Bu ders doğrusal cebir ikili dizinin birincisidir. Doğrusal uzaylar kavramı, doğrusal işlemciler, matris gösterimleri ve denklem sistemlerinin hesaplanabilmesi için temel araçlar vb. konuları içermektedir. Ders gerçek yaşamdan gelen uygulamaları da tanıtmaya önem veren “içerikli yaklaşımla” tasarlanmıştır. Bölümler: Bölüm 1: Doğrusal Cebirin Matematikdeki Yeri ve Kapsamı Bölüm 2: Düzlemdeki Vektörlerin Öğrettikleri Bölüm 3: İki Bilinmeyenli Denklemlerin Öğrettikleri Bölüm 4: Doğrusal Uzaylar Bölüm 5: Fonksiyon Uzayları ve Fourier Serileri Bölüm 6: Doğrusal İşlemciler ve Dönüşümler Bölüm 7: Doğrusal İşlemcilerden Matrislere Geçiş Bölüm 8: Matris İşlemleri ----------- This is the first of the sequence of two courses. It develops the fundamental concepts in linear spaces, linear operators, matrix representations and basic tools for calculations with systems of equations. The course is designed with a “content based” emphasis, answering the “why” and “where“ of the topics, as much as the traditional “what” and “how” leading to “definitions” and “proofs”. Chapters: Chapter 1: Place and Contents of Linear Algebra Cebirin Chapter 2: Learning From Vectors in the Plane Chapter 3: Learning From Equations For Two Unknowns Chapter 4: Linear Spaces Chapter 5: Function Spaces and Fourier Series Chapter 6: Linear Operators and Transformations Chapter 7: From Linear Operators to Matrices Chapter 8: Matrix Operations ----------- Kaynak: Attila Aşkar, “Doğrusal cebir”. Bu kitap dört ciltlik dizinin üçüncü cildidir. Dizinin diğer kitapları Cilt 1 “Tek değişkenli fonksiyonlarda türev ve entegral”, Cilt 2: "Çok değişkenli fonksiyonlarda türev ve entegral" ve Cilt 4: “Diferansiyel denklemler” dir. Source: Attila Aşkar, Linear Algebra, Volume 3 of the set of Vol1: Calculus of Single Variable Functions, Volume 2: Calculus of Multivariable Functions and Volume 4: Differential Equations.
Çözümlü Problem 1: Kutu Fonksiyonun Fourier Serileri / Solved Problem 1: Fourier Series Representation of "Box" Functions
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Na lição
Fonksiyon Uzayları ve Fourier Serileri / Function Spaces and Fourier Series
Ministrado por
Attila Aşkar
Prof. Dr.
Hello.
We have achieved our overall structure of the previous session.
Now if we do a variety of examples.
Our first example; This "box function" We'll see just how well we said,
A function defined in JOIN with 0 to 1;
such that 0 1/4 up,
From 0 to 2.1 4.1 0.25
value between 1 and 1 1/2 in value between 0.
Now we have seen four options; this function directly
right sinus and express my kosinusl general angle
or that range by extending symmetrically range
symmetric function and therefore can be illustrated by taking only the cosine,
it's almost the equivalent of taking the sine antisymmetric extension
be shown, as well as a general representation of my true,
this time as the sine and cosine functions with real
JOIN with 0 to 1 range of complex exponential representation again.
Now let them function means
it; 0 to 0.25, wherein the first,
up to 1 and a half after the 0.
Why do we have such a simple function?
Two reasons, actually functions in this way digital communications
You are creating the rectangle so arranging them side by side, this range also
If you choose a very small well, a function can get.
The second reason, of course, under a simple function Let the account
I do not miss it reviews how she drowned.
We'll do four kinds of complex here, because there are three here
it is also not possible to show the real value of the overall representation of my wife.
Now when we get to see that this function between 0 and 1,
at the same thing from all of the symmetrical JOIN with 0 to 1,
JOIN with 0 to 1 in the same thing, in the same complex.
But we do not only with the general representation of this range only
If we do so without stretching any periodicity of sine and cosine
We found ötelenmiş of this function because devinimlilig out of it.
But this function is always the same in this region.
But we're doing it that our symmetric extension of me
JOIN with 0 to 1 between the part where we take care of artificially -1'l 0
we have the symmetry between it, it's such a function.
No illusion that it is going beyond what we can deal with it other than to say the region
We're not interested in this function mostly outside of it
whereas here the displacement of a two-period that period, this important difference.
We're doing the same thing in the antisymmetric extension, again between the JOIN with 0 1
in the same function, but it -1'l 0
antisimetrig so instead we extend the plus and minus values,
where each x value on the value of the function here comes the same,
The minus sign here that the function of each x-value -x'li
We arrive in the coming antisymmetric function -1'l 1 for him.
Other than that, we think that what is happening so antisymmetric function
but this is happening shifted all three functions in all three display also
We have the same function between the JOIN with 0 to 1.
In this same complex functions but do not draw it here
We work going on, but the complex variable.
Therefore, this option will see the opening of four series of this function.
First this.
JOIN with 0 to 1 range, never without something like an extension etc.
period, therefore p = 1 2p = 1 1/2 192 look here.
the general formula of pages, the general formula for the Fourier series
We need to show that we look for the function.
Something to be aware of where 2p / p were below 1/2
It comes up because it is below 2,
We will see in others p = 1 will not be here in two important
This accounts for the difference, but the results are always going to the same place.
What we do in this f (x) we take the integral f (x) to kosinüsl
We're getting hit the integral f (x) we are getting hit the sine integral.
Because we have chosen a very simple function from 0 to 1/4
JOIN with 0 of 0 means that part falls as 1 / up to 4
1 / 4th of a hard sheer size between 1/2
See you say f (0), such as the fact that we chose 1,
it is of course that integration of the integral of x gives x 1/2 -1/4 1/4 data.
0 threw the EC again this integral in the place where we keep the non-0,
in this way a sinus tactics.
Now here's become attached to them just the way you'll see a lot of numbers
important to see.
F we're getting things done for the integration itself, we're getting hit with kosinüsl,
We're getting hit the sinus.
You will remember that in the beginning and if you look at this page 2
1 / p has p = 1/2 is a 1 / p the above 2
while coming in here 1 / 2p 2p 1 because it had one staying here,
See these formulas will be visible, but it can also feel the need.
Integral to this is very simple functions we take off
We can make this integration is coming this way.
Sine cosine integral, since the sine integral minus cosine
See here comes obviously the denominator coefficients.
This is an important thing because it grows short denominator for the coefficient k that is getting smaller,
already already shrinking in the series to converge
We have to k increase, and it shows.
Now there's benefit in the regulations as an account.
Let's look at the current structure where we have little EC
When a 1 / K characteristics of the term has a pin in it there is also a separate denominator
They are always here in order to keep the numbers are only 1 sinuses.
We take these sinuses because of this 2 1/2 above
k p x 1/2 time sine we provide remains,
also take a short 1/4 sine pi / 2 remains.
As you can see here the sine function k 1 2 3 4
He brought five stops each time you come to the neck and over again.
With the number 4 is composed as follows:
If data k 1
sinus p 0 but sine pi / 2 1; said
If we draw a circle here so
Let's draw a circle,
this apartment pi / 2
As we can say that there is one value, but also in front of the North Pole
because it is minus 1, k 2, when we give here is 0,
sine sine pi where he 2p 0 0 0 see here,
3 When we give again this term 0
but 3p / 2-circle back when you come to the South Pole that values -1,
1 because it is a minus in front.
4 time sine we provide perimeter 4p 0
You start wandering twice came to his sinus 0.
Here is a time when we decided to circle because of 2p 4
You still wandering sinus catheter 2p 0, 0 again, as you see here.
When we look at B'li cosine are here, there
numbers also give you time k 1
cosine of pi -1 on the west side of the circle again.
Cosine of pi / 2 in the north, the northern point of the cosine of 0,
so here comes 0, -1 happening here comes one.
See when we decided k 2
cosine of pi comes from here, cos p -1,
We have k 2, cos 2 p 1,
here there is a -1, -1 so there's a sign on the front here; -two.
3 When we arrived the cosine 3p / 2 at the south pole 0
cosine of pi three
If a minus because once you turned,
You went to another pin, and welcome again to the west, it also becomes a value.
When you give four, where two cosine of pi, circle once
When you go around the circle eastern ucund here, a cosine.
Four pi circle you wander twice,
Welcome to the eastern end there is also a gene comes from a negative one to zero.
These coefficients are standing longitudinal repeated.
But this brings short reducing the current denominator.
Hani enough to get air to the account details.
Now we know that the coefficients, we found.
Coefficients here is this great little chord
It is obtained by dividing the current KPIs.
see also see this great divided by KPIs.
Now, after we put them in, see where the numbers minus one, zero,
one meant zero.
See here minus one, zero, one, zero.
Likewise, the negative one, zero, one, zero being repeated.
There were times the k p in the denominator, we got out of the common pie.
k one, two, zero multiplied, but it is wise to put it systematically.
Two, three, four, five, six, seven, eight going this way.
If we look at similar situations in the sine-sine.
There coefficient of one, minus two, one and zero is coming.
Now when we look at here are a sine,
minus two, one, zero, one, minus two, one,
Go this way, we see that at zero but each KPI is still divided.
Taking Pier common k1, we divided one.
k2, we split, we split three, we divided into four, we divide by five, we divide by six,
We split into seven, were divided into eight, going this way, it goes on forever.
These are compiled and they sum up some zeros falling.
Here we organize properly coefficients we get such a number.
We will see this chart.
This account is sufficient to follow the outline.
When there is a need to sit or be made ready formula,
software is used, based on the fast Fourier series.
But it's important to find out whether it is what lies beneath,
You can not receive more than one benefit or closed box.
It opened without knowing how the series.
The second representation, between us, we had a reset but now we
If we extend it symmetrically remains only a minus cosine Fourier series.
So we can only work with cosine, this is an advantage.
Because not only you will find both a both a and b.
Important change is happening.
In a previous pin, a split, two were, because we're getting a reset in between.
a b, b eksi a, 2pi'ydi.
That one, here also note here is that if you pay attention and pi'y
You have not got a factor of two.
Because p is a divide that the denominator for two, up comes as the two share.
Here it extends the period therefore p is equal to two.
As you can see this because no two front where k is pi,
If you divide p p a.
This series happens.
This means that there is a significant difference between the two.
Yet in the same way that we take for the integration of the FX.
We stood with FX cosine, we're getting the integration.
No sinuses.
Because we extend symmetrically see coefficient was zero.
Here again, because the function of easy integration we chose to do here is very simple.
A reset function is zero divided among the four,
means that fell integral part of it.
One of the four was split up into two parts divided by a one or f0'l also multiplied, then he entered.
One half a pardon from zero up to zero, no part of it.
So this is a very simple integration integration was one of the integral.
We do.
Here likewise integral f FX zero zero
threw one over the remains of up to two x's part of a split of the four.
This fx fixed it together.
Only remains cosine.
We do now integral to this cosine, sine tactics.
Limits of four to one half of a split.
No b, so this advantage.
I would like to repeat a previous,
k is 2k where I had here.
These flows are still sinuses see here the same way.
They counted on the order and the periodic
It will be for a small number of them.
So when we put one half instead of sine x
k x pi divided by two instead of four, divided by a trailing four when we put k p.
For example, when we get a k, pi divided by four to 45 degrees,
The sine and cosine of 45 degrees at the root of two divided by two.
That occurs where these coefficients and
When you grew up, this time 45 k
degree is not 135 degrees,
225 degrees, 315 degrees, such as value goes.
So it comes minus a trailing minus root of two divided by two plus sometimes.
If you look at them calmly sit down, draw a circle like this.
Keep track of these numbers available.
Cons easy to write these numbers on a missing term negative alpha,
If a cross hair of the term beta again here in this time period is eight.
When you come here because they see it four times for four but is coming.
Here's coefficient alpha, minus one, minus the beta so it goes that way.
If we arrange them in the series we get a series like this.
These are all clear, the alpha, beta is defined here to mean the series
We're going to have obtained.
Again, the function given to us again, but this time it functions
If we extend between a reset from a minus, but we reset the artificial,
antisymmetric this time we extend the series will be composed of only the sinuses.
This again is an advantage both in general a'la B'La to work instead of just
You are trying to be.
Is there an advantage compared to the previous one?
Not much of an advantage but it can be a feature of the FX function.
He may benefit from maintenance here.
It is useful here as just like the previous series of the cosine.
Asymmetries will function itself, then there is nothing you will ever make.
Directly in this manner, working in this range.
See k sinuses are still here.
There were also short cosine.
But there was general 2k series.
This is important because these are the pin one here, between minus one,
between a general display zero.
Because of this difference comes from him.
Now this far, check account of things can follow this interim.
We also will see the çizitl.
Because here comes 45 degrees multiples.
To make a circle because it is eight multiples of 45 °
I need time to raise k.
He then stops himself lengthwise again.
This accounts for an important embodiment to perform properly.
I've already done them in a very simple way.
It was performed using Excel program.
No need to go to a fast Fourier transform.
Because these functions are very simple.
Considering the complex representation of the function to the forces there,
f you will multiply by pi to over supply.
Here again, there need to pay attention to these things and so on,
We have to look at the details of these accounts, rather than the course.
eg was.
FX we get a split among four reset to zero again.
We take a divide between four and one half.
This multiplication yapınca we get these things.
Now let's try to figure out where the result here.
One of those obtained from this first seen way cosine sine series.
Taking a single term that is a sine,
Taking becomes a cosine function in this way.
As you can see something quite rude.
Zero, one divided by four, two and one-to-one split
It is the first of two terms as trying to emulate a little bit, but quite distant.
If we take four terms.
See gitttik she began to approach a little more.
Taking eight terms, also see quite a few simulations.
Discontinuity here because it is also an important thing here,
It's called the Gibbs phenomenon.
It also passes through the discontinuity in a good way,
He is trying to zoom in.
See here is not exactly zero, but soon begins to be exceptional as well.
Learn more here see as well reduce the term here, where small
little things happening and continues to converge, as you see more and more.
You take more terms, when you get 252 terms here
I did it, I enjoy it, this is almost the same.
How much you get for that term will approach this series is so good.
These sine and obtained from kosinüslü.
If we look at others,
obtained from the sine and kosinüslü and periodically moved.
This is because the kosinüslü see only moderate zero,
this is a divided four symmetrically divided by a four again.
There is a split when you move back four but a split comes four here
Or, this is divided by four and a half here, a trailing four again,
half and a covering, but it looks like the two you see the differences,
This symmetrical, the starting point of this is it.
Here antisymmetric getirelim,
display and six or 16 terms here
see also increasingly taken by converging that you see
reset here and in fact we are interested in.
This is done in order to highlight the difference between something that just happened.
We see that similar to each other Side rapidly up and down.
There are minor differences, but you taşdıg out of time on the difference there.
Here, coefficients are given.
Blue aka coefficients, red also see coefficients.
As you can see the numbers growing, k where k is increasing steadily.
one of k, two, three, four,
five goes this way, we show here just to 16.
As you can see factor k is going down due to the denominator.
This is quite typical behavior.
Now I want to take a break here.
After that, we will examine it and discuss different functions of a second.
But the transactions also more or less the same.
Again we get the integral, integral to hit the kosinüsl,
We will hit the integration with the integral hit the sinus or exponential function.
These are the issues we are interested in an additional exercise.
Bye now.
In a second sample, and following him on the course this course a
real life example of a sound wave from a machine part,
We could put the vibrations of an earthquake part.
Fourier transform how the vibration of a molecule of
we will examine it and see how important information out of here.
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