The standard forrmat necessitates that endogenous variables would be on one side and exogenous (or constant) variables on the other, to have the form Ax=d. There needs to be 2 systems of equations, both with the same variables on exactly yhe same side.
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Text box
We evaluate at x=0 simply to get the precise values of the coefficients.
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The taylor expansion is used to say something about consumer behaviour and utility expansion. This can be done by expanding around the mean value, and thus, x-xo, and (x-xo)^2 would be the standard deviation and variance, respectively. We would therefore see behaviour under risk and uncertainty, and the effect on welfare and utility.
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Text box
The Remainder is extremly important here. The function of an arbitrary number is equal to the sum of successive derivatives plus a remainder term. Why? Because the arbitrary number is always less than n, which is the highest order derivative. Thus, any number less than n can only give an approximation of the function.
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Contribution to growth
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Remember that dz = fxdx + fydy. This is put in function 11.6. Thus: d(dz) = d(fxdx+fydy) which is the sum of two partial derivatives.
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Discriminant is the b^2 - 4ac
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The leading principal minor of A of order k is the minor of order k obtained by deleting the last n - k rows and columns.
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First: delete rows 3 and 2, coloumns 3 and 2 Second: delete row 3 and coloumn 3 Third: don't delete anything
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Note that we have d11-r and d22-r on the diagnol of the matrix only since the characteristic root is multiplied by the identity matrix. Thus: [d11 d12] [d21 d22] - r[1 0] [0 1] = [d11 - r d12 ] [d21 d22 - r ]
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Imagine a hollow bell. Take two points from its bottom which correspond to 2 vectors. It is like putting a small flat metal piece at the bottom of the bell. Now, push it up until it hits the points of the surface. This is now the line which correspond to the vectors points of the surface (MN). Above it there is an arc.
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We say this is the height because M corresponds to the function of u, and N to the function of v. M and N are reached vertically upwards since they are the points as a result of the vectors at u and v. Therefore, the line between them is the weighted average of the line at u and v, which are again reached vertically upwards, so its the height of the line segment
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Trust me: IT MAKES SENSE!!!
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I worked this out with exercises
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f(v)-f(u) / v-u < f '(u) multiply by v-u: f(v)-f(u) < f '(u)*(v-u) deduct f(u) throughout: f(v) < f(u) + f '(u)*(v-u) which is the same as 11.24
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This is very much the same as the first-order and second-order derivative test. However, these are tests that work only when the function is at a stationary point. These tests are however used at points other than the stationary point
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Showing linear combination (see pg. 69)
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Note that here we usurp the jacobian determinant, because this will only tell us whether we have a set of fucntionally in/dependant functions, that will give us a unique solution. We are assuming that this unqinueness (ie J is not equal to 0) already exists. The Hessian (second-order partial derivatives) will tell us whether we have a maximum or a minimum.
Note that we are here using: V(m) = A(1+i/m)^-mt since i is here defined per quater, we don't have to divide by m. Also, since t = 1/4 (it takes only one quarter) -mt = -1
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a and b are the endogenous variables, and Po, Pbo, r, t, are the exogenous variables
The main point here is that dy is treated as a variable not an arbitrary constant, and thus we should find its derivative just like any other variable.
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Makes sense: we are never interested in all the values of the function because we are concerned with the range and domain of the constrained part. So in terms of second order necessary and sufficient conditions we are only interested in the dx and dy satisfying the constraint.
![[Alpha C. Chiang, Kevin Wainwright] Fundamental Methods of Mathematical Economics](https://img.pdfslide.us/doc/110x75/563dba14550346aa9aa28294/alpha-c-chiang-kevin-wainwright-fundamental-methods-of-mathematical-economics-569150ac2359b.jpg)
