2.8. Equation of a circumference. Smooth curves.

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Из курса от партнера National Research University Higher School of Economics

Mathematics for economists

1 оценки

National Research University Higher School of Economics

Mathematics for economists

1 оценки

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.

Из урока

Differentiation. Gradient. Hessian.

Week 2 of the Course is devoted to the main concepts of differentiation, gradient and Hessian. Of special attention is the chain rule. Also students will understand economic applications of the gradient.

2.1. Differentiation. Partial differentiation.7:38

2.2. Example of differentiation. Cobb-Douglas function.9:32

2.3. Tangent plane.7:46

2.4. Total differential.7:05

2.5. Chain rule for multivariate functions.8:05

2.6. Gradient of the function.9:58

2.7. Economic applications of the gradient.9:31

2.8. Equation of a circumference. Smooth curves.13:42

2.9. Chain rule for differentiation.10:07

2.10. Linear approximation. Example of tangent plane for particular function.10:40

2.11. Second-order derivatives.10:34

2.12. Young's Theorem.5:20

2.13. Hessian matrix.5:24

Познакомьтесь с преподавателями

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

We considered straight lines in n-dimensional space

and we define them using parametric equations.

But this also can be done with the general curves.

For example, let's consider

equation of a circumference in R2,

how to write it down.

That's drawing. The circumference is the boundary of the unit circle as we know,

x and y axis.

So, what we're doing,

we're trying to figure out how the points which belong to the circumference

can be written as functions of parameter t. It can be done

in many ways but this is the most common way of writing down.

So, x is cosine t,

and y is sine t. T,

a real valued variable can take any values from zero up to two pi.

Well actually, t can take any real values,

but if we choose a zero value for t where exactly at this point.

As t grows we are moving in counterclockwise direction,

and we rotate the full circle and we return back to this point when t approaches two pi.

So, if there are no limits on t values from below and from above,

then with a groove of parameter t,

we make infinitely many rotations about the origin.

So, this is a curve which is represented in the parametric form.

Now, we can extend this concept of equation of a curve to n-dimensional space.

So, we can consider curve in

n-dimensional space represented by a system of functions.

So, this curve is represented in the form of ith coordinates or the point which

belongs to the curve is some function using Greek letter pi of t,

where t can take any value or is restricted to lie within some interval.

We'll be interested in so-called smooth curves, smooth curves.

We can call a curve smooth if each of these functions pi is

a continuously differentiable function of

parameter t. Let me remind you that in single variable calculus,

we called a function f of x

a function which belongs to the class of continuously differentiable functions.

So, f of x belongs to what?

To the class of C1.

If its derivative,

f prime is a continuous function.

Then, we'll call a curve in n-dimensional space which is defined

with the help of the collection of the functions pi of t if all of them-.

So, this curve is smooth if is a continuous and differentiable function.

Moreover, when we find derivatives,

all these coordinate functions,

and we write them down,

we'll avoid the case

when all these derivatives turns zero simultaneously.

So, that will be constraint is not a zero vector.

Then we'll call it a smooth curve.

Now, the tangent vector

to such a smooth curve is given exactly with this vector made of derivatives.

So this is a tangent vector

to a smooth curve.

It's exactly this vector.

Going back to the chain rule, remember,

that was the title of this topic.

If we are interested in finding

derivative with respect to t,

when here, we construct

the composite function made of the function which was defined earlier, f of x.

We substitute for x coordinates the corresponding pi functions,

that's how we get the composite function.

So, the total derivative

with respect to t can be found in a similar fashion as we did with the straight line.

So here, we get a sum multiply by pi prime.

But we haven't exhausted this topic yet,

about the chain rule.

Let's consider a function of two variables.

So, given a function f of x, y,

let's suppose that x and y are no longer independent variables,

they're functions of some other variables.

So, let's suppose x is a function of u and v,

and y is a function of u and v. So,

if we substitute both functions into the variables of the original f,

we get a composite function again.

So after substituting, we'll get a composite function,

it becomes x (y,u) and y (u, v).

Now, the question is how to find derivatives of

this composite function with respect to the independent variables,

u and v. So,

we need to derive the formula.

So the question is, how to find df over du,

and df over dv.

Actually, this is a theorem which we're not going to prove,

but I'll show you the idea with this prove along with the rule,

how to memorize the formula which is used for calculating these derivatives.

Well, this is a chain that was as earlier.

So, how to memorize that rule?

First of all, we can write total differential of the given function as early,

df by definition is df over

dx times dx plus df over dy times dy.

Now, in order to calculate the first derivative df over du,

we keep v fixed,

and we divide both parts of

this formula by the increment in u.

So that becomes du in the denominator,

and we continue to divide both terms.

Now, if we take the increment in u which is the same as du,

it's the same as du,

and we turn it to zero,

so, it's becoming smaller and smaller,

then first of all,

we can replace straight d's, here and there,

and also in the right-hand part of the formula

with a corresponding around d's everywhere,

and we have the right to do so,

because v value is fixed,

then the derivative in the left hand side becomes exactly what we're looking for.

So, we get df over du which equals df

over dx times dx over du

plus df over dy times dy over du,

and we've got the important formula.

Also, if we're interested in finding df over dv, we'll do the same.

So, we will take the total differential formula,

we divide both sides by the increments in v. Now,

this increment will turn to zero.

If limit exists, then we get a formula

which very similar to the previous one with only difference that

we have df over dv equals df over

dx times dx over dv

plus df over dy times dy over dv.

Also, I'll enclose it in a box and quite

soon we'll use both formulas in order to find some interesting results.

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