) = f(1+1) = f(1) + f(1) ???

36. Ishaan94 Group Title

I don't understand it. Can you give me an example which proves f(2)f(2) isn't necessarily f(4). Sorry If I sound dumb.

37. KingGeorge Group Title

@experimentX Not necessarily. This is not necessarily a homomorphism. @Ishaan94 Suppose $$f(3) =10$$ and $$f(6)=f(2)f(3)=20$$. $$f(4)$$ could be anywhere in between. The function is given as multiplicative (see first post for definition), so if the two arguments don't have a gcd of 1, you can't say anything about the product.

38. KingGeorge Group Title

@experimentX It would probably be simpler to say that the function is just multiplicative. Not additive, and what you posted was the definition of additive.

39. Ishaan94 Group Title

$f(3)>f(2) \implies f(3) \ge f(2) + 1 \implies f(3) \ge 3$$f(4) \ge 4 \implies f(n) \ge n$Can we assume $$f(n)=n+k$$? but k may not be necessarily a constant term right?

40. Ishaan94 Group Title

But this doesn't help either :/

41. KingGeorge Group Title

That's the right direction. If you follow that direction correctly, you'll be able to get the lower bound you need. To get the upper bound you need, you'll need to change things a little bit first.

42. KingGeorge Group Title

I'll post a second, more helpful hint in half an hour or so if you think you need it.

43. Ishaan94 Group Title

$f(3) =3 +k, \quad f(6) =f(3)f(2) = 6+2k, \quad f(5)< 6 +2k, \quad f(4)> 3+k$$f(5) \le 5 + 2k, \quad f(4) \ge 4+ k$$f(5) \ge f(4) + 1 \implies f(4)+1\le f(5)\le5+2k$ If I prove $$f(4) +1 = 5+2k$$ then this might work out, but $$f(4) +1 \ge 5+k$$. It's just more and more and more inequalities. :/

44. Ishaan94 Group Title

From the above inequalities, $5+ k\le f(5) \le 5+2k$

45. KingGeorge Group Title

Have posted 2nd hint. I would also suggest letting $$f(3)=k$$. It makes the manipulations easier

46. Ishaan94 Group Title

wait... $$f(4) + 1 \le 5+ 2k \implies f(4) \le 4 +2k \implies f(3) <(4) \le 4+2k$$ $$\implies 4 + k \le f(4) \le 4 + 2k$$ $\implies n + k \le f(n) \le n + 2k$ woow, nice. 4 + 2k< 5 + k k< 1 Woow, Wow. K<1. Yay! I am so happy! K can not be less than Zero as f(n) is supposed to be an increasing function. So, K has to be Zero. Yay! Yayyy! But I am not sure if it's right. :/

47. KingGeorge Group Title

I don't think that's quite right. It's almost correct, but in your last expression, 4+2k>5+k, I'm pretty sure the second k is not the same k.

48. KingGeorge Group Title

If you could convince me that the k's are in fact the same, you would indeed have a proof.

49. Ishaan94 Group Title

No, it's same. Really?$f(3) >2 \implies f(3) \ge f(2) + 1 \implies f(3) \ge 3$$\implies f(n) \ge n$ Let $$f(3) = 3 + k$$. $f(6) = f(2) \cdot f(3) = 6 + 2k$ $f(5) < f(6) \implies f(5) < 6+ 2k \implies f(5) \le 5 + 2k \tag1$ $f(4) < f(5) \implies f(4) + 1 \le f(5) \tag2$ From1 and 2.$f(4) +1 \le f(5) \le 5+2k \tag3$ $f(3) < f(4) \implies f(3) + 1 \le f(4) \implies 4 + k \le f(4) \tag 4$ From 4 and 3 $5 + k \le f(5) \le 5 + 2k \tag5$Also from 3$f(4) + 1\le 5+2k \implies f(4) \le 4 + 2k\tag6$From 4 and 6 $4+k \le f(4) \le 4+ 2k$ $\implies n+k \le f(n) \le n +2k\tag7$ From 5 and 6 $4+2k < 5+k \implies k<1 \implies k=0\tag8$From 7 and 8 $n\le f(n)\le n \implies f(n) = n$

50. Ishaan94 Group Title

Can I type QED now? :D

51. KingGeorge Group Title

I think I see the actual problem now. You're trying to compare an upper bound for 4, and a lower bound for 5. Your claim is that those bounds can't overlap, but I don't see anything that shows those bounds can't overlap. It is true of course, that the best bounds don't overlap, but these bounds are not as small as possible. If $$f(4)=4+2k$$, then of course $$4+2k<f(5)<6+2k$$. And we're done. But what if $$f(4)<4+2k$$? That is still possible and we no longer get the same implication.

52. KingGeorge Group Title

53. Ishaan94 Group Title

Ohh hmm :( I will try again.

54. Ishaan94 Group Title

Hmm but I still do have the bounds hehehe $n+k \le f(n) \le n+2k$ Wait... for $$n=2$$,$2+k \le f(2) \le 2+2k \implies 2+ k \le 2\le 2+2k \implies k=0$

55. Ishaan94 Group Title

56. Ishaan94 Group Title

Shall I type QED now? :D

57. KingGeorge Group Title

You have those bounds for $$n \geq3$$ so you can't use it for $$n=2$$. For $$n=2$$, you're given that it's 2. You're method is great, you just need to refine it to get an upper and lower bound for a single number and show that the bounds are the same.

58. Ishaan94 Group Title

Oh I knew something was wrong. hmm I will try again. (Y) :D

59. KingGeorge Group Title

It's very helpful to set $$f(3)=a$$. From there, get a lower and upper bound for $$f(18)$$. If you do it just right, you'll get a quadratic that you can solve for a.

60. Ishaan94 Group Title

I can't take $$f(18)=f(3)f(6)$$. Can I? I think $$f(2)\cdot f(9)$$ is right. Hmm $$2f(9)$$.

61. KingGeorge Group Title

That's the right track. Now what's the upper bound of$$f(9)$$ in terms of $$f(3)$$?

62. KingGeorge Group Title

Hint: $f(9)\leq f(10)-1$

63. Ishaan94 Group Title

4k-3 or 4a -3?

64. KingGeorge Group Title

Right. So this means that $$f(18) \leq \text{____}$$

65. Ishaan94 Group Title

8k - 6 $$\ge$$ f(18) How do I solve it for f(17)?

66. KingGeorge Group Title

Ignore f(17). You found the upper bound of f(18). Now we need to find the lower bound for f(18). Once again, you want to start at f(3)=k.

67. KingGeorge Group Title

In this part, you'll want to get a quadratic. You'll want to look at f(3) and f(5) and use these to get to f(18).

68. Ishaan94 Group Title

I can only get f(15) using f(3) and f(5)

69. KingGeorge Group Title

And from f(15) you can get f(18)

70. Ishaan94 Group Title

$2k + k^2 \le f(15) \le 2k^2 -k$

71. KingGeorge Group Title

Just use $2k + k^2 \le f(15)$Now this means that $\text{______}\le f(18)$

72. Ishaan94 Group Title

$$(k+1)^2 + 2 \le f(18) \le 8k+6$$

73. KingGeorge Group Title

$(k+1)^2 + 2 \le 8k+6$Now just solve for k

74. Ishaan94 Group Title

$$(k+1)^2 + 2 \le 8k + 6 \implies k^2 + 2k+3 \le 8k+6 \implies k^2 -6k - 3\le0$$ $\implies 3 -2\sqrt3 \le k \le 3 + 2\sqrt3$ But is this what I am supposed to get?

75. KingGeorge Group Title

There was a typo that was making this rather hard. It should be $(k+1)^2 + 2 \le f(18) \le 8k-6$You had this right originally, there was just a typo that was perpetuated. Try and solve this one.

76. Ishaan94 Group Title

Ohh I can't concentrate at all :( I see now, $3\le k \le 3 \implies k=3 \implies f(3) =3$

77. Ishaan94 Group Title

So, we finally have f(3)=3. But I don't get it why did we have to go all the way from 4 and 5 to 18. Why a quadratic?

78. KingGeorge Group Title

We can get the relation $k^2-6k+9 \le 0\implies (k-3)^2 \le 0$There is only one value of $$k$$ that solves this, $$k=3$$. That's why we need a quadratic. It allows us to get an inequality with only 1 solution. If it were linear, we would have a line with infinite solutions satisfying the inequality.

79. Ishaan94 Group Title

eh the proof still isn't complete I will have to show it for f(n). I think I will have to use induction. $f(3) =3$ Lets assume it works up from 3 to 2k-1. $f(k) = 2\cdot f\left(k\right) = 2k$ I am not sure if it's right. I can only recall strong induction from your previous problem.

80. KingGeorge Group Title

To show it for all n, just take $$f(2)f(3)=f(6)=2*3=6$$ So $$f(4)=4$$ and $$f(5)=5$$ since it has to be strictly increasing. We can just continue this process up to infinity.

81. Ishaan94 Group Title

and why wasn't f(1) enough for us to use induction? why did we need to solve it for f(3).

82. KingGeorge Group Title

$$f(1)$$ just let us get $$f(2)$$. We needed a value greater than 2 to generate larger numbers using the multiplicative rule.

83. Ishaan94 Group Title

ohh, thanks. i couldn't have done this without your help.

84. KingGeorge Group Title

You did very well. =D

85. KingGeorge Group Title

I appreciate the amount of time you spent on this. Thanks for actually doing this.


text:
# proving an intersection empty set

by rbzima
Tags: intersection, proving
 P: 86 I'm trying to prove $$\bigcap^{\infty}_{n=1}(0,1/n) =$$ EMPTY SET. One thing that I can't seem to bypass is getting past the closed intervals as stated in the Nested Interval Property, which states "For each $$n\in$$ N, assume we are given a closed interval $$I_n = [a_n, b_n] = {x \in$$ R $$: a_n \leq x \leq b_n}$$. Assume also that each $$I_n$$ contains $$I_n_+_1$$. Therefore, $$\bigcap^{\infty}_{n=1}I_n \neq$$ EMPTY SET. Basically, I'm stuck at taking care of the open interval part of this proof. The original for R was proved by showing sup A = x, then x is between the closed interval. Any advice would be great!
 P: 1,076 Obviously the intersection is a subset of (0,1), Why? So it will suffice to show that for any x in (0,1) that there exists a set of the form (0, 1/n) to which x does not belong. So basically given x in (0,1) can you figure out how to find n such that x does not belong to (0, 1/n)?

 Related Discussions Calculus 5 Calculus & Beyond Homework 2 General Math 3 Introductory Physics Homework 4 Introductory Physics Homework 1


text:
The main objective of finite group theory is to classify all possible finite groups up to isomorphism. This problem is very difficult even if we try to classify the groups of order less than or equal to $60$. However, we can break the problem down into several intermediate problems. This is a challenging project that requires a working knowledge of the group theory you have learned up to this point. Even if you do not complete it, it will teach you a great deal about finite groups. You can use Table 15.21 as a guide.

##### 1

Find all simple groups $G$ ( $|G| \leq 60$). Do not use the Odd Order Theorem unless you are prepared to prove it.

##### 2

Find the number of distinct groups $G$, where the order of $G$ is $n$ for $n = 1, \ldots, 60$.

##### 3

Find the actual groups (up to isomorphism) for each $n$.


text:
# TECHNICAL LANGUAGE

A statement is an assertion which is either true or false, e.g. "London is the capital of Britain" and "The Earth is hotter than the Sun" are statements: the one true, the other false. On the other hand, the following are not statements:

• How old are you?
• Wake up!
• Capital the Britain London of is,
they are, respectively, a question, a command and nonsense.

Mathematics is concerned with statements expressed in precise and technical terms: indeed we can speak of a mathematical language with its own syntax; e.g.

• $1+5>2+3,$
• $5^2+12^2=3^2,$
• $(-1)^3=1,$
• $2(3+4=(2+3)+4.$
Of these, only the first three are statements; the fourth is nonsense (although it is expressed using mathematical symbols, it does not obey the rules of syntax). Often we deal with assertions containing variables $x,y,\ldots,$ etc., e.g. $x\geq10,x^2-y^3=z^2,-1<xy<1.$ These become statements if the variables are replaced by particular and appropriate mathematical entities in this case numbers. We shall say that these are "statements about $x,y,\ldots".$

Throughout this document all variables are understood to refer to /integers:/

$\ldots{\quad},-2,{\quad}-1,{\quad}0,{\quad}1,{\quad}2,{\quad}\ldots$

## "There exists" and "For all"

In common language we make statements asserting that something exists or that all things of a certain type have some attribute, e.g.

• "There is a man in Britain over 100 years old"
• "All women in Britain are less than 200 years old"
These notions also arise in mathematical statements, e.g. There is an integer $x$ satisfying $x^2=49.$

All integers x satisfy $(x-1)(x+1)=x^2-1.$

Three useful abbreviations are s.t. (such that) and the quantifiers: $\exists$ (there exists) and $\forall$ (for all). Thus the two previous statements could be re-written as follows:
 $\exists{x}$ s.t. $x^2=49$
and
 $\forall{x;}\quad{(x-1)(x+1)=x^2-1}$
the context here dictating that x is integer. $\exists$ and $\forall$ can appear together - for example:

• $\forall{\quad}n\geq{2},{\quad}\exists{\quad}primes{\quad}p,q{\quad}s.t.{\quad}p+q=2n.$
Incidentally, this statement is known as GoldbachsConjecture: it is not known whether it is true or false!

## AND, OR and NOT

We can form composite statements by using and, or and not (sometimes symbolized by $\wedge$ $\vee$ and $\neg$ ), e.g.

• $2>1$ and $3>2$
• $5^3=100$ or $7-3=4$
• not ( $4\leq{3}$ and $2+3=5$ ).
All three of these statements are true. The term or in mathematics is always inclusive, i.e. it includes the possibility of "or both", e.g. ( $1+1=2$ or $2+2=4$ ) is true.

If $P$ is a statement then not P is called the negation of P. Not P is true if and only if P is false. Note that:

• not( P or Q ) is the same as ( not P ) and ( Q )
• not( P and Q ) is the same as ( not P ) or ( not Q )
• not( not P ) is the same as P.
Exercise. Find the negations of the following statements.

• $x\leq{4};$ or $x$ is even,
• $\exists$ x s.t. $x^2$ =51.

## Conditional Statements

A conditional statement takes the form "if ... then ...". For example, "if Liverpool win, then Liverpool score."

This means, of course, that when the statement "Liverpool win" is true, the statement "Liverpool score" is inevitably true.

The conditional statement "if P then Q ", where P and Q are statements, can be written $P\Rightarrow{Q}$ ( $\Rightarrow$ means "implies").

In this case we also say Q if P or P only if Q. For example, $x^2=4$ if $x=2;$ but $x^2=4$ only if $x=2$ or $x=-2.$

By convention, the statement $P\Rightarrow{Q}$ is deemed to be true when P is false, e.g. both of the implications below are true(!)

• $1+1=3\Rightarrow{2+2=4},$
• $1+1=3\Rightarrow{2+2=5}.$

## Converse and Contrapositive

Consider the implication ${P}\Rightarrow{Q}.$ The reverse implication ${Q}\Rightarrow{P}$ is called the converse, e.g. $(x=1)\Rightarrow(x^2=1)$ has converse $(x^2=1)\Rightarrow(x=1)$ Note that here the first implication is true, but the converse is false: the two statements are completely different.

The converse must not be confused with the contrapositive of $P\Rightarrow{Q},$ which is the implication $not{\quad}Q\Rightarrow{\quad}{not}{\quad}P.$ For example, "Liverpool win $\Rightarrow$ Liverpool score" has contrapositive "Liverpool don't score $\Rightarrow$ Liverpool don't win."

It is IMPORTANT to note that the contrapositive is logically identical to the original implication: they say the same thing in different ways.

## Necessary and Sufficient

If $P\Rightarrow{Q}$ we say that P is sufficient for Q or that Q is necessary for P, the idea being that for Q to be true it is enough that P is true; alternatively, if P is true then Q must be true, e.g.

• "For S to be a quadrilateral, it is sufficient for S to be a square."
• "For S to be a square, it is necessary for S to be a quadrilateral."
If $P\Rightarrow{Q}$ and also $Q\Rightarrow{P}$ we write $P\Leftrightarrow{Q}$ and say that P is necessary and sufficient for Q, or that P if and only if Q, (sometimes abbreviated P iff Q or that P is equivalent to Q.

It is a commonly recurring theme in mathematics to show that two different-looking statements are in fact equivalent, e.g.

• x is odd $\Leftrightarrow$ x+1 is even,
• x+y=0 is equivalent to x=-y,
• xy=0 iff x=0 or y=0,
• x^2=1 if and only if x=1 or x=-1.
It is IMPORTANT to remember that an "if and only if" statement is really two separate statements.

## Types of Statement

Mathematicians deal in Axioms, Definitions, Theorems and Lemmas.

### Axioms

An /axiom/ is a foundational statement that we simply decree is true, e.g.

• Axiom 1. $\forall$ $x,$ $x+0=x.$

### Definitions

A definition is a statement which states precisely what a technical term is to mean, e.g.

### Theorem

A /theorem/ is any statement which is a consequence of our axioms, definitions and logical rules (though we often reserve the term for important statements), e.g.

### Lemma

A lemma is a minor theorem which is a key step to proving a more major theorem.

• Lemma 1. The sum of two odd numbers is even.

CategoryMaths


text:
# MIT PRIMES Question, polynomial satisfying conditions

There was an MIT PRIMES application problem that goes like this: (don't worry, the admission ended on Dec 1, so I'm not cheating or anything)

For all $d\geq 0$, determine if there is a polynomial $p(x)$ with degree $d$ such that for all $n$ in $1\ldots 99$ inclusive satisfies $$p(n)=n+\frac1n$$ and if so, what is $p(x)$ and find the value of $p(100)$.

In English terms: find a polynomial with degree $d$ such that $$p(1)=1+\frac11=2$$ $$p(2)=2+\frac12=2.5$$ and so on. What values of $d$ work? What are the polynomials for each of those $d$? What is $p(100)$ for each of those polynomials?

My solution was to brute force it by solving a matrix (see Finding polynomials from huge sets of points), but Java wasn't good enough to do all that math (couldn't contain huge numbers with good enough precision).

Is there a better, more elegant way to do the problem? If not, how would I brute force it?

If $p(x)$ satisfies the given conditions, then $$f(x) = xp(x)-x^2-1$$ is a polynomial of degree $d+1$ (assuming $d\geq 1$) with zeros at $x = 1,2,\ldots,99$.

Note that $f$ satisfies $f(0) = -1$, so it cannot be the polynomial that is identically zero-valued. Thus $f$ can have at most $d+1$ distinct zeros, so we must have $d\geq 98$.

If $d = 98$, then the location of zeros of $f(x)$ implies $$f(x) = k(x-1)(x-2)\cdots(x-99)$$ for some constant $k$. But since $$-1 = f(0) = k\cdot(-1)^{99}(99!),$$ we may solve for $k = \frac1{99!}$. In this case $f(100) = k\cdot(99!) = 1$, so then $$p(100) = \frac1{100} (f(100) + 100^2 + 1) = 100 + \frac2{100}.$$

If $d\geq 99$, then we have $$f(x) = \frac1{99!}(x-1)(x-2)\cdots(x-99)g(x)$$ for some (arbitrary) polynomial $g(x)$ of degree $d - 98$, normalized to $g(0) = 1$. In this case $f(100) = g(100)$, and $$p(100) = \frac1{100} (g(100) + 100^2 + 1 ) = 100 + \frac{1 + g(100)}{100}.$$

From the given condition $nP(n)=n^2+1$ for $n=1,2,\cdots,99$. So the polynomial $P^{\star}(x)=xP(x)-x^2-1$ of degree $d+1$, has 99 zeros. If $d<98$ then $P^{\star}(x)$, despite of being of degree $d+1$, would have $>d+1$ roots, and hence should be identically equal to zero for every $x$. Then $P(100)$ would be simply $\frac{P^{\star}(100)+(100^2+1)}{100}=100+\frac1{100}$.

Now if $d=98$ then we can represent $$P^{\star}(x)=(x-1)(x-2)\cdots(x-99)$$ assuming $P$ to be a monic polynomial. Then $P(100)=\frac{P^{\star}(100)+(100^2+1)}{100}=\frac{99!+10001}{100}$.

If $d>98$ then we can represent $$P^{\star}(x)=(x-1)(x-2)\cdots(x-99)g(x)$$ With $g(x)$ being a polynomial of degree $\ge1$. But then due to indeterminacy of $g(x)$ we can't calculate the exact value of $P(100)$.

• That's wrong: $P^{\star}(x)=xP(x)-x^2-1$ doesn't only have to have zeros at $x=1,2,\ldots,99$, it also must satisfy $P^{\star}(0)=0\cdot P(0)-0^2-1=-1$. – Professor Vector Dec 14 '17 at 5:34
• You can’t assume $P^*$ is monic, since it necessarily has constant term equal to $-1$. Similarly, the hypothetical case where $P^*$ is identically zero cannot actually occur. – Erick Wong Dec 14 '17 at 5:35


text:
# zbMATH — the first resource for mathematics

Every compact set in $$\mathbb{C}^ n$$ is a good compact set. (English) Zbl 0188.39003

##### Keywords:
complex functions
Full Text:
##### References:
 [1] A. MARTINEAU, Sur LES fonctionnelles analytiques et la transformation de Fourier-Borel, J. Analyse Math. 9, 1-164 (1963). · Zbl 0124.31804 [2] GUNNIG-ROSSI, Analytic functions of several complex variables, Prentice Hall (1965). · Zbl 0141.08601
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# Reversion of Series

Suppose we have two quantities, $x$ and $y$, and we are given that $y$ may be expressed as an infinite power series in $x$:
$$y = a_1 x + a_2x^2 + a_3x^3 + \cdots$$


text:
### Home > CCG > Chapter 6 > Lesson 6.2.2 > Problem6-64

6-64.

Write a converse for each conditional statement below. Then, assuming the original statement is true, decide if the converse must be true or not. Homework Help ✎

1. If the cat knocks over the lamp, then it will run away frightened.

To write a converse, switch the hypothesis and conclusion of the conditional statement.

Converse: If the cat runs away frightened, then it knocked over the lamp.
This is not always true. There are lots of reasons cats run away.

2. If a standard $6$-sided die is rolled, then the chances of getting a $3$ are $\frac { 1 } { 6 }$.

What if I had a $12$-sided die?

3. If a triangle has a $90º$ angle, then it is a right triangle.

Make sure the converse is $100\%$ true for all cases.


text:
The Solar System

The total distance D_{total} = 238,857 miles.

Distance from the Earth's center to the barycenter (point O to point X) is defined as:

D_{barycenter} = \tfrac{M_{moon}}{ M_{moon} + M_{Earth}} \times D_{total}

Where if the mass of the Earth were one, the mass of the Moon relative to the earth would be 0.012. Substituting for values, we get the distance from Earth's center to the barycenter of the Earth-Moon system.

D_{barycenter} = 2,832.30 miles.

Since the radius of Earth is 3,959 miles, it means that the barycenter is located inside of Earth at about 1121 miles below the surface. Could anything weird happen in there at the level of barycenter? Well, it depends. Gravitational forces may loosen rocks a bit, push magmatic material up and that in turn could trigger an earthquake — but this is a conjecture not fully understood yet.


text: Net Fixed Assets are a metric that represents the net book value of all fixed assets on our balance sheet and allows us to calculate the residual value of the asset. Each year the depreciation is charged on the asset and that is then added to accumulated depreciation account. Net fixed assets are useful for a company to keep track of what may need to be replaced in the future. The accounting department is to calculate the fixed assets to net worth ratio. effectively property, plant and equipment after depreciation. Fixed assets are usually reported on the balance sheet as property, plant and equipment. The net fixed assets is calculated by deducting accumulated depreciation from total fixed assets:eval(ez_write_tag([[728,90],'studyfinance_com-banner-1','ezslot_2',109,'0','0'])); The net worth of the company is the value that is left after all of the liabilities have been paid off. supposes a reduction of the NFA that will be reflected in the P&L as an expense. Accelerated depreciation must be taken into account, which will allow some assets to be depreciated faster than normal, so we must take into account the differences between accounting and tax depreciation. In the merger and acquisitions, this metric mostly use. Net fixed assets are used by small business owners to figure out how much their total fixed assets are really worth or how much liability they have. The calculation of net fixed assets is: The net fixed assets calculation is useful for someone evaluating the fixed assets of an acquisition candidate, and who must rely on financial information to … Pro members can track their course progress and get access to exclusive downloads, quizzes and more! Fixed assets refer to the long-term, tangible business assets that are classified as property, plant, and equipment. To calculate net worth, we use the following formula:eval(ez_write_tag([[580,400],'studyfinance_com-large-leaderboard-2','ezslot_4',110,'0','0'])); Note: net worth is also known as shareholders’ equity and the formula above is the basic accounting equation rearranged to give the value of shareholders’ equity. In this article, we are going to see what Net Fixed Assets are and the concepts of CAPEX and Depreciation, and we will learn how to calculate each one. Some examples of fixed assets include: real estate; furniture; manufacturing equipment; antiques; Real estate is one of the more common items brought to mind when one thinks about fixed assets. eval(ez_write_tag([[300,250],'studyfinance_com-medrectangle-3','ezslot_3',108,'0','0'])); The ratio shows how much of the owner’s cash (net worth) is tied up in the form of fixed assets such as property, plants and equipment. A fully amortized asset does not mean that it is no longer useful, in fact, they usually continue to be used once it is fully amortized. How to calculate Net Fixed Assets? 2. \text{Fixed Assets to Net Worth} = \dfrac{Net\: Fixed\: Assets}{Net\: Worth}, Net\: Fixed\: Assets = \text{Total Fixed Assets} - Accumulated\: Depreciation, Net\: Worth = Total\: Assets - Total\: Liabilities, \text{Fixed Assets to Net Worth} = \dfrac{100{,}000}{300{,}000} = 33.33\%, Fixed Assets to Net Worth Ratio Conclusion, Fixed Assets to Net Worth Ratio Calculator. For instance, an influx of customer deposits or progress payments with corresponding current liabilities might lead to lower fixed ratio whereas using most of the available cash to acquire a new asset at the end of the financial year might distort the ratio for the year. Fixed assets to net worth, also known as the non-current assets to net worth ratio, is a financial ratio used to measure the solvency of a company. The net fixed asset formula is calculated by subtracting all accumulated depreciation and impairments from the total purchase price and improvement cost of all fixed assets reported on the balance sheet.Net Fixed Assets = Total Fixed Assets – Accumulated DepreciationThis is a pretty simple equation with all of these assets are reported on the face of the balance sheet. they are the sum of the three previous concepts and will be directly integrated into our Balance. A fixed asset is a long-term tangible piece of property or equipment that a firm owns and uses in its operations to generate income. You can use the fixed assets to net worth ratio calculator below to quickly calculate the fixed assets to net worth ratio of a company by entering the required numbers. Fixed assets consist of plants, properties or equipment. Fixed assets are those long term, tangible assets held by the firm for business use, and which the firm does not plan to convert to cash in the current or upcoming fiscal years. Our microtraining contains many more topics, easy and difficult to teach you financial modeling, 15 Southampton Buildings,London,WC2A 1AL,UK. The concept is used to determine the residual fixed asset or liability amount for a business. For example, On 1st April 2016 Furniture worth $100,000 was purchased. Generally, a higher fixed asset ratio implies more effective utilization of investments in fixed assets to generate revenue. … If we calculate the fixed assets turnover ratio for ABC firm, it comes out to be 2.5. Fixed-assets-to-net-worth ratio can be calculated by dividing the value of all fixed assets by net worth. When the company has a negative Cash Flow due to the acquisition of fixed assets, it will indicate that it is in a growth phase. Definition: Net of fixed assets are the net of gross value of fixed assets in balance sheet after the elimination of accumulated depreciation expenses, accumulated impairment expenses, and the debt or liabilities that entity used to acquire fixed assets. Net fixed assets are calculated with the following formula: fixed asset purchase price + additions to existing assets - accumulated depreciation - accumulated asset impairment - liabilities associated with the fixed asset. Accumulated depreciation expenses are the total depreciation expenses of assets from the beginning to the reporting date. The useful life of plant & machinery is 15 years and say its residual value is 10% of the cost of the asset. Study Finance is an educational platform to help you learn fundamental finance, accounting, and business concepts. Consider their net revenue is 50 lakhs. Basic Net Fixed Assets Formula. is given by the NFA at the end of the previous exercise. 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text:
A Domain object is a representation of a single dimension of a ParamSet. Domain objects are used to construct ParamSets, either through the ps() short form, or through the ParamSet$search_space() mechanism (see to_tune()). Domain corresponds to a Param object, except it does not have an $id, and it does have a trafo and dependencies (depends) associated with it. For each of the basic Param classes (ParamInt, ParamDbl, ParamLgl, ParamFct, and ParamUty) there is a function constructing a Domain object (p_int(), p_dbl(), p_lgl(), p_fct(), p_uty()). They each have the same arguments as the corresponding Param $new() function, except without the id argument, and with the the additional parameters trafo, and depends. Domain objects are representations of parameter ranges and are intermediate objects to be used in short form constructions in to_tune() and ps(). Because of their nature, they should not be modified by the user. The Domain object's internals are subject to change and should not be relid upon. p_int( lower = -Inf, upper = Inf, special_vals = list(), default = NO_DEF, tags = character(), depends = NULL, trafo = NULL, logscale = FALSE ) p_dbl( lower = -Inf, upper = Inf, special_vals = list(), default = NO_DEF, tags = character(), tolerance = sqrt(.Machine$double.eps),
depends = NULL,
trafo = NULL,
logscale = FALSE
)

p_uty(
default = NO_DEF,
tags = character(),
custom_check = NULL,
depends = NULL,
trafo = NULL
)

p_lgl(
special_vals = list(),
default = NO_DEF,
tags = character(),
depends = NULL,
trafo = NULL
)

p_fct(
levels,
special_vals = list(),
default = NO_DEF,
tags = character(),
depends = NULL,
trafo = NULL
)

## Arguments

lower (numeric(1)) Lower bound, can be -Inf. (numeric(1)) Upper bound can be +Inf. (list()) Arbitrary special values this parameter is allowed to take, to make it feasible. This allows extending the domain of the parameter. Note that these values are only used in feasibility checks, neither in generating designs nor sampling. (any) Default value. Can be from the domain of the parameter or an element of special_vals. Has value NO_DEF if no default exists. NULL can be a valid default. The value has no effect on ParamSet$values or the behavior of ParamSet$check(), $test() or $assert(). The default is intended to be used for documentation purposes.  (character()) Arbitrary tags to group and subset parameters. Some tags serve a special purpose: "required" implies that the parameters has to be given when setting values in ParamSet. (call | expression) An expression indicating a requirement for the parameter that will be constructed from this. Can be given as an expression (using quote()), or the expression can be entered directly and will be parsed using NSE (see examples). The expression may be of the form  ==  or  %in% , which will result in dependencies according to ParamSet$add_dep(on = "", cond = CondEqual$new()) or ParamSet$add_dep(on = "", cond = CondAnyOf$new()), respectively (see CondEqual, CondAnyOf). The expression may also contain multiple conditions separated by &&. (function) Single argument function performing the transformation of a parameter. When the Domain is used to construct a ParamSet, this transformation will be applied to the corresponding parameter as part of the $trafo function. Note that the trafo is not inherited by TuneTokens! Defining a parameter with e.g. p_dbl(..., trafo = ...) will not automatically give the to_tune() assigned to it a transformation. trafo only makes sense for ParamSets that get used as search spaces for optimization or tuning, it is not useful when defining domains or hyperparameter ranges of learning algorithms, because these do not use trafos. (logical(1)) Put numeric domains on a log scale. Default FALSE. Log-scale Domains represent parameter ranges where lower and upper bounds are logarithmized, and where a trafo is added that exponentiates sampled values to the original scale. This is not the same as setting trafo = exp, because logscale = TRUE will handle parameter bounds internally: a p_dbl(1, 10, logscale = TRUE) results in a ParamDbl that has lower bound 0, upper bound log(10), and uses exp transformation on these. Therefore, the given bounds represent the bounds after the transformation. (see examples). p_int() with logscale = TRUE results in a ParamDbl, not a ParamInt, but with bounds log(max(lower, 0.5)) ... log(upper + 1) and a trafo similar to "as.integer(exp(x))" (with additional bounds correction). The lower bound is lifted to 0.5 if lower 0 to handle the lower == 0 case. The upper bound is increased to log(upper + 1) because the trafo would otherwise almost never generate a value of upper. When logscale is TRUE, then upper bounds may be infinite, but lower bounds should be greater than 0 for p_dbl() or greater or equal 0 for p_int(). Note that "logscale" is not inherited by TuneTokens! Defining a parameter with p_dbl(... logscale = TRUE) will not automatically give the to_tune() assigned to it log-scale. logscale only makes sense for ParamSets that get used as search spaces for optimization or tuning, it is not useful when defining domains or hyperparameter ranges of learning algorithms, because these do not use trafos. logscale happens on a natural (e == 2.718282...) basis. Be aware that using a different base (log10()/10^, log2()/2^) is completely equivalent and does not change the values being sampled after transformation. (numeric(1)) Initializes the $tolerance field that determines the (function()) Custom function to check the feasibility. Function which checks the input. Must return 'TRUE' if the input is valid and a character(1) with the error message otherwise. This function should not throw an error. Defaults to NULL, which means that no check is performed. (character | atomic | list) Allowed categorical values of the parameter. If this is not a character, then a trafo is generated that converts the names (if not given: as.character() of the values) of the levels argument to the values. This trafo is then performed before the function given as the trafo argument.

## Value

A Domain object.

The p_fct function admits a levels argument that goes beyond the levels accepted by ParamFct$new(). Instead of a character vector, any atomic vector or list (optionally named) may be given. (If the value is a list that is not named, the names are inferred using as.character() on the values.) The resulting Domain will correspond to a range of values given by the names of the levels argument with a trafo that maps the character names to the arbitrary values of the levels argument. ## See also Other ParamSet construction helpers: ps(), to_tune() ## Examples params = ps( unbounded_integer = p_int(), bounded_double = p_dbl(0, 10), half_bounded_integer = p_dbl(1), half_bounded_double = p_dbl(upper = 1), double_with_trafo = p_dbl(-1, 1, trafo = exp), extra_double = p_dbl(0, 1, special_vals = list("xxx"), tags = "tagged"), factor_param = p_fct(c("a", "b", "c")), factor_param_with_implicit_trafo = p_fct(list(a = 1, b = 2, c = list())) ) print(params) #> <ParamSet> #> id class lower upper nlevels default #> 1: unbounded_integer ParamInt -Inf Inf Inf <NoDefault[3]> #> 2: bounded_double ParamDbl 0 10 Inf <NoDefault[3]> #> 3: half_bounded_integer ParamDbl 1 Inf Inf <NoDefault[3]> #> 4: half_bounded_double ParamDbl -Inf 1 Inf <NoDefault[3]> #> 5: double_with_trafo ParamDbl -1 1 Inf <NoDefault[3]> #> 6: extra_double ParamDbl 0 1 Inf <NoDefault[3]> #> 7: factor_param ParamFct NA NA 3 <NoDefault[3]> #> 8: factor_param_with_implicit_trafo ParamFct NA NA 3 <NoDefault[3]> #> value #> 1: #> 2: #> 3: #> 4: #> 5: #> 6: #> 7: #> 8: #> Trafo is set. params$trafo(list(
bounded_double = 1,
double_with_trafo = 1,
factor_param = "c",
factor_param_with_implicit_trafo = "c"
))
#> $bounded_double #> [1] 1 #> #>$double_with_trafo
#> [1] 2.718282
#>
#> $factor_param #> [1] "c" #> #>$factor_param_with_implicit_trafo
#> list()
#>
# logscale:
params = ps(x = p_dbl(1, 100, logscale = TRUE))

# The ParamSet has bounds log(1) .. log(100):
print(params)
#> <ParamSet>
#>    id    class lower   upper nlevels        default value
#> 1:  x ParamDbl     0 4.60517     Inf <NoDefault[3]>
#> Trafo is set.
# When generating a equidistant grid, it is equidistant within log values
grid = generate_design_grid(params, 3)
print(grid)
#> <Design> with 3 rows:
#>           x
#> 1: 0.000000
#> 2: 2.302585
#> 3: 4.605170
# But the values are on a log scale with desired bounds after trafo
print(grid$transpose()) #> [[1]] #> [[1]]$x
#> [1] 1
#>
#>
#> [[2]]
#> [[2]]$x #> [1] 10 #> #> #> [[3]] #> [[3]]$x
#> [1] 100
#>
#>
# Integer parameters with logscale are ParamDbls pre-trafo
params = ps(x = p_int(0, 10, logscale = TRUE))
print(params)
#> <ParamSet>
#>    id    class      lower    upper nlevels        default value
#> 1:  x ParamDbl -0.6931472 2.397895     Inf <NoDefault[3]>
#> Trafo is set.
grid = generate_design_grid(params, 4)
print(grid)
#> <Design> with 4 rows:
#>             x
#> 1: -0.6931472
#> 2:  0.3372003
#> 3:  1.3675478
#> 4:  2.3978953
# ... but get transformed to integers.
print(grid$transpose()) #> [[1]] #> [[1]]$x
#> [1] 0
#>
#>
#> [[2]]
#> [[2]]$x #> [1] 1 #> #> #> [[3]] #> [[3]]$x
#> [1] 3
#>
#>
#> [[4]]
#> [[4]]\$x
#> [1] 10
#>
#>
`