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

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Del curso dictado por National Research University Higher School of Economics

Mathematics for economists

1 calificación

National Research University Higher School of Economics

Mathematics for economists

1 calificación

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

So let us check that the conditions of IFT are met.

So we apply the conditions to the first order conditions

provided by this equation.

What we need to check?

In terms of differentiability and

continuity of the functions involved we see that since

demand function is twice continuous differentiable this equation.

Depending on y and c is differentiable with

respect to both variables y and c.

This is a second order condition which we'll use a bit later.

Now, let's check whether this condition.

Remember from IFT, we need to check that

the derivative with respect to y is not

zero where y is exactly this equation.

So let us differentiate with respect to y.

So when we differentiate we get exactly pi double prime.

So this is pi double prime and a condition is set that them

sign over this second over derivative should be negative, so this is a negative.

So the condition claiming that such a derivative shouldn't be zero.

This condition is met.

So the implicit

function ym, the output of the monopolist

exists as a function of c the value of the marginal cost.

Moreover, we can find derivative we're looking for.

According to IFT we use the formula where

in the denominator replace exactly the f with the y which equals

p double prime y plus 2 p prime.

And what goes up there in the numerator we need to put

the derivative of this equation with respect to C which is simply negative one.

So then by crossing out the minuses together we get 1 over.

And this is negative because denominator is negative.

So what's the idea behind it.

When the marginal costs of the monopoly increases their output

of this monopoly goes down.

And if we would like to find another derivative

derivative of the monopoly price with respect

to module cause we simply apply the chain rule.

So we need to differentiate our demand

function at this value and multiply by.

So all in all.

We have the formula.

Well, both derivatives in the denominator and y value are taken at ym of course.

Now, we can get an interesting conclusion.

We can draw conclusion from this formula.

Let's suppose that the government imposes an excise tax on this monopoly.

So the value of this text is t tax is imposed on the output.

And it can be interpreted as the growth of the marginal costs.

So initially there.

Of course function was c times y.

And now one is becomes greater earlier,

so we add up all the value of that imposed text.

How will it affect the final price.

A charge by the monopolist.

It's hard to tell without knowledge of the demand function, but quite often.

The demand function is a linear function.

So let's suppose that this is exactly the case,

so this is a linear function where a and

b are positive numbers and we plug into this final

formula in order to calculate the derivative

Since, it's linear its second derivative equals 0.

So we can cross out this term and

when we cross it out as a result we get 1 over 2.

It's interesting it means that if a monopoly charges firstly.

Some price which is pm then after

the government imposes a tax the monopoly

price grows by the amount of t over 2.

So initially it was n.

And this prize becomes BM plus t over 2

because this derivative gives rise to the price in exactly this amount.

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