Free Essay

# Differential Equations

In: Other Topics

Submitted By tarungehlot
Words 2717
Pages 11
Differential Equations by tarungehlots

Concepts of Differential Equation Consider a variable that might denote the per capita capital stock level in an economy. When convenient, we can recognize that the level of capital depends upon time by addition the time argument. Doing so, we would write the per capita capital stock level as , rather than just writing . If we think of time as unfolding continuously, we can also think of capital as being a continuous function of time, and we can assume that the function has derivate . This derivative is the instantaneous change in per capita capital. In many presentations, is written as , but we will here use the more familiar notation . By writing the derivate as , so that we include the time argument, we are emphasizing that the value of the derivative may change with time. Because it is typically cumbersome to repeatedly write down the argument, we can also write the derivative as just , while remembering that its value may change with time. A differential equation is an equation that relates the time derivative of a variable to its level. An example is the equation (1) . The variable is called a state variable because it gives the state of the system at any given point in time. In our example, gives the state or level of per capita capital stock.

2. A Dynamic System

The basic dynamic principle is the idea that “the way things are determines the way things change.” A differential equation is one way of modeling the basic dynamic principle. The state variable indicates the way things are, while the time derivative indicates how things change. The differential equation itself is what relates the two. More precisely, a differential equation typically presents a functional relationship, showing how depends upon . When we want to emphasize the functional relationship, we can write the differential equation as , where the general functional form is the rule that tells us how the value of the time derivative is determined from the level of the state variable .

A difference equation can also be used to represent the basic dynamic principle. When a difference equation is used, time unfolds as sequential time periods, rather than unfolding continuously. We will focus on using the differential equation here.

A solution to a differential equation is a function that satisfies the differential equation for all points in time that are of interest. Some differential equations can be solved. However, most differential equations that are of interest in economics cannot be solved. Nonetheless, we can still learn about the path followed by the state variable over time by using a phase diagram technique.

Assuming the function in equation (1) is nonlinear, we can describe equation (1) as a single state variable system, and equation (1) is a nonlinear and non-autonomous differential equation. As a dynamic system, equation (1) is a single state variable system because there is only one time derivative present, which is . If the system were a two state variable system, you would see two time derivatives and two different state variables. The equation is labeled nonlinear because the functional relationship between and is not linear. Typically, nonlinear differential equations cannot be solved. However, the presence of the general functional form in equation (1) assures us that we cannot solve it. Finally, equation (1) is labeled non-autonomous because it includes variables other than state variables and time derivatives. The variables , , , and in equation (1) are exogenous variables, which are variables that describe the environment impacting the system. Usually exogenous variables are constants, which is the case in equation (1).

Classifying the variables of a differential equation system helps clarify how you the model builder believe the system works. Because our system (1) is a single equation system, we can only have one endogenous variable. The endogenous variables a differential equation system are always the time derivatives. So, for our system, is endogenous. The state variables of a system are always classified as predetermined, so is predetermined for our system. The level of the variable is predetermined at point in time because the instantaneous change in the level of we just determined by the model in the previous instant of time. Non state variables are exogenous, so , , , and are exogenous. The last value that needs to be specified for the system is the initial condition for the state variable , which is the value of the variable at the initial point in time . Thus, we can summarize the classification of variables as follows:

Classification of Variables
Endogenous (1): Exogenous (4):, , , Predetermined (1): Initial Conditions (1): It is common that differential equation models contain auxiliary equations, which are equations that endogenized (i.e., determine) the values for other variables by relating them to the state variable. Using standard techniques for analyzing a differential equation of this type, we can learn how the path for state variable will evolve over time. Once we know the path followed by the variable , we can use the following “auxiliary” equations to determine the paths followed by the other variables of interest. To illustrate, let us add the endogenous variables , , , and to our system, and assume that they are determined from the following four equations: (2) (3) (4) (5)

The variables, , and are directly related to the variable , and only related to . Thus, if we know the path followed by , we can use equations (2)-(4) to determine the paths followed by , , and . Equation (5) indicates that we need to know the path followed by , along with the value of the exogenous variable , to determine the path followed by . So we can determine the path followed by by first using equation (2) to determine the path followed by . This discussion indicates that we can classify the variables of the dynamic system (1)-(5) as follows.

Classification of Variables
Endogenous (5): , , , , Exogenous (4):, , , Predetermined (1): Initial Conditions (1):

The core of this model is the path followed by the state variable . Once you understand the core of a model, you can use that knowledge to then understand the auxiliary aspects of the model. Notice that adding the auxiliary equations only changes the number of endogenous variables. We can think of the core of the model, which is equation (1) and the first classification of variables as a subsystem, a system upon which the auxiliary aspects of the model depend.

3. Analyzing the Steady State Finding the Steady States The steady state for the state variable is the state where the variable is not changing, or where . Setting in (1), we know SS1 . There are five variables in equation SS1, the model’s four exogenous variables and the state variable. Equation SS1 determines the steady state values of the state variable as a function of the exogenous variables. Assuming , one value of for which equation SS1 holds is . This steady state is not so interesting in that this result is simply telling us that if no capital initially exists, then none will accumulate over time, (which is because capital is assumed to be essential in the production process). If there is a non zero value for that satisfies SS1, let it be denoted by . To illustrate a situation where there is a unique value for the steady state , consider Figure 2 below. The right side of equation SS1 is linear in , with a zero intercept, as shown in the figure. The assumption implies the left side of equation SS1 also has a zero intercept. The conditions and would ensure that the production function increases at a decreasing rate as increases, as shown in the figure. If it were true that, then the only steady state would be associated with , because the two curves and would only intersect at the point in Figure 2. The Inada condition ensures that the curve initial rises above the curve, and the Inada condition ensures that the curve eventually intersects the curve. Because and for all , we know that this intersection at a capital level only occurs once, which implies is unique, as shown in Figure 2. Figure 2: The steady state for the Solow-Swan growth model

Stability of the Steady State The Inada conditions and also ensure that the steady tate associated with is stable. A steady state is locally stable if the state variable moves toward the steady state value for the variable when it is in the “neighborhood” of the steady state. Define the function so . Notice that, if , then the steady state is locally stable. This is because implies when and when . For our model given by SS1, . So, . Because, our steady state is locally stable if . Notice, in Figure 2, that the slope of the curve, at the steady state , is less than the slope of the curve. That is, . So, we find, meaning our steady state is locally stable. Examining Figure 2, we can also see that the steady state is locally unstable. In the neighborhood of , with positive, notice in Figure 2 that the slope of the curve is greater than the slope of the curve. This implies , which implies . If in the neighborhood of , then is increasing in that neighborhood, meaning it increasing away from the steady state value of zero. Figure 3 is a phase diagram. The traditional phase diagram for a single state variable differential equation is constructed by plotting the state variable along the horizontal axis and plotting the time derivative of the state variable along the vertical axis. So, for our model is plotted horizontally and is plotted vertically. The function shows how depends upon . The Inada conditions tell us that the function is positive and diverging to infinity as approaches zero, and they tell us decreases as increases, eventually becoming negative, as shown in Figure 3. The steady state value for is where equals zero. The arrows along the horizontal axis are presented to emphasize is increasing when but decreasing when . Figure 3: Phase diagram for the Solow-Swan growth model

To summarize, we know that the steady state is locally unstable and the steady state is locally stable. The Inada conditions indicate that the path taken over time by the state variable is as shown in Figure 3. If the initial capital stock level is between zero and , then increases over time and approaches . Alternatively, if is greater than , then decreases over time and approaches . Figure 4 shows these two potential time paths for by plotting as it depends upon time.

Figure 4: Two possible time paths for the state variable in the Solow-Swan growth model

4. A System of Two Differential Equations Suppose a dynamic system is represented by the following system of equations and the following classification of variables (A)

(B) Endogenous (2): , Exogenous (2): , ,
Predetermined (2): ,
Initial Conditions (2): ,

How can we characterize the paths followed by the two state variables?

5. Using a Phase Diagram to Characterize the Two Dimensional Differential Equation System

We now characterize the path followed by and diagrammatically in a “phase diagram.” The first step in constructing the phase diagram is to plot the “nullclines” associated with the steady state conditions and . From (B), we can see that implies . The usual restrictions on the production function imply there is a unique value such that . (Notice that we need the Inada conditions and to conclude this. The restrictions and alone are not enough.) When , the assumption implies, so which implies the right side of (B) is negative. Because and are typically assumed for the utility function, we know must be positive for the left side of (B) to be negative. Thus, when , . Using analogous reasoning, when , . This is shown by the arrows in Figure 1.

We obtain the nullcline by setting in condition (A), which implies
. Because , we know when . The Inada conditions and allow us to draw conclusions about the shape of this nullcline consumption level . The assumption ensures that when is near zero, which implies when is near zero. Moreover, ensures increases faster than as increases from a number near zero, which implies is increasing as increases from zero. The assumption ensures that, when is large enough, increases faster than as increases. This implies that, as increases, must reach a peak and then decline thereafter. This nullcline separates the region where is increasing from where it is decreasing. Condition (A) indicates that, when is greater than its nullcline value for a given level , then . Conversely, when is less than its nullcline value for a given level , then . This is shown by the arrows in Figure 1.

Figure 1: Dynamics implied by the Euler Equation and Capital Accumulation Constraint for the Neoclassical Model of Optimal Growth

Starting at any point in the of Figure 1, the arrows tell us the direction of movement. This allows us to trace the path from any starting point . The starting point for is given to us in the problem. The starting point for is a choice. Without additional restrictions, we cannot determine which value of will occur. However, proceeding with some analysis can provide an indication of what those restrictions might be.

Figure 2 is drawn under the assumption that . (After understanding this analysis, it should be apparent how to present the case for .) Path 1 is associated with a relatively high initial consumption choice . Because is in the upper left quadrant of the phase diagram, the direction of movement must be up and to the left ( increasing and decreasing). . Path 2 is still associated with a relatively high initial consumption choice , but one where falls in the lower left quadrant. Initially, the direction of movement must be up and to the right ( increasing and increasing). However, the nullcline is eventually reached, and the direction of movement of is reversed. Thus, the paths of type 2 eventually become like those of type 1. For all paths of type 1 and type 2, must eventually occur. To rule out such paths as equilibrium paths, we need only add the reasonable condition to the problem.

Figure 2: Phase Diagram for the Neoclassical Model of Optimal Growth

3
2
1
4

Path 3 is associated with a relatively low initial consumption choice . Because is in the lower left quadrant of the phase diagram, the direction of movement must initially be up and to the right ( increasing and increasing). However, on this path, the nullcline is reached, and the direction of movement of is reversed. Eventually, such a path must involve a negative consumption level. One way to rule out such a path is to recognize that, at any given point in time , when the capital-consumption combination is , the consumer can discontinuously increase or decrease the consumption level, for such an action is equivalent to the choosing when . On path 3, by choosing to jump paths by discontinuously increase , (by choosing to reduce the saving level ), the consumer can move to a new path that will provide a higher consumption level in every period, for all time, which is clearly preferable. Thus, it is not optimal to choose a consumption level that generates a path of type 3.

Having ruled out paths of types 1 and 2 because they are not feasible, and paths of type 3 because they are clearly not optimal, we can conclude that the unique optimal choice for , given the initial capital level is the level for that generates path 4, or the level shown in Figure 2. On this path, consumption and capital each increase en route to the steady state , never crossing a nullcline.

It is worth noting that, if the consumer believes that the economy would end at a particular point in time , rather than assuming the economy will never end, then it would be rational to choose a consumption path such that . That is, the consumer should choose the consumption path where capital is just used up when the economy terminates. (The Euler equation and capital accumulation equation would be the same; only the “boundary condition” condition would change to .) This implies, rather than choosing path 4 in Figure 2, the consumer would choose a path either of Type 1 or Type 2, so that vertical axis where is eventually reached. Which path is chosen would depend upon the length of time T. The optimal initial consumption level becomes lower as the length of time T becomes longer.

### Similar Documents

#### Differential Equation

...Exercise 1 – Find the first derivative and the second derivative of the following functions Answer: Applying constant function and power function rule (A) Y = 3 + 10X + 5X2 dY/dX = 0 + 1.10.X1-1 +2.5.X2-1 dY/dX = 10 + 10X d2Y/ dX2 = 0 + 1.10.X1-1 d2Y/ dX2 = 10 (B) Y = 2X (4 + X3 ) Y = 8X + 2X4 dY/dX = 1.8.X1-1 + 4.2.X4-1 dY/dX = 8 + 8X3 d2Y/ dX2 = 0 + 3.8.X3-1 d2Y/ dX2 = 24X2 (C) Y = 3 /X2 Y = 3X-2 dY/dX = -2.3.X-2-1 dY/dX = -6X-3 dY/dX = -6/X3 d2Y/ dX2 = -3.-6X-3-1 d2Y/ dX2 = 18X-4 d2Y/ dX2 = 18/X4 (D) Y = 18T – 2T2 dY/dT = 1.18.T1-1 – 2.2.T2-1 dY/dT = 18 – 4T d2Y/ dT2 = 0 – 1.4.T1-1 d2Y/ dT2 = - 4 Exercise 2 - Find the partial Derivative of Y with respect to X Answer (A) Y = 10 + 3Z + 2X ∂Y/∂X = 0 + 0 + 1.2.X1-1 ∂Y/∂X = 2 (B) Y= 18Z + X2 + Z.X ∂Y/∂X = 0 + 2.1.X2-1 + Z ∂Y/∂X = 2X + Z Application - The nursing home industry is growing rapidly because the aging of American population. According to the study of an economist, the average cost per patient day of a nursing home can be approximated by C = A – 0.16B + 0.002B2 Where, B is the nursing home’s number of patient days per year ( in thousands) and A is the number that depends on the location and other factors but not on B. Based on the information , how big must a nursing home be ( in terms of patient – days) to minimize the cost per patient day ? Answer – C= ƒ (A, B) Where C is Avg. Cost per patient day A is variable depends on location and other......

Words: 404 - Pages: 2

Free Essay

#### Ordinary Differential Equations

...MATH 364A: Ordinary Diﬀerential Equations (Midterm 1) Name: Student ID: Signature: Question 1 (40 points) Solve the following initial value problems. (a) y + t3 y = t3 y(0) = 0. (b) y = − (1+x) y y(−1) = 1. Question 2 (40 points) Solve the second-order initial value problem 2y − 3y − 5y = 0 y(0) = 0 2 y (0) = 1. Question 3 (40 points) For each equation below, ﬁrst determine whether the equation is exact or not exact. If the equation is exact, ﬁnd the solution. (a) cos(y) + (2y − x sin(y)) dy = 0. dx (b) 3xy 2 + (y 3 + 3x2 ) 3 dy = 0. dx Question 4 (40 points) Consider the following ﬁrst-order diﬀerential equation y = (y 2 − 1)(4 − y 2 ). Find all critical (equilibrium) solutions and classify their stability. 4 Question 5 (40 points) (True/False) In each of the following, determine whether the given function solves the given diﬀerential equation or initial value problem. (a) If k > 0 denotes any real constant, then the function y(t) = e−kt solves the initial value problem y = −ky y(0) = 1. (b) The function y(t) = (1 − t)−1 = 1/(1 − t) solves the initial value problem y = y2 y(0) = 1. (c) The function y(t) = e2t solves the second-order initial value problem y = 2y y(0) = 1 y (0) = 2. (d) Any function y(x) deﬁned by the implicit equation x4 + 2x2 y 2 + y 4 = C solves the diﬀerential equation x3 + xy 2 + (x2 y + y 3 )y = 0. 5 Question 6 (Extra Credit, 10 points) An object with......

Words: 324 - Pages: 2

Free Essay

#### Classification of Differential Equations

...CLASSIFICATION OF FIRST ORDER DIFFERENTIAL EQUATIONS SUPPLIMENTARY PROBLEMS 1. Write the given equation in standard form 3.16. exy'-x=y' solution exy'-x=y' =exy'-y'=x ex-1y'=x y'=xex-1 ANS 3.25 dy+dx=0 Solution. dy+dx=0 dydx+1=0 y'=-1 2. The differential equations are given in both standard and differential form. Determine whether the equation in standard form are homogeneous and/or linear, if not linear, whether they are Bernoulli: determine whether the equations in differential form as given are separable and/or exact. 3.28 y'=xy+1: xy+1dx-dy=0 Solution. y'=xy+1 = y'-xy=1 this is in the form y'+pxy=qx for px=-x , qx1 thus this is linear next xy+1dx-dy=0 is not separable si9nce the variables cannot be separeted. To check for exactness take Mx=xy+1 and Nx=-1 for an exact equation, ∂M∂y=∂N∂x ∂M∂y=x and ∂n∂y=0 thus ∂M∂y≠∂N∂x so it is not exact 3.30 y'=x2y2: -x2dx+y2dy=0 Solution. y'=x2y2 =≫ y'+0y=x21y2 The equation is of the form y'+pxy=qxyn where px=0, qx=x2 and n=-2 so it is a Bernoulli equation NEXT For -x2dx+y2dy=0 it is separable since the variables are separated. To check for exactness take Mx=-x2 and Ny=y2 thus ∂M∂y=0 and ∂N∂x=0 hence its EXACT. 3.35 y'=2xy+x: 2xye-x2+xe-x2dx-e-x2dy=0 Solution. For y'=2xy+x =≫y'-2xy=x This is in the form y'+pxy=qx for px=-2x and qx=x Thus it’s a linear equation NEXT 2xye-x2+xe-x2dx-e-x2dy=0 this is not separable since the variables...

Words: 336 - Pages: 2

#### To Sir with Love

...Engineering and Science subjects. Here our intention is to make the students acquainted with the concept of basic topics from Mathematics, which they need to pursue their Engineering degree in different disciplines. Course Contents: Module I: Differential Calculus Successive differentiation, Leibnitz’s theorem (without proof), Mean value theorem, Taylor’s theorem (proof), Remainder terms, Asymptote & Curvature, Partial derivatives, Chain rule, Differentiation of Implicit functions, Exact differentials, Tangents and Normals, Maxima, Approximations, Differentiation under integral sign, Jacobians and transformations of coordinates. Module II: Integral Calculus Fundamental theorems, Reduction formulae, Properties of definite integrals, Applications to length, area, volume, surface of revolution, improper integrals, Multiple Integrals-Double integrals, Applications to areas, volumes. Module III: Ordinary Differential Equations Formation of ODEs, Definition of order, degree & solutions, ODE of first order : Method of separation of variables, homogeneous and non homogeneous equations, Exactness & integrating factors, Linear equations & Bernoulli equations, General linear ODE of nth order, Solution of homogeneous equations, Operator method, Method of undetermined coefficients, Solution of simple simultaneous ODE. Module IV:  Vector Calculus                                                               Scalar and Vector Field, Derivative of a Vector, Gradient,......

Words: 310 - Pages: 2

#### Econ60081

...REMOVED FROM THE EXAMINATION ROOM Electronic calculators may be used provided that they cannot store text. 1 of 3 P.T.O. 2 of 3 ECON60081 SECTION A Answer ALL Questions. Each question carries equal weight. Question 1. Consider the problem Ax = b where ⎛ ⎞ ⎞ ⎛ ⎛ ⎞ 2 1 3 1 1 A = ⎝ 3 2 5 ⎠  x = ⎝ 2 ⎠  b = ⎝ 3 ⎠  1 1 2 3 2 Determine the degrees of freedom and the number of redundant equations of this system. Further, determine the solution(s) if solutions exists. [10 marks] Question 2. Give formal deﬁnitions for the following: (a) a convex function, (b) a strictly con­ vex set, (c) a differentiable function. Further, give an example of a concave function that is not differentiable. [10 marks] Question 3. Find the solution of the following differential equation  = 1 + 3 − 2  ˙ where (0) = 5. [10 marks] Question 4. Find the general solution of the following second order differential equation  + 4 + 10 =  ¨ ˙ [10 marks] Question 5. Consider the following system of nonlinear difference equations ½ 1+1 = 31 − 2  2 2+1 = 2 + 1  Find the equilibria and classify them as sink, source or saddle. [10 marks] Continued 3 of 3 ECON60081 SECTION B Answer ALL Questions. Question 6. Assume  ≥ 4. For   −1 and   −1, solve the utility maximization problem (a) max ( ) = 1 ln(1 + ) + 1 ln(1 + ) subject to the constraint 2 + 3 = .[15 marks] 2 4 (b) Let (∗......

Words: 347 - Pages: 2

Free Essay

#### Abstract Model

...deterministic or probabilistic decision models. As deterministic models, decisions to bring good final outcomes. A deterministic model is “you get what you expect” risk-free model, which determines the outcome. It also depends on the influence of the uncontrollable the factors that determine the outcome of a decision and the information the decision-maker input as a predicting factor (Arsham, 1996). According to Schrodt (2004), deterministic models was widely used in the early 18th century to study physical processes to develop differential equations by many mathematicians. These differential equation allow values of a variable as function of its value at any point in time and as a common form of the deterministic concept. The differential equations apply to a variety of astronomical and mechanical phenomena and have produced crucial scientific literature (Schrodt, 2004). The model is important in astronomical and mechanical phenomena because of the finite equations...

Words: 1025 - Pages: 5

Free Essay

#### Chocolate

...Syllabus Cambridge International A Level Further Mathematics Syllabus code 9231 For examination in June and November 2013 Contents Cambridge A Level Further Mathematics Syllabus code 9231 1. Introduction ..................................................................................... 2 1.1 1.2 1.3 1.4 Why choose Cambridge? Why choose Cambridge International A Level Further Mathematics? Cambridge Advanced International Certificate of Education (AICE) How can I find out more? 2. Assessment at a glance .................................................................. 5 3. Syllabus aims and objectives ........................................................... 7 4. Curriculum content .......................................................................... 8 4.1 Paper 1 4.2 Paper 2 5. Mathematical notation................................................................... 17 6. Resource list .................................................................................. 22 7 Additional information.................................................................... 26 . 7 .1 7 .2 7 .3 7 .4 7 .5 7 .6 Guided learning hours Recommended prior learning Progression Component codes Grading and reporting Resources Cambridge A Level Further Mathematics 9231. Examination in June and November 2013. © UCLES 2010 1. Introduction 1.1 Why choose Cambridge? University of Cambridge International Examinations (CIE) is the world’s largest provider of......

Words: 7018 - Pages: 29

Free Essay

#### Structure

...Beams in this chapter, we describe methods for determining the equation of the deflection curve of beams and finding deflection and slope at specific points along the axis of the beam 9.2 Differential Equations of the Deflection Curve consider a cantilever beam with a concentrated load acting upward at the free end the deflection in the y v is the displacement direction of the axis the angle of rotation (also called slope) is the angle between the x axis and the tangent to the deflection curve point m1 is located at distance x point m2 is located at distance x + dx slope at slope at m1 m2 is is +d denote O' the center of curvature and the radius of curvature, then d = ds is and the curvature 1 = 1 C = d C ds the sign convention is pictured in figure slope of the deflection curve dv C dx for = tan or ds j dx d C dx d C = dx = cos j 1 and d 2v CC dx2 = dv tan-1 C dx tan dv C dx j , then small = 1 C = 1 C = = if the materials of the beam is linear elastic = 1 C = M C EI [chapter 5] then the differential equation of the deflection curve is obtained d C dx d2v = CC dx2 = M C EI and v dV CC dx d 4v CC dx4 = -q q -C EI it can be integrated to find ∵ dM CC dx d 3v CC dx3 = V then V = C EI = 2 sign conventions for M, V and q are shown the above equations can be written in a simple form EIv" = M EIv"' = V EIv"" = -q this equations are valid only when Hooke's law applies and when the slope and......

Words: 5105 - Pages: 21

Free Essay

#### Matlab and Ode Solvers

...Lab 0: MATLAB and ODE Solvers | ME 4173 Robot Kinematics | | | | Introduction The following report will display the results and conclusions of an experiment to simulate the output of an inverse pendulum system in MATLAB. The objectives of this experiment were to review MATLAB programming and using MATLAB to simulate ODEs and systems. Objectives * Examine the basics of MATLAB * Use MATLAB to simulate a system * Use ODE solvers to numerically integrate the system over a set time period Apparatus The apparatus used in this experiment was MATLAB. It was used to provide a simulation environment to analyze the inverse pendulum’s motion. Experiments and Results There were six components of this experiment. This experiment was mostly familiarizing with MATLAB. All code used is illustrated in the Appendix – Code. The first part consisted of learning commands within the MATLAB environment. It was a brief overview of how commands work in MATLAB. There was no code used in the part of the experiment. The second part of the experiment examined how arrays were created and used in MATLAB. The first step was to create a matrix. This matrix was then subjected to various commands including eye( ), zeroes( ), and ones( ). Indexing was also used to access various parts of the matrix. Matrix operations such as transpose, inverse, size and length were also shown. Part three of the experiment explained how a script was created and what it was...

Words: 828 - Pages: 4

Free Essay

#### Personal Statement

...entrepreneur. Accordingly, I have been interested in business courses and I decide to take financial related courses, especially Actuarial Studies as the direction of my master’s study. I have been interested in numbers since I was a high school student. I felt satisfied even though I had to contribute more than one hour to solve a mathematic problem. I often spent time to think about other methods to solve mathematic problems that my teacher had provided answers. My enthusiasm about mathematics was inspired again when I began my college study. I took some basic mathematics concepts, such as limit, series, calculus and differential coefficient. I also learned some basic theories and the application of related concepts, such as differential coefficient of function of one variable, calculus, partial derivative of function of many variables, differential equation, and Taylor's formula, intermediate value theorem and infinite series which help me to know the nature of function, and the independent vector algebra and space analytic geometry. To be honest, I even made more efforts in the study of mathematics than that in my academic courses. Therefore, I believe that my mathematics achievements are pretty competitive in pursuing Master of Actuarial Studies. I took multi-directed development when I was in college since I believe that enhancing my learning and surviving ability accounts even...

Words: 578 - Pages: 3

Free Essay

#### Final

... What is called as the setting device? 15. What is called as a function chart and of what it consists? 16. In what difference of a signal from physical size? 17. In what an essence of a principle of the opened management? 18. In what an essence of a principle of indemnification? 19. In what an essence of a principle of feedback? 20. List merits and demerits of principles of management? 21. What special case of management is called as regulation? 22. In what difference of systems of direct and indirect regulation? 23. List and give the short characteristic of principal views CS? 24. What is called as static mode CS? 25. What is called as static characteristics CS? 26. What is called as the equation of statics CS? 27. What difference from strengthening factor is called in transfer factor, in what? 28. In what difference of nonlinear links from the linear? 29. How to construct the static characteristic of several links? 30. In what difference of astatic links from the static? 31. In what difference of astatic regulation from the static? 32. How to make static CS astatic? 33. What is called as a static error of a regulator how it to reduce? 34. What is called statizm of CS? 35....

Words: 2580 - Pages: 11

Free Essay

#### Maths

...COORDINATE GEOMETRY. EQUATION OF A STRAIGHT LINE SOLVED EXAMPLES. 1. ( ) ( Solution. ( Now, using the formula ) we have: ( ( ). ) . . ( 2. ) Solution. ( ) ( ) ( ( ) ), ( ) ) ( ) ( ) Solution. ( ) ( ) ( ) ( ) EQUATION OF A CIRCLE. The general equation of a circle is of the form .Where (– √ is the centre of the circle and the radius is: Finding the equation of a circle of a circle given its radius and centre Let ( ) be any point on a circle whose centre is ( circle is given by : ( ) ( ) ) and ( r ( ( ) ) ) 0 If the centre is at the origin ( ( ) ( ) ) then the equation becomes: ‘ the equation of a ) Solved Examples. 1. Find the equation of the circle with centre ( ) and radius Solution. Using ( ) ( , ( ( )- ) we have, ) ( ) ( ) 2. Find the equation of a circle with centre ( ) which passes through the point( Solution. ( ) ( ( ) ) ( ) √ ( ) ( ( ) ( , ) ) (√ ) 3. Find the centre and radius of the circle Solution. ( ) Comparing (1) with the general equation of a circle The radius of the circle is: √ The centre is ( √ √ ) ( ) ) EXERCISE. 1. Find the centre and radius of the circle 2. Find the centre and radius of the circle 3. Find the equation of the circle which......

Words: 700 - Pages: 3

Free Essay

#### Laplace Transform

...transformation called the Laplace transform. It is very effective in the study of initial value problem involving linear differential equation with constant coefficient. Laplace transform was first introduced by a French mathematician called Pierre Simon Marquis de Laplace about 1780’s. This method associated with the isolation of the original problem that is function ƒ(t) of a real variable and some function ƒ(s) of a complex variable so that the ordinary differential equation for the function ƒ(t) is transformed into an algebraic equation for ƒ(s) which in most cases can readily be solved. The solution of the original differential equation can be arrived at by obtaining the inverse transformation. The transformation and its inverse can be derived by consulting already prepared table of transform. This method is particularly useful in the solution of differential equation and has more application in various fields of technology e.g. electrical network, mechanical vibrations, structural problems, control systems. Meanwhile in this research work, I shall look into the Laplace transform, the properties of the Laplace transform and the use of this technique in solving delay differential equation will be looked into. 1.2 Statement of the Problem There are so many engineering and other related problems that can be expressed in the form of ordinary differential equations. But such problems cannot easily be solved using the elementary method of solution. In such cases, the......

Words: 3775 - Pages: 16

Free Essay

#### Schaltalgebra

...Diese Form der Algebra beschäftigt sich mit mathematischen Aussagen und ihren Verknüpfungen. Dabei gelten Aussagen als mathematisch, wenn sie klar als wahr (w) bzw. falsch (f) zu bezeichnen sind. Wahre Aussagen werden durch einer "1" dargestellt, falsche durch eine "0". Beispiel: Eine mathematische Aussage wäre: 1. Innsbruck liegt in Tirol. (wahre Aussage) 2. Das Münchner Gymnasium hat 10000 Schüler. (falsche Aussage) Dagegen aber keine mathematische Aussage wäre: 1. Schule ist toll. 2. Mathematik macht Spaß. Mit bestimmten Verknüpfungen können mehrere mathematische Aussagen verbunden werden. Diese wären dann: 1. Und ( ) 2. Oder ( ) 3. Nicht ( ) Die Fach-Bezeichnung für die Nicht-Verknüpfung ist Negation, für die Und-Verknüpfung Konjunktion und für die Oder-Verknüpfung Disjunktion. Während und zwei Aussagen Verknüpfen bezieht sich nur auf eine einzelne Aussage oder das "Ergebnis" mehrer Aussagen. Dabei wird dieses Ergebnis folgendermaßen ermittelt: • Eine wahre und eine falsche Aussage durch "Und" verknüpft ergeben eine falsche Aussage. • Zwei wahre Aussagen durch "Und" verknüpft ergeben eine wahre Aussage. • Zwei falsche Aussagen durch "Und" verknüpft ergeben eine falsche Aussage. Beispiele dazu wären: 1. Innsbruck liegt in Tirol (w) und ( ) das Münchner Gymnasium hat 10000 Schüler (f), ist insgesamt eine falsche Aussage. 2. Österreich liegt in Europa (w) und ( ) hat eine Grenze zu Deutschland (w), ist eine......

Words: 422 - Pages: 2