 first class, we have discussed that, what is the quadratic form of a function. That quadratic form of the function may be positive definite, positive definite function, positive semidefinite function, negative definite function and negative semi-definite functions. Then, we have discussed that what we mean by that positive definite matrix, positive semi-definite matrix, negative definite matrix and negative semi-definite matrix. Then how to test a matrix is a positive definite matrix or negative definite matrix or positive semi-definite matrix or negative semi-definite matrix or function, quantity function is positive definite function or not that can be tested through a Sylvester criteria. And Sylvester criteria requirement is there, the matrix P must be symmetric. So, we have shown it, if a function is expressed in quadratic form, again if a function is in quadratic form that can be expressed x transpose P x. And the P always we can select that from the quadratic function is symmetric that we have shown it. And we have if you recollect that we have considered a function of this, which is function of x 1, x 2 and x 3. Our job is to whether this what is the nature of this quadratic function, whether it is a positive definite function means for all values of x x 1, x 2, x 3 the function value is positive definite or not nature of the function. It may be positive definite, may be negative definite, may be positive semi-definite, may be negative semi-definite. So, we have to check, we have to test that whether what is the nature of this function. With this one we have form x transpose P form and where from this of quadratic function we have written P is symmetric matrix. And we have used after that we have used the what is called Sylvester inuclidic Sylvester criteria for finding out the positive definite matrix. So, all leading principal minus of order 1, order 2, order 3 and so on should be greater than 0. That is now next is suppose a matrix is whether a matrix is negative definite matrix or not how to test. The same idea what we have considered for testing for positive definite matrix, same idea can be extended here. Let us take test for negative definite matrix and when we are doing this test using the Sylvester criteria we assume that matrix is symmetric matrix. That means q is equal, let us call that q matrix whether it is a negative definite matrix how to test it. So, and let us call this dimension is n cross n. If you recollect the positive definite matrix positive definite matrix test that P is our earlier discussion P is greater than 0. This indicates that P is a positive definite matrix. In other words you can say if you form a quadratic function with P and x, x is any vector of dimension compatible dimension with P. If you P multiplied by x transpose, post multiply by x and it will form a quadratic function and this value if P is a positive definite, this value which is a scalar quantity is always greater than 0 when x not equal to null vector of dimension n cross 1. That is the by definition of positive definite matrix. Now, if I multiply by since this is a scalar quantity scalar, if I multiply by both sides by minus let us call it as 5 multiplied by minus 1. So, I can write once you multiply by this I can write P of x is less than 0. So, this is the important step see what we can do then minus sign I can just push inside with P that is also less than 0. So, this quantity is less than 0 if P minus P is a negative definite matrix by the definition of quadratic function. So, minus P I am considering as a let us call new matrix Q whose dimension is n cross n because dimension of P and q must be same difference between P and q is minus P is equal to q. So, now I have to check whether this will be less than 0 provided q is negative definite matrix just now we have seen P is a positive if multiplied by this one minus P this matrix will be a negative definite matrix, but how to test it that whether it is a negative definite matrix or not. So, if you proceed in the same manner for positive definite matrix we will see what the changes are there. So, for our case Q we have considered here minus P will dimension n cross 1 this constant. So, P I multiplied by minus that means all the elements of P is multiplied by minus 1. So, what we consider P is a positive definite all the diagonal elements are positive now since it is multiplied by minus 1 P. So, all the diagonal elements will be negative and non-zero number. So, our test for this one all the diagonal elements of Q or Q is what this are negative and non-zero numbers. So, this less than 0 means Q must be a negative definite and Q is nothing but a P. P is a positive definite we have started from this one multiplied by minus 1. So, all the diagonal elements previously was plus since it is multiplied by minus 1. So, all the diagonal elements will be minus that means all the diagonal elements of Q agree are negative and non-zero numbers. Next is you see this one the test for positive definite by using the Sylvester matrix is leading principle minus of order 2 order 1 we have seen the diagonal element in case of negative must be negative order 1. So, order 2 what will be there before that I just tell you these things you just if the determinant of note this results I am using determinant of minus a is equal to minus 1 whole to the power of n where n is the dimension of this matrix a I multiply by minus 1 and finding out the determinant of minus 1 is equal to determinant of a. So, when n is equal to odd when n is equal to odd the determinant of the odd order dynamic odd order matrix determinant of odd order matrix with negative sign is equal to minus determinant of a. So, I can write it this equal to minus determinant of a when n is odd this equal to determinant of a determinant minus a is equal to determinant a when n is even. So, this results I will apply when I will check the test when will test the negative definiteness of this matrix Q. So, you see Q is nothing but a minus p that all the elements of p are now multiplied by minus. So, now you when will find out the what is called leading principal minors there are two steps now the all leading principal minors of principal minors with even order are positive all leading please remember all leading principal minors with even order means I know how to find out the leading principal minors. So, even order means the dimension of the matrix which will extract from the original matrix Q of either 2 by 2 or 4 by 4 or 6 by 6 or even dimension order of this matrix we have to generate from the Q matrix. So, the even order the principal leading principal minors determinant with even order are positive not only positive it is non zero numbers and non zero it cannot be zero the determinant non zero number again all leading principal minors with odd order are negative and non zero numbers. So, determinant of odd numbers will be negative or it cannot be zero number non negative numbers agree. So, now this is the test of this one. So, alternative first order will be negative second order leading principal minors second order will be a positive third order leading principal third order will be a negative fourth order leading principal minors fourth order will be a positive and so on until unless you reach to the original system order that means n cross n. So, let us call the earlier example if I multiply by earlier example we have shown it that quadratic form is a what is whatever the given is quadratic form is a positive, definite quadratic form that function is a positive definite function agree. So, I multiply it by both I multiply it by this function by minus 1. So, naturally this function will be a negative definite function. So, if you check this negative definite means of this one check the nature of the quadratic function and if you see the earlier example I just multiplied by that example by minus 1 and I earlier example we have seen that is the positive definite function that means for any value of x this quadratic function value is greater than zero when x is not equal to a null vector. So, that quantity I now multiplied by minus 1. So, that indicates that function value will be less than 0 forward. Now, how to test it that function is a quadratic the function which is given is a negative definite function by the what are the steps we have mentioned you can check it with this one. So, I have just written this one x 1 7 x 1 square minus 4 x 1 x 2 minus 10 x 2 sorry x 1 x 3 then minus 5 x 2 square minus 8 x 2 x 3 minus 9 x 3 square for this example we check it that that by using the silvestre matrix silvestre criteria check what is the nature of this quadratic function whether it is it is it should be a negative definite function because previously you was a positive definite function since I multiplied by minus 1 it should be negative definite you check this thing whatever I have discussed this one. So, next is your what is called test for semi definite matrix using silvestre criteria test for positive semi definite matrix and that test you can do. So, semi definite matrix P which is equal to n cross 1 and that matrix is symmetric matrix see matrix. I repeat once again that you can test the silvestre criteria whether the matrix is positive definite semi definite positive semi definite negative definite or negative semi definite you can test it if the matrix is positive definite. If it is not there I can easily convert into a positive what is called symmetric matrix because by definition of this one I can write that P is a if it is a semi definite the symbol is this one if it indicates P is greater than 0 means P is positive semi definite means. In other words you form a quadratic function with P and any vector x and it is value this scalar value is always greater than equal to 0 it may be some value of x may be 0 and other than x is equal to null vector and other values of x it is greater than 0 for x not equal to null vector. So, this test how you do it the silvestre criteria tells that let A is P is n cross n is a symmetric matrix. Then what you have to do you check first the matrix is given all the diagonal elements it is a positive semi definite must be positive or some may be 0 also, but it cannot be negative agree. So, it should be all the diagonal elements in other words should be a non negative numbers. So, all diagonal elements of matrix P elements of matrix P must be non negative that once the matrix is given immediately I can say whether it is non negative if it is a non negative then we can proceed further many non negative means it is may be positive some element may be 0. Next is all the now I am not using the leading principle only principle minors all the principle minors means determinant are non negative agree this non negative that means all principle minors will be the determinant will be a positive and some may be 0, but it cannot be negative then we will call this is a positive semi definite matrix. So, I must know what is principle minors how to find out the principle minors of a matrix Q matrix P which is a symmetric matrix. So, let us take let us explain this way with an example that how we can find out the principle minors of a matrix P which is symmetric matrix find the find all principle minors of the matrix. Let us call Q or some let us call P is now how to find out P let us call that matrix is 1, 2, 3, 4, 5, 6, there is 2 6, 7, 8, 9, but when you find out the principle minors is not necessary that how to find out the principle minors of this one that not necessary P should be a symmetric matrix agree, but when you check it that whether the matrix is positive semi definite matrix cannot even it is not a symmetric matrix I can always convert into a symmetric matrix that I told by definition of this one P plus P transpose by 2 if you multiply by X transpose of X the value of will be remain same. So, this let us call it is not a symmetric matrix, but my interest is to find out the all principle minors of this matrix is not a symmetric matrix let us call this I am writing 4 agree. So, our aim is to find out the principle minors of this matrix, so what will do it first find out principle minors of order one. So, principle minors of order one you have to make all the 1 1 element that means a matrix out of this from this matrix you pick up the matrix of size 1 by 1, so I have a 9 elements. So, you have to select only that elements which are the diagonal elements also, so this will come 1 5 and 9 these are the principle minors of order one you see I have a 9 elements 1 by 1 matrix out of this I am considering only 1 5 9, because these are the diagonal elements of the original matrix. Similarly, principle minors of order 2, so you have to pick up a matrix of size 2 by 2 from the matrix of whose size is 3 by 3. So, 2 by 2 you have a different combination you see this is 2 by 2 and this and this also 2 by 2 this this this this 2 by 2, but you have to pick up those 2 by 2 matrix whose diagonal elements are the diagonal elements of the original matrix. In other words you can see that first row first column you delete and what were the matrix is what is the elements are left that is the principle minors what is called principle minors of order 2. So, again second row second column if you mean it is it is a 1 9 3 7 and 1 and 9 is the diagonal elements are same as the original matrix diagonal elements part of this one. So, principle minors of order is determinant of 1 2 4 5 this is also 1 principle minors of order 2 that means third row third column you delete it that what is left is this one. Another principle minors of order 2 is determinant of first row first column if you consider a delete it then it is a 5 8 5 6 8 9 determinant of this min another is if you consider that is what we have considered first row first column if you separate this now second row second column second row second column if you do it this will be a determinant of 1 9 1 9 then 3 7 to 3 7. So, this is the determinant there is no other choices there which size of the matrix is 2 by 2 and not only that is diagonal elements are the diagonal elements of this matrix. So, now is left is principle minors of order 3. So, principle minors of order 3 is the matrix itself minors of order 3 is the determinant of first row first column. So, I know how to find out the leading sorry not leading principle minors of a matrix Q. So, now let us take an one example and check how to find out the what is called that a quadratic function example. So, consider the quadratic function and check function f of x is the 1 and x 2 I can write in short is a function of x is a vector whose dimension is 2 cross 1 1 element 1 variable is x 1 another variable is x 2. So, this is 4 x 1 square plus 4 x 1 x 2 plus x 2 square. So, your problem is check the positive semi definite function check the definiteness of this function or check this this quadratic function is positive semi definite. So, that you can write show that the quadratic function f of x is positive semi definite that is our problem. So, this function if you see this function I can always write into matrix and vector form this matrix is positive semi definite means is nothing but a x x 2 this is equal to I can write in matrix and vector form as I told you that x 1 and x 2 and this I can represent in different ways, but I will prefer to represent this in symmetric matrix. So, that I can test I can use the silvestre criteria for checking this matrix is positive semi definite or not. So, I will write it 4 1 4 is there 4 x 1 x 2 here x 2 x 1. So, I will divide into 2 parts 2 2. So, it is a symmetric matrix this p is a symmetric matrix. So, this will be positive what is positive semi definite matrix in the sense the function value is always greater than equal to 0 for all values of x when x is not equal to null vector. In other words I can write x transpose p of this will this quantity is greater than 0 when x is not equal to null means this will be greater than 0 the function value provided p is a provided p is positive semi definite matrix. So, we know if you use the silvestre criteria for p p p is a symmetric matrix then you can say our p is like this way what we got it here 4 to sorry this is you see this quantity is minus this quantity is minus. So, this is minus this is minus. So, I mistake here this is minus quantity this is minus because minus 4 I will minus 2 minus 2 here. So, I am writing p is this one then matrix this one is 1 if you see this p. So, I have to check it this matrix is positive semi definite matrix according to the silvestre criteria look at this one positive semi definite matrix I have to prove it that function means p must be positive definite. All the diagonal elements all the diagonal elements will be greater than equal to 0. So, it is a positive it may be 0, but here in this case one element, but it cannot be negative non negative. So, it test is our satisfy first checking that principal minus of order 1 are 4 which is greater than 0 and 1 is greater than 0. So, it is a next I have to go for order 2. So, principal minor of order 2 when it is a we have to check semi definite positive semi definite negative semi definite we have to consider the principal minors when only positive definite matrix or negative definite matrix we have to test it using silvestre criteria we have to consider leading principal minors we know how to find out the leading principal minors. So, this order is determinant of this this size is 2 by 2 determinant of our original matrix p. And if you see the determinant of this p is 4 minus 4 is 0. So, one case is getting positive another is 0. So, the matrix is positive semi definite matrix therefore, the matrix test using the matrix p whose dimension 2 cross 2 is positive semi definite matrix. And hence the function given function is positive semi definite matrix and hence the function f of this is positive semi definite. In other words, so any value of x you put it here the infinite number of x you will see the function value will be either positive or some value of x it may be negative 0, but not negative that is sure. So, this is we know, but if we ask to test for negative what is called semi definite matrix then how will you do it test. That means, if you multiply by p matrix by what is called minus then this matrix p will be a negative definite. And using the same logic that what we have arrived with a form positive definite matrix to a negative definite matrix same logic you can apply it here. Only the things same thing in the sense your that first that elements of all leading all diagonal elements will be either negative or 0. Then the even the principal minus of even order will be positive or some value will be 0. And negative that means what is called the odd order of principal minus will be what is called negative or 0 that is one the same thing what you did it here you can do it this one. So, I am I am leaving this in an exercise to check it this one this function what we have considered this one if you see this example what you have considered you multiply by minus 1. So, that this function will be negative definite function. So, check please do it this one please. So, check the what is called definite miss of check the definite miss of the quadratic functions f of x 1 x 2 is equal to minus 4 x 1 square plus 4 x 1 x 2 minus x 2 square you see this this function I am multiplied by minus 1. So, previously we have proved it is a positive semi definite. So, you have multiplied by minus. So, it will be negative definite. So, it applies Sylvester theorem only the what is called even order values and odd order pencil minus values you check it all these things you can easily prove it this Sylvester inequality criteria. So, I leave this in an exercise for this one. Now, let us that next is our optimality conditions optimality conditions for a function f of x 1 x 2 only two variable function that can be extended from n variable case function f of x of two variables. So, let us consider we have the function this optimality condition can be obtained from the by using the Taylor series expansion of the function f which is the function of x 1 and x 2. Let us call f 1 of the function f which is the function of x 1 and x 2 is we have the point here x 1 star and around this epsilon 1 x 2 star plus epsilon 2. Let us assume that x 1 star and x 2 star are the optimal function of x 1 point. That means when the value of x 1 is x star x 2 is x star x 2 is x 2 star we get the function value function f of x which is the function of two variable case optimum value of the function either minimum or maximum. Then around this we given a perturbations from x 1 star and x 2 star. So, let us see the Taylor series expansion of this function. So, it will be a Taylor series expansion x 1 star comma x 2 star this then this is a function of two variables. So, if you see this one I can write it this one del f of x del x 1 again del f of x del x 1 x is a you can say function of x 1 and x 2 into that you are doing the Taylor series expansion around x 1 star and x 2 star. So, write x 1 is equal to x 1 star x 2 is equal to x 2 star and what is this incremental is epsilon 1 again plus another function is del f just Taylor series expansion now we did it we are doing it now this at x 2 is equal to x 1 is equal to x 1 star x 2 is equal to x 2 star multiplied by the incremental delta epsilon 2 this is the first order approximate then second order terms is what half factorial. Then what is this one second derivative of f x 1 x 2 differentiate this with respect to x 1 square twice with respect to x 1 twice then it is a epsilon put this value you find out this value x 1 is equal to x 1 star and x 2 is equal to x 2 star because around this point you are doing. So, this is will be a epsilon 1 square half bracket then next term is your half is common. So, next term is twice del square f x 1 x 2 and differentiation of with respect to x 1 then with respect to x 2 this is equal to at what point x 1 is equal to x 1 star. And x 2 is equal to x 2 star plus plus. So, I am writing here plus then del square f x 1 x 2 and this is differentiate with respect to x 2 square that you evaluate this value at x 1 is equal to x 1 star and x 2 is equal to x 2 star multiplied by here I missed it epsilon 1 and epsilon 2. So, here you have multiplied by epsilon 2 square plus plus some other terms and let us call some other terms are what is this some other terms what we consider is capital R rest of the series term third order fourth order order of this what is called Taylor series expansion of this one is R. So, R is the where you can write where R is the where R is the is rest of the terms this term is much smaller since this term that epsilon 1 and epsilon 2 the perturbation what we have given around the point x 1 star and x 2 are very small. So, this contribution is small compared to the first term and second term. So, now if you investigate the terms each individual terms then we will see that what we will get it that one. Now, see this one let us call this one this point this slide I am keeping it here just to understand. Let us call f of x which is a scalar variable one variable only at a single variable this and we have f of function is like this way again. Then x is equal to x star you see at this point you got the maximum value of the function again for scalar case we know the derivative at this point slope will be 0. So, we are getting the maximum value of this function here and this is the minimum value of function at this point that slope is again 0. So, the first scalar case the derivative of the function if the function value at that point is minimum or maximum or optimum the derivative of the function value is 0. What is the saddle point of this one a saddle point is a point that does not carry any information about the local minimum or local maximum or optimum point you see this does not shows any local minimum or local optimum points. So, this point is called the saddle point again. So, our definition now this is for simple scalar function function which is function of only single variable. So, our definition for stationary point you know whether the function will be maximum or minimum for scalar case the derivative of this one must be 0. So, a stationary point is a point x star at which the derivative of the function is 0. But when this function is a scalar one, but in case of a function which is a function of multivariable case x 1 x 2 dot dot x n then we have we will show you the gradient of this function must be assigned to null vector then only we will get at that point if that maximum or minimum value of the function agree. So, this is the and this figure shows that function is a this function f is a function of x 1 and x 2 and the function value we have plotted in y z directions and this is the function it is shown. So, you see it has a minimum value maximum value at this maximum value here here and minimum value is here. So, let us see this one at this moment that if you recollect we had a function is f which is a function of x 1 and x 2 and if you assume that x 1 star x 2 the optimum value may be maximum value of or minimum value of the function at this point x 1 is x star from there we have given a some perturbations. Then by Taylor series expansion we have retained the function value at x is equal to x 1 star and x 2 is equal to x 2 star and the first order derivative and second order derivative plus higher order derivative whose sum of the rest of the series I have kept is r. This quantity r is very small provided that a perturbations of from the what is called x 1 star or x 2 star which is the optimum point of the function is small. Then this is small compared to this terms first order, second order terms of this thing. So, now let us see this thing. So, we are writing the original function x 1 star plus epsilon 1 x 2 star plus epsilon 2 star and this f of 1 x I am keeping in the left term side the function value at x is equal to x star and x 2 is equal to x star this part is equal to what is left in the right hand side. This part I have taken this side and what is left first order derivative and second order derivative functions here. So, I am writing it this you see carefully what I am writing into matrix and vector form only is a row vector x 1 star x 2 differentiate with respect to f with respect to x 2 star x 1 star x 1 then all of this x 1 x 2 then with respect to x 2 this multiplied by x 1 and x 2. So, this is the first order part of derivative is coming to the now second part is half factorial this you see what I am writing, but this have to evaluate this have to evaluate I am writing x 1 is equal to x 1 star and x 2 is equal to x 2 star. So, this quantity is known to you, but this quantity this may be positive this may be sorry this is the epsilon 1 this is the epsilon 2 see this expression this is the epsilon 1 this is the epsilon 2 this not x this is the epsilon 1 and this is the epsilon 2 again this now this quantity when you put x 1 is equal to x star x 2 is equal to x 2 star this quantity you do not know it may be positive may be negative may be 0 all these things can be, but what about epsilon perturbations epsilon 1 epsilon 2 it can be positive side or it can be negative also. So, if you just multiply this into this the resultant quantity you are not sure whether it will be plus or minus, because once you assume that this is a plus let us call multiplied by epsilon, epsilon can be negative may be positive, because it is the either side of this optimal point x 1 is equal to x star perturbation may be in either side. So, this also if you assume this is a negative and this epsilon 1 and epsilon 2 may be positive or negative. So, you this whole product you cannot say confidently resultant will be a positive or negative. So, this things if you keep in mind I am now seeing the what is the second order derivative part what you can write it. So, this is a delta square f x 1 x 2 delta x 1 square this find out this value x 1 is equal to x 1 star and x 2 is equal to x 2 star multiplied by epsilon square. And this value you differentiate the what is called the function of f with respect to x 1 twice that and that put the value of x 1 is equal to x star x 2 that multiplied by epsilon square. Now, next is 2 then this x 1 x 2 then del of x 1 del of x 2 put this values x 1 is equal to x 1 star x 2 is equal to x 2 star and multiplied by epsilon 1 and epsilon 2 plus del square f del x 1 is into del x 2 differentiate with respect to x 2 twice put the value x 1 is equal to x 1 star and x 2 is equal to x 2 star multiplied by epsilon 2 square plus the rest of the terms which is denoted by r. Now, in this expression you see this minus this this 2 quantities can be positive can be negative you do not know because if it is a optimum point is it is like this way it is the x star agree if it is this side agree and if it is this side that x plus epsilon this one agree that means this is a positive quantity difference if it is this side this will be positive side positive. Then I can say if it is positive that this is the optimum point of this one if it is like this way from this and this agree then this value will be a negative. So, now I cannot say that one with whether it will be positive or negative because due to this factors this factor I told you if it is a let us call this is a positive due to epsilon it may be positive negative epsilon may epsilon 1 can be positive epsilon 1 can be negative similarly this time if it is a negative this can be a positive negative total resultant whether it will be positive or negative we cannot say. So, similar to that one this I can write it this is nothing but a gradient transpose. So, assign gradient of this vector f of x assigned to 0. So, this part is vanished now. So, this ambiguity because we know at this point slope is 0 if it is a multiplicative case that gradient will be 0 assigned this one. So, only this one is left and this I can write it if you see this I can write it now into a that form x 1 star plus epsilon x 2 star plus epsilon 2. So, this minus f of x star x 2 star is equal to I can write it this gradient of f of x 1 x 2 transpose x 1 is equal to x 1 transpose x 2 is equal to x 2 star into what I can write it that one your epsilon 1 epsilon 2 see this one this I have written is gradient transpose epsilon 1 epsilon 2. And what we can write it this one you see something like this is the square quantity you can say a epsilon square 2 b epsilon 1 epsilon 2 c is equal epsilon 2 square. So, it is a quadratic form I can always write that one into this form we have shown earlier that what we can write it I can write always epsilon 1 epsilon 2 transpose this is this form then I can write it gradient that is x 1 x 2 del x 1 square del square x 1 x 2 del x 1 square del square f x 1 x 2 del x 1 del x 2 and del square f x 1 x 2 del x 1 del x 2. And this one is del square f x 1 x 2 and x 2 and del x 2 square and that is that one and this value you have to calculate x 1 is equal to x 1 star and x 2 is equal to x 2 star into this into this is into epsilon 1 epsilon 2 and what is the term is left plus r. So, if you consider this is a vector of epsilon ultimately I am writing because you see this quantity will be positive or negative I cannot take decision with this one because of it can be anything. So, I assigned this quantity that this whole quantity is assigned to a null vector. So, if it is so the right hand side now I can write it half epsilon transpose epsilon transpose into this is the Hessian matrix h you calculate h is x 1 is equal to x 1 star x 2 is equal to x 2 star into epsilon plus r. So, now you see I just mentioned it that r is sufficiently small compared to this previous terms because that epsilon 1 epsilon 2 because higher order derivatives will be negligible because of epsilon 1 epsilon 2 are very small around the x 1 star x 2. So, if you assign this is 0 this term is not there only term is left this plus this. So, I can easily now tell that what is called this quantity will be positive provided h is a positive definite matrix quadratic function. So, if it is a positive definite function then what does it mean if you give a perturbation from x 1 star and x 2 star that quantity is positive means we have obtained the what is called local minimum. If this quantity is negative when it will be negative h is negative definite matrix h is negative definite matrix then this quantity will be negative, but that contribution is negligible because of epsilon 1 epsilon 2 are very close to x 1 star x 2 star. So, this will be negative the function below difference is function below negative means it has reached to the local maximum the difference is coming. So, today I will stop here and next class we just continue that what is our final conclusion. Thank you.