The post How to solve equations of 1st, 2nd and 3rd degree appeared first on Math Academy.
]]>Table of content:
First, we need to understand what are the possible solutions for solving equations, here they are:
We define as the set of the possible solutions for the equation.
1-No solution: some equations don’t have a solution, i.e. there is no value possible for the variable to make the equation verified or true. Here is an example:
We simplify the equation by multiplying through the parentheses we get
and by subtracting from both sides we get 6=10 which is false and therefore we deduce that there is no solution for this equation i.e. is empty or .
2- Unique solution: equations may have a unique solution that verifies it, meaning that there is one and only one value for the variable to take to make the equation true. Here are some examples:
Example 1:
by subtracting 5 from both sides, we get
and by dividing by 3 both sides, we get
There it is the solution we are looking for and it is unique. is the only value for to make the equation true, .
Example 2:
By subtracting from both sides to eliminate the on the right side of the equation we get
and by adding 14 to both sides we get , by dividing by 4 we get , we get one and only one value for i.e. .
3- Multiple solutions: Equations may have multiple solutions, where there are several values for to verify the equation, here is an example of that:
we can do the multiplication of the parentheses and get
Using the factored writing of the equation, for the right side to be equal to 0, one of the two parentheses must equal 0 and for that we have two cases:
Either by subtracting 5 we get and by dividing by 2 we get ,
or and by adding 3 for both sides we get .
So we have two possible solutions for this equation .
4- Infinite solutions: An equation with infinite solutions is an equation always verified no matter the value of , let’s take a look at the following example:
by simplifying both sides, we get
and then
by subtracting from both sides we get .
By simplifying the equation we got to which is true all the time, it doesn’t depend on the value of , so no matter the value of the equation is always true, and since has infinite possible values we have infinite solutions for this equation.
Now, after seeing the different cases of the number of possible solutions, let’s see how to solve equations of first degree, second degree, and third degree.
We call an equation of first degree every equation written like follows:
Where is the variable and and are real numbers with different than 0.
Note that some equations of first degree are not on this form but after simplifying it always ends up with the form above.
We call it equation of first degree due to the variable begin up to power 1, and it is the highest power of the variable on the equation, meaning .
To solve an equation of first degree we first simplify it if it is not simplified to get to the form and then all we need is pass b to the other side and divide by a, i.e.
and there it is the solution of the equation
Example:
We simplify the equation
then we get the form
and the solution is i.e. .
We can solve the equation geometrically by considering both sides of the equation as a straight line equation meaning the left side is a line equation and the right side is a line equation.
Then we can draw both lines in an orthometric plane and we draw the line and the line which is equivalent to the x-axis (because the x-axis is the line with )
(i.e. ) means the point where the two lines intersect, so by drawing the line and take the where it intersects with the x-axis we get our solution and it is the same as the algebraic method .
Example:
Let’s solve the equation
We draw the line with the equation
We pick 2 values of and get the respective value of and then we graph the two points in a plane and the new draw the line passing by the two points, and the coordinate of the point of intersection of the line and the x-axis is the solution of the equation.
We call an equation of second degree, every equation with the standard form with , and being real numbers and different than zero. It is called a second degree equation because the highest power of in this equation is 2 (i.e. ).
A method to solve a second degree equation is to write it in the form of multiplication of two first degree equations and solving it by finding the solution for the two first degree equations.
How to factor a second degree equation?
If we consider a second degree equation like follow
So, to get from the right side form to the left side factored form we need to figure out the values of and knowing the value of and from the right side form. Let’s try an example:
We need to divide by 2 to remove the factor of and to get the form
so, we get:
Now with this form, we know that and .
So, we need to find the two numbers and that their sum is equal to 10 and their product is equal to 21.
We have and 21 can be written as product as or , and since must equal 10 we have , so the values of and are the ones that make and .
After that, all we have to do is to write the equation in the form .
so, we get:
Now the resolution is straightforward since we have the product of two first degree equal to zero then we know for sure that either the first term of the product is equal to zero or the second is equal to zero, meaning either or , we solve each first degree term of the left side, we get:
and therefore we have the two solutions of the second degree equation , .
We can verify by giving the value or , as follow:
We call the discriminant of the equation the expression , usually it is represented with the letter , i.e. .
Depending on the sign of discriminant we can determine the number and the value -if there is any- of the solutions, and the possible cases are the following:
1- If the discriminant is strictly positive (), then the equation has two different solutions, and the solutions are:
.
Example:
Let’s determine the solutions for the equation:
We evaluate :
so we have
We conclude that the equation has two distinct solutions, and they are the following:
2- If the discriminant is equal to zero, then the equation has one double root, meaning that the equation has two solutions that are the same so one repeated (or doubled) solution. The solution is given as follow:
Example:
Let’s determine the solutions for the equation:
by evaluating we get:
We conclude that the equation has one solution, and the solution is: ; .
The reason why we call this solution a double root or repeated solution because the equation in fact can be written as the product of the same first degree polynomial, and therefore the same solution for the two first degree polynomials.
If we take the previous example, we have:
3-If the discriminant is strictly negative (), then the equation has no solutions.
Example:
Let’s solve the equation
by evaluating we get:
We have , so we conclude that the equation has no solution (Because we can’t take the square root of delta since it’s negative).
In this method, we use the algebraic identity
,
where is variable and is a real number.
To solve the equation , we follow these steps:
1- We divide both sides by , we get: .
2- We subtract from each side, we get: .
3- We add the value (i.e. the square of one half of ) to both sides, and we get:
.
4- Now we have the left side is written as Expansion of the algebraic identity , so we can write the left side as follow:
.
5- We square root both sides and solve the equation in hand.
For a better explanation, let’s use this method with an example:
First, we divide by 3, we get: .
Second, we subtract 4 from both sides, we get: .
Third, we add the to both sides, we get: .
We simplify the right side:
,
,
.
Forth, we write the left side as the algebraic identity, we get: .
Fifth, we take the square root of both sides, we get: .
Sixth, we subtract from both sides, we get: .
And therefore we have the two solutions of the second degree equation , the solutions are:
.
we can graph the function and we look for which values using graphing software or Graphing calculator.
By graphing the function , the graph we get is a parabola, solving the equation is equivalent to determine the value of for the intersecting points of the graph with the x-axis. There are three cases:
The following figure shows the three possible cases:
We call equation of third degree or cubic equation every equation when simplified has the following standard form:
where ,,, and are real numbers and different than 0.
It is called a third degree equation because the highest power of in this equation is 3 (i.e. ).
For the number of possible solutions, unlike first and second degree equations, a third degree equation has at least one solution. Algebraically the reason is that the term with the highest power of , i.e. overgrow the rest of the terms and tend to infinity in both sides depending on the sign, meaning that for very small negative values for () tend to , and for very big positive values for () tend to (or the opposite is depending on the sign of the coefficient of the term ), that means going from one infinity to the other it passed by zero at least one time. There are three possible cases: one, two, or three solutions.
To solve a third degree equation, it would be helpful if we know one solution (or root) to start with. By knowing one solution (remember every cubic equation has at least one solution) we proceed by factoring the third degree equation into a product of a first degree polynomial (using the solution we know) with a second degree polynomial. At this point, we don’t know the coefficients of the second degree polynomial, so we find out their value and then we solve the second degree equation, and therefore we get the solutions of the third degree equation.
For a better understanding let’s try solving this equation:
knowing that is a solution.
Since is a solution then the left side of the third degree equation can be factored into a product of first degree polynomial by a second degree polynomial , meaning that we can write the equation in the form:
Now we need to find the values of and , in order to do that we use the first expanded form of the third degree polynomial, i.e.
by expanding the left side, we get:
Since both sides now are in the standard form, to figure out the values of , and . All we need to do is to equal each coefficient from the left to the correspondent coefficient from the right, in other words:
Now we determine the values of and :
So we have
also so
we have , by replacing a with 1 we find so
Therefore, we have the values , , .
so the factored form is now as follow
Now all that is left is to solve the second degree equation
using any one of the methods we saw previously, we get two solutions
Therefore, the third degree equation has three distinct solutions , and the equation can be written in the factored form
.
As we mentioned before, there are three possible cases for the number of solutions: one, two, or three solutions, and since we start with a known solution, it comes to the second degree polynomial to determine the number of solutions, and it goes as follow:
Notice that in case the constant in the standard third equation form is zero, meaning the equation is in the form
we know that is a solution since each term has
therefore, we don’t need to go through all the process to determine the second degree coefficients, we just take as a factor and we get our factored form as follow
with , and already known and no need to determine them, so we proceed directly to solving the second degree equation.
Example:
let’s solve the equation
since there is no constant, we take as a factor
We know is a solution, so we proceed to solve the second degree equation
using one of the methods shown before we get two solutions or .
We conclude that the equation has three solutions , and the factored form is .
Geometrically, the reason why third degree equation has at least one solution is that the graph pass from to or the opposite (from to ), and therefore we are sure that the graph intersects with the x-axis at least one time.
To solve a third degree equation, we can graph the function and we look for which values using graphing software or graphing calculator.
Solving the equation is equivalent to determine the value of for the intersection point of the graph and the x-axis. There are three possible cases:
The following figure shows the different possible cases.
In conclusion knowing these solving methods can make the process of solving equations of first, second, and third degree equations straightforward, easier, and simpler with clear steps.
You want to have more fun! Check the graph below and see how the graphs change depending on the values of the coefficients. Check the type of equation you want to show (one or multiple), then slide to change the values of ,,, and and watch the graphs change dynamically. Enjoy!
The post How to solve equations of 1st, 2nd and 3rd degree appeared first on Math Academy.
]]>The post Operations Research: The Science Of Doing Better appeared first on Math Academy.
]]>Operations research is the art of problem-solving and decision-making. In this article, we will learn more about “The Science of Doing Better” and its endless applications.
Table Of Content:
If ever you have been in any of these situations or similar problems, you may be interested in knowing about the operations research field. In this article, we will take a journey into the world of operations research, where we strive for the OPTIMUM.
The Operations Research (shortly O.R.) has many names, it is also named Operational Research or Operations Management, usually, the “Operational research” term is used in British English, while the “Operations research” is the name widely used in American English.
Operations research is the discipline of using various and advanced mathematical techniques and methods in order to determine the best solution for a given real-life decision-making problem. It is considered as a branch of the Applied Mathematics field, and sometimes considered the same as decision-making science and management science.
Operations research focuses on practical and real-world applications, applying multiple mathematical methods to obtain optimal (when using exact methods) or near-optimal (when using approximative methods).
Some sources may suggest that operations research started way before (it goes back to the 17th and 18th century), but the beginning of the modern operations research took place during World War I when science was used as a way to ameliorate military operations in England, when the British army needed help to reinforce their air and navy forces. In 1917 and as Germany increased the attacks on Britain’s supply lines resulting in destroying an average of one ten ships, causing big losses in lives and supply line disruptions, and to face this situation, Britain’s army gathered scientists to help avoid or reduce the damage caused by the Germans.
After analyzing all the data available to them and modeling the problem, the scientists were able to optimize the size of the convoys, the speed, and the timing for the safest sailing, and only six months later the average losses were reduced to an impressive one in 200 rate, which made Germany’s assaults inefficient.
After World War I, in 1934 and as a measure to reinforce their defense in case of an air attack from Germany, scientists’ simulations showed that in case of war, the German air force can inflict massive damages and takedown vital targets easily, and by 1938 this fatal attack be delivered in only 24 hours window. In order to prevent that, a set of scientists were assigned by the British government to study the practicality of using radio waves against aircraft, the scientists proved that radio waves could detect, track down, and range aircraft, resulting in the invention of the radar.
This invention was not enough, so by merging Scientific Research with Operational Experience (the reason for the name Operational Research) the team was able to determine the best placement and positioning of the radars, resulting in an exceptional air defense system. Therefore, operations research allowed the earliest possible detection and interception of enemy aircraft. Some sources suggest that by the battle of Britain, radar has increased fighter command’s defensive capability by a factor of 10, and operations research doubled this to a factor of 20.
After World War II, operations research was integrated into all kinds of sectors and widely applied to solve problems in infrastructures, health care systems, transport and supply, business, industry, and banking to name a few. The use of operations research quickly increased because of its incredible effectiveness with real-life decision-making problems.
The applications of operations research are endless, helping to solve different problems and backing decisions with science, here is a general list of the main fields using it:
The first step of the O.R process is to formulate the problem, and it goes without saying that problems in the real world don’t present themselves ready, clearly described, and with well-put information and details in front of the eyes at first sight. The problem statement is often ambiguous and incomplete, usually, the problem contains information that need to be extracted then rewritten in a way that helps get the best use of it, in addition sometimes a problem may embody several other sub-problems to deal with which is may make the problem that seems from far easy to describe and formulate, more complex.
An important goal of this step is to determine the objectives that we want to achieve like minimizing the costs, delivery time or maximizing the profits …etc. Seeing and being able to formulate the problem the right way is a critical step in order to assure that the rest of the work build on this formulation is done on a correct and solid base since it is the first step of the O.R process, also helping in achieving a correct, relevant and useful solution provided for the real problem in hand, not for just a superficial undetailed version of it.
The second step of the O.R. process is modeling the problem mathematically, in other terms passing from the word description of the problem in hand to the mathematical version using mathematical modeling methods, the importance of this step is that it prepares the problem for the resolution phase and making it ready to apply solving methods on it. There are many mathematical modeling methods here is a shortlist of the most used and well-known ones:
It is very important to keep in mind that the solving methods and algorithms depend on the chosen modeling method, i.e. choosing a specific modeling method implies that we use for solving the problem the correspondent solving techniques and changing the model obligate a change in the resolving technique. Every modeling method opens a door for its solving approaches and therefore choosing the best model for a problem is of the essence. A problem can be modeled in several ways but some ways can grant us a better solution with better precision while saving time.
The next step is selecting the method to use, and in this matter, the menu is nothing less than diverse, through time mathematicians and scientists developed a lot of methods, algorithms, and techniques to resolve the different problems. The modeling step will determine in general the set of methods we are able to use but that doesn’t limit the choices at all since every modeling method have plenty of resolution techniques. For example, if we choose to model a given problem using graph theory, now for the resolution we need to choose the methods and algorithms that are developed for graphs, like the Bellman-Ford algorithm or Dijkstra’s algorithm for finding the shortest path in a graph.
Now, and knowing that O.R. solving methods have a great deal of variety, let’s go through the major classifications of these methods:
In this step, we apply the chosen method to the problem with its data, variables, the relation between them, the restrictions, and the objectives and targets of the optimization process. Depending on the selected approach, the time required for the resolution may vary a lot depending on the level of accuracy wanted for the solution.
For example, after establishing all the possible solutions also called the solution space or the search space which usually contains an infinite number of possible solutions or a finite but immense number of solutions, we have to decide how many iterations we want to perform in the solution space to accept the best solution found by then as the solution to take, since for endless or giant number of possible solutions it’s impossible to try them all and choose the best one.
This step is a very good demonstration of the power of mathematics combined with computing using optimized algorithms and powerful hardware. In addition, very helpful softwares are available to help with the resolution step, from simulation, planning, and scheduling to optimization, forecasting, and computation for discrete combinatorial problems. Some notable software: IBM ILOG CPLEX Optimization Studio, Excel solver, Simulink.
The final step of the O.R. process is the implementation, after we got the solution, we proceed to execute the solution in the real world, knowing the specific conditions needed if any, under which the solution can be applied to achieve the wanted result. Also bearing in mind the limitations of the solution if there are any, and keep the implementation and the real execution within the limits, or else the expected results won’t be met and may result in deteriorating the situation instead of improving it.
The step allows us to detect any issues made during the other steps or any unrealistic or impractical assumptions of the variables of the problem or the execution potential in real-life or unexpected complications. In this phase, we validate the results to determine their accuracy and validity or if they need some tuning to overcome some overlooked issues, in addition, reporting any recommendations to be made aiming to ameliorate the results and the implementation process.
Let’s keep in mind that this process divided into 5 steps is just one of many other ways to describe the operations research process, you may find several ones using only 3 steps or some other ones detailing into 7 steps depending on the level of details intended for describing the process.
After introducing operations research, its development, applications, and its process, we will discover in this section the possible opportunities and careers that operations research opens, the demand for O.R. in the market, the salaries, and some forecasts into the future. For this, we will take the USA as a study case and the statistics and results we are about to study here are from the US Bureau of Labor Statistics Occupational Employment Statistics (2019 Data).
The job of Operations research analyst can also have many other similar, usually overlaying or intersecting, occupations like Logisticians, Management Analysts, Market Research Analysts …etc.
In the following statistics, we will take operations research analysts as the study case.
The statistics show the growth of the base salary during 20 years (from 1999 to 2019), we can see a big rising in salaries from $53,850 in 1999 to $90,600 in 2019 which means that the wages grew by $36,750 in 20 years or by an average of $1,837 per year.
Note that the salaries do not include bonuses, signing bonuses, benefits, overtime, tips, or commissions.
This chart represents the average base salary estimation for a window of 10 years (2019-2029), it shows that by 2021 the average salary is estimated to be $94,275 and by 2029 it would rise to $108,975.
The map shows a remarkable difference in base salaries between different states, where the darker blue the state is on the map the higher is the base revenue.
The top 5 highest paying states are:
The top 5 lowest paying states are:
We can see that there is a $57,600 difference between the highest and the lowest paying state.
We notice that the Finance & Insurance sector and the Professional Services sector are incorporating the most operations research in their environments, followed by the Federal Government and Manufacturing sectors. We also see a big portion of 32% for the other sectors since operations research has multiple applications in almost every field.
Also, the statistics give us a view on how the 2019 pay breaks down by work environment:
We can see that the Federal Government sector is paying higher with a remarkable margin from the other work environments.
Reading the chart, we conclude that taking the average base pay as a reference, operations research analysts outearn logisticians and market research analysts with a remarkable margin. Also, operations research analysts are outearned with a small difference by industrial engineers and management analysts and outearned with a big difference of $26,030 by economists.
As we can see in the chart, operations research analysts are more and more in demand, and we expect that to be the case for the long future since decision-making and optimization are not a luxury for any field and for its uses in all the sectors including the super growing ones like Artificial Intelligence and Machine Learning. At the end of the decade, we predict to reach 124600 new employed operations research analysts.
The chart shows that even operations research analysts that only have a bachelor’s degree have a decent portion with 30% of the total occupations, but we also see that operations research analysts with master’s degree have better chances and representing 70% of the total occupations, meaning that having a bachelor’s degree good and opens several opportunities and you can choose to widen your knowledge with a master degree and enter the job market with more chances at hand.
Are you interested in knowing more about operations research and its endless applications? Here are some links to some resources that you may want to check:
Operations research is an important discipline and science to incorporate into every business and so many fields, and since decision-making is necessary, crucial, and a key step of today’s world, then comes into view the operations research earning the title of the art of problem-solving and decision making, letting the world reap the benefits by integrating the science of doing better in every aspect possible.
The post Operations Research: The Science Of Doing Better appeared first on Math Academy.
]]>The post 20 Professional Roles Requiring Statistics appeared first on Math Academy.
]]>Ever since the cheapening of computer data storage in the 1990s, data has been collected at a rate faster than the development of interpretive schemas from which to analyze this data into useful decision-making results, causing a surplus of data in relation to interpretations. Thus, “Data Science” evolved as a science and role for the application of statistical methods to these large unstructured databases where the required hypothesis to test is ambiguous by using specific techniques and technologies for these datasets, such as Hadoop.
3. Data Analyst
The post 20 Professional Roles Requiring Statistics appeared first on Math Academy.
]]>The post Research Colloquium: SEIR Model of COVID-19 using Economic Functionality for Transmission Coefficient Dynamics appeared first on Math Academy.
]]>Within the short-term, it is not possible to restructure employment in order to accommodate for the loss of proximate-communicative jobs, which must necessarily be halted in order to stop the spread of the infection. Economic-shutdown, as shown in the below figure (3) as total employment loss, was followed within a month with a decrease in the infection-rate. The positive effect of the economic shutdown was shown more clearly in NYC, where the infection-peak was much higher and the subsequent shutdown also more intensive, but nationally the job-loss was followed by a flattening of the infection-rate curve.
For both scales, NYC and nationally, a decrease in infection-rate was followed by a reopening of the economy after an apparent equilibrium was reached with 1/3 dysfunctionality (i.e. labor-side shut-down) in NYC and 1/4 dysfunctionality nationally. These may have been premature and the effects buffered by the fall moratorium on evictions.
I follow a simplified and modified version of the SEIR model presented by Yang and Wang [2], using many of the parameter values discerned by them, but with the viral transmission rate functions based upon the continuous measurement of the economic shut-down, as given by employment-rate changes since January, 2020. In addition to the standard SEIR model, Yang and Wang’s model includes a novel environment variable of viral density to model the indirect pathway of the infection by surface contact and aerosol inhalation. Here I collapse the hospitalized subpopulation (H) into the infected subpopulation (I), which is thus any individual who has tested positive for coronavirus. The compartments in this model thus considered are Susceptible (S), Exposed (E), Infected (I), Recovered (R), and the Environment (V), thus forming a SEIRV model.
The Exposed population is an unmeasured quantity of those who have been exposed to the virus and are thus `infected’ but have not been tested, either because they are still in the incubation period ( days) or do not show any symptoms due to strong immune and respiratory systems. After this incubation period, a proportion of the exposed population () [4] develop symptoms and so get tested, while those of the post-incubation non-symptomatic population, , get tested at the same probability as the general population. Thus, given a testing rate of for the whole population, the post-incubation non-symptomatic get tested at a rate of . This is summarized as a post-incubation exposed testing rate of .
The Environment is included as a compartment since it holds viral density in the air as aerosol and in surfaces from those exposed or infected and can therefore transmit to the susceptible population via .
In this model, the demographic parameters are the population flux rate () and the natural birth-death rate . The birth rate (from 2017) for NYC is (per day) and the death rate for NYC is (per day) giving a natural birth-death rate of [5]. The population of NYC at the start of 2020 was . The population flux from NYC was negative this past year, (people per day) [6]. The 2020 birth-rate for the USA was and the death rate was [7], so , and the migration rate was . The population of the USA at the start of 2020 was .
The parameter estimations for the recovery rates and the environmental shedding/removal rates are from Yang, 2020. The recovery rate of exposed individuals is per day, while of infected individuals is per day. The viral shedding rates () are the rates at which viral counts accumulate in the environment due to infected individuals. These are for exposed individuals and for infected individuals. The rate at which the virus is removed from the environment due to decay is .
The functions are the direct human-to-human transmission rates between the exposed and susceptible individuals and between the infected and susceptible individuals, respectively, while represents the indirect environment-to-human transmission rate between the environment and susceptible individuals. For each transmission coefficient we separate them into their biological invariant coefficients measured from the ‘normal’ conditions of viral propagation, , although these may have possibly increased over time due to mutations, and their social dynamic components based upon the functional communicativity of the populations’ sociality given mostly by health policies and their compliance. Most likely, although the virus takes some time to replicate during the incubation period until the person is fully contagious. For the betas of the 3 transmission compartments, E, I, & V, we can formulate them functionally through the parallel expressions:
In order to test the economic shut-down hypothesis, it remains thus to test how well it fits the data, which will be the subject of the next research colloquium.
The post Research Colloquium: SEIR Model of COVID-19 using Economic Functionality for Transmission Coefficient Dynamics appeared first on Math Academy.
]]>The post How to Hire the Right Math Tutor appeared first on Math Academy.
]]>When searching for a tutor there are three major things you need to have in place to ensure a good experience. Having a good tutor can be a great benefit far more bang for your buck than classroom education, but there are three common problems that can interfere with your getting your money’s worth. These problems are inability to communicate, lack of real knowledge of the subject matter, and plain irresponsibility. On the other side of the equation, you can (if you are deliberate in your choosing) find a tutor who doesn’t merely avoid these problems but gives far greater benefit than all the teachers you’ve ever had combined.
Brilliant mathematicians are notoriously bad communicators. (Those of us who are not downright schizophrenic, like John Nash, famously portrayed in A Beautiful Mind, often love the world of ideas especially because of its removal from real life!). If you have an absent-minded- professor type for a tutor, you have an abundance of knowledge but no way to access it. If you have a present-minded tutor who knows how to nurture the students’ learning process, you have an invaluable aid to your learning.
What a good tutor and good tutoring company do to communicate clearly:
People who apply to work as tutors at many tutoring companies have studied at some of the top universities in the country. Yet there are some candidates who shouldn’t be hired even with the kind of credentials any student of mathematics would envy, simply because they do not have listening or teaching abilities. Many students and their parents believe that if they get a tutor from an ivy league school, then that tutor is immediately knowledgeable and possesses the necessary teaching and communication skills, and can also build rapport with them. This is a misconception that needs to be dispelled. Many tutoring companies actually use this approach for their marketing strategy:
“All our tutors are from ivy league schools”
does NOT necessarily translate into:
“Our tutors can help you, they are knowledgeable, and have good teaching and communications skills”
Therefore, a successful tutoring company that actually yields significant results with its students, seeks talent everywhere. Whether it’s from an ivy league school or from a small college with a strong math program. It can be a brilliant student that immigrated to the United States from abroad or a talented math professor from a public college.
In matching you with your tutor, the right tutoring company listens carefully and gets a sense of what the best kind of match will be. And if they really has class, then if the tutor isn’t a match for you, they won’t pair you with a tutor who can’t really help your needs, they may simply refer you to another company which may help.
What YOU can do to ensure your needs:
When you speak with a tutoring company the first time, be sure to ask about the communication skills of the tutors. Then, when you get matched up, have a conversation with the tutor yourself, and ask questions. Ask the tutor to explain a mathematical concept to you. If you don’t understand the explanation, or if you find you’re having to work hard to understand or keep up, this might not be worth pursuing.
What to look for:
A tutor needs to know the material not just well enough to pass the exam himself/herself, but know it when on the spot, and be able to explain it as well as doing it. Real knowledge of subject matter is actually a different skill set from the ability to solve a problem and get a grade on a exam.
A tutor needs to have teaching-knowledge as well as learning knowledge. Concepts often seem to slip out of some people’s heads when they’re asked to explain them. A real understanding of the context for a mathematical concept is necessary, and of the fundamental principles involved, not merely a pat method for solving a problem. This is particularly true on the SAT or ACT, tests of reasoning rather than subject matter, and on higher level math courses such as college and graduate courses.
For higher-level courses: if you’re a college or graduate student, you need a tutoring company that specializes in tutoring for college graduate level math. There are few companies in New York City and around the country who are advanced enough in mathematics to be able to tutor these subjects. The best math tutors know the best places to work, so ask them which companies are good.
Some tutoring companies are geared toward their bottom line, not your learning. Their corporate structure legally requires them to maximize their profits, not your learning. They use rote teaching methods, formulas to ensure that quality is controlled. The downside of this is that, since the tutors aren’t free to depart from the tried-and-true methods, the improvement in students is very slight. The tutoring isn’t individualized or tailored to the student’s particular needs. As a business model it makes sense: put a large quantity of students through a rote program and ensure they’re all slightly satisfied. But the great additional gains that could have been garnered are all lost.
How a good tutoring company ensures knowledge of the subject matter:
Good companies test everyone who works for them. Potential tutors have to get a score of %80 or better on their assessment test. A good tutoring company doesn’t go by references alone. They ask tutors to explain a concept aloud, on the spot, in order to see directly how they teach. Many good companies give you a chance to “test-drive” a tutor for one session and guarantee a full refund if you’re not completely satisfied. (A good company will also have %95 or more of their clients satisfied and continuing lessons.)
How you can ensure knowledge of subject matter:
When you approach a tutoring company, ask what credentials the tutors have. Look at the tutor profiles page. When talking with a tutor ask the tutor to explain why something works as well as how to do the math.
The reality is some tutors, especially in fast-paced New York City, just don’t show up, or don’t return phone calls for weeks at a time.
What A Good Company Does to Prevent Run-away Tutors:
We’ve had a few come through our company even. We were astonished that some people could be so unprofessional. We learned. Now, though our default attitude continues to be trust-based, we understand it is possible for this to occur, and if you came to us for tutoring you’d need to know about us, that a tutor who does this is removed from her/his jobs and replaced, period. They don’t work for our company again. (Who knows, they may be working for some other company now, so be sure to do some due diligence!). We expect this is how any good company will handle this. But not all companies do have a backup tutor at the ready, one who’s fully competent to take over in the middle of the process and get your ready for your exams.
This kind of thing happens very rarely with tutoring companies, and is by far the most infrequent of the three problems we’ve outlined here. But when it does happen it can be very distressing for students. So it’s good to know you won’t run into that problem.
How to Do Your Due Diligence:
Ask a tutoring company if they’ve ever had this occur, and how they handled it. Ask for a guarantee that not only will your problem be addressed but that a backup tutor will be available immediately to ensure you’re ready for your exams in time. And that this backup tutor be at least as good as the original.
In summary, communication skill, real knowledge of the principles behind the material, and guarantees of responsiveness are the three most important things to ensure you’re getting a good tutoring experience that brings you to where you want to be in your studies and your grades. Don’t just take our word for it, ask us, ask whatever other company you look at out there.
Take the time to get it right the first time, since it will waste valuable time and make things that much more frantic if you have to do it all again later, plus adding more confusion to your studies or those of your child. And also, if you get a really good tutor, a really great match, you won’t just get the benefit of repairing what’s not been working in your education previously, you may find you are learning worlds more than you were before.
What to ask the tutor or tutoring director:
What to look for:
The post How to Hire the Right Math Tutor appeared first on Math Academy.
]]>The post Absolute Value and Square Roots appeared first on Math Academy.
]]>Example 1: Simplify
This problem looks deceptively simple. Many students would say the answer is and move on. However, that is only true for positive values of .
Try
Left-hand side:
Right-hand-side:
The equality holds for (and actually for all values of that are ).
Try .
Left-hand side:
Right-hand-side:
Since , this equation does not hold for . Notice that when , the left hand side is actually equal to . Therefore, for all
To summarize, for all , , and for all .
We can write this as a piecewise function as follows:
Our assignment was to simplify, and a piecewise function is surely not simpler than . However, this piece-wise function is actually the exact definition of absolute value. Therefore, we can write , and our simplification is complete. This holds true, for any expression in the square root that has a even exponent. Even though is positive, it can be product of two negatives ex: or two positives . Therefore, the absolute value ensures that both cases are addressed in the solution.
Example 2: Solve for
To solve for , we first take the square root of both sides.
As we discussed earlier, , so
The equation isn’t quite solved for yet. To remove the absolute value, we write:
, and our work is done.
When working on problems involving square roots, remember to always check the positive and negative cases and be careful that you don’t miss the absolute value.
The post Absolute Value and Square Roots appeared first on Math Academy.
]]>The post Absolute Value and Logarithms appeared first on Math Academy.
]]>Absolute values often turn up unexpectedly in problems involving logarithms. That’s because you can’t take the log of a negative number.
Let’s first review the definition of the logarithm function:
Log_{b }x = y ⇔ b^{y} = x
(The double arrow is a bi-conditional, which means that one side is true if and only if the other side is true).
There are several restrictions on this definition when using real numbers (those not involving i = √-1 ).
If any of these conditions are not met, we risk introducing imaginary solutions or an infinite number of solutions. To prevent this from happening, we must follow these rules when working with real numbers, and to do so we often need to introduce the absolute value.
Example 1. Expand the function f(x,y) = log(xy).
We must always have a positive number inside the log function, and since x and y are being multiplied together, they must then always have the same sign. Therefore, the domain of our function is xy>0 which means that f(x,y) lies in Quadrants I and III on the x-y plane.
To expand, we must follow the above rules, with emphasis in this case on Rule 3. To make sure that the values inside the log are positive, we introduce the absolute value.
log(xy) = log|x| + log|y|
If we had not included the absolute values, log(x) + log(y) would only be defined when x and y are both positive (Quadrant I). Since the domain is different than that of log(xy), they cannot be equivalent functions.
Example 2. Simplify the function g(x,y) = log|x| + log|y|
It is important to notice that the above solution is not true in the reverse direction. g(x,y) = log|x| + log|y| is defined for all real, nonzero values of x and y. Therefore, to simplify this function, we must make sure the domain remains all real nonzero numbers. To do so, we use the absolute value. Simplifying,
log|x| + log|y| = log|xy|
Example 3. Evaluate the indefinite integral ∫ 1⁄x dx
This common integral can be found in many integral tables. It is equal to ln|x| + C. The absolute value is important because this is an indefinite integral, which means x might range through the entire real number line (There is a singularity at x=0, but log(0) is undefined too). We introduce the absolute value into the log to ensure that the antiderivative is defined everywhere the integral is.
When working in the real numbers with log functions, remember to always follow the three rules listed above. For many expressions involving logs, it is often necessary to introduce absolute values to ensure that these rules are met.
The post Absolute Value and Logarithms appeared first on Math Academy.
]]>The post L’Hospital’s Rule appeared first on Math Academy.
]]>If and are differentiable on some interval around the number (or if , and are differentiable for all for some ), and , then
We can take the derivatives of the numerator and the denominator and then take the limit and it will be the same as the limit of the original ratio. Indeterminate form includes limits of the forms , , , , , and .
Example 1.
In this example, we see that taking the limit of the numerator and denominator results in , which is an indeterminate form. And since there is not an obvious way to simplify the fraction in order to make the limit easier to compute, it is a good candidate for L’Hospital’s Rule. We take the limit of the numerator and denominator as follows:
Example 2.
Here we see that attempting to take the limit as is results in . So again we use L’Hospital’s Rule:
Taking the limit at this point still results in , so we apply L’Hospital’s Rule again.
is not an indeterminate form and so that is our final answer.
Example 3.
If we plug zero into the limit, we get the indeterminant form . We can’t apply L’Hospital’s Rule right away because we do not have a ratio. Instead, we must manipulate the function so that it becomes a ratio. We can move to the denominator by rewriting as .
Now we have a ratio and the indeterminate form . Applying L’Hospital’s Rule,
We still have an indeterminate form , but before applying L’Hospital’s Rule again, we should notice that we can cancel the in the top and the bottom. It is always best to simplify after applying L’Hospital’s Rule if possible.
If we had tried to use L’Hospital’s Rule before canceling, our ratio would become more complicated with each application of the rule.
Example 4.
Although L’Hospital’s Rule can always be applied to functions that meet the criteria of being differentiable around an interval, it does not always yield a useful answer, even if we see indeterminate forms. Here we will try to use L’Hospital’s Rule.
Noticing that this ratio isn’t any simpler, we can try to apply L’Hospital’s Rule again.
This is the original limit we started with, so we are not getting anywhere using L’Hospital’s Rule. Instead, we should try to simplify our problem by dividing everything by the element with the highest power, in this case .
Noticing that approaches as approaches infinity, we can now take the limit
L’Hospital’s Rule is often useful to evaluate limits of indeterminate form. It is most useful when it makes the numerator and denominator simpler than they were before. Remember to always simplify after applying the rule. L’Hospital’s Rule is applicable to all limits of ratios of differentiable functions, but is not guaranteed to make the limit easier to find.
Although the rule was named for Guillaume de l’Hôpital, it was actually discovered by his employee Johann Bernoulli
The post L’Hospital’s Rule appeared first on Math Academy.
]]>The post Algebraic Groups appeared first on Math Academy.
]]>One of the most fundamental algebraic structures in mathematics is the group.
A group is a set of elements paired with an operation that satisfies the following four conditions:
I. It is closed under an operation (represented here by “+”, although it does not necessarily mean addition):
For all elements a and b in the set S, a+b is also in S.
II. It contains an identity element (often written as “e”):
There is some element e in the set S such that for every element a in S, a+e = e+a = a.
III. The operation is associative:
For all a, b, and c in the set S, (a+b)+c = a+(b+c).
IV. Inverses exist:
For every element a in the set S, there is an a^{-1} in S such that a+a^{-1} = a^{-1}+a = e We will look at some examples of groups and sets that aren’t groups.
Example 1. Integers under addition, (ℤ, +).
I. The group operation is addition. For any two integers n and m, n+m is also an integer.
II. The identity element is 0 because for any integer n, n+0 = 0+n = n.
III. Addition is associative.
IV. Inverses exist. The inverse of any integer n is -n, and n + (-n) = (-n) + n = 0. Therefore, (ℤ, +) is a group.
Example 2. Integers under multiplication, (ℤ, ⋅)
I. The group is closed under multiplication, because for any two integers n and m, n⋅m is also an integer.
II. The identity element is 1. For any integer n, n⋅1 = 1⋅n = n.
III. Multiplication is associative.
IV. Inverses do NOT exist. For any integer n, 1/n is not an integer except when n=1. Since the the set does not meet all four criteria, (ℤ, ⋅) is not a group.
Example 3. The set of rational numbers not including zero, under multiplication (ℚ – 0, ⋅)
I. The group is closed under multiplication.
II. 1 is the identity element.
III. Multiplication is associative.
IV. Inverses exist in this group.
If q is a rational number, 1/q is also a rational number. And q⋅ (1/q) = (1/q) ⋅ q = 1.
Notice that if 0 were in this set, it would not be a group because 0 has no inverse. Therefore, ℚ – 0 is a group.
Example 4. The set {0, 1, 2} under addition
I. This set is not closed under addition because 1 + 2 = 3, and 3 is not part of the set. Therefore, the set cannot be a group.
Example 5. The trivial group {e} This group consists only of the identity element. We don’t need to specify the operation here because it works for both multiplication and addition.
For addition, e=0. 0+0=0 so the group is closed under addition, has an identity element, and is closed under inverses (and addition is associative).
For multiplication, e=1 and similarly, 1⋅1 = 1.
The examples here involved only numbers, but there are many different types of groups.
For example, the ways to transform a triangle is called a Dihedral Group, with the group operation being the act of rotating or reflecting the shape around an axis. The possible permutations of a Rubik’s Cube are also a group, with the operation being a sequence of moves. It’s powerful to know that a set is a group because it gives you an understanding of how the elements will always behave. Groups have many other properties that are useful to mathematicians, and there is a whole field of study built off of this knowledge called Group Theory.
The set of permutations of a Rubik’s Cube is considered an algebraic group.
The post Algebraic Groups appeared first on Math Academy.
]]>The post Algebraic Rings appeared first on Math Academy.
]]>An algebraic ring is one of the most fundamental algebraic structures. It builds off of the idea of algebraic groups by adding a second operation (For more information please review our article on groups).
For rings we often use the notation of addition and multiplication because the integers are a good analogy for a ring, but the operations “+” and “⋅” do not necessarily represent addition and multiplication in the ring.
A ring is defined as a set S combined with two operations (“+” and “⋅”) that has the following properties:
I. The set S is a group under one operation (we will call this operation “+”):
a. It is closed under this operation: For all elements a and b in the set S, a+b is also in S.
b. It contains an additive identity element (which we will call “0”, although it does not necessarily represent the integer 0): There is some element 0 in the set S such that for every element a in S, a+0 = 0+a = a.
c. The operation is associative: For all a, b, and c in the set S, (a+b)+c = a+(b+c).
d. Inverses exist: For every element a in the set S, there is a (-a) in S such that a+(-a)= (-a)+a = 0.
II. The group operation is commutative: For all a and b in the set S, a + b = b + a.
III. The second operation (written as “⋅”)
Follows these rules:
a. It is closed under this operation.
b. The set contains a multiplicative identity element (which we call “1”, although it does not necessarily represent the integer 1): such that for all elements a in the set S, a⋅1 = 1⋅a = a. c.
It is associative: For all a, b, and c in the set S, (a⋅b)⋅c = a⋅(b⋅c). d. It is distributive over “+” on both the right and left side: For all elements a, b, and c in the set S, a⋅(b+c) = a⋅b + a⋅c and (b+c)⋅a = b⋅a + c⋅a.
Note that inverses are not required for the “⋅” operation. If inverses do exist under “⋅”, we have a special kind of ring called a field.
We will now look at some examples of rings and sets that aren’t rings.
Example 1.
The integers under addition and multiplication.
I. The integers are a group under addition (see our group article). I. Addition can be carried out in any order so it is commutative. III. Multiplication follows the additional rules laid out for rings. The integers are closed under multiplication. The identity element is 1. Multiplication is associative and it is also distributive.
Therefore, the integers are a ring under addition and multiplication. ∎ This is the reason addition, multiplication, 0, and 1 show up in our notation. If a set “acts like” the integers by following the ring axioms, it is a ring.
Example 2.
The set of 2×2 matrices under addition and multiplication.
I. The set of 2×2 matrices form a group under addition.
a. For any 2×2 matrices A and B, A+B is also a 2×2 matrix.
b. The additive identity “0” is the zero matrix .
c. Matrix addition is associative.
d. The additive inverse of a matrix A is just -A. Observe .
II. The addition of 2×2 matrices is commutative, A+B = B+A. III. Multiplication of 2×2 matrices satisfies the following conditions:
a. It is closed under the operation. The multiplication of two 2×2 matrices results in another 2×2 matrix.
b. The identity matrix is I = and for any 2×2 matrix A, A⋅I = I⋅A = A.
c. Matrix multiplication is associative. d. Matrix multiplication distributes over addition on the left and right side (This can be easily proven, but to save space it will not be proven here). Therefore, the set of 2×2 matrices is a ring.
Example 3.
The set of 2×2 matrices with non-negative entries under addition and multiplication. This set satisfies all of the requirements for the multiplication operation, but it is not a group under addition. Without negative entries, most matrices in this set do not have additive inverses.
Therefore, this set is not a ring. Like groups, rings do not have to be composed of real numbers or matrices. Some examples include sets of polynomials, the complex numbers, and integers modulo n. The word “ring” comes from the German word “Zahlring,” which means a collection of numbers that circle back on themselves.
However, now that the field has been studied more extensively, we know that cyclic rings are just one type of rings. The symbol ℤ for the integers is also named after a German word for numbers “Zahlen.”
The post Algebraic Rings appeared first on Math Academy.
]]>