An Introduction to Applied Mathematics: An Iterative Process

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1 An Introduction to Applied Mathematics: An Iterative Process Applied mathematics seeks to make predictions about some topic such as weather prediction, future value of an investment, the speed of a falling mass, epidemic population as well as many others. One must have a math model, a method of solution, usually a computing tool and an evaluation or assessment of the above steps. The assessment most often leads to variations and enhancements to the model. This creates an iteration of the stages (application, model, method, implementation and assessment) so that applied mathematics in a continuing process as is depicted in the following figure. Application Model Method Implementation Assessment Applied Math as a Process Applications include motion of a mass, price prediction or time-value of money. Models might be algebraic equations, differential equation, discrete systems or probability descriptions. Methods of solution could be by-hand calculations, algorithms such as Newton s method or computer calculation. Implementations of method might be done using spreadsheets, Maple, Java or Matlab. The assessments may deal with the derivation of the model, accuracy of methods and computer calculations as well as comparison of predicted values with observed values. The assessment is a critical part of the applied mathematics process as it points the way to more accurate models. Example One. Consider a projectile, which has been observed several times. The applied problem is to determine where the projectile is going and where it came from. A first model might be Newton's law of motion in two directions. If only gravitational

2 force acting on the projectile is used, then the problem can be solved exactly by basic algebra methods, provided one has exact observation data and exact computing tools. (x(t ),y(t )) 1 1 (x(t ),y(t )) 2 2 Trajectory of a Projectile The simplest model of motion is in 2D with the only force being gravitation in the vertical direction. Newton s law of motion applied to both the horizontal and vertical directions give a pair of differential equations mx = 0 and my = -32. In this case the path can be described in terms of the initial position (x(0),y(0)) = (x 0,y 0 ) and initial velocity (u(0),v(0) = (u 0,v 0 ) as x(t) = u 0 t + x 0 and y(t) = v 0 t + y 0-16t 2. These four parameters are the unknowns, and they will require four equations, which can be gotten from two observations of the position. Let (x 1,y 1 ) and (x 2,y 2 ) be the observed positions at times t 1 and t 2 as illustrated in the above figure. Then u 0 t 1 + x 0 = x 1, 2 v 0 t 1 + y 0 16 t 1 = y 1, u 0 t 2 + x 0 = x 2 and 2 v 0 t 2 + y 0 16 t 2 = y 2. The method of solution is from basic algebra. The first and third equations decouple, and by subtracting the first equation from the third equation we have the smaller system for (x 0,u 0 ) u 0 t 1 + x 0 = x 1, u 0 (t 2 - t 1 ) + 0 = x 2 - x 1.

3 The implementation stage is a simple calculation to solve this system. Here it is possible to get different answers depending upon the computing tool being used. An assessment of the problem indicates some difficulties. If the mass is moving fast, the observations will have to be close together which means the t 1 and t 2 will be close. If t 2 - t 1 is small, then significant round-off errors can occur in the calculation of u 0. If the projectile is moving very fast and the observations are close together, then there may also be significant observation errors, that is, we may have uncertain data for x 1 and x 2. The first model is very primitive because it neglects air resistance, earth's rotation, motion in three directions and other possible complications. Therefore, one is inclined to repeat and enhance the above stages so as to be able to make a more accurate prediction. Example Two. Consider the price data given in the following table for the past three months. Month Price of HD TV The applied problem is to predict the price in the future months. The prices decrease by 50 and then by 40. This suggests a model for the price data might be a linear function of time, y = mt + c, where the slope m and the intercept c are to be chosen so that the line is close to the data. price Price Data and Linear Model time

4 For the method of finding m and c, one could eye ball the data and various lines as is illustrated in the above figure where the center line seems to be closer to the data. A more price method is to choose m and c so that the least square function is a minimum F(m,c) = (m1 + c 2000) 2 + (m2 + c 1950) 2 + (m1 + c 1910) 2. The minimum of this function can be found by either using calculus or matrix algebra. If there is a great deal of price data, the computation will be tedious. Here the implementations of the methods are best done using a computational tool. A computational tool could refer the either the software (program for algorithm or method) or the hardware (calculator or PC). Once the m and c are found, predictions about future prices can be computed by evaluating the linear function at the desired time. Comparing the predicted prices with any additional price data can do the assessment. There may be inaccurate data, and perhaps the linear model should be replaced by a quadratic function of time! Example Three. Consider the amount of money is an account. The applied problem is to predict the amount in the account based on interest rates, initial amount and monthly deposits and withdrawals. The model is for the amount at each month, A k, where k is month from the initial deposit, A 0. The change in the amount is A k+1 A k = (r/12) A k + D where D is a deposit and r is the interest rate compounded monthly. Thus, the model is an iterative form y k+1 = a y k + b where a = 1 + r/12 and b = D. One method of solution is an iterative calculation: y 0+1 = a y 0 + b y 1+1 = a y 1 + b. y K+1 = a y K + b. Another approach is to use the closed form of the solution, which is generated by use the partial sums from the geometric series, y K+1 = a K+1 y 0 + b (a K+1 1)/(a 1).

5 The implementation of either method of solution requires some computation tool. The iterative approach will require a programming loop, and the closed form solution requires the evaluation of an exponential function. The assessment involves changing interest rates, ability to make deposits and unpredictable withdrawals. That is, the values of a and b may depend on k, and there may be some uncertainty associated with these parameters. In these cases the closed form solution is no longer valid, and here one must rely on the iterative approach. The first example was derived from differential equations whose solution is a continuous function of time. The third example is discrete as the amounts change at the end of every month. A generalization of the third example has the vector form y k+1 = A y k + b where A is an nxn matrix and y k, a and b are nx1 vectors. An example can be derived from the discrete version of the heat equation ρcu t (Ku x ) x = f where u = u(x,t) is the temperature and y k equal to the vector temperatures. The i th component will be an approximation of the temperature at time step k and space position i. In more complicated models n can easily be several million! Each iteration of the applied mathematics process requires some mental and computational effort as well as money. Another way of depicting the applied mathematics process is given in the following algorithmic format: Get $ Do While $ > 0 Applied Problem Model Method Implementation Assessment $ = $ - $ End Do

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