3.3. Applications of the Implicit Function Theorem (part 1).

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Del curso dictado por National Research University Higher School of Economics

Mathematics for economists

10 calificaciones

National Research University Higher School of Economics

Mathematics for economists

10 calificaciones

This course is an important part of the undergraduate stage in education for future economists. It's also useful for graduate students who would like to gain knowledge and skills in an important part of math. It gives students skills for implementation of the mathematical knowledge and expertise to the problems of economics. Its prerequisites are both the knowledge of the single variable calculus and the foundations of linear algebra including operations on matrices and the general theory of systems of simultaneous equations. Some knowledge of vector spaces would be beneficial for a student. The course covers several variable calculus, both constrained and unconstrained optimization. The course is aimed at teaching students to master comparative statics problems, optimization problems using the acquired mathematical tools. Home assignments will be provided on a weekly basis. The objective of the course is to acquire the students’ knowledge in the field of mathematics and to make them ready to analyze simulated as well as real economic situations. Students learn how to use and apply mathematics by working with concrete examples and exercises. Moreover this course is aimed at showing what constitutes a solid proof. The ability to present proofs can be trained and improved and in that respect the course is helpful. It will be shown that math is not reduced just to “cookbook recipes”. On the contrary the deep knowledge of math concepts helps to understand real life situations.

De la lección

Implicit Function Theorems and their applications.

Week 3 of the Course is devoted to implicit function theorems. In this week three different implicit function theorems are explained. This week students will grasp how to apply IFT concept to solve different problems.

3.1. Implicit functions.7:05

3.2. Implicit Function Theorem.5:33

3.3. Applications of the Implicit Function Theorem (part 1).10:33

3.4. Applications of the Implicit Function Theorem (part 2).7:01

3.5. Gradient is perpendicular to a level curve of a function.11:41

3.6. Implicit function theorem for the function of many variables.6:20

3.7. Example of application of the IFT for the function of many variables.8:45

3.8. Implicit Function Theorem for the system of implicit functions. Jacobian matrix.10:53

3.9. Example of application IFT for the system of implicit functions (part 1).8:41

3.10. Example of application IFT for the system of implicit functions (part 2).7:05

3.11. Example of application in microeconomics.6:26

3.12. Cramer's rule.2:45

Conoce a los instructores

Kirill Bukin
Associate Professor, Candidate of sciences (phys.-math.)
Faculty of Economic Sciences, Department of Theoretical Economics

Let us apply this Implicit Function Theorem or IFT for

short, for our example with the unit circle equation.

The same point as earlier.

Okay, let's check whether it's applicable, the theorem is applicable to this

particular equation considered at this point.

First of all, the function, capital F, is x squared plus y squared -1.

And it is, continues to differentiable for all x, y values.

Secondly, when we substitute the coordinates into the equation,

they satisfy this equation, easy to check.

Now, what's df over dy.

This is 2y, and when we substitute

this number, It's not 0 for

y equals 1 over square root of 2.

So the theorem is applicable and it claims,

as we have learned, that implicit function exist and

this function on some interval is continuous differentiable.

When we substitute 1 over the square root of 2, it takes the same value.

And when we substitute this function into equation, so

this equation is satisfied identically within the same interval.

So, as for the interval, this theorem doesn't give a clue,

simply because this theorem is of local nature,

we can guarantee that such an interval exists.

What about the derivative?

At this point.

So according to the formula, we need to substitute into the denominator.

2y taken at this value, which becomes the square root of 2.

What about dF over dX?

dF over dX, at this point is also square root of 2.

So we have -1 and clearly,

if we draw the tangent line to the unit

circle at this point, this angle,

this is an acute angle whose value

is definitely 45 degrees.

And this is an obtuse, so we can easily check, this is a correct result.

A word of caution should be said here.

What if we tried to apply IFT to the point x0=1,

y0=0?

Well first of all, the condition, or the IFT's not met,

because df over dy at this point

is 0, because y0 is 0.

But, we need to understand what was the problem here.

The problem here is as follows.

If we're trying to solve the equation in terms of y as a function of x,

it's impossible to do so within the neighborhood of the given point.

x0=1, y0=0, because what part

of the unit circle should we choose

as the graph of the implicit function.

Should it lie in the upper half plane or lower half plane?

Impossible to tell, and

also if we can see the points which lie to the right of this point.

There is no graph there at all.

And when we're looking for the function, the function

should be defined in some neighborhood around the point.

A question, whether it's possible to find implicit function,

even if dF over dy is zero to given point?

If dF over dy is zero to given point,

probably another derivative dF over dX at this point is not 0.

This is exactly what happens here at a given point,

where derivative equals to x and x is 1, so it's not 0.

We can reformulate the theorem and

we claim the results that there is a function, also implicitly defined.

A function X as a function of Y.

So here Y is an independent variable and X is dependent variable.

And it exists in some neighborhood of all the given point.

Although the interval where this function is defined is along y axis,

and more over we can try to find derivative.

This is x prime with respect to y.

And well, if we consider this particular problem

we need to Use the formula.

Here we have number 2 in the denominator, and here we have 0.

So we have 0 here.

This is a result for this particular unit circle.

But we can try to do the same in case then the gradient of the function,

which defines this equation, doesn't turn zero at a given point.

Now, let's consider an example, or other problem, from microeconomic theory.

Let us suppose, there is a monopolist

Whose profit can be found by the formula, this is the inverse demand function,

p(y) where y is the output of the monopolist.

So that's, this product provides the revenue and

the marginal cost or the monopoly is constant and this the value is C.

So, we multiply in order to find the revenue fixed cost,

in order to find the total cost function we multiply the marginal costs by Y.

And this is a problem of maximization of profit with respect to the output.

What's given?

What information is provided about demand?

The demand function p(y) Belongs to c2 class of functions.

That means that it has the second other derivative which is also continuous.

And what is the problem we are going to set and solve?

So the monopoly maximizes its profit.

So, in order to find the output of the monopoly,

we need to find the first derivative of the profit and

that first order condition looks like

And equals 0.

Also let us assume that the second order condition is also valid.

Meaning that whether we find the solution or this equation.

I will be using notation y, and

with upper script M, indicating the output of the monopolist.

So let's suppose that second order derivative of the profit at this value ym.

Where m is negative.

That provides the maximization of the profit at a given point.

Now the question is, how to find derivative

of this output with respect to the value of them?

Marginal cost which is given by C.

So that, this is the question number one and the question number two,

how to find the derivative over a monopoly price?

Which will be denoted by pm

with the respect to the same level of marginal cost dc.

So, that's question number two.

And, in order to find answers to these questions,

once again we'll apply IFT.

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