Ders çok değişkenli fonksiyonlardaki ikili dizinin birincisidir. Burada çok değişkenli fonksiyonlardaki temel türev ve entegral kavramlarını geliştirmek ve bu konulardaki problemleri çözmekteki temel yöntemleri sunmaktadır. Ders gerçek yaşamdan gelen uygulamaları da tanıtmaya önem veren “içerikli yaklaşımla” tasarlanmıştır. Bölümler Bölüm 1: Genel Konular ve Düzlemdeki Vektörler Bölüm 2: Uzayda Vektörler, Doğrular ve Düzlemler; Vektör Fonksiyonları Bölüm 3: Düzlem Eğrilerinden Hatırlatmalar ve Uzay Eğrileri, İki Değişkenli ve İkinci Derece Fonksiyonlar ve Karşıt Gelen Yüzeyler Bölüm 4: Özel Yapıdaki İki Değişkenli Olarak Karmaşık Fonksiyonlar, İki Değişkenli Fonksiyonlarda Kısmi Türev ve İki Katlı Entegralin Temel Tanımları; Limit Kavramının Gerekliliği ve Anlatımı Bölüm 5: Türev Hesaplama Yöntemleri Bölüm 6: Türev Uygulamaları Bölüm 7: İki Katlı Entegraller ve Uygulamaları ----------- The course is the first of the sequence of calculus of multivariable functions. It develops the fundamental concepts of derivatives and integrals of functions of several variables, and the basic tools for doing the relevant calculations. The course is designed with a “content-based” approach, i. e. by solving examples, as many as possible from real life situations. Chapters Chapters 1: General Topics and Vectors in the Plane Chapters 2: Vectors in Space, Lines and Planes; Vector Functions Chapters 3: Reminders of Plane Curves and Space Curves, Quadratic Functions and Variables, Surfaces Chapters 4: Special Two Variables Complex Functions, the Basic Definition of Partial Derivatives and Two Storey Integrals in Two Unknown Functions ; Necessity and Details of Limits Chapters 5: Methods of Derivative Calculations Chapters 6: Application of Derivatives Chapters 7: Two Storey Integrals and Applications ----------- Kaynak: Attila Aşkar, “Çok değişkenli fonksiyonlarda türev ve entegral”. Bu kitap dört ciltlik dizinin ikinci cildidir. Dizinin diğer kitapları Cilt 1 “Tek değişkenli fonksiyonlarda türev ve entegral”, Cilt 3: “Doğrusal cebir” ve Cilt 4: “Diferansiyel denklemler” dir. Source: Attila Aşkar, Calculus of Multivariable Functions, Volume 2 of the set of Vol1: Calculus of Single Variable Functions, Volume 3: Linear Algebra and Volume 4: Differential Equations. All available online starting on January 6, 2014
Çözümlü Örnekler ve Ödevler
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來自 Koç University 的課程
多变量微积分 I：概念
16 個評分
從本節課中
Türev Uygulamaları
İki değişkenli sayısal açık fonksiyonlarla tanımlanan yüzeyde teğet düzlem ve diferansiyel. Zincirleme türev yöntemi ve tam türev. Yöne göre türev. Gradyan. Koordinat dönüşümü ve Jakobiyan. Taylor serileri. Kritik noktalar, en büyük ve en küçük değerler. Türev hesaplamalarının üç ve “n” değişkenli fonksiyonlara genellenmesi.
與講師見面
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Hello.
Our previous session of the Taylor series 've seen.
Starting from there, the main one variable We remember the concepts.
Two variables, even the three variables variable
Taylor series obtained in the same structure as I have seen can be.
Here I want to start with an example.
Whether a function like e Euler's number, to the exponential function
e to the minus x squared plus you get three years.
It is equal to x is zero, y is equal to
Taylor series achieve near zero We want to.
This is actually the McLaurin series
You can call it zero zero Area expansion.
Here I offer three options.
It using one of the main formula various partial derivatives
calculate zero at zero, the coefficients to achieve using the series.
This has some advantages, but some As in everything there are burdens.
A second method of separating these two exponents,
With so e to the minus x squared minus E y as the product of three
taking over, but we know u e to We know that opening.
If you say where u or u to x squared minus If negative at three years
their single variable denominated We know expansions,
therefore obtained by multiplying this series We can.
Again, a single variable function taking advantage of those in
e to make the definition, but this time u all these exponents
In terms selecting e to u again, ie, exponential
expansion of functions of one variable using this series can produce.
As you can see there are some future product.
We make them naturally, as expected such as
two methods give the same expansion will be.
Now let's do the calculation with partial derivatives.
Here, you know, given function.
Function that we have chosen zero zero the value at the
derivative of the function with respect to x is zero again at zero
i.e., the first derivative value to y order
partial derivatives, the second order partial derivatives are passing.
These three.
x-x, x-y, y-y by the partial derivatives basis.
Then we go to the third order.
derivatives with respect to x three times, two times x a time
According to the partial derivative of y, x in a time of two time to y
partial differentiation by three times to y partial derivative and
in all their zero zero
values to account numbers we find.
Now you see a few of them simply do it.
mean is that the partial derivative with respect to x; E over u to say
the term, these compounds function derivative, partial derivatives, chain in derivatives
As we have seen above to u by u derivative of u with respect to x derivatives in revenue.
derivative of e to u again to u the above, with respect to x
derivatives minus two x, as you can see minus two x's coming here.
You will find zero value at zero.
e to the x is zero, regardless of Whether the term is reduced.
Similarly, the derivative with respect to y to get compounds via this email will say again
function in the chain derivative rule As we have seen, to the next higher.
u is the derivative of y.
There are cons as you can see here three years.
x to y doing the hard task of partial For derivatives that we received.
Here comes a minus three.
When the account to zero over zero is zero, be one, because we find minus three.
When we come to the second derivative is now wherein the term minus
leaving out or how the two x If you want to get.
derivatives with respect to x, minus two x with respect to x derivative minus two.
Here you have a product as you can see.
Then the minus sign to those of u with respect to x over partial derivative, he brings a minus two x's.
Because we do here in this u x It was derivative, minus two x sounded.
He will give four x squared.
You can see it when he calculated at zero zero As we get minus two.
In this way, it's a nice exercise, the various order
To get derivatives, I would recommend Try them for yourself who do not.
Here are some others will be zero zero will come.
We take all of these places When there is one constant term.
There are two first-order terms.
But be zero partial derivatives with respect to x is fx for
X, the term, not only have y'l term, minus three times a year.
In the next term You will recall one half
factorial, a divided three-factorial The term comes from.
When we opened this term here divided
minus two times two factorial, see where two
There, x the square, but it was next to zero xy because it is falling.
After nine YY value.
This year is multiplied by the square.
After that one-third factorial We put here for terms divided
xxx'l in three factorial terms, ie x diced No, because these coefficients are zero.
x squared term y'l that you find here, As you can see six going on.
But I have three of them.
His eighteen come here.
second power with respect to x, y, first by force.
There are six such terms, but twice before xa the derivatives according to y
As we first x then y then We can get a more derivatives with respect to x.
Derivatives of x to y then twice before We can get and
they are equal in this The term is three times the future.
x according to the first, the second derivative of y
zero, so this term will fall, i.e. x No y squared term.
y we get the third order derivatives There's one of them.
How the third order derivatives with respect to x is a varieties are obtained.
Before you take the derivative with respect to x.
with respect to x, with respect to x.
Here, too, according to y to y to y.
No other kinds of options here.
Coming coefficient of minus twenty-seven.
This year the couple multiplied.
Other terms in this way then is going to be edited.
As the second method.
We know that the expansion of e to u.
That's a plus, plus a square divided into two factorial
plus the cube divided by three factorial, such that respectively, they are going.
If we say before the minus x squared, then ua
If we say that e to the minus x squared minus three of the We find the opening.
expansion to minus three y we can find.
Because here, the term e to the minus x squared times
In three years he multiplied to minus We have seen that we can show.
Therefore, the first expansion to hit here where y stands for the product minus three.
Now we need to edit them, of course.
Now to the other advantage of this There was no need for editing.
But there is also a lot to take partial derivatives I had to.
Here, as in real life, a
You can earn anywhere but in this price you need to pay.
There are more processes are doing At the beginning then going straight.
Easy to start here alone this I need to buy the product.
How will this product, constant terms will be.
There are only a constant term.
What else you çarpsa because this is the first X, will braces terms.
Then, linear, first-order term future, here are the only minus three years.
However, when only one hit We bring multiplying minus three y.
Then have squared terms.
See squared minus x squared here have the future.
And here are nine divided by the square of two years, he product coming from the merger.
These two will come.
So this is actually secondary shock, but as course in secondary school
because it can not do so quickly account They do not even do.
But here these terms a little patience You can sit.
Initially, human error can also through the eyes of a
escapes them, and then you look again, 'll get.
And when you make this initiative a natural As such it should be
Taylor coefficients of partial taking formula directly
we found using the obtained expansion We are.
There is a third way.
Again, the only variable u e to expansion of the exponential function
but here u using this open we want
cool so u accept the exponential minus x square
plus three years of taking u take here We put them in place.
Refer came here.
Where u came.
the x squared plus y plus a three Eskil
divided by the square of the two factorial, As you can see the square came here.
Then it came, but here's cube minus that came minus the front.
In this way we are moving.
Now you can stop it but it's full bi bi structure does not seriously.
Complicated because the terms here.
Here squared with the term bi-linear terms way.
If you look here, when you open it x The term will quadruple.
will be y'l to square one in terms of x y The term will be checked.
As you can see here squared term, diced
mer and fourth terms, very terms 's.
Here you will edit them and that's
When you edit achieve the same expansion You will.
Generally, of course, somewhere in this twentieth order terms,
you're going to order derivatives of the twentieth remain overwhelmed everyone.
However, the computer can be done, but maybe in general there is no need.
Did you know the most used third-order and the third order term are required.
Because we usually do investigations We examine around the point.
You will also see an application.
For that reason, there is nothing to be scared now so because,
such as taking too many terms will not be necessary.
The term another one we take three variables
X,, and this is roughly y'l and z'l just similar.
As a convenience wherein three I dropped.
X, with a term of y'l z'l term bi product.
Gene expansion around the zero point Consider
and how it will achieve them Let's see.
Gene servant three methods, we can get.
Partial derivatives can get.
So x and y, z The first several second order derivatives of the third order
derivatives thereof instead of opening them you'll get when positioned these terms.
To you, for the sake of convenience we accounts
I'm giving for the sake of providing these expansions is here.
Again earlier E, E as in opening u can open via a separate series.
A separate series of cosine of z You can open.
We know them individually expansions e to u stands for a
plus, plus, plus a one-half square
but you can see the cube divided by three factorial where the minus sign here as
u'l terms for the only forces that negative sign for the future.
Double the length minus the square of forces,
with future increases to the fourth power future.
Such an initiative would achieve.
If you do not know precisely what is in part z'l an opportunity to remember to do it
Have you seen a single variable functions cosine of z stands for something like that.
Here's the same result by multiplying them enunciating is obtained.
Gene could make something else.
After the old x squared minus y u
Select the first terms
e to the minus x squared the series, as A second group
terms as the expansion of e to the minus y, A third group of terms
whether the cosine of z obtained as expansions Edit your hit them when you see
them as more than simple algebraic operations deep bi
No, but this is a very useful concept method.
As I said a little while ago now ep, but to the many problems now,
second moment or did you know would be needed to force the third degree.
Twenty-order term here if you receive As you can see
very rapidly growing number of terms such occurs.
These are not very useful manual In work to be done.
No need computer as bi If it is not usually
Such a simple computer program takes bi, typing these coefficients are calculated.
What is important lies behind thereof The concept is to know.
Now here's to do with Taylor series I want the presentation ends.
The second critical points bi basic applications and local extreme values.
You will recall only the extreme values not local
There are parts of so-called absolute extreme values.
Under certain restrictions extreme values is there.
One of the extreme values of some integral There exist.
The second part of the course, which they very value,
derivatives of functions of several variables and integration at two we'll see.
After this seven-week course
also interested to follow that course of will open.
While there are more advanced applications
in other optimization problems will be seen.
Bye for now.
Thereafter to discuss.
