 So, we will resume our discussion of the various methods of solving problems. The first important method of solving problems is to make a good representation of the problem. So there are many different ways you can represent a problem. So appropriate quotations here, well begun is half done. So if you make a good representation of a problem, you have come 50% close. I mean, you have solved 50% of the problem. Because of the uncertainties involved in ill-defined problems, the problem solver must often devote a great deal of time to both the step of forming and initial representation of the problem and the step of reformulating it. So in fact, reformulation and initial representation, all these can be regarded as some sort of representations of the problem. So use of diagrams or pictures is very much encouraged. So a picture is worth a thousand words, that is another quotation. So what are the various techniques of representation? So you use symbols to represent a problem. Very often we write equations using symbols. This is a very simple and common example of using symbolic representation. Another strategy is reformulation. So there is a statement of the problem. You can modify the statement without changing its meaning. And this modification of the statement helps you to get a solution easily. So there is one statement of the problem, where you try to get a solution from that statement. It may be somewhat longer or difficult, but you change the statement without changing the meaning. This is important. And then the solution comes very quickly. That is the meaning of reformulation. You can use of tables. There are different types of tables such as a list form of table or a matrix form of table. Then you can draw graphs to illustrate the problem. You can use Venn diagrams. And very important, you must draw diagrams to scale. This we would have learnt even while solving our puzzle of dividing a square into five equal parts. If you don't draw a diagram to scale, you have a feeling that your approach will lead to a solution. But if you draw a diagram to scale, you show the various lengths properly to scale, then you can quickly see that it will not lead to a solution. So let us start with the first representation technique. That is symbolic representation. How it is important in solving problems. Now here is a long statement. This particular statement is actually a literal translation of a Sanskrit verse. This verse is by a mathematician named Madhava. He is an Indian mathematician, a few hundred year old. Let us read this statement. Then we will appreciate why if you use symbols to represent situations, things can be much easier. So what does this statement say? It says it is telling you how a solution to a particular problem can be obtained. It is stating the problem as well as the solution. So to give you a hint, he is trying to give a series expansion for a trigonometric term. Statement is the first term is a product of the given sign and the radius of the desired arc divided by the cosine of the arc. The succeeding terms are obtained by a process of iteration when the first term is repeatedly multiplied by the square of the sign and divided by the square of the cosine. All the terms are then divided by the odd numbers 1, 3, 5 and so on. The arc is obtained by adding and subtracting respectively the terms of odd rank and those of even rank. It is laid down that the sign of the arc or that of its complement whichever is the smaller should be taken here as the given sign. Otherwise, the terms obtained by the above iteration will not tend to the vanishing magnitude. Now this is a long statement. Now if you want to figure out what this is trying to say, it takes you a long time. You can try it. Actually this statement is nothing but if you write it symbolically it is an equation. The equation is as follows. This is what the entire statement is meaning. So it means r into theta is equal to r into tan theta and then you have a number of terms in brackets 1 minus tan square theta by 3 plus tan power 4 theta by 5 and so on. If you substitute theta equal to tan inverse x or tan theta equal to x, it gives you a series for tan inverse x. So tan inverse x is equal to x minus x cube by 3 plus x power 5 by 5 and so on. So when you write a series like this, it appears to be a very simple representation. Now what you have done is you have used symbols. Because we are using symbols to write equations so commonly, we do not appreciate the extent of simplification that has happened in problem solving. But use of symbols is of recent origin. As you can see, earlier days the entire thing was stated in words. Now if you carefully observe in many situations when we encounter new problems, we do not use symbols to represent the problem and as a result we take longer time to solve. So we should remember the strategy of solving problems that is symbolic representation. We try to use symbols to represent various elements of the problem and then show the interrelation between them. Now let us take another problem. How can you become productive? That is a statement. Let me explain to you the context. One of the Japanese companies, it found that its productivity was going down and it was losing out to other companies in competition. So the management thought that it must revise its entire process of producing the goods and they thought a good way of coming up with a revision or a new process would be to ask the workers who participate in production to give suggestions. Let us invite suggestions from the workers. So they called a meeting of all the workers and they presented them this particular question. How can you become more productive? And they said that all of you think over it and please write your suggestions on a piece of paper and drop these in a box kept here. We will look at all the suggestions and if a suggestion is good we will implement it and if it improves productivity then the person who has made the suggestion will get an award. So they thought this award can be a motivation for people to suggest new ways of doing production. Surprisingly they found that there were very few suggestions and they were surprised why are the people not coming up with suggestions. So one of the persons in the management did a silent investigation to find out what is the cause of this. And then he found that the workers were not suggesting because they thought it is a management trick to extract more work out of the workers where you involve them in making the suggestion so that later on you can always say this is your own suggestion which we are implementing and then they will make them work more because how can you become more productive unless you work more? It is not possible unless you work more you cannot become more productive. So this is not good. Let us not get involved in this trick. Let us not be exploited. This was the reason why they did not come up with suggestions. So what I am describing is a practical situation, something that has happened in practice it has been reported. It is a case study. Now once they realize this is the reason why people are not coming up with suggestions then someone thought of an idea. He said let me rephrase this statement. The statement how can you become more productive does not seem to be very convenient for the workers for them to respond. So they gave some time so after six months they called another meeting and this time they made a new statement. They said please give us suggestions on how can you make your job easier. Now actually if you see the implications of the two statements are the same because if you want to be more productive then if you will be more productive if your job is made easier then you can do more jobs in a given time. If every job is made easier which means you will be able to do the job in lesser time then in a given time you will be able to do more jobs. So after all the company is not going to reduce the working hours. It is going to tell you that you have to work from this time to that time. But this statement of the problem appears much more friendly towards the workers. So it appears as though you are trying to make the job of the worker easy and then it invited suggestions. The management invited suggestions from the workers and it practically happened that they got hundreds of suggestions on how the job can be made easy and they actually implemented some of the suggestions and their productivity went up. So it has been reported in a place how companies have improved their productivity. So here you can see the formulation of the statement is very crucial. In fact in research it has been found that in many cases it is the formulation of your research problem that is very critical and therefore as you go on doing research you start with a particular statement and then you go on reformulating it until you find that it is in a form in which you can solve it. So this reformulation is a very important strategy of solving problems. So whenever we formulate any statement of our research we should carefully see whether there are different ways of formulating it so that you can get a solution of the problem easily. In mathematics the transform techniques that we employ are nothing but sophisticated examples of this strategy of reformulation. So all of us know that when you use a binary number system then you can implement various mathematical operations easily in practice. All our computers work based on the binary system, all operations become easy. Why? Because in binary system an operation such as addition or subtraction becomes a logical operation. The operation of addition or subtraction in a decimal system becomes a logical operation in binary system and of course you know that multiplication is repetitive addition and so every mathematical operation can be reduced to a logical operation by using a binary number system. So this is how a binary number system has made various mathematical operations easily implementable in practice. Logarithm is another example. You know that now of course we have computers, calculators but in our days I am talking about 1970s we used to carry log tables to the exam because how do you calculate various quantities multiplication and division? You take the logarithm and then you add the logs and then you take the anti-log. So logarithm converts multiplication into addition, division into subtraction. So that is how it is simplifying the operation. It is again an example of transformation. So you transform, some information is available, you put the same information in a different form. It is easier to process. Let's take another example of a problem. I mean this is the name given Tirupati temple problem. What it means is it is about a person climbing up and going to the temple and coming down. So statement of the problem is like this. Exactly at sunrise one morning you set out to climb the Tirupati temple. The path wound around the mountain, you climb the path at varying rates of speed. You stop many times along the way to rest and to eat the fruit you carried with you. You reach the temple just before sunset. So you started in the morning and you reach the top. So after fasting and meditating for several days, you began your journey down along the same winding path, starting at sunrise and walking as before at variable speeds. Your average speed down the hill was more than your average climbing speed. Obviously when you come down, you will come down faster. There must be a spot along the path that you will pass on both trips at exactly the same time of the day. Finally that is the problem. So you are climbing up and then you are rested there at the top and then you are coming down. So what you have to prove is there is some spot along your path that you will pass both during onward and return trips at the same time of the day. So there is some particular spot which you will pass at the same time of the day both in onward and return trips. The statement of the problem is that you went up, you waited or you rested there for some time and then you came down. But the fact that you waited there and all it does not really matter. So what you have done is you have reformulated the problem and you have said one person is going up and other person is coming down the way you would have come down from the top. Now it is very easy to see that the two people will meet at some point. One person is going up, the other person is coming down and that is the point where which you will pass at the same time of the day. So simple restatement of the problem has given you a solution. Instead of one person going up and then climbing down you say two people are there, one person going up, other person is coming down. You can also use a graphical approach to solve the problem. Now we will see the graphical approach along with some other examples which also can be shown to be solvable by graphical approach. Now let's take another problem. Derive the trend in the behavior of a plating adhesion on silicon substrate from the measured data as a function of substrate area and doping level. Adhesion is measured for 0.51 and 2 centimeter square area and P plus Pn and n plus doping levels. Each measurement is repeated twice. Now some technical terms have been used because I have taken the problem as it is but the technical terms are not very important. So let me phrase the problem for those who may not understand it, rephrase. So I will draw a diagram. So the situation is that you have substrate and this is your plating, metal coating. Plating is nothing but a metal coating. So don't bother the fact that it is a silicon substrate that is not important. It is a substrate that is what is important and on this you are plating. You have developed a procedure of plating the substrate and you are trying to study this procedure. So what is the adhesion between this metal that you have plated and substrate? So when you measure the adhesion then you will know how good is your process. If your adhesion is good as compared to another process, better. Then you say your process is good that is what this problem is about. So you are measuring the adhesion between the metal and the substrate for two variables. So one variable is area of the substrate. So I will draw the top view on the top here. So this is the top view. So this is the area of the substrate. So if I show on the top view, I am going to take substrates of different size that is 0.5 by 1 centimeter, then 1 centimeter by 1 centimeter, this is different size again and then 1 centimeter by 2 centimeters and I am going to measure adhesion in each case. Then I am going to change this condition of the substrate. There are four different conditions. Basically P plus means a substrate which has a high number of positive charge carriers. P is a substrate which has positive charge carriers but not that high concentration. So you do not bother about these descriptions. I am just giving this so that you become comfortable with the problem. So basically the condition of the substrate is being changed. There are four different conditions. P plus P, N, N means that the substrate has more number of negative charge carriers and N plus means it has very high number of negative charge carriers. So the result of this experiment is that you are going to measure adhesion. This is the variable being measured for three different areas and for each area you will have four different substrate conditions. In other words, how many readings you would have taken? So you would have taken 12 readings. Now you know that any experiment should be repeated at least twice. We will discuss more about how many readings you must take in an experiment so that your errors are minimized. That will come in the experimental skill. Right now let us assume that we have at least repeated each experiment twice. So we have total of 12 into 2, 24 readings. Now from these 24 readings we want to extract a pattern. What happens to the adhesion? Does it change with area? If it changes, is there a regular pattern, increase or decrease? Does it change with substrate conditions? Is there any regularity in this behavior? That is what we want to extract. So what will you do? One impulse would be to draw a table and put all these readings in a table. Now here I must point out though this looks like the thing to do, that is at least put the data in a table form, I have seen many scholars do not even draw a table when they come with the readings. So a typical method that a scholar will use to record the readings would be he will write area equal to 0.5 centimeter square. Let me write and show you because I am quoting all practical instances. So the scholar comes like this, area equal to 0.5 centimeter square substrate P plus adhesion 10 into 10 power 6 Newton per meter square in the first experiment and second experiment it is 10.2 into 10 power 6 Newton per meter square, not centimeter square meter square. Then 2, area equal to 1 centimeter square substrate P plus adhesion. So when the scholar comes to you, he comes with 4 or 5 pages like this one by one values whereas an impulse, first impulse should be to at least draw a table. So whenever you have a large amount of data you have to organize it, this is the first important thing if you want to derive a trend you have to organize it. Now it is interesting to note there are several ways of drawing the tables, arranging the tables. This is perhaps the simplest and most common way, this particular method of arranging the data in a table is called a list form, LISD list. What you are doing is you are doing a straight forward thing, the first column is doping the substrate condition. So you have 4 different substrate conditions. Then for each condition you have 3 different areas 0.5, 1, 2. For each area you are doing 2 experiments, these 2 experiments are shown as 2 columns experiment 1 and 2 and then you are going to enter all your readings of adhesion. This you will repeat for other substrate conditions. Now if you want to derive the behavior of the adhesion as a function of let us say the substrate condition. So what you should do when you change the substrate condition your area should remain constant right normally whenever you want to derive trends how does one variable influence the other variable then except for these 2 variables all other variables should be maintained constant that is the way it is. So if you want to do that here what are you going to do you are going to take let us say 0.5 centimeter square area and then you will see the 2 readings perhaps you will average the 2 to generate 1 reading corresponding to 1 area. Then you will go to next surface condition or substrate condition again look at the 0.5 centimeter square area reading here and then average the 2 then you skip the next 2 now you go to the third substrate condition and again you choose the 0.5. Now when you want to derive the trend by looking at the numbers what is happening is that you have to skip a few numbers so you should take the first number then you should skip the 2 numbers then take the fourth one then skip 2 numbers then take the next one and then look at all these numbers to derive a trend then you should repeat it for the next area so you see this is possible but the other numbers are interfering right and you to derive a trend to see whether addition is decreasing or increasing it is a little bit confusing so what a person will do is he will prepare another table where only for 0.5 centimeter square area what is the variation then 1 centimeter square area what is the variation okay then if you want to see how the things change as a function of area now you should keep the substrate condition constant and see how things vary with area that is done easily here because all that I have to do is let us say I keep this substrate condition constant I see just these three consecutive numbers and then I find that as the area is increasing the addition is decreasing so this is done easily here so this particular approach of listing the readings in this case it is enabling you to derive the variation of addition with area easily but if you want to derive the variation of the addition with substrate condition then it is a little bit involved now you look at another method of representing the same information this is called matrix representation of the data this is also a table here what is done is that one variable is shown in the vertical direction that is substrate condition the other variable is shown in the horizontal direction okay and the two experiments corresponding to each combination of substrate condition and area are shown in adjacent columns the advantage of this approach is I can see both variations of addition as a function of area and variation of addition as a function of substrate condition very easily if I want the variation as a function of substrate condition I move in the vertical direction area is constant so I can take for 0.5 centimeter square area or 1 centimeter square area or 2 centimeters square area if I want variation as a function of area I keep substrate condition constant that means I move in horizontal direction okay so this is matrix form of representation of the data people rarely use matrix form of representation okay but one must be encouraged to use matrix form of representation if it is possible in a given situation because it enables you to derive trends much more easily than the list form okay there is another method of representing the same information this is a graphical form let us see what are the advantages of this representation so here a graph has been plotted as follows so the four substrate conditions are shown in the horizontal axis n plus n P P plus the three different areas readings for three different areas are shown adjacent to each other okay corresponding to any particular substrate condition the vertical axis shows the adhesion now the two different readings obtained in two experiments for the same condition are shown in the vertical along the vertical axis and they are joined together by a line so whenever you join these two readings together by line it shows that the substrate condition and area is the same for both of these and the two readings are obtained by repeating the experiment and then now you have represented the complete data that you have taken in this particular case the condition for 2 centimeter square is not shown because no reading could be obtained so it is not shown here now what is the advantage of this form you can see the trends are much more easy to derive in this graphical representation than in even matrix form of representation of the data for example if I say the take the P plus condition I clearly see that as area increases the adhesion decreases this is true for P also so for P or P plus this trend is clearly seen it is also seen that for n plus or n you cannot derive any particular trend as a function of area okay you also find that while for P and P plus conditions repetitions of the experiment give more or less close readings for n and n plus the dispersion between two readings is rather large particularly when the area is high area is more so in such cases you cannot really believe these readings right if the dispersion is too much so you do not know whether the experiment was done not done correctly or it is the characteristic of that subset condition that things cannot be well controlled the point however is that all these different aspects of the results are very nicely seen in a graphical representation than in a table representation although they are present in the table representation if you want to derive this information you know you have to do mathematical operations and subtract this number from that number and write it and so on whereas all these various operations are done intuitively when you draw on a graph that is why people should be encouraged to draw graphs whenever they want to represent their results okay you must try to see whether you can draw a graph of the given results now in this context let us take up the problem that we were considering the so-called Tirupati temple problem a graphical solution for that so this is the time and the horizontal direction and this is the height now you are starting in the morning this is the bottom of the mountain and then you are going up and you have reached the top now when you are climbing down you are starting from the same point at t equal to 0 that is how you are drawing it and you are coming down now the point is it is not necessary that you will reach the bottom of the mountain at the same time because you are going to walk down faster so this is not a correct representation in fact correct representing may be like this because you take less time to come down so while going up this is the way it is while coming down now it is easy to see that there is one point of intersection so you can find out at what time of the day and at what height which spot you will pass at the same time on both your onward and return so this is a graphical solution to the problem so a problem can have you can get a solution by multiple approaches since this graphical approach is rather powerful we will consider one more problem and show how you can get a graphical solution not one more in fact we will see two more problems let us say you want to solve the equation x equal to e power minus x I can use a tabular form of representation of the solution process so what do I do I take the left hand side and right hand side left hand side is x right hand side is e power minus x and then I go on varying x and I show what is the left hand side what is the right hand side value x is equal to 0 left hand side is 0 right hand side is e power minus 0 that is 1 can I take x equal to 0.4 left hand side is 0.4 right hand side is e power minus 0.4 0.67 I go on writing the numbers like this now what I find is when I increase x in the beginning left hand side is less than right hand side the difference goes on decreasing and then between 0.5 and 0.6 I find the left hand side has become more than right hand side left hand side is 0.6 right hand side is 0.55 for 0.5 left hand side is 0.5 right hand side is 0.61 so I know that the solution lies somewhere between 0.5 and 0.6 so now I focus on this interval again I prepare a table 0.5 0.55 or 0.5355 and so on then I will find some two numbers between which the change occurs then again I will focus on that interval this is one way of solving this equation it is an iterative process but it is not a very efficient iterative process fine now supposing we want to show the this iterative process on a graph why because our aim is to get the solution and we want to see how we can get the solution quickly now graphical solution for this would be to draw line y equal to x and y equal to e power minus x and find the point of intersection on the same graph if you want to represent the method that we just used preparing a table of values it amounts to doing the following so you take one value of x you see what is the value on y equal to x what is the value on y equal to minus x you find that the difference is large so I move to the right 0.5 it is small then I move to 0.6 now y equal to x has become more than y equal to minus x so I know solution lies between these two then I repeat the same operation this is a graphical method of representing the operation that I have just done but we know that the iterative process we normally employ is not this the iterative process we employ is we substitute a value of x on the left hand side get a new value of x this new value of x we substitute back and get a further new value of x and we go on doing like this until two constitutive values of x are as close as possible as close as we want now how do you know that this approach gives you a solution why should it converge to a solution if you want to explain this again you have to use a graph when you show this mathematical operation on a graph as many of us would have already known this operation is shown by this particular spiral shown by dotted line this operation of starting with a value of x I guess generating a new value then putting this new value back then generating another value is equivalent to moving along a spiral on this graph so I start with x is equal to 0 so I put x equal to 0 here xn is x equal to 0 xn is equal to 0 so e power minus 0 is 1 so xn plus 1 is 1 how do I get that one graphically so I put the value x equal to 0 for that e power minus 0 is 1 that one I make it as the x new x so I move horizontally on the y equal to x line I have located this new value of x now so for this value of x I try to see what is the new value of e power minus x that means I am moving down along this dotted line and this operation is being repeated so now I move again horizontally in this direction to get the new value of x then I move vertically so like this by using a combination of horizontal and vertical segments I find that I am moving along a spiral which ultimately leads you to the solution you can see that the spiral is ultimately converging and it will give you a solution so if you want to explain how the iterative process works then it is a graphical approach that helps you to explain very easily is another example where drawing a graph for the given situation helps you to explain the solution process in fact this graphical approach is the one that was used to develop the iteration iterative process of getting the solution let us take one more problem to illustrate how we can have a graphical solution let me describe the situation but the situation is that you have two classrooms in which teaching is going on let us assume that the same topic is being taught in both the classrooms and you have the same number of students approximately and the level of students is also same now you are asked to observe from outside the classroom and then at the end of the class give a comparison as to in which class the teaching was going on better okay suppose this is what you are told so you to observe from outside two classrooms where the same thing is being taught and you are asked to find out in which classroom the teaching is going on better now what would you do so what you would do is you look at a number of activities which are going on in the class so can you give me some indices which you will use to determine which class the teaching is better class participation so when you say class participation what form of participation so one form of participation is student asking questions okay that is one thing you will observe yes pardon facial expressions okay whether people are excited or not use of very good point teaching eggs so is the teacher using slides or any other form of audio visual presentation and so on okay that is another point because it is seen that if you use these eggs then the teaching is better then what else whether blackboard teacher is writing on the blackboard using the blackboard or not so yes ha number of questions being asked also you can also see whether if the teacher ask a question students are answering them okay effective utilization of the backboard so now you list all these activities now the point is you should also know for every activity what is the intensity as he said number of questions also is important so both classes questions may be going on I mean people may be asking questions but question point is in which class the number of questions is better okay now the point is that all these activities are going on and against each of the activity you can have a measure okay and then you will have to take an overall view right Samara is all this and then say which is better now it is interesting to see how you could represent this information on a graph so this is the graph of the teaching learning process going on in a class in this graph the y axis is the activities the x axis is time so as you see here the y axis is all the activities which are going on in the class and the x axis is time if you see carefully the activities have been arranged in a particular order this is very important the activities which are common in the class such as teacher talking and teacher using the chalkboard are shown very close to the teaching baseline timeline time axis the activities which are relatively rare but which show more involvement in the teaching learning process are shown further and further away from the axis for example it is more common to have a teacher asking a question than a student asking question so student asking question is shown further away from this line student action if the student is going to the board and trying to explain something that is even rare so that is shown even further away so this is how the carefully the activities are arranged similarly use of audio visuals and so on is shown on the negative side but again what is rare is shown further away for example use of multimedia it is rare but it is supposed to be very effective in teaching learning process so that is shown much further away from the timeline now you plot a graph of what is happening in the class the graph would look something like this so here teacher is talking for some of time then teacher has thrown a question to the class some student is responding then the teacher has gone to the board and started using the board at which point some student has thrown a question so the graph moves up here to this point the teacher is trying to answer the question teacher response then teacher talking teacher using some projection and so on right so this is how you plot the activities on a graph you compare the graphs for the two classrooms by looking at the graphs one can easily determine in which classroom the teaching and learning is going on better whichever classroom the graph has a high amplitude and more frequency okay because the high amplitude indicates more student involvement and greater use of the teaching aids and more frequency shows all the various activities going on if the number of activities is less then you will not have things changing okay not have more frequency so like this by measuring the frequency and the amplitude you can easily compare the two classrooms so this is an example where you can use a graphical representation of the information to derive the solution for the problem so like this one can give any number of examples to show how a graphical approach helps you to solve problems and you may get up in the you may get innovative solutions new solutions by representing the information in a graphical form so one of the assignments that I suggest to students is that in your own area of research you take up a book and from beginning to end just see what are the various graphs which have been drawn there in the book okay and not only that I even recommend that whenever you enter a library if you find a new arrival a book you please open the book and only look at the graphs in the book doesn't matter whether it is in your area or not it is interesting how in different subjects people use different types of graphs to represent information okay who knows there is a particular method of representing information that someone else has used which if you employ in your situation right you will get a solution so this assignment given to you right you can look at the books and look at only the graphs in the book and see how the graphs have been drawn for what situations the graph have been drawn how many different ways graphs have been drawn for example those who are from electrical engineering might have heard of what is called the Smith chart which is used for analysis of microwave circuits the Smith chart is nothing but a graphical representation of the information in the situation you see this graph is so useful for solution that it has been named after the person who suggested the graph Smith chart Smith is the name of the person who suggested the graph so like this graphical representation is a very very important method of solving problems and coming up with innovative solutions