 We are going to conclude the discussion of least square problems static deterministic least square problems linear non-linear with the discussion of examples. I am interested in couple of essentially three examples to be precise. I would like to be able to recover the vertical temperature profile of an atmosphere from satellite radiance measurements that is a linear problem. I am going to talk about 1D and 2D spatial interpolation that is again a linear problem. I am going to deal with a non-linear least square problems again related to vertical temperature profile recovery. So, what is the idea here? Recovery of that vertical temperature profile. The problem is to retrieve the vertical temperature profile of the atmosphere from satellite radiance measurements. Problem 1 thus radiance received by the satellite in the infrared domain. Let R be the energy received in a frequency F by a satellite and that is F is related to the vertical temperature of the but this RF is related to the vertical temperature profile TP where P is the pressure. So, I am going to talk about vertical coordinate at pressure levels of the atmosphere. So, you can think of RFF which is the energy received by the satellite. The physical model for this is given by the equation 1. Gamma F is a constant that depends on the frequency of the channel in which the energy is received. WP gamma F which is the kernel in the integral in equation 1 is given explicitly by P times gamma F exponential of gamma F times P. P is the pressure level. So, you can think of the pressure level like this at the sea surface level pressure is equal to 1 I am going to go to the atmosphere where the pressure is 0. So, I am interested in trying to recover the temperature profile T of P how T varies as a function of P when I go from sea surface level to where at different heights the heights is indirectly measured through pressure. So, the weight function is given by P times the exponential the energy is related by this equation and this equation constitute the model the mathematical model that we are going to be concerned with T P is the temperature profile. So, you can readily see the energy received is related to the temperature distribution with respect to pressure and the formula is derived from the radiation physics. Let us not go into the derivation of this formula let us take this formula to be granted. So, what is the basic idea here what is the inverse problem W is known this quantity is known because gamma F is known for a particular channel. So, gamma F is known P is known. So, the W weight function is known and RF is also known from the satellite measurements I would like to be able to recover the temperature I would like to be able to recover the temperature namely. So, knowing RFF and knowing all the other things I would like to be able to recover the temperature. So, that is what is called temperature retrieval problem. Please understand what is the forward problem if I know T P I can recover RF that is the forward problem what is the inverse problem I know RFF I want to find T P that is the inverse problem satellite. So, this kind of problem is very routinely done. For example, we are now talking about El Nino what is RF RF is the radiance measured by the satellite just above the equatorial Pacific. So, I would like to be able to understand or recover the temperatures of the equatorial Pacific waters based on the satellite measurement temperature and from the current estimate tells you we are very warm Pacific ocean that has led to this notion of what is called we are in the El Nino strong El Nino regime. So, these kinds of problems arise very routinely and meteorologists all these problems just about every day. So, here are some basic values relevant to the problem. So, F I I is 1 2 3 4 5 that means I have 5 channels with 5 different frequencies the frequencies are given in some scale let us not worry about the scale. So, the frequency ranges are from 0.9 to 1.3 increasing values the frequency this is some scaled values are a normalized frequency gamma F F gamma F F I these are all constants these are all radiation related constant physicists have already estimated these constants. So, for the first channel is 1 over 0.9 second channel 1 over 0.7 for the fifth channel is 1 over 0.2. So, gamma I gamma F is known F I is known frequencies and the corresponding radiation constants are known to us. Now, so I have a problem where I am concerned with measurement of energy in 5 channels. So, I have measured energy radiated and observed in 5 channels. So, based on the energy observed from 5 channels I want to recover the temperature distribution please understand temperature is a continuous function of the pressure continuous function of the height any continuous function is an is an infinite tree object it is extremely difficult to be able to recover that function, but I can however computationally discretize the system. So, for the sake of illustration I am going to discretize the discretize the atmosphere into a 3 layered system layer 1 layer 2 layer 3 as shown in figure the layers are marketed by demarcated by the pressure levels. So, 1 to 0.5 0.5 to 0.2 to 0 I am assuming the temperature in layer 1 is T 1 temperature in layer 2 is T 2 temperature in layer 3 is T 3 T 1 T 2 T 3 are constants. You can readily see instead of 3 layers I can have 30 layers or I can have 50 layers which are defined by the pressure levels. So, the conceptually the problem is not very different whether I consider 3 layer or 30 layers or 300 layers. The number of layers is simply a question of convenience and the accuracy of representation. So, without loss of generality let us try to illustrate it using 3 layers. Let T not be the temperature of the surface of the F, T I be the average temperature at level I the layers are bounded by isophobic surfaces at 1.5 0.2 and 0. So, I would like to be able to go back to the first equation. So, equation 1 is a continuous representation as a function of pressure I am going to have to discretize this integral. So, I am going to express this integral look at this now Zi which is the so, Zi is going to be RF what is RF? RF is the energy measured by the satellite in the ith channel. I am going to call that observation as Zi in our notation. So, Zi so, RF is we already know RF is equal to exponential of gamma F plus the integral I am taking the exponential of the left hand side. So, I am going to call RFI minus this because I can compute this quantity this comes from the satellite. So, the difference between the two if I am going to call it ZFI my observation the integral can be now represented as a sum of three simple integrals one goes from see T P comes within the integrand if I assume that is the constant T P gets out T 1 the constant temperature in the first layer 0.5 to 1 T 2 0.5 to 0.2 to 0.5 T 3 0 to 0.2 I can now evaluate this integral I can evaluate this integral I can evaluate this integral I know P this is the integration with respect to P I know gamma I know this exponential function. So, by doing a very simple integration in this domain this integral value is going to be A I 1 ith observation A I 1 this integral is going to be A I 2 this integral is going to be A I 3. So, A I 1 A I 2 A I 3 are known constants T 1 T 2 T 3 are unknowns. So, I get the my first equation Zi is equal to T 1 A I 1 plus T 2 A I 2 plus T 3 A I 3 if I change I from 1 to 5 I have 5 such equation Z 1 Z 2 Z 3 Z 4 Z 5. So, 5 channels 3 layers. So, there are 3 unknowns there are 5 equations the number of channels equal to number of equations the number of layers is equal to number of unknowns. So, the constants are simply numerical values of the respective integrals and all these things can be calculated using the table. Therefore, by collating this 5 linear equations T 1 T 2 T 3 are unknowns T 1 T 2 T 3 are unknowns Zi's are knowns A I 1 A I 2 A I 3 A 2 1 A 5 1 A 5 2 A 5 3 this is the matrix H this is the vector x and that is the vector z therefore, I have z is equal to h of x with z belonging to R 5 h belonging to R 5 by 3 and t belonging to R 3. So, if I invert this problem I get t. So, this is a linear least square problem I can solve this problem. So, what is my xls is equal to tls which is equal to h transpose h inverse h transpose z. So, I can solve this linear least square problem very easily we can solve this linear least square problem very easily. So, now you see how our simple linear least square problem can be used for satellite retrieval measurements all that is needed is a mathematical model which is given by the radiation physics once you have the radiation physics problem I can discretize it by levels. So, the number of levels refers to the number of unknowns the number of channels refers to the number of knowns I can evaluate this matrix H so you can z z is equal to h of x is a very simple very simple problem. So, I am now going to go further I am going to give you typical values let us pretend t1 is 0.9 t2 is 0.85 t3 is 0.875. So, what am I what am I going to do now I am going to solve the forward problem first I am going to assume t1 t2 t3 that is going to be the set x bar I have already evaluated all the a functions from the integration so that gives the matrix H. So, using t1 t2 t3 assumed value and the computed matrix H I am going to compute z bar is equal to h of x bar I am going to use this z bar as my observation I want to be able to use the model itself to generate the observation and then use that observation so generated to be able to do the trivial this aspect of using the model to generate the observation and then to solve the inverse problem is called twin experiment. So, I am not going to wait for me to get the real data from satellite measurements I am developing methods when I am developing methods I do not have to worry about the actual observation I am going to have to generate observation I am going to illustrate the methodology by using this artificially generated observation and that is the goal of what is called twin experiments. So, now what is that we need to do we need to generate the actual observation to do the actual observation I have to create a noise vector V. So, I am going to have to create a noise vector V whose covariance is sigma square I phi I phi is the identity matrix of order phi sigma square is the common variance. So, what is that I am assuming all the channels that measure the energy radiated in the satellite the instruments that measure them they are equally good or equally bad. So, they have a common variance for example, if you buy a voltmeter from a shop the voltmeter specification will tell you it can measure voltmeter from 0 to 200 with an accuracy of plus or minus 5%. So, that gives you the standard error in the measurements. So, all the so by talking to people who design these instruments in satellite one can very easily compute the variance. So, sigma square is the common variance of the instrument that measure the radiance. So, Z bar I have actually calculated synthetically V is the noise. So, I would like to add Z to Z bar to get Z. So, now I am going to consider this Z which is considered to be Z bar plus V as my noisy observations. So, if I use Z in my calculations and recover I solve the inverse problem I should get a T which is very close to T1.9 T2 is equal to 0.85 T3 is equal to 0.085 and who is going to control the difference between the retrieved value and the original value the observation noise covariance sigma square. If sigma square is 0 I should be able to recover them precisely if sigma square is not equal to 0 I will recover them with some error the error is largely due to the measurement noise. So, using noisy observations of the vector Z we now solve the overdetermined least square problem Z is equal to h of x and recover x recover x. Now, I would like you to compute the residual I would like you to be able to compute the residual I think this is this is this should be R of I think this expression is wrong I want to be able to compute Z minus h x l of s now this is the actual residual Z minus h l of x now that is the one that goes in here. So, if I measure this now I would like to be able to plot that RLS so I would like to RLS is equal to Z minus h of x l s and I would like to be able to plot against the variance and I would like you to see when this is the claim if the variance is 0 the residual will be 0 if the variance is not equal to 0 then the error will be more. So, I would like to I would like you to be able to plot the variance and understand the impact of variance on the recovery and I would like you to be able to solve the problem and comment on it this is a computer based computer based homework problem I would like all of you to be able to do this problem enjoy and understand the impact of variance on the optimal residual so what is this this is the optimal residual so that is the first problem. Now I am going to go to problem 2 where I am going to do a spatial interpolation problem so consider a uniform spatial grid in one dimension I have 8 grid points I have 7 grid intervals so I have 8 unknowns at each of the grid point x 1 to x n n in this particular example is 8 all the grid intervals are assumed to be uniform and equal to unity let there be m measurements of a scalar field so what is what is that it can be I am measuring temperature pressure or concentration of a pollutant to name a few in the only difference here is that I am simply considering a spatial expanse of 1d why 1d 1d is not practical 2d 3d are more practical but to go to 2d and 3d I would like to be able to solve a simple problem of 1d so let us pretend that I have an observation of z1 the observation is located in the interval in the in the grid from 2 to 3 observation z2 is located from 4 to 5 observation z5 is located from 5 to 6 so there are 3 observation stations the observation station z1 is at a distance a1 from the grid point 2 z2 is the grid a2 from grid point 4 and z3 is the distance a3 from the grid point 5 so all these things are given in given information n the number of grid points m the number of observations what is being observed is a scalar field the scalar field such as temperature concentration of a pollutant or whatever that be I am assuming m is smaller than n m is smaller than n that means I have more number of grid points less number of observations therefore this is an undetermined case in fact this is the problem that occurs very naturally in the context of pollution estimation we would like to be able to give alerts for days in which where the pollutant are very strong we do not have measurements measuring system for pollutants at every place we have a fixed number of locations where we measure the pollutants but we would like to be able to extrapolate those measurement to a larger domain so that we can say how the concentration of the pollutant varies spatially and to be able to extrapolate I need a matrix so extrapolation interpolation these are very similar problems this I am talking about spatial interpolation problem in here. So let there be m locations where there are m observations of a scalar field let there be n unknowns I would like to be able to estimate the n unknowns using m knowns so this is the standard problem I am going to formulate these problems the linear problem the jth observation so I am going to have some basic basic notations let the jth observation let the jth observation be contained in the ith interval the ith grid point is the space from I to I plus 1 so Z j is the I so what does it mean if this is I if this is I plus 1 this is Z j Z j so let the jth observation be contained in the ith interval referring to the figure in the previous page I had 4 observations I have 8 unknowns Z 1 is in the interval 2 3 Z 2 in the interval 4 5 Z 3 in the interval 5 6 and Z 4 in the interval 7 8. So given 4 observations I would like to find 8 unknowns the unknown refers to again temperature concentration pressure so on and so forth some scalar field so I hope the problem is clear now I am going to talk about how do we take this 4 observations and create estimates of the 8 unknowns this is done by simple linear interpolation that we learn in a first course in numerical analysis. So let us consider the ith sub interval I to the sub interval I to I plus 1 let jth observation be contained in this interval let the jth observation be located at distance a j from the end I since the distance between I and I plus 1 is 1 the distance of Z j from I plus 1 is a j bar where a j bar and a j is 1 so a j bar is 1 minus a j in other words I have a I know exactly where the observation location is with respect to the computational grid. So I and j are the computational grid point Z j the observation location at I the value of the unknown is X I at j I plus 1 the value of the unknown is I plus 1 the value of the observation is Z j Z j is known I do not know X I do not know X I plus 1. Now I am going to relate the known to the unknown Z j is the known X I X I plus 1 are not known in order to relate X I plus 1 X I and Z j I am going to use a simple linear relation. So what is the relation the value X I say this is the value of X I this is the value of X I plus 1 this is the value of Z j I am going to assume the line joining X I Z j and X I plus 1 has a constant slope. So what does it mean Z j minus X I by a j is equal to X I plus 1 minus Z j by a a j bar. So I am simply trying to express this constancy of the slope which if I simplify this I get the relation a j bar X I plus a j X I plus 1. So this is the linear relation this relation trans this relation connects the unknown to the known the X I's and X I plus 1 are the unknown Z j's are known they are related by the parameters a j and a j bar. Now if I can do this for jth observation I should be able to do this for every observation if I did that I get a matrix. So I have M observations I have 8 unknowns so this is the unknown vector this is the known thing my problem is linear so this is the matrix H the H matrix is 4 rows and 8 columns. For example Z 1 is located in the interval between 2 and 3. So only 2 and 3 are affected by Z 1 rest of all 0 Z 2 is affected only by X 4 and X 5 because it lies in the interval 4 5 Z 3 in the interval 5 6 Z 4 in the interval 7 8 and you can see in every row there are more 0's and non 0's further the sum of the non 0 elements in each row is 1 it need not be the case a 1 a 1 need not be equal to a 2 need not be equal to a 3 need not be equal to a 4 a 1 a 2 a 3 are numbers in the interval 0 to 1 when ai is 1 it lies on the grid point when ai is 0 also the observation lies in the grid point. So we are simply assuming for generality the observation location and the grid point locations are not the same if the observation location the grid point location coincide there is no need to interpolate X i will be equal to either X i will be equal to Z j or X i plus 1 will be equal to Z j depending on which grid point the Z j lies on that is an easy case that is why we are considering a very general case where the observations are not located at the grid points. So by simple concept of constancy of the slope we have we have been able to derive this relation Z is equal to H of X please remember we have now converted the problem of spatial interpolation to a linear least square problems. The H has 4 rows and 8 columns H is such that the rows is equal to 1 I can solve the problem Z is equal to H of X by the least square solution which we have already obtained now we can estimate the temperature concentration and the computational grid. So I can estimate the value by 8 points knowing the value only at 4 points this ability to be able to extrapolate the observation in a smaller subset to a larger region using the notion of spatial interpolation by converting the problem to a linear least square problem I am able to estimate the distribution of concentration the distribution of temperature the distribution of pressure or whatever quantity that is being observed. Now I would like to be able to extend this to spatial 2D interpolation. So in this case I am solving the same problem except that the space domain is 2D instead of 1D. So this is the 2D version there are n number of grid points nx is the number of grid points in the x axis ny is the number of grid points in the y axis I am giving an example where nx is 4 ny is 4 there are n by n n is 16 nx times ny I can so this grid has 16 locations I can label the grid points 1 through 16 in this snake like order that is that is one way of notation another notation would be 11121314212222324 using the standard way of numbering in geometry x coordinate y coordinate the left the left numbering system is called the row major order the right numbering system is the standard ij notation these two notations are related. So if k is the value of the grid point in the row major order if ij is the coordinate in the two dimensional representation k and ij are related by this relation k is equal to i minus 1 times the nx plus j for example when nx is 4 the node label 7 corresponds to 23 since 7 is equal to 2 minus 1 times 4 plus 3. So you can think of the two dimensional grid labelled in two different ways row major order you can use column major order too we will get a similar formula so you can see there are several ways of numbering and each way of numbering we need to know the relation I have related two distinct ways of numbering. So I have now the number of unknowns the number of unknowns are 16 the number of unknowns are 16 now I am going to go back to statement of the problem there are 16 unknowns the z's are known there are only 4 z's that are given in here you can really see there is a z1 there is a z2 there is a z3 there is a z4. So given 4 observations of concentration I have to evaluate the concentration at 16 points that is the problem we are going to be concerned with again I need to develop a relation between the known the unknown the z's are known x's are not known. So I am now going to consider let the jth observation be contained in a square whose origin is i if this is i this is i plus 1 if there are nx points in the row major order the label for this is i plus nx the label for this i plus nx plus 1 therefore I am going to consider the square as 1 with the origin i. So let the jth observation be located in the ith grid where i is the origin. Let aj be the distance of the observation along the x axis let bj be the distance of this from the origin along the y axis. So the coordinate of zj or aj bj aj bar is 1-aj bj bar is 1-bj so we have the standard relation therefore we can now compute the distance of zj from each of the corners. So what are we first going to do I have to relate zj to xi xi plus 1 xi plus nx xi plus nx plus 1. So I am going to do this as two applications of linear interpolation we have already seen. So I am now going to relate zj to eta i and eta i plus 1 using a 1D interpolation and once eta i is computed I am then going to relate this to xi and xi and i plus nx xi plus nx and this. So I will do this first and then do this next I will do this next so that way I am going to spread zj into 4 of these quantities. Using the 1D interpolation you can readily see zj is equal to a bar eta i and aj eta i plus 1. Again using the same linear interpolation eta i is equal to xi bj bar and xi nx i plus nx bj likewise eta i plus 1 I can now substitute 8 and 9 into 7 if I substitute that I get a formula that relates zj so the right hand side has all the grid points xi xi plus 1 xi n i plus 1 all the coefficients are distances that are known so I can write this in the form of a matrix. There are 4 observations so there are 4 relations so this is the observation vector there are 16 unknowns the 16 unknowns are naturally partitioned into vector of 4 segments. So z1 now lies so let us go back to the picture now z1 is in the grid rooted at with all origin 1 so z1 will affect only 1 2 5 and 6 that can be readily seen 1 I am sorry 1 so this is non-zero this is non-zero 5x is non-zero 6 is non-zero these stars represents the coefficient which are given by products of aj aj bar bj bj bar you can fill them in I do not want to put all of them in to make it more complex I simply talked about their value as a star non-zero value so we can again see z1 affects only 4 neighbors likewise z2 affects again 4 neighbors 3 4 and 7 8 z3 affects 6 7 and 10 and 11 c4 affects 11 12 and 15 16 so this is essentially a summary of the relation between the knowns and the unknowns. So this gives raise to a problem z is equal to hfx this is the matrix hfx this relation becomes z is equal to hfx h is a vector of size 4 x is a vector of size 16 you can readily see this is an under-determined problem under-determined problem I want to emphasize in the case of satellite measurements vertical temperature profile distribution we have an over-determined system in the case of spatial interpolation like this we have an under-determined system I am illustrating all the aspects of my least square theory the 2D interpolation matrix is such that the rosa is always 1 I want to be able to solve z is equal to hfx zls is equal to h transpose so this is the under-determined case I want to remember the under-determined solution is different from the over-determined solution this is the formula for the optimal estimate for the under-determined system again this comes from the previous results that we have already seen. Now before I go to the next nonlinear problem I would like to be able to summarize the second problem so the second problem is simply a spatial problem which is which occurs again and again in many different facets of geophysical applications. So I have a sparse set of observation I have a larger domain I have a computational domain embedded and I would like to be able to carry over the observation information to a larger domain that is where the spatial interpolation comes into being. Now I am going to talk about the last of the illustration using a nonlinear problem I am again going to go back to the atmosphere the vertical temperature retrieval problem I am still going to consider 3 layer problem the pressures are given the last pressure I think there is an a last pressure this is not 0 this is 1.0 that is the C level pressure decreases 3 layers. Now I am going to assume an empirical relation for the variation of temperature with pressure a quadratic relation. The previous one was arrived at by a physical argument from radiation physics that will give us a linear problem. Now let us try to conjure up an atmosphere where the temperature at pressure p is given by this nonlinear function where x1, x2, x3 are the unknowns p is the known the temperature is the measure of the pressure is the nonlinear function. So the unknowns are x1, x2, x3 this is the nonlinear problem because of the nonlinear relation. Now I am going to talk about how to develop the mathematical relation for the observation to the temperature the observations are measures of overlapping fractions of the area under the curve. So Tp is the curve so if this is p I am sorry if this is p is decreasing Tp is going to be defined over this so this is p this is Tp. So I have pressure levels if I am going from pi to pj if I integrate it I am going to get an observation which is z bar ij. So 10 refers to the model this is the observation this is the unknown the observation unknown are related to the integral and Tp is a nonlinear function of the unknown parameters x1, x2, x3. So if I substitute for Tp from the previous thing I can relate the parameters to the observations the pressure level I am going to assume are 0 to 2.5, 2 to 5, 3 to 0.7, 0.6 to 0.8. So that is the pressure level from which observations are going to be coming they are overlapping regions. So this is another simple formulation of the problem. So I am going to now relate my unknown to the known I am going to derive my model this is the unknown this is the known this is the temperature profile if I integrate this I will get this quantity I would like you to verify the results of this integration. Now you can see this is the observation this is the model equation the observation and the model equations are nonlinear related p's are known x's are unknowns observations are known. So referring to the table on slide 19 I know the values of p, i and p, j let us go back. So this is the p, i and p, j so if I substitute these numerical values in here I get the first relation 0, 1 is equal to 0.25 and that is equal to this function that is h1 of x and what is my x? My x is equal to x1, x2, x3 because I have a 3 layer problem my t's are x. So z2 is equal given by this z3 is given by this it should be z4 is given by this. So z's are given so again what am I going to do these are the values again I want you to understand I have been given the z's in here I would like to go back. So pressure level the z's the observations are given okay so let us go to the next one. So this is where the z's are coming into play so z1 is equal to h1 of x, z2 is equal to h2 of x, z3 is equal to h3 of x, z4 is equal to h4 of x. So you can now z the whole problem is z is equal to the whole problem now reduces to z is equal to h of x the whole problem reduces to z is equal to h of x. So in this case m is 4, n is 3 so this is the nonlinear problem and the nonlinear problem essentially comes from the fact the model relate the temperature to the pressure using a nonlinear function of the unknown parameters. Therefore z is equal to h of x therefore z is equal to h of x, z is given by z1 to z4, h is given by this I have to compute my residual z – h of x I have to compute my f of x which is the sum of the square of the residual I have to compute the gradient to 0 I have to solve for the Hessian and verify this is a positive definite function. So I would like you to compute the optimal solution as a homework problem I have to formulate the problem it is simply you need to be able to solve this problem numerically if you solve this problem numerically you will understand the methods of nonlinear solutions extremely well. And what are the methods we have seen we can use the first order approximation we can use the second order approximation we have described all these methods extremely well in detail. So I have formulated the problem I have already described the algorithm I would like you to be able to combine the algorithms with the problem on these simple cases each of them are derived from chemical setup atmospheric temperature retrieval or spatial estimation of concentration of certain pollutant these are all very simple problems of great interest in applications. So by solving these problems you can master the techniques behind solving linear least squares nonlinear least squares static deterministic problems. So I am going to now illustrate some of the major steps here so what is that we need to do you know H so you need to be able to compute the Jacobian you need to compute the Hessian term you can build the first order as well as second order approximation then you can do the minimization arising from the first order and as well as the second order approximation and also I would like you to utilize this problem to be able to compare the quality of the solution for the first order and the second order. So doing all these will help you complete doing all these will help you to complete the estimation problem of interest in here and this also help you to look at an actual problem which is related to or which is derived from practical considerations in meteorology you will tell you how to build your H function how to do the Jacobian where the Jacobian Hessian gradient all these calculations comes into play where the minimization what is the role of the minimization what is the role of the first order approximation what is the role of second order approximation. So if you complete all the three problems numerically you will have a total understanding of not only the algorithms but also many of the mathematical principles that we have reviewed in the previous lectures with this we conclude our discussion of the static deterministic inverse problems of both linear and non-linear type both well posed and well posed type both offline as well as online type. So I would like to be able to now draw a little picture of least square problems so if you consider least square problems least square problems can be classified as linear non-linear well posed ill posed offline online and we can also think of deterministic and stochastic. So we have talked about linear problems we have talked about non-linear problems we have talked about well posed problems we have talked about ill posed problems we have talked about online problems we have talked about offline problems we have talked about deterministic problems. We also can relate to static and I am sorry we can also relate this to static and dynamic so one is static so we talked about static. So in these lectures so far we have covered static deterministic online offline well posed well posed linear and non-linear and we have talked about all the associated mathematical principles as well as we have derived algorithms to solve all these problems and that is the this picture provides a summary of what we have done so far. I would like to encourage you to further continue your solution process by working with these examples that are given in page 20 slide 23 I would like to refer to chapters 5 through 7 these slides this particular module where the 3 examples are illustrated they are taken from chapters 5 through 7, 5, 6 and 7 of our book Lakshmi Rahan, Louis Lakshmi Rahan and all 2006 Cambridge University Press book so the module essentially is a summary of what is happening in many of these chapters. So if you work out all the problems which are part of the development as well as exercises you will gain a very thorough and good working knowledge of solving static inverse problems thank you.