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# Applications of Integrals in Mathematics

Applications of integrals can be tricky- it can be applied to math, science, and engineering, economics etc. If you don’t have the right teacher or textbook survival guide, getting lost before exam can be easy.

In this guide you’re going to learn how integration is applied in various scenarios, the different types of integrals, and some examples.

## Applications of Integrals: What does it mean?

Integrals can be applied in many different areas like math, science, and engineering, statistics,and economics.

In math, we use it to find:

• The center of mass (Centroid) of an area with curved sides
• The area between two curves
• The area under a curve
• The average value of a curve
• Integral transforms like Laplace and Fourier transforms.

When speaking about integrals, there are two different types: definite integrals (also called Riemann Integral) and indefinite integrals.

In simple terms, a definite integral has definite limits ( defined upper and lower limits/bounds) which can be represented by:

In contrast to definite integrals, an indefinite integral is an integral which has no limits. This can be represented by:

(C is the constant of integration)

## Solved: Example Problem of Applications of Integrals

The given function is a type of parabola, the graph is as shown below:

First, take a look at our notes below:

We will integrate the given function to get . We know that using which we will compute the value of the constant of integration. Once we know the value of constant of integration, the next step is to find f(4).

1) Knowing all that, let’s integrate .To do that, we need to apply the power rule of integration which would be:

2)  Using the power rule,

……………(1)

Thus we got the result of integration.

3) Now we will find the value of the constant of integration,

We know that , hence applying this value to equation(1) we get

Therefore we found the value of the constant of integration which is

(4) Now we will apply this value of integration constant to

## Conclusion

This is just one example of many different scenarios. We have to remember in math alone there are numerous applications and in physics even more! Applications of integrals is a topic that takes practice with the right survival guides and experts to help you with solutions– you’ll be on the right track.


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Outlook: MVB Financial Corp. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Hold
Time series to forecast n: 27 Feb 2023 for (n+16 weeks)
Methodology : Modular Neural Network (Speculative Sentiment Analysis)

## Abstract

MVB Financial Corp. Common Stock prediction model is evaluated with Modular Neural Network (Speculative Sentiment Analysis) and Logistic Regression1,2,3,4 and it is concluded that the MVBF stock is predictable in the short/long term. According to price forecasts for (n+16 weeks) period, the dominant strategy among neural network is: Hold

## Key Points

1. Market Outlook
2. Short/Long Term Stocks
3. What are buy sell or hold recommendations?

## MVBF Target Price Prediction Modeling Methodology

We consider MVB Financial Corp. Common Stock Decision Process with Modular Neural Network (Speculative Sentiment Analysis) where A is the set of discrete actions of MVBF stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4

F(Logistic Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Modular Neural Network (Speculative Sentiment Analysis)) X S(n):→ (n+16 weeks) $R=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$

n:Time series to forecast

p:Price signals of MVBF stock

j:Nash equilibria (Neural Network)

k:Dominated move

a:Best response for target price

For further technical information as per how our model work we invite you to visit the article below:

How do AC Investment Research machine learning (predictive) algorithms actually work?

## MVBF Stock Forecast (Buy or Sell) for (n+16 weeks)

Sample Set: Neural Network
Stock/Index: MVBF MVB Financial Corp. Common Stock
Time series to forecast n: 27 Feb 2023 for (n+16 weeks)

According to price forecasts for (n+16 weeks) period, the dominant strategy among neural network is: Hold

X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)

Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)

Z axis (Grey to Black): *Technical Analysis%

## IFRS Reconciliation Adjustments for MVB Financial Corp. Common Stock

1. If items are hedged together as a group in a cash flow hedge, they might affect different line items in the statement of profit or loss and other comprehensive income. The presentation of hedging gains or losses in that statement depends on the group of items
2. If there is a hedging relationship between a non-derivative monetary asset and a non-derivative monetary liability, changes in the foreign currency component of those financial instruments are presented in profit or loss.
3. For example, when the critical terms (such as the nominal amount, maturity and underlying) of the hedging instrument and the hedged item match or are closely aligned, it might be possible for an entity to conclude on the basis of a qualitative assessment of those critical terms that the hedging instrument and the hedged item have values that will generally move in the opposite direction because of the same risk and hence that an economic relationship exists between the hedged item and the hedging instrument (see paragraphs B6.4.4–B6.4.6).
4. The purpose of estimating expected credit losses is neither to estimate a worstcase scenario nor to estimate the best-case scenario. Instead, an estimate of expected credit losses shall always reflect the possibility that a credit loss occurs and the possibility that no credit loss occurs even if the most likely outcome is no credit loss.

*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.

## Conclusions

MVB Financial Corp. Common Stock is assigned short-term Ba1 & long-term Ba1 estimated rating. MVB Financial Corp. Common Stock prediction model is evaluated with Modular Neural Network (Speculative Sentiment Analysis) and Logistic Regression1,2,3,4 and it is concluded that the MVBF stock is predictable in the short/long term. According to price forecasts for (n+16 weeks) period, the dominant strategy among neural network is: Hold

### MVBF MVB Financial Corp. Common Stock Financial Analysis*

Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementBaa2B3
Balance SheetBa2Baa2
Leverage RatiosBaa2C
Cash FlowB2Caa2
Rates of Return and ProfitabilityB2Baa2

*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?

### Prediction Confidence Score

Trust metric by Neural Network: 90 out of 100 with 624 signals.

## References

1. A. Tamar, Y. Glassner, and S. Mannor. Policy gradients beyond expectations: Conditional value-at-risk. In AAAI, 2015
2. Bessler, D. A. S. W. Fuller (1993), "Cointegration between U.S. wheat markets," Journal of Regional Science, 33, 481–501.
3. Varian HR. 2014. Big data: new tricks for econometrics. J. Econ. Perspect. 28:3–28
4. Wan M, Wang D, Goldman M, Taddy M, Rao J, et al. 2017. Modeling consumer preferences and price sensitiv- ities from large-scale grocery shopping transaction logs. In Proceedings of the 26th International Conference on the World Wide Web, pp. 1103–12. New York: ACM
5. Çetinkaya, A., Zhang, Y.Z., Hao, Y.M. and Ma, X.Y., When to Sell and When to Hold AQN Stock. AC Investment Research Journal, 101(3).
6. Bera, A. M. L. Higgins (1997), "ARCH and bilinearity as competing models for nonlinear dependence," Journal of Business Economic Statistics, 15, 43–50.
7. K. Tuyls and G. Weiss. Multiagent learning: Basics, challenges, and prospects. AI Magazine, 33(3): 41–52, 2012
Frequently Asked QuestionsQ: What is the prediction methodology for MVBF stock?
A: MVBF stock prediction methodology: We evaluate the prediction models Modular Neural Network (Speculative Sentiment Analysis) and Logistic Regression
Q: Is MVBF stock a buy or sell?
A: The dominant strategy among neural network is to Hold MVBF Stock.
Q: Is MVB Financial Corp. Common Stock stock a good investment?
A: The consensus rating for MVB Financial Corp. Common Stock is Hold and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of MVBF stock?
A: The consensus rating for MVBF is Hold.
Q: What is the prediction period for MVBF stock?
A: The prediction period for MVBF is (n+16 weeks)

## People also ask

What are the top stocks to invest in right now?

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### Summary of Wave Digital Elements

From Eq.(F.1), we have that the general reflectance of impedance with respect to the reference impedance in the wave variable formulation is given by

 (F.13)

In WDF construction, the free constant in the bilinear transform is taken to be . Thus we obtain . When is first order, it is possible to choose the reference impedance so as to eliminate the delay-free path in the digital reflectance , and so its value depends on the actual physical element being digitized.

Next Section:
Wave Digital Mass
Previous Section:
A Physical Derivation of Wave Digital Elements


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Chromatic polynomial

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All non-isomorphic graphs on 3 vertices and their chromatic polynomials, clockwise from the top. The independent 3-set: ${\displaystyle k^{3}}$. An edge and a single vertex: ${\displaystyle k^{2}(k-1)}$. The 3-path: ${\displaystyle k(k-1)^{2}}$. The 3-clique: ${\displaystyle k(k-1)(k-2)}$.

The chromatic polynomial is a graph polynomial studied in algebraic graph theory, a branch of mathematics. It counts the number of graph colorings as a function of the number of colors and was originally defined by George David Birkhoff to study the four color problem. It was generalised to the Tutte polynomial by Hassler Whitney and W. T. Tutte, linking it to the Potts model of statistical physics.

History

George David Birkhoff introduced the chromatic polynomial in 1912, defining it only for planar graphs, in an attempt to prove the four color theorem. If ${\displaystyle P(G,k)}$ denotes the number of proper colorings of G with k colors then one could establish the four color theorem by showing ${\displaystyle P(G,4)>0}$ for all planar graphs G. In this way he hoped to apply the powerful tools of analysis and algebra for studying the roots of polynomials to the combinatorial coloring problem.

Hassler Whitney generalised Birkhoff’s polynomial from the planar case to general graphs in 1932. In 1968, Read asked which polynomials are the chromatic polynomials of some graph, a question that remains open, and introduced the concept of chromatically equivalent graphs.[1] Today, chromatic polynomials are one of the central objects of algebraic graph theory.[2]

Definition

All proper vertex colorings of vertex graphs with 3 vertices using k colors for ${\displaystyle k=0,1,2,3}$. The chromatic polynomial of each graph interpolates through the number of proper colorings.

For a graph G, ${\displaystyle P(G,k)}$ counts the number of its (proper) vertex k-colorings. Other commonly used notations include ${\displaystyle P_{G}(k)}$, ${\displaystyle \chi _{G}(k)}$, or ${\displaystyle \pi _{G}(k)}$. There is a unique polynomial ${\displaystyle P(G,x)}$ which evaluated at any integer k ≥ 0 coincides with ${\displaystyle P(G,k)}$; it is called the chromatic polynomial of G.

For example, to color the path graph ${\displaystyle P_{3}}$ on 3 vertices with k colors, one may choose any of the k colors for the first vertex, any of the ${\displaystyle k-1}$ remaining colors for the second vertex, and lastly for the third vertex, any of the ${\displaystyle k-1}$ colors that are different from the second vertex's choice. Therefore, ${\displaystyle P(P_{3},k)=k\cdot (k-1)\cdot (k-1)}$ is the number of k-colorings of ${\displaystyle P_{3}}$. For a variable x (not necessarily integer), we thus have ${\displaystyle P(P_{3},x)=x(x-1)^{2}=x^{3}-2x^{2}+x}$. (Colorings which differ only by permuting colors or by automorphisms of G are still counted as different.)

Deletion–contraction

The fact that the number of k-colorings is a polynomial in k follows from a recurrence relation called the deletion–contraction recurrence or Fundamental Reduction Theorem.[3] It is based on edge contraction: for a pair of vertices ${\displaystyle u}$ and ${\displaystyle v}$ the graph ${\displaystyle G/uv}$ is obtained by merging the two vertices and removing any edges between them. If ${\displaystyle u}$ and ${\displaystyle v}$ are adjacent in G, let ${\displaystyle G-uv}$ denote the graph obtained by removing the edge ${\displaystyle uv}$. Then the numbers of k-colorings of these graphs satisfy:

${\displaystyle P(G,k)=P(G-uv,k)-P(G/uv,k)}$

Equivalently, if ${\displaystyle u}$ and ${\displaystyle v}$ are not adjacent in G and ${\displaystyle G+uv}$ is the graph with the edge ${\displaystyle uv}$ added, then

${\displaystyle P(G,k)=P(G+uv,k)+P(G/uv,k)}$

This follows from the observation that every k-coloring of G either gives different colors to ${\displaystyle u}$ and ${\displaystyle v}$, or the same colors. In the first case this gives a (proper) k-coloring of ${\displaystyle G+uv}$, while in the second case it gives a coloring of ${\displaystyle G/uv}$. Conversely, every k-coloring of G can be uniquely obtained from a k-coloring of ${\displaystyle G+uv}$ or ${\displaystyle G/uv}$ (if ${\displaystyle u}$ and ${\displaystyle v}$ are not adjacent in G).

The chromatic polynomial can hence be recursively defined as

${\displaystyle P(G,x)=x^{n}}$ for the edgeless graph on n vertices, and
${\displaystyle P(G,x)=P(G-uv,x)-P(G/uv,x)}$ for a graph G with an edge ${\displaystyle uv}$ (arbitrarily chosen).

Since the number of k-colorings of the edgeless graph is indeed ${\displaystyle k^{n}}$, it follows by induction on the number of edges that for all G, the polynomial ${\displaystyle P(G,x)}$ coincides with the number of k-colorings at every integer point x=k. In particular, the chromatic polynomial is the unique interpolating polynomial of degree at most n through the points

${\displaystyle \left\{(0,P(G,0)),(1,P(G,1)),\cdots ,(n,P(G,n))\right\}.}$

Tutte’s curiosity about which other graph invariants satisfied such recurrences led him to discover a bivariate generalization of the chromatic polynomial, the Tutte polynomial ${\displaystyle T_{G}(x,y)}$.

Examples

 Triangle ${\displaystyle K_{3}}$ ${\displaystyle x(x-1)(x-2)}$ Complete graph ${\displaystyle K_{n}}$ ${\displaystyle x(x-1)(x-2)\cdots (x-(n-1))}$ Edgeless graph ${\displaystyle {\overline {K}}_{n}}$ ${\displaystyle x^{n}}$ Path graph ${\displaystyle P_{n}}$ ${\displaystyle x(x-1)^{n-1}}$ Any tree on n vertices ${\displaystyle x(x-1)^{n-1}}$ Cycle ${\displaystyle C_{n}}$ ${\displaystyle (x-1)^{n}+(-1)^{n}(x-1)}$ Petersen graph ${\displaystyle x(x-1)(x-2)\left(x^{7}-12x^{6}+67x^{5}-230x^{4}+529x^{3}-814x^{2}+775x-352\right)}$

Properties

For fixed G on n vertices, the chromatic polynomial ${\displaystyle P(G,x)}$ is a monic polynomial of degree exactly n, with integer coefficients.

The chromatic polynomial includes at least as much information about the colorability of G as does the chromatic number. Indeed, the chromatic number is the smallest positive integer that is not a zero of the chromatic polynomial,

${\displaystyle \chi (G)=\min\{k\in \mathbb {N} :P(G,k)>0\}.}$

The polynomial evaluated at ${\displaystyle -1}$, that is ${\displaystyle P(G,-1)}$, yields ${\displaystyle (-1)^{|V(G)|}}$ times the number of acyclic orientations of G.[4]

The derivative evaluated at 1, ${\displaystyle P'(G,1)}$ equals the chromatic invariant ${\displaystyle \theta (G)}$ up to sign.

If G has n vertices and c components ${\displaystyle G_{1},\cdots ,G_{c}}$, then

• The coefficients of ${\displaystyle x^{0},\cdots ,x^{c-1}}$ are zeros.
• The coefficients of ${\displaystyle x^{c},\cdots ,x^{n}}$ are all non-zero and alternate in signs.
• The coefficient of ${\displaystyle x^{n}}$ is 1 (the polynomial is monic).
• The coefficient of ${\displaystyle x^{n-1}}$ is ${\displaystyle -|E(G)|}$.
• The coefficient of ${\displaystyle x^{1}}$ is ${\displaystyle (-1)^{n-1}}$ times the number of acyclic orientations that have a unique sink, at a specified, arbitrarily chosen vertex.[5]
• The absolute values of coefficients of every chromatic polynomial form a log-concave sequence.[6]
• ${\displaystyle \scriptstyle P(G,x)=P(G_{1},x)P(G_{2},x)\cdots P(G_{c},x)}$

The last property is generalized by the fact that if G is a k-clique-sum of ${\displaystyle G_{1}}$ and ${\displaystyle G_{2}}$ (i.e., a graph obtained by gluing the two at a clique on k vertices), then

${\displaystyle P(G,x)={\frac {P(G_{1},x)P(G_{2},x)}{x(x-1)\cdots (x-k+1)}}.}$

A graph G with n vertices is a tree if and only if

${\displaystyle P(G,x)=x(x-1)^{n-1}.}$

Chromatic equivalence

The three graphs with a chromatic polynomial equal to ${\displaystyle (x-2)(x-1)^{3}x}$.

Two graphs are said to be chromatically equivalent if they have the same chromatic polynomial. Isomorphic graphs have the same chromatic polynomial, but non-isomorphic graphs can be chromatically equivalent. For example, all trees on n vertices have the same chromatic polynomial. In particular, ${\displaystyle (x-1)^{3}x}$ is the chromatic polynomial of both the claw graph and the path graph on 4 vertices.

A graph is chromatically unique if it is determined by its chromatic polynomial, up to isomorphism. In other words, G is chromatically unique, then ${\displaystyle P(G,x)=P(H,x)}$ would imply that G and H are isomorphic. All cycle graphs are chromatically unique.[7]

Chromatic roots

A root (or zero) of a chromatic polynomial, called a “chromatic root”, is a value x where ${\displaystyle P(G,x)=0}$. Chromatic roots have been very well studied, in fact, Birkhoff’s original motivation for defining the chromatic polynomial was to show that for planar graphs, ${\displaystyle P(G,x)>0}$ for x ≥ 4. This would have established the four color theorem.

No graph can be 0-colored, so 0 is always a chromatic root. Only edgeless graphs can be 1-colored, so 1 is a chromatic root of every graph with at least one edge. On the other hand, except for these two points, no graph can have a chromatic root at a real number smaller than or equal to 32/27.[8] A result of Tutte connects the golden ratio ${\displaystyle \phi }$ with the study of chromatic roots, showing that chromatic roots exist very close to ${\displaystyle \phi ^{2}}$: If ${\displaystyle G_{n}}$ is a planar triangulation of a sphere then

${\displaystyle P(G_{n},\phi ^{2})\leq \phi ^{5-n}.}$

While the real line thus has large parts that contain no chromatic roots for any graph, every point in the complex plane is arbitrarily close to a chromatic root in the sense that there exists an infinite family of graphs whose chromatic roots are dense in the complex plane.[9]

Colorings using all colors

For a graph G on n vertices, let ${\displaystyle e_{k}}$ denote the number of colorings using exactly k colors up to renaming colors (so colorings that can be obtained from one another by permuting colors are counted as one; colorings obtained by automorphisms of G are still counted separately). In other words, ${\displaystyle e_{k}}$ counts the number of partitions of the vertex set into k (non-empty) independent sets. Then ${\displaystyle k!\cdot e_{k}}$ counts the number of colorings using exactly k colors (with distinguishable colors). For an integer x, all x-colorings of G can be uniquely obtained by choosing an integer k ≤ x, choosing k colors to be used out of x available, and a coloring using exactly those k (distinguishable) colors. Therefore:

${\displaystyle P(G,x)=\sum _{k=0}^{x}{\binom {x}{k}}k!\cdot e_{k}=\sum _{k=0}^{x}(x)_{k}\cdot e_{k}}$,

where ${\displaystyle (x)_{k}=x(x-1)(x-2)\cdots (x-k+1)}$ denotes the falling factorial. Thus the numbers ${\displaystyle e_{k}}$ are the coefficients of the polynomial ${\displaystyle P(G,x)}$ in the basis ${\displaystyle 1,(x)_{1},(x)_{2},(x)_{3},\dots }$ of falling factorials.

Let ${\displaystyle a_{k}}$ be the k-th coefficient of ${\displaystyle P(G,x)}$ in the standard basis ${\displaystyle 1,x,x^{2},x^{3},\dots }$, that is:

${\displaystyle P(G,x)=\sum _{k=0}^{n}a_{k}x^{k}}$

Stirling numbers give a change of basis between the standard basis and the basis of falling factorials. This implies:

${\displaystyle a_{k}=\sum _{j=0}^{n}(-1)^{j-k}{\begin{bmatrix}j\\k\end{bmatrix}}e_{j}}$   and   ${\displaystyle e_{k}=\sum _{j=0}^{n}{\begin{Bmatrix}j\\k\end{Bmatrix}}a_{j}}$.

Categorification

The chromatic polynomial is categorified by a homology theory closely related to Khovanov homology.[10]

Algorithms

Chromatic polynomial
InputGraph G with n vertices.
OutputCoefficients of ${\displaystyle P(G,x)}$
Running time${\displaystyle O(2^{n}n^{r})}$ for some constant ${\displaystyle r}$
Complexity#P-hard
Reduction from#3SAT
#k-colorings
InputGraph G with n vertices.
Output${\displaystyle P(G,k)}$
Running timeIn P for ${\displaystyle k=0,1,2}$. ${\displaystyle O(1.6262^{n})}$ for ${\displaystyle k=3}$. Otherwise ${\displaystyle O(2^{n}n^{r})}$ for some constant ${\displaystyle r}$
Complexity#P-hard unless ${\displaystyle k=0,1,2}$
ApproximabilityNo FPRAS for ${\displaystyle k>2}$

Computational problems associated with the chromatic polynomial include

• finding the chromatic polynomial ${\displaystyle P(G,x)}$ of a given graph G;
• evaluating ${\displaystyle P(G,x)}$ at a fixed point x for given G.

The first problem is more general because if we knew the coefficients of ${\displaystyle P(G,x)}$ we could evaluate it at any point in polynomial time because the degree is n. The difficulty of the second type of problem depends strongly on the value of x and has been intensively studied in computational complexity. When x is a natural number, this problem is normally viewed as computing the number of x-colorings of a given graph. For example, this includes the problem #3-coloring of counting the number of 3-colorings, a canonical problem in the study of complexity of counting, complete for the counting class #P.

Efficient algorithms

For some basic graph classes, closed formulas for the chromatic polynomial are known. For instance this is true for trees and cliques, as listed in the table above.

Polynomial time algorithms are known for computing the chromatic polynomial for wider classes of graphs, including chordal graphs[11] and graphs of bounded clique-width.[12] The latter class includes cographs and graphs of bounded tree-width, such as outerplanar graphs.

Deletion–contraction

The deletion-contraction recurrence gives a way of computing the chromatic polynomial, called the deletion–contraction algorithm. In the first form (with a minus), the recurrence terminates in a collection of empty graphs. In the second form (with a plus), it terminates in a collection of complete graphs. This forms the basis of many algorithms for graph coloring. The ChromaticPolynomial function in the Combinatorica package of the computer algebra system Mathematica uses the second recurrence if the graph is dense, and the first recurrence if the graph is sparse.[13] The worst case running time of either formula satisfies the same recurrence relation as the Fibonacci numbers, so in the worst case, the algorithm runs in time within a polynomial factor of

${\displaystyle \phi ^{n+m}=\left({\frac {1+{\sqrt {5}}}{2}}\right)^{n+m}\in O\left(1.62^{n+m}\right),}$

on a graph with n vertices and m edges.[14] The analysis can be improved to within a polynomial factor of the number ${\displaystyle t(G)}$ of spanning trees of the input graph.[15] In practice, branch and bound strategies and graph isomorphism rejection are employed to avoid some recursive calls, the running time depends on the heuristic used to pick the vertex pair.

Cube Method

There is a natural geometric perspective on graph colorings by observing that, as an assignment of natural numbers to each vertex, a graph coloring is a vector in the integer lattice. Since two vertices ${\displaystyle i}$ and ${\displaystyle j}$ being given the same color is equivalent to the ${\displaystyle i}$’th and ${\displaystyle j}$’th coordinate in the coloring vector being equal, each edge can be associated with a hyperplane of the form ${\displaystyle \{x\in R^{d}:x_{i}=x_{j}\}}$. The collection of such hyperplanes for a given graph is called its graphic arrangement. The proper colorings of a graph are those lattice points which avoid forbidden hyperplanes. Restricting to a set of ${\displaystyle k}$ colors, the lattice points are contained in the cube ${\displaystyle [0,k]^{n}}$. In this context the chromatic polynomial counts the number of lattice points in the ${\displaystyle [0,k]}$-cube that avoid the graphic arrangement.

Computational complexity

The problem of computing the number of 3-colorings of a given graph is a canonical example of a #P-complete problem, so the problem of computing the coefficients of the chromatic polynomial is #P-hard. Similarly, evaluating ${\displaystyle P(G,3)}$ for given G is #P-complete. On the other hand, for ${\displaystyle k=0,1,2}$ it is easy to compute ${\displaystyle P(G,k)}$, so the corresponding problems are polynomial-time computable. For integers ${\displaystyle k>3}$ the problem is #P-hard, which is established similar to the case ${\displaystyle k=3}$. In fact, it is known that ${\displaystyle P(G,x)}$ is #P-hard for all x (including negative integers and even all complex numbers) except for the three “easy points”.[16] Thus, from the perspective of #P-hardness, the complexity of computing the chromatic polynomial is completely understood.

In the expansion

${\displaystyle P(G,x)=a_{1}x+a_{2}x^{2}+\dots +a_{n}x^{n},}$

the coefficient ${\displaystyle a_{n}}$ is always equal to 1, and several other properties of the coefficients are known. This raises the question if some of the coefficients are easy to compute. However the computational problem of computing ar for a fixed r ≥ 1 and a given graph G is #P-hard, even for bipartite planar graphs.[17]

No approximation algorithms for computing ${\displaystyle P(G,x)}$ are known for any x except for the three easy points. At the integer points ${\displaystyle k=3,4,\dots }$, the corresponding decision problem of deciding if a given graph can be k-colored is NP-hard. Such problems cannot be approximated to any multiplicative factor by a bounded-error probabilistic algorithm unless NP = RP, because any multiplicative approximation would distinguish the values 0 and 1, effectively solving the decision version in bounded-error probabilistic polynomial time. In particular, under the same assumption, this rules out the possibility of a fully polynomial time randomised approximation scheme (FPRAS). For other points, more complicated arguments are needed, and the question is the focus of active research. As of 2008, it is known that there is no FPRAS for computing ${\displaystyle P(G,x)}$ for any x > 2, unless NP = RP holds.[18]

References

• Biggs, N. (1993), Algebraic Graph Theory, Cambridge University Press, ISBN 978-0-521-45897-9
• Chao, C.-Y.; Whitehead, E. G. (1978), "On chromatic equivalence of graphs", Theory and Applications of Graphs, Lecture Notes in Mathematics, 642, Springer, pp. 121–131, ISBN 978-3-540-08666-6
• Dong, F. M.; Koh, K. M.; Teo, K. L. (2005), Chromatic polynomials and chromaticity of graphs, World Scientific Publishing Company, ISBN 978-981-256-317-0
• Giménez, O.; Hliněný, P.; Noy, M. (2005), "Computing the Tutte polynomial on graphs of bounded clique-width", Proc. 31st Int. Worksh. Graph-Theoretic Concepts in Computer Science (WG 2005), Lecture Notes in Computer Science, 3787, Springer-Verlag, pp. 59–68, CiteSeerX 10.1.1.353.6385, doi:10.1007/11604686_6, ISBN 978-3-540-31000-6
• Goldberg, L.A.; Jerrum, M. (2008), "Inapproximability of the Tutte polynomial", Information and Computation, 206 (7): 908, arXiv:cs/0605140, doi:10.1016/j.ic.2008.04.003
• Helme-Guizon, Laure; Rong, Yongwu (2005), "A categorification of the chromatic polynomial", Algebraic & Geometric Topology, 5 (4): 1365–1388, arXiv:math/0412264, doi:10.2140/agt.2005.5.1365
• Huh, J. (2012), "Milnor numbers of projective hypersurfaces and the chromatic polynomial of graphs", arXiv:1008.4749v3 [math.AG]
• Jackson, B. (1993), "A Zero-Free Interval for Chromatic Polynomials of Graphs", Combinatorics, Probability and Computing, 2 (3): 325–336, doi:10.1017/S0963548300000705
• Jaeger, F.; Vertigan, D. L.; Welsh, D. J. A. (1990), "On the computational complexity of the Jones and Tutte polynomials", Mathematical Proceedings of the Cambridge Philosophical Society, 108 (1): 35–53, Bibcode:1990MPCPS.108...35J, doi:10.1017/S0305004100068936
• Linial, N. (1986), "Hard enumeration problems in geometry and combinatorics", SIAM J. Algebr. Discrete Methods, 7 (2): 331–335, doi:10.1137/0607036
• Makowsky, J. A.; Rotics, U.; Averbouch, I.; Godlin, B. (2006), "Computing graph polynomials on graphs of bounded clique-width", Proc. 32nd Int. Worksh. Graph-Theoretic Concepts in Computer Science (WG 2006), Lecture Notes in Computer Science, 4271, Springer-Verlag, pp. 191–204, CiteSeerX 10.1.1.76.2320, doi:10.1007/11917496_18, ISBN 978-3-540-48381-6
• Naor, J.; Naor, M.; Schaffer, A. (1987), "Fast parallel algorithms for chordal graphs", Proc. 19th ACM Symp. Theory of Computing (STOC '87), pp. 355–364, doi:10.1145/28395.28433, ISBN 978-0897912211.
• Oxley, J. G.; Welsh, D. J. A. (2002), "Chromatic, flow and reliability polynomials: The complexity of their coefficients.", Combinatorics, Probability and Computing, 11 (4): 403–426, doi:10.1017/S0963548302005175
• Pemmaraju, S.; Skiena, S. (2003), Computational Discrete Mathematics: Combinatorics and Graph Theory with Mathematica, Cambridge University Press, section 7.4.2, ISBN 978-0-521-80686-2
• Read, R. C. (1968), "An introduction to chromatic polynomials" (PDF), Journal of Combinatorial Theory, 4: 52–71, doi:10.1016/S0021-9800(68)80087-0
• Sekine, K.; Imai, H.; Tani, S. (1995), "Computing the Tutte polynomial of a graph of moderate size", Algorithms and Computation, 6th International Symposium, Lecture Notes in Computer Science 1004, Cairns, Australia, December 4–6, 1995: Springer, pp. 224–233
• Sokal, A. D. (2004), "Chromatic Roots are Dense in the Whole Complex Plane", Combinatorics, Probability and Computing, 13 (2): 221–261, arXiv:cond-mat/0012369, doi:10.1017/S0963548303006023
• Stanley, R. P. (1973), "Acyclic orientations of graphs", Discrete Math., 5 (2): 171–178, doi:10.1016/0012-365X(73)90108-8
• Voloshin, Vitaly I. (2002), Coloring Mixed Hypergraphs: Theory, Algorithms and Applications., American Mathematical Society, ISBN 978-0-8218-2812-0
• Wilf, H. S. (1986), Algorithms and Complexity, Prentice–Hall, ISBN 978-0-13-021973-2
• Ellis-Monaghan, Joanna A.; Merino, Criel (2011). "Graph polynomials and their applications I: The Tutte polynomial". In Dehmer, Matthias (ed.). Structural Analysis of Complex Networks. arXiv:0803.3079. doi:10.1007/978-0-8176-4789-6. ISBN 978-0-8176-4789-6.


text:
# Properties

 Label 5040.bm Number of curves $4$ Conductor $5040$ CM no Rank $1$ Graph

# Related objects

Show commands for: SageMath
sage: E = EllipticCurve("5040.bm1")

sage: E.isogeny_class()

## Elliptic curves in class 5040.bm

sage: E.isogeny_class().curves

LMFDB label Cremona label Weierstrass coefficients Torsion structure Modular degree Optimality
5040.bm1 5040bn3 [0, 0, 0, -38547, -2912814] [2] 12288
5040.bm2 5040bn4 [0, 0, 0, -12627, 510354] [2] 12288
5040.bm3 5040bn2 [0, 0, 0, -2547, -40014] [2, 2] 6144
5040.bm4 5040bn1 [0, 0, 0, 333, -3726] [2] 3072 $$\Gamma_0(N)$$-optimal

## Rank

sage: E.rank()

The elliptic curves in class 5040.bm have rank $$1$$.

## Modular form5040.2.a.bm

sage: E.q_eigenform(10)

$$q + q^{5} + q^{7} + 4q^{11} - 6q^{13} - 2q^{17} + O(q^{20})$$

## Isogeny matrix

sage: E.isogeny_class().matrix()

The $$i,j$$ entry is the smallest degree of a cyclic isogeny between the $$i$$-th and $$j$$-th curve in the isogeny class, in the LMFDB numbering.

$$\left(\begin{array}{rrrr} 1 & 4 & 2 & 4 \\ 4 & 1 & 2 & 4 \\ 2 & 2 & 1 & 2 \\ 4 & 4 & 2 & 1 \end{array}\right)$$

## Isogeny graph

sage: E.isogeny_graph().plot(edge_labels=True)

The vertices are labelled with LMFDB labels.


text:
# American Institute of Mathematical Sciences

January  1998, 4(1): 73-90. doi: 10.3934/dcds.1998.4.73

## Remarks on resolvent positive operators and their perturbation

 1 Department of Mathematics, Arizona State University, Tempe, AZ 85287-1804, United States

Received  January 1996 Revised  January 1997 Published  October 1997

We consider positive perturbations $A = B+ C$ of resolvent positive operators $B$ by positive operators $C: D(A) \to X$ and in particular study their spectral properties. We characterize the spectral bound of $A$, $s(A)$, in terms of the resolvent outputs $F(\lambda) = C (\lambda - B)^{-1}$ and derive conditions for $s(A)$ to be an eigenvalue of $A$ and a (first order) pole of the resolvent of $A$. On our way we show that the spectral radii of a completely monotonic operator family form a superconvex function. Our results will be used in forthcoming publications to study the spectral and large-time properties of positive operator semigroups.
Citation: Horst R. Thieme. Remarks on resolvent positive operators and their perturbation. Discrete & Continuous Dynamical Systems, 1998, 4 (1) : 73-90. doi: 10.3934/dcds.1998.4.73
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2019 Impact Factor: 1.338


text:
# Eigenvalues of $\ddot {y} +\dot{y}-6y = 0$

I want to write $\ddot {y} +\dot{y}-6y = 0$ in the form

$$\dot{x} = \cdots$$

$$\dot{y} = \cdots$$

then create a matrix from which I can calculate $\lambda_1,~\lambda_2$.

But I keep getting the wrong eigenvalues if I let $x = \dot{y}$.

• Just solve the characteristic equation. – Rodrigo de Azevedo Aug 7 '17 at 23:11
• "But I keep getting the wrong eigenvalues if I let x=y˙" This is the part we ought to see if this was to become an interesting question. – Did Oct 14 '17 at 13:48
• @Did How did you even find this? I posted this about ten weeks ago! – Alt-Rock Oct 14 '17 at 13:57

We let:

$$x_1 = y \\ x_2 = y'$$

Next

$$x_1' = y' = x_2 \\ x_2' = y'' = -y' + 6 y = - x_2 + 6 x_1$$

We can now write

$$X' = \begin{bmatrix} 0 & 1 \\ 6 & -1 \end{bmatrix}x$$

Can you take it from here?

Note: as an alternate approach, we can find the roots of the characteristic equation $\lambda^2 + \lambda - 6 = 0 \implies \lambda_1 = -3, \lambda_2 = 2.$


text:
# 5. Superimposed Rotation

In preparation for the constitutive formulations in later chapters, this short chapter will derive how different tensors behave under superimposed rigid body motion. This will enable us, for example, to show that is a function of in the beginning of the next chapter. The idea of “work conjugate” pairs will be introduced in this chapter. We will see that the Cauchy stress tensor, , for example, changes with rigid body rotation, while does not. Thus, and are not work-conjugate, and therefore it would not be appropriate to use them as a pair when forming a constitutive relationship. This was first recognized by Hill [Hill].

Let’s define the vector as follows:

, where is a superimposed rigid body motion, sometimes called a Euclidean transformation (see Fig).

We know that

Consider the following square body (see Fig), paying careful attention to the heavily bolded edge in order to track the motion of the body.

Clearly, if (which is what we’re going to want), and, say, ,
then (this should be clear from the above Fig).

Since we want to find such that , we can conclude that .

What we do know is the following:

; ;

So, , which gives:

(1)

Eq. 1 is what we wanted to find, and it should be expected. Recall that is physically understood to be a deformation (axial strains and shear), , followed by a rotation, . Thus, is the total deformation + rotation, , followed by an explicit rigid body rotation, , – i.e. a rotation that is superimposed on .

Now, we can see how different measures of strain and stress behave under superimposed rigid body motion, by substituting the above expression for .

(2)

Thus, is invariant to rigid body motion.

note: always ; since is an orthogonal tensor

note:
informal proof: ;

or

NOT = which does not match either of the above expressions for
(or which is similarly )

matches expression for
or
matches expression for

formal proof: see Appendix A.1

(3)

What about , , and ?

Recall,

Insert into the above expression

So, ; ( is symmetric ; is orthogonal)

Also, (skipped work)

Recall also,

So,

note: ( since is orthogonal)

(4)

note: is antisymmetric, and indeed, the two terms that make up are each antisymmetric

Since , we can say that (1)

We also know the following three expressions to be true:
(2)
(3)
(4)

(4) (1) (3) (5)

(2) (5) ()

(5)

note: Since and , we can say that and are a “work-conjugate pair.” A more rigorous proof, starting with the “balance of mechanical energy” can be found in [Holzapfel].

How about the nominal stress and the Second Piola-Kirchhoff Stress?

Again, we will need to use

Since depends on , we will need . It turns out, though, that

note: the determinant of an orthogonal tensor is “1”

So,

(6)

is a good measure of stress when there are small (infinitesimal) strains. It will also be used as a “work-conjugate pair” with in the following chapters on constitutive relationships.

• G. A. Holzapfel, Nonlinear Solid Mechanics: A Continuum Approach for Engineering, Baffins Lane, Chichester, West Sussex PO19 1UD, England: John Wiley & Sons Ltd., 2000.
[Bibtex]
@book{holzapfel,

address = {Baffins Lane, Chichester, West Sussex PO19 1UD, England},

author = {Holzapfel, Gerhard A.},
publisher = {John Wiley \& Sons Ltd.},

title = {{Nonlinear Solid Mechanics: A Continuum Approach for Engineering}},

year = {2000}
}
• R. Hill, “On constitutive inequalities for simple materials I,” Journal of the Mechanics and Physics of Solids, vol. 16, iss. 4, pp. 229-242, 1968.
[Bibtex]
@article{Hill,
title={On constitutive inequalities for simple materials I},
author={Hill, R},
journal={Journal of the Mechanics and Physics of Solids},
volume={16},
number={4},
pages={229--242},
year={1968},
publisher={Elsevier}
}


text:
Find the Antiderivative 6/( square root of 1-x^2)

Write as a function.
The function can be found by finding the indefinite integral of the derivative .
Set up the integral to solve.
Since is constant with respect to , move out of the integral.
Rewrite as .
The integral of with respect to is
Simplify.
The answer is the antiderivative of the function .
Find the Antiderivative 6/( square root of 1-x^2)


text:
# The Limitations of Algebra Tiles

(Click aqui para leer este articulo en español.)

Algebra Tiles can be extremely useful in developing early algebraic thinking and modeling certain types of problems. However, they have very definite limitations and should be left behind at a certain point in the learning process.

Algebra tiles are limited by their 2-dimensional nature to modeling second degree or lower polynomials. Their physical nature also limits the number of variables that can appear in polynomials they model as well as the depth of understanding of negation and subtraction.

Algebra Tiles are limited by their dimensionality. Using only two dimensions, Algebra Tiles limit the algebra teacher and student to second-degree (quadratic) or lower (lines or constants) polynomials. Secondly, only some sets of algebra tiles come with more than one variable, and almost never with more than two variables, further limiting the polynomials they can model. Further, once negative coefficients and subtraction are introduced, the physical interpretation of algebra tiles as area becomes suspicious for two reasons: 1) Area can’t be negative, and 2) $x^2-3x$ should not cover the same amount of area as $x^2+3x$, yet in algebra tiles, they do. Whenever language around the “negative” tiles is not carefully used, algebra tiles can also exacerbate misunderstandings around whether $-x$ is always negative. Finally, algebra tiles, as many manipulatives, while great for building understanding and being able to envision and interpret, can also become cumbersome and less efficient than other methods of performing algebraic manipulations.

How can you overcome these? Don’t abandon the tiles altogether. Instead, always use the tiles simultaneously with the symbolic manipulations. Use both at first, and make sure you understand each step of the symbolic representations. Gradually transition to using the tiles less and less and the symbolic strategies more and more. Let’s explore each of these limitations so that we understand why this is important.

This post contains affiliate links. If you purchase through these links, I may earn a small commission at no additional cost to you. You can read more about how I choose affiliates and products at my affiliate page.

## Two Dimensions Limit Polynomials That Can Be Modeled

You may have noticed already that the “biggest” algebra tiles represent $x^2$ (or maybe the $xy$ or $y^2$ tile if you’re lucky enough to have a set with a second variable). In order to expand further to $x^3$, we would need to deal with 3-dimensional algebra cubes, not 2D tiles. Without that third dimension, we are stuck with second-degree (quadratic) or lower polynomials. Thus, the algebra tiles (affiliate link) won’t help us when we have any terms with $x^3$ or higher powers.

## Lack of Variables Limit Polynomials That Can Be Modeled

If you’re lucky, your set of tiles might have a second variable (unknown side length). Almost none will have more than 2. Algebra Tiles are limited to representing polynomials with only the number of variables that they have unknown side lengths — usually just one variable, sometimes two.

## Introducing “Negative” Tiles Throws Off Area Interpretation

There are multiple problems with reading the algebra tiles as area covered when we introduce negative coefficients and subtraction. Since area in this context measures a physical length times width, and since length and width won’t ever be negative, the whole idea of negative area breaks the rules of measurement. Even if we thought of these negative coefficients as a “taking away” kind of subtraction, we don’t actually take away using the algebra tiles. Instead, we appear to add the same way we would if the coefficient was positive, just with a contrasting color instead. Which leads polynomials with opposite coefficients to cover the same area.

## “Negative” Tiles May Reinforce Misunderstandings of Sign of $-x$

If we are at all lazy with language, which often happens, students frequently gain a misconception that $-x$ is always a negative number. But this isn’t the case. $x$ might take on either positive or negative values. $-x$ should always be read as “the opposite of $x$.” Thus, if $x$ was positive, then $-x$ is negative (opposite of positive). But whenever $x$ takes on a negative value, then $-x$ will be the opposite: positive. This is a common misunderstanding, reinforced by the casual (and wrong) language “negative $x$.” Algebra tiles could be used to reinforce either interpretation, and care must be taken to know how the “negative” tiles are being used.

## Algebra Tiles Can be Cumbersome and Inefficient

You never want to rely so heavily on a tool that it becomes a crutch. Use algebra tiles to gain understanding, but don’t neglect the symbolic manipulations. Practice those, as ultimately they will be quicker, more efficient, have a wider range of uses (such as higher degree polynomials), and be more readily available even when you don’t have access to the tiles. Use the tiles to understand, but don’t let them hold you back. And above all, remember…

You’ve Got This!

Barbra

Barbie has taught math, supported students, and volunteered in classrooms for over 20 years. Her daughter is currently learning math in a Common Core state.


text:
Tag Info

2

Keyword arguments are not interpreted by the parser of TXS. Qrrbrbirlbel has already indicated possible workarounds (recommended) Define a new command \ctablewithlabel{<label>}[<other options>]{<arg 1>}{<arg 2>}{<arg 3>} and declare it in a cwl file as label Trick the parser with an additional \iffalse\label{tab:somelabel}\fi ...

2

We have different versions of pst-plot (how it is possible?!), so I can't copy your beginning version. However, it seems that showorigin=false should give you the desired result. If no, I will remove my answer. \documentclass{article} \usepackage{pst-plot} \begin{document} \begin{figure} \centering \def\dx{2.5} \def\dy{1} \def\dy{1.0} \makeatletter ...

6

The label information is all in the aux file, so you can extract it something like this: \documentclass{article} \usepackage{hyperref} \long\def\findlabel#1#2\findlabel{% \ifx\newlabel#1\lablist\expandafter{\the\lablist\showlabel#2}\fi} %hyerref has 4 felds in each label culd use them but don't here \def\showlabel#1#2{% \par \noindent Label: ...

5

Insert the label on a node inside the segment line. Remove the label 1,2,3 from node (b). \documentclass{article} \usepackage{tikz} \begin{document} \begin{tikzpicture} \tikzstyle{every node}=[draw,circle,fill=white,minimum size=18pt, inner sep=0pt] \def \x{2} \def \y{1.75} %% edited here for better view \draw (1 * \x, 0 * \y) node (leer) ...

Top 50 recent answers are included


text:
# Representing any number using only 0 [closed]

I am afraid to even ask this question, but this is something that was asked over a breakfast table by a friend of mine.

How do I represent a number, say 4, by using just zeros as a numeral in tandem with any function or mathematical operator(s)?

Now one way to do this is by using a factorial, since $0!=1$: $$0! + 0! + 0! + 0!$$ Another way uses $\cos 0=1$: $$\cos 0+\cos 0+\cos 0+\cos 0$$ Also, the topmost $1$ on Pascal’s Triangle is the zero$^{th}$ row, zero$^{th}$ entry, i.e., $0C$$_0 = 1. The Partition function , P(n) is , by convention , defined as 1 for n=0, i.e. , P(0) = 1. Is there any other way I could possibly achieve this?I would be more interested to know the functions which when applied on zero produce non-zero results , like the ones mentioned earlier in the question, rather than semantic manipulations or visual representations. ## closed as off-topic by user21820, Matthew Towers, user133281, Jean-Claude Arbaut, Ali CaglayanOct 16 '16 at 14:13 This question appears to be off-topic. The users who voted to close gave this specific reason: • "This question is not about mathematics, within the scope defined in the help center." – user21820, Matthew Towers, user133281, Jean-Claude Arbaut, Ali Caglayan If this question can be reworded to fit the rules in the help center, please edit the question. • Why the hell would you and your friend want this? – Billy Rubina Oct 16 '16 at 1:26 • It started off as more of a philosophical debate that zero holds up infinity and then took rather this silly turn. Besides imagine designing a scientific calculator with just a 0 digit-button. – naveen dankal Oct 16 '16 at 1:31 • Although not "0", I like |\mathcal{P}(\mathcal{P}(\mathcal{P}(\emptyset)))| – Bonnaduck Oct 16 '16 at 4:42 • The problem with your question is the meaning of "just zeros". None of your examples consists of ""just zeros". So you would have to define what you mean. – Martin Argerami Oct 16 '16 at 14:17 • @naveendankal, you can click on the "reopen" button just below your question in order to have it reopened. – Mikhail Katz Oct 19 '16 at 13:40 ## 11 Answers Peano says: with the successor function.$$4=S(S(S(S(0))))$$You could try using unary with 0. So 4_{10} is 0000 in unary. Also, this usually works:  0 0 0 0 0 0 0 0000000000 0 0 0  • Your examples separate the symbol 0 from its standard meaning: the value of zero. In a similar way we could introduce bogo numerals (as opposed to Arabic numerals) with the symbol 0 meaning four, then just write 0. – Kamil Maciorowski Oct 16 '16 at 0:19 • OP did ask to use only zero, so even if this 0 isn't the same as the one with the usual arithmetic properties, it wouldn't take anyone more than couple of guesses to tell what 0000 means. I think. Also, I took the idea of only zero literally. So the first example is only zeros representing 4. – Cehhiro Oct 16 '16 at 0:22 • OK, your ways are imaginative and they as well may be what OP wants. I'm just pointing out they are semantic tricks more than math tricks. My feeling is it should be explicitly noted here on Mathematics Stack Exchange. – Kamil Maciorowski Oct 16 '16 at 0:37 • @ O. Von Seckendorff : your answer is creative but I am more interested in the Mathematical representation , maybe some function that translates 0 into 1 or something like the Cos and factorial. – naveen dankal Oct 16 '16 at 0:48 • If "four" is represented by 0000, then "one" is represented as '0'. Which makes it impossible to distinguish between "zero" and "one" as both are written as '0'. Unless one would like to write zero as '' (empty). However, inventing a symbol for zero usually is considered as a milestone, and this would revert it. TL;DR -> I think that 4 should be written down as '00000' (five zeros). – Sjoerd Oct 16 '16 at 3:05$$\Large{\vert\{}\normalsize{ 0,\{0\},\{\{0\}\},\large{\{}\normalsize{\{\{0\}\}}}\large{\}}\Large{\} \vert}$$• Is that what they call a family tree? – Mateen Ulhaq Oct 16 '16 at 7:00 • @MateenUlhaq I've never heard that phrase in math. – GFauxPas Oct 16 '16 at 16:16 Well 4 = (0!+0!)^{(0!+0!)} And we could take this notation:$$x_n=\sum_{n=0!}^n 0!$$as we are not explicitly using the number n , but rather this is just short hand. So now we may express e in terms of 0 as$$lim_{n\rightarrow \infty}\displaystyle (0!+\frac{0!}{x_n})^{x_n} $$or if you prefer:$$ e = (0!+0!)+\cfrac{0!}{0!+\cfrac{0!}{(0!+0!)+\cfrac{0!}{0!+\cfrac{0!}{0!+ \cfrac{0!}{(0!+0!)^{(0!+0!)}+\cdots }}}} } $$but of course this is all rather ridiculous! Of course assigning x_n in this way, is essentially the same way as assigning a number such as 4 or 300 -- it is arbitrary, which opens all such thought on the proper existence of numbers ... but this is starting to dig far deeper into your silly question!!! • That's exactly what has been troubling me too with regards to this question. It sounds both stupid and silly and deep and profound at the same time, maybe less mathematically but more philosophically. It leads to a the vulnerability of the existence of numbers and raises a slight doubt whether they are essential or inconsequential. I am glad that you added that last paragraph to your answer because this is exactly what my fear was when I stated this question. An extra up for that para!! – naveen dankal Oct 16 '16 at 5:28 • Thus the unreasonable effectiveness of mathematics follows! – Hugh Entwistle Oct 16 '16 at 5:32 A solution only using "0" once is$$\large\sqrt{\left\lceil\exp(\exp(\exp(0)))\right\rceil}=4$$But I admit, that it is not very aesthetic. • Use Mathjax with HUGE. It could be nice ! – Claude Leibovici Oct 15 '16 at 18:23 • @ClaudeLeibovici Decided to use \large instead, feeling that \huge is overkill. – Simply Beautiful Art Oct 16 '16 at 0:54 • It might be a good idea to use$\exp^n$or$\exp \circ \dots\circ\exp$to represent successive iterations of$\exp$, and$\lceil\cdot\rceil$for ceiling. – Henricus V. Oct 16 '16 at 6:00 • @Henry W$1)$I do not know how to produce the ceil braces. (If you know it, you can edit the answer)$2)$"$exp^n(x)$would contain other digits. We are only allowed to use "0" as a digit here. – Peter Oct 16 '16 at 8:46 Arguably the von-Neumann construction does exactly that: It identifies each natural number with the set of natural numbers below (where$0$is considered a natural number). In that construction, of course$1=\{0\}$: The only natural number below$1$is$0$. Obviously this only uses$0$(and the set construction, but then, you were obviously willing to use extra constructions as long as they are not themselves$0$). Furthermore,$2=\{0,1\}$. Now, this obviously uses$1$besides of$0$. But we've just seen that$1=\{0\}$. So$2=\{0,\{0\}\}$. Then$3=\{0,1,2\}$. Using the previous results, we get$3=\{0,\{0\},\{0,\{0\}\}\}$. And$4=\{0,1,2,3\} = \{0,\{0\},\{0,\{0\}\},\{0,\{0\},\{0,\{0\}\}\}\}$So clearly$4$(and any other natural number) can be represented using only zero. But it's even better: I omitted the question: What is$0$? Well,$0$is the set of natural numbers less than$0$. But there are no natural numebrs less than$0$. Therefore$0$is the empty set,$0=\emptyset$. So actually the von-Neumann construction creates all natural numbers literally out of nothing. Given the number$n$, you simply get the next natural number by appending the number$n$as new element to itself:$n+1=n\cup\{n\}$So$0$is the empty set. So you get the next number,$1$, by appending$0$as element; that is, you've got$1=\{0\}$. To get from$1$to$2$, you append the element$1$, so$2=\{0,1\}$, and so on. Moreover, you mention that the question arose from a discussion about infinity. And it turns out that the von-Neumann construction is exactly the way to get to infinity (and beyond!). That's because you can now ask, what is the set of all natural numbers. Well, if we want to consider it a number, it must be a number large enough that all natural numbers. That is, it is infinite. This specific number is commonly called$\omega$. So it is infinity, right? Well, wait. If it is a number in its own right, then we can make again a set of numbers containing all natural numbers and$\omega$. This set by construction contains all numbers$\le\omega$, so it cannot have any "holes". But then, according to the rule, it should be another number, that is also larger than every natural number, and furthermore larger than$\omega$! Indeed, since we've just appended the element$\omega$to$\omega$, so what we get should be$\omega+1$. And we can repeat that construction to get$\omega+2$,$\omega+3$, … So we get actually infinitely many additional infinities. The numbers whose construction I just describes (well, strictly speaking, I only described the construction of the lowest ones) is called the ordinal numbers. Now you may ask: What does the complete set of ordinal numbers look like? Well, here comes the next surprise: The set of ordinal numbers does not exist! That's seen directly from the construction. Remember how we arrived at$\omega$? Well, we had the set of all natural numbers, and we decided that the very fact that we have that set makes it a new number, that's larger than any of them. Now imagine we'd have the set of ordinal numbers. Then by the very construction, that set itself should be an ordinal number larger than any ordinal number. But then it cannot be in that set, and therefore that set cannot be the set of ordinal numbers, as we've just found an ordinal number that's not in that set, namely the very set itself! So you see, we've started with$0$(actually, with literally nothing), and ended up with so many numbers that they don't even fit in a set! My guess is$0^0 =1$thus$0^0 + 0^0+0^0+0^0 =4$. •$0^0$is undefined – Bernard Masse Oct 15 '16 at 17:49 • I like this because it uses only the zero symbol. While an analyst might feel$0^0$is undefined in some contexts, a combinatorialist will likely be happy with a value of one assigned. – hardmath Oct 15 '16 at 17:56 • Isn't$0^0 = \prod_\text{false}0 = 1$? What else would it mean? – GFauxPas Oct 16 '16 at 3:35 • @Peter That limit is definitely not why$0^0$is conventionally defined to be$1$. The reason is twofold: there is precisely one function$\varnothing\to\varnothing$, and$0^0=1$allows us to plug$x=0$into$\sum_n a_n x^n$reasonably. – arctic tern Oct 16 '16 at 5:17 •$0^0=1$when the exponent is the integer 0.$0^0$is undefined when the exponent is the real 0. – Meni Rosenfeld Oct 16 '16 at 7:47 Every real number admits a binary expansion. For every occurrence of$1$in the expansion, write$1=0^0$. • 1. You've mis-spelled "write". 2.$0^0$is undefined. Perhaps you meant$1=\lim\limits_{x\to0}x^x$. – barak manos Oct 15 '16 at 17:45 • @barakmanos Thanks for pointing out the spelling error. Your other point is wrong.$0^0$is$1$. This is the standard convention. – MathematicsStudent1122 Oct 15 '16 at 17:46 • I've never heard of a unified "standard convention". What I have seen is that in particular settings where such undefined expressions arise, they are defined situationally as the limit of some expression as the parameter in question approaches$0$. – AJY Oct 15 '16 at 18:26 • @AJY Do you accept the binomial theorem? Does it hold for$(x+0)^n$? This isn't really up for debate. – MathematicsStudent1122 Oct 15 '16 at 18:56 • @AJY Or how about the power series of$\exp$,$\sum_{n=0}^{\infty} \frac{x^n}{n!}$at$x=0\$. Is that also disputed? – MathematicsStudent1122 Oct 15 '16 at 18:57

Add zero to a number let say x so now the new number is x0 , x0-x = yz , now u= sum y+z , now sqrt u + u to the power 0 will be 4. Explanation u will always be 9 so sqrt 9 will be 3 and power 0 will be one so 3+1=4

In some languages...

i = 0, i++, i++, i++, i++, i

0 = 0
1 = 0++
2 = (0++)++
3 = ((0++)++)++

and so on. This uses the syntax of C-like languages.

• Maybe you could explain your notation? – snulty Oct 16 '16 at 11:47


text: The Hidden Markov Model or HMM is all about learning sequences. Stock prices are sequences of â¦ hmmlearn implements the Hidden Markov Models (HMMs). Problem 1 in Python. Machine Learning using Python. A Hidden Markov Model (HMM) is a statistical signal model. The observation set include Food, Home, Outdoor & Recreation and Arts & Entertainment. In the part of speech tagging problem, the observations are the words themselves in the given sequence. A lot of the data that would be very useful for us to model is in sequences. 3. emission probability using hmmlearn package in python. Tutorial¶. 1. Stock prices are sequences of prices. We can impelement this model with Hidden Markov Model. 5. We also went through the introduction of the three main problems of HMM (Evaluation, Learning and Decoding).In this Understanding Forward and Backward Algorithm in Hidden Markov Model article we will dive deep into the Evaluation Problem.We will go through the mathematical â¦ The Overflow Blog Podcast 288: Tim Berners-Lee wants to put you in a pod. Introduction to Hidden Markov Model article provided basic understanding of the Hidden Markov Model. 53. A lot of the data that would be very useful for us to model is in sequences. Stock prices are sequences of prices. The Hidden Markov Model or HMM is all about learning sequences.. A lot of the data that would be very useful for us to model is in sequences. Problem with k-means used to initialize HMM. In simple words, it is a Markov model where the agent has some hidden states. English It you guys are welcome to unsupervised machine learning Hidden Markov models in Python. Today, we've learned a bit how to use R (a programming language) to do very basic tasks. My program is first to train the HMM based on the observation sequence (Baum-Welch algorithm). â¦ In Python, that typically clean means putting all the data â¦ together in a class which we'll call H-M-M. â¦ Featured on Meta New Feature: Table Support. Credit scoring involves sequences of borrowing and repaying money, and we can use those sequences to predict whether or not youâre going to default. Featured on Meta Responding to the â¦ This code implements a non-parametric Bayesian Hidden Markov model, sometimes referred to as a Hierarchical Dirichlet Process Hidden Markov Model (HDP-HMM), or an Infinite Hidden Markov Model (iHMM). IPython Notebook Tutorial; IPython Notebook Sequence Alignment Tutorial; Hidden Markov models (HMMs) are a structured probabilistic model that forms a probability distribution of sequences, as opposed to individual symbols. So the time dependency involves the speed, pressure and coordinates of the pen moving around to form a letter. This model is based on the statistical Markov model, where a system being modeled follows the Markov process with some hidden states. You'll also learn about the components that are needed to build a (Discrete-time) Markov chain model and some of its common properties. Be comfortable with Python and Numpy; Description. The HMM is a generative probabilistic model, in which a sequence of observable variable is generated by a sequence of internal hidden state .The hidden states can not be observed directly. This short sentence is actually loaded with insight! Figure 1 from Wikipedia: Hidden Markov Model. Python Hidden Markov Model Library ===== This library is a pure Python implementation of Hidden Markov Models (HMMs). Installation To install this package, clone thisrepoand from the root directory run: $python setup.py install An alternative way to install the package hidden_markov, is to use pip or easy_install, i.e. A lot of the data that would be very useful for us to model is in sequences. The following will show some R code and then some Python code for the same basic tasks. sklearn.hmm implements the Hidden Markov Models (HMMs). NumPy, Matplotlib, scikit-learn (Only the function sklearn.model_selection.KFold for splitting the training set is used.) We know that to model any problem using a Hidden Markov Model we need a set of observations and a set of possible states. Hands-On Markov Models with Python helps you get to grips with HMMs and different inference algorithms by working on real-world problems. You will also learn some of the ways to represent a Markov chain like a state diagram and transition matrix. Improve database performance with connection pooling. Descriptions. Credit scoring involves sequences of borrowing and repaying money, and we can use those sequences to predict whether or not youâre going to default. Browse other questions tagged python hidden-markov-models unsupervised-learning markov or ask your own question. Description. The Hidden Markov Model or HMM is all about learning sequences. The Hidden Markov Model (HMM) was introduced by Baum and Petrie [4] in 1966 and can be described as a Markov Chain that embeds another underlying hidden chain. Part of speech tagging is a fully-supervised learning task, because we have a corpus of words labeled with the correct part-of-speech tag. Related. In short, sequences are everywhere, and being able to analyze them is an important skill in â¦ Hidden Markov Models¶. A Hidden Markov Models Chapter 8 introduced the Hidden Markov Model and applied it to part of speech tagging. The resulting process is called a Hidden Markov Model (HMM), and a generic schema is shown in the following diagram: Structure of a generic Hidden Markov Model For each hidden state s i , we need to define a transition probability P(i â j) , normally represented as a matrix if the variable is discrete. The transitions between hidden states are assumed to have the form of a (first-order) Markov chain. A statistical model estimates parameters like mean and variance and class probability ratios from the data and uses these parameters to mimic what is going on in the data. A Hidden Markov Model (HMM) is a specific case of the state space model in which the latent variables are discrete and multinomial variables.From the graphical representation, you can consider an HMM to be a double stochastic process consisting of a hidden stochastic Markov process (of latent variables) that you cannot observe directly and another stochastic process that produces a â¦ A lot of the data that would be very useful for us to model is in sequences. Prior to the creation of a regime detection filter it is necessary to fit the Hidden Markov Model to a set of returns data. Since your friends are Python developers, when they talk about work, they talk about Python 80% of the time. Best Python library for statistical inference. Next, you'll implement one such simple model with Python using its numpy and random libraries. 3. Browse other questions tagged python hidden-markov-model or ask your own question. Training the Hidden Markov Model. Stock prices are sequences of prices. Be comfortable with Python and Numpy; Description. Write a Hidden Markov Model in Code; Write a Hidden Markov Model using Theano; Understand how gradient descent, which is normally used in deep learning, can be used for HMMs; Requirements. This package has capability for a standard non-parametric Bayesian HMM, as well as a sticky HDPHMM (see references). One way to model on how to get the answer, is by: Hidden Markov Model using Pomegranate. The Hidden Markov Model or HMM is all about learning sequences.. A lot of the data that would be very useful for us to model is in sequences. Simple Markov chain weather model. The mathematical development of an HMM can be studied in Rabiner's paper [6] and in the papers [5] and [7] it is studied how to use an HMM to make forecasts in the stock market. Credit scoring involves sequences of borrowing and repaying money, and we can use those sequences to predict whether or not youâre going to default. Language is a sequence of words. As an example, I'll use reproduction. The hidden states include Hungry, Rest, Exercise and Movie. As for the states, which are hidden, these would be the POS tags for the words. The 3rd and final problem in Hidden Markov Model is the Decoding Problem.In this article we will implement Viterbi Algorithm in Hidden Markov Model using Python and R. Viterbi Algorithm is dynamic programming and computationally very efficient. Language is a sequence of words. You only hear distinctively the words python or bear, and try to guess the context of the sentence. Gesture recognition with HMM. Language is a sequence of words. Swag is coming back! For this the Python hmmlearn library will be used. The transitions between hidden states are assumed to have the form of a (first-order) Markov chain. ... We can define what we call the Hidden Markov Model for this situation : The Overflow Blog How to put machine learning models into production. - [Narrator] A hidden Markov model consists of â¦ a few different pieces of data â¦ that we can represent in code. A Markov Model is a stochastic state space model involving random transitions between states where the probability of the jump is only dependent upon the â¦ In our case this means, that a signature is written from left to right with one letter after another. How can I predict the post popularity of reddit.com with hidden markov model(HMM)? The project structure is quite simple:: Help on module Markov: NAME Markov - Library to implement hidden Markov Models FILE Markov.py CLASSES __builtin__.object BayesianModel HMM Distribution PoissonDistribution Probability The data is categorical. Hidden Markov Model (HMM) is a statistical Markov model in which the system being modeled is assumed to be a Markov process with unobserved (i.e. But many applications donât have labeled data. hidden) states. For this experiment, I will use pomegranate library instead of developing on our own code like on the post before. Package hidden_markov is tested with Python version 2.7 and Python version 3.5. Prior to the discussion on Hidden Markov Models it is necessary to consider the broader concept of a Markov Model. The API is exceedingly simple, which makes it straightforward to fit and store the model for later use. Related. 2. Hidden Markov Model is a partially observable model, where the agent partially observes the states. Hidden Markov Model (HMM) is a statistical model based on the Markov chain concept. The standard functions in a homogeneous multinomial hidden Markov model with discrete state spaces are implmented. I would like to predict hidden states using Hidden Markov Model (decoding problem). Multi-class classification metrics in R and Pythonâ¦ The HMM is a generative probabilistic model, in which a sequence of observable $$\mathbf{X}$$ variables is generated by a sequence of internal hidden states $$\mathbf{Z}$$.The hidden states are not observed directly. Browse other questions tagged python markov-hidden-model or ask your own question. The Hidden Markov Model or HMM is all about learning sequences. Language is a sequence of words. 1. The states in an HMM are hidden. I am taking a course about markov chains this semester. A Tutorial on Hidden Markov Model with a Stock Price Example â Part 1 On September 15, 2016 September 20, 2016 By Elena In Machine Learning , Python Programming This tutorial is on a Hidden Markov Model. Stock prices are sequences of prices. It will enable us to construct the model faster and with more intuitive definition. 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text:
# Conditional Probability Word Problem - Drug Testing

I've been having some difficulty with this homework problem related to conditional probabilities. I think part a is correct but the answer to part b seems wrong to me.

Professional athletes who play in the NFL are subject to random drug tests. Suppose the probability that the test shows a positive result given that the athlete is using drugs is 0.93 and the probability that the test shows a negative result given that the athlete is not using drugs is 0.98. Assume that three percent of the NFL players use drugs.

Let $S=\{TP, TN, FP, FN \}$ where TP stands for True Positive, TN fo True Negative, FP for False Positive, and FN for false negative. All of these events are mutually exclusive.

a) What is the probability that a NFL player chosen at random will test negative for drug use?

$P(TN\cup FN)=P(TN)+P(FN)=0.97*0.98+0.03*0.07=0.9527$

b) What is the probability that the player is actually using drugs given that he has a positive test?

First, note that the probability of a positive test is $1-0.9527=0.0473$ since this event is the complement of the event in problem a. Also the probability that a player is using drugs is given at 0.03.

By the conditional probability formula: $$P(\text{using drugs }\vert \text{ positive test})=P(TP)\backslash P(\text{positive test})=P(TP)\backslash 0.0473$$ To find $P(TP)$, note that: $$P(\text{positive test }\vert \text{ using drugs})=P(TP)\backslash P(\text{using drugs})=P(TP)\backslash 0.03$$ This was give to be 0.93. Solving yields $P(TP)=0.0279$.

Hence the solution is $0.0279\backslash 0.0473=0.5899$

I am encouraged that my solution is between 0 and 1, but it seems off to me. Can anyone tell me where I made a mistake?

If the athlete has a positive test, he falls either into TP or into FP.

Given that

$$P(\text{positive test}|\text{using drugs}) = 0.93$$

we know

$$P(\text{negative test}|\text{using drugs}) = 0.07$$.

I assumed your (a) is correct, I check only (b) on a different way.

We need to find $P(\text{using drugs}|\text{positive test})$. So, we use the Bayes-theorem:

$P(\text{using drugs}|\text{positive test}) = P(\text{positive test}|\text{using drugs}) \cdot \frac{P(\text{using drugs})}{P(\text{positive test})} = 0.93 \cdot \frac{0.03}{1-P(\text{negative test})} = 0.93 \cdot \frac{0.03}{1-0.9527} = 0.589852$.

What you've found, is a known surprising result of the probability theory. It is caused by the fact, that after a positive result, the human thinking is often minded to ignore the fact, that only a small part of the athletes are using drugs.

Thus, between the actual positive results, we will have much more false positives.

• This "paradox" has also a name, unfortunately I forget it, but it is a well-known phenomenon. – peterh May 21 '18 at 16:07


text:
# Extras: Bash and Shell Scripting

## Bash Shell

There are often several different shells installed on a Linux system. bash is probably the most common, and is typically the default for new users on any system. bash is running in the terminal, and it interprets and executes the commands we type in.

### Environment

#### User Configuration

On startup, such as when a new terminal is opened, or you connect to a remote system over the network, bash will read several configuration files depending on how it detects it is running.

• For login shells (shells where you are prompted to login when you start it), bash will read configuration options in ~/.bash_profile or ~/.profile if these exist. Note ~ is interpreted by the shell as the users home directory.
• For non-login shells (such as when you open the terminal emulator) bash will read configuration options in ~/.bashrc.
• A lot of the time, for simplicity, many people will put all their configuration options in ~/.bashrc and leave their ~/.profile empty except for a command to read the ~/.bashrc file.

These files can be used to configure many aspects of the shell, such as how the prompt looks in the terminal.

#### Environment Variables

Many important aspects of how bash behaves are governed by environment variables. These can be modified to your preference in your configuration files. To see the current value of one of these variables you can do, e.g. echo $PATH. A $ symbol is used when referring to the data stored in a variable, and echo is a built-in bash command that outputs something to the terminal.

• PATH: this tells bash which directories to look in for executables so they can be run without typing in their full path. For example, we can just type in gedit in the terminal to launch the program, as the PATH variable contains the directory holding an executable of this name. Different directories are separated by “:“.
• To add the directory ~/.bin to your path, you can add the line export PATH="~/bin:$PATH" to the .bashrc file. Here the export statement is used to allow the variable to be usable by any child processes rather than local to the script. Note when we are naming the variable we want to assign a value to we don’t use a $. Also we just want to add a directory to the existing PATH so we add :$PATH to keep all the existing directories. ### Scripting Shell scripts allow you easily string together sets of commands in a useful way. For example, you could write a script that runs a calculation, parses the important results to another file, and then generates a plot. This course will only be able to briefly cover scripting. If you’re interested in developing more advanced scripts, I strongly suggest referring to the Advanced Bash-Scripting Guide. #### The basics A simple example script to output “Hello World!” is as follows: #!/bin/bash # Simple example script that outputs "Hello World!" echo "Hello World!"  • The first line tells the system what command is used to interpret the contents of the script. This line should always begin with #! and then the path to the executable. For bash the executable will almost always be /bin/bash. • Comments being with #. • Commands can be entered otherwise as you would enter them to the terminal. • Try creating a file called hello.sh with the contents listed above. • You will need to make the script executable with chmod u+x hello.sh. • Then you can run it by typing ./hello.sh. #### Variables Variables don’t need to be declared in any way, you can assign a value and start using them. For example: #!/bin/bash # Example using variables. var1="Hello" var2="World!" echo$var1 $var2 # Note the use of the$ symbol when we want to use the value stored in the
# variable.

# Any command can use these variables
dirname="tmpdir"
mkdir $dirname  You can also read a value from stdin using the read command as follows: #!/bin/bash # Example showing how the read command is used. echo "Please enter some text:" read user_text echo "The text you entered was:"$user_text

##### Capitalization

Note: variable names are case sensitive in bash. The usual convention used is that environment variables (such as PATH), and internal shell variables (such as BASH_VERSION) are capitalized, while other variables are in lowercase. Adopting this convention will ensure that you don’t accidentally overwrite any important variables in your scripts.

#### Command Substitution

Command substitution allows the result of a command to replace the command itself, acting much like a variable in practice. For example:

#!/bin/bash
# Example of command substitution.

# This can be done by enclosing the command in $( ) echo "It is now$(date)."
# Note: the "date" command outputs the current date and time.

# This can also be done by enclosing the command in backtics
echo "The files in this directory are:" ls

#### Conditional Statements

if statements can be used in bash, and many types of tests are possible. You can test if the value stored in a variable equals something as follows:

#!/bin/bash
# Example using if statatements

echo "Please enter a yes or no: "

# Note the spaces following [ and before ] are important.
if [ $user_response = yes ] then echo "You entered yes." elif [$user_response = no ]
# "elif" is the same as "else if"
then
echo "You entered no."
else
echo "You didn't enter yes or no."
fi
# "if" statements are always ended with "fi".

We can also check, e.g., if a file or directory exists:

#!/bin/bash
# Check if the directory "tmpdir" exists, and if not, create it, then check
# if the file "tmpdir/testfile" exists, and if not, create it.

dirname="tmpdir"
filename="testfile"

if [ ! -d "$dirname" ] # The "!" is logical negation # -d tests that a file exists and is a directory. then mkdir$dirname
fi

if [ ! -f "$dirname/$filename" ]
# -d tests that a file exists and is a regular file (i.e. not a directory).
then
touch "$dirname/$filename"
fi

Many other tests of this form are possible.

Note, bash can also use the [[ ... ]] test construction rather than [ ... ], with the former offering some more options and functioning more like other programming languages, though it won’t work in many other (non-bash, or older bash version) shells.

#### Arithmetic and Testing

To test numeric values, the (( ... )) construction can be used. For example:

#!/bin/bash
# Example showing the use of the (( )) construction

var1=4
var2=5
var3=8

if (( var1 + var2 > var3 ))
# Note within the (( )) don't use $symbols before variables. then echo "$var1 + $var2 >$var3"
else
echo "$var1 +$var2 <= $var3" fi # We can also use the construction to perform basic arithmetic var4=$(( var1 * var3 ))
echo "$var1 *$var3 = $var4"  #### For Loops For loops iterate a variable over a range of values. For example: #!/bin/bash # Simple for loop example. # This construction loops over a space separated list of values. A variable # whose contents contains spaces would work in the same way. for i in 1 2 3 do echo "Iteration number$i"
done

A for loops can be used to apply a command repeatedly to a set of files. For example:

#!/bin/bash
# Example showing a simple for loop over a list of arguments used to convert
# a set of data files with comma separted columns to tab separated columns.

for csvfile in "$(ls *.csv)" do datfile="$(basename $csvfile .csv).dat" # basename is a tool to strip the suffix from a file. Here we use it # to construct a new filename with the .dat extenstion. sed 's/,/\t/g'$csvfile > $datfile done  Bash also supports numeric loop ranges using the syntax {start..finish..increment). For example #!/bin/bash # Example of for loop with numeric increments for i in {0..6..2} do echo "The value of i is$i"
done

#### While Loops

Bash also supports while loops, where a set of commands are repeated continuously while a given condition is true. For example:

#!/bin/bash
# Example of while loop in bash

counter=0

# Here -lt means less than
while [ $counter -lt 10 ] do echo$counter
counter=$((counter + 1)) done  It is often convenient to use while loops to operate on every line of a file. For example: #!/bin/bash # Example of while loop reading lines from a file linenumber=0 while read line do linenumber=$((linenumber + 1))
echo "$linenumber:$line"
# This will prepend each line of a file with its line number.
done < test.txt

#### Arguments

We can also pass arguments to bash scripts from the command line. For example:

#!/bin/bash
# Example using command line arguments. Call this script with several
# arguments.

echo "The first argument is $1" echo "The second argument is$2"
echo "The number of arguments is $#" echo "The full list of arguments is$@"

We can loop over arguments using a for loop as follows:

#!/bin/bash
# Example using a for loop to iterate over command line arguments.

for arg in $@ do echo$arg
done

#### Functions

We can also define functions in bash as we would in other languages. These allow sections of code to be reused as needed and can help make a script easier to follow.

A simple example is as follows:

#!/bin/bash
# Simple example of a function in a bash script

# First we define a function called "hello", that will output "Hello World!"
# when called.
hello() {
echo "Hello World!"
}

# To call the function, we just use its name.
# Note function names in the script will take precedence over executable names
# from the path in the script.
hello


text:
# Revision history [back]

### Interactive question in notebooks

When trying to find some solution to the double integral problem and poking around with Sage (4.7.2), I stumbled upon this behavior:

x,y,u,v,p,k=var('x,y,u,v,p,k')
integrate(x+y^k, y)

output (resembles maxima interaction):

Traceback (click to the left of this block for traceback)
...
Is  k+1  zero or nonzero?

How can I answer to this (with nonzero)?

### Interactive question in notebooks

When trying to find some solution to the double integral problem and poking around with Sage (4.7.2), I stumbled upon this behavior:

x,y,u,v,p,k=var('x,y,u,v,p,k')
integrate(x+y^k, y)

output (resembles maxima interaction):

Traceback (click to the left of this block for traceback)
...
Is  k+1  zero or nonzero?

How can I answer to this (with nonzero)?

Update: I accept @god.one solution below as there seems no way as for now to interactively answer those maxima questions.

### Interactive question in notebooks

When trying to find some solution to the double integral problem and poking around with Sage (4.7.2), I stumbled upon this behavior:

x,y,u,v,p,k=var('x,y,u,v,p,k')
integrate(x+y^k, y)

output (resembles maxima interaction):

Traceback (click to the left of this block for traceback)
...
Is  k+1  zero or nonzero?

How can I answer to this (with nonzero)?

Update: Update: I accept @god.one solution below as there seems no way as for now to interactively answer those maxima questions.