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.
Gibbs Olayı, Yakınsama ve Taylor Serileriyle Karşılaştırma / Gibbs Phenomenon and Convergence in the Norm
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Doğrusal Cebir I: Uzaylar ve İşlemciler / Linear Algebra I: Spaces and Operators
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Koç University
25 classificações
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Fonksiyon Uzayları ve Fourier Serileri / Function Spaces and Fourier Series
Conheça os instrutores
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Hello.
So far in our study,
We have seen how to do calculations with Fourier function.
We saw where they came from, a variety of Fourier series that
sine and cosine both general ones, are only cosine,
We have only sinus or exponential calculations with complex functions.
We have also seen several examples.
However, until now we did not ask the following questions: Does every
Is there a function of the Fourier series?
Fourier series will be an important issue by the convergence of infinite series.
Is not no setbacks in Fourier series?
Some of the examples we have had in discontinuities.
Discontinuities in a function that is really desirable,
Not something cute, but sometimes we can not avoid.
We selected the Fourier series on discontinuities related
an artificial jump will occur.
But that's not the strangest thing too, because you already have a jump in discontinuity.
Fourier series, Gibbs also added a splash of these things are happening more
says emissions.
After seeing them, nevertheless after seeing an example
Taylor Serileri'yl a fundamental Fourier series
section with a series of analogies comparing the differences
Fourier series will finish the bet.
This will be our last session on Fourier series.
As I said just now, if a discontinuity function
if, for example if there is a discontinuity at a point x equals,
function will find a limit when approaching from the left,
We will find the right approaching the limit.
Fourier series and the continuous sine
the cosine trying to produce this discontinuity.
It will have an observation he passes through this discontinuity.
General cosine and sine series that asks, should the only kosinüsl
or just get sine series or series exponential always
the median will pass before these discontinuities as a visual
We will see, we will see later in an analytical way.
We examined a sample of this event ever on a
Take the example.
You will recall in this box from scratch, split up to four functions
the value of the function is zero, one to jump on a trailing four.
Half a split up to a value of up to two hard going, after a again
splashed with drops to zero and remains at zero until the end of the range.
That is a function that resets a line between the two periods.
These black lines as we have seen again
the function of this series, the discontinuity
including how to approach this function, it shows how close.
n equal terms refer only to remember if you get a little bit of box function
There are many differences, but one approach works because it is a very small term.
See if we can function a little better approaching four terms.
Taking eight terms till she gets closer.
We have 16 pretty in terms of approaching the box function.
We stopped here because the main idea
conveys but if you get too close this time to more thoroughly term
wherein these oscillations, errors will be well reduced.
Splash it goes like this: See, over there more specific.
4 x 1 split of the function when approaching continuous
cool the value generated from roughly minus zero function,
less than a comma as two zero two, making a splash with a comma.
After doing a bit of such a release,
When we get a lot more smaller release terms.
Then again, in this case a trailing one point,
one point two zero to minus one,
zero point two huge decline, going down to release.
This event is called Gibbs.
Obviously this discontinuity near discontinuities
completely not able to show continuous function,
some extra jumps, consists oscillations.
I'm a little more irdeliyel this instance.
We have examined in detail before it 210'unc page.
Fourier series representation of the overall function of this box,
so that both state and cosine and sine on.
We have seen that this series came the following structure there.
Of course, you do not need us to take here again, we see copying from there.
A scratch from a split of a four,
After a split pin had kosinüsl terms of this infinite series.
Again, there goes an infinite sine series with a split pu.
As you see here, there are two works,
not four, because four coefficient of zero, you will see if you look back to this series.
Three that come three times two pi comes six, it's worth it.
So when he went four times to eight zero two pie again.
This series here, as we have seen,
It's like before we give a lot of detail in 210'unc a page,
zero, one, zero, but a split in the denominator
k is a split two for one split of its three, one divided by four, one over five,
A split is going under, but that does not seem to zero in some yellows here.
Similar situation in sine.
See here, two there, four here, six there, not eight.
That is four times two pi coefficient is zero.
After five five times, there are six, seven have coefficients of eight zero again.
In such a series,
This situation brought about something that we can not give account of our decision.
Now here we will do the following: Put one half instead of x
We know that a split in the two discontinuities.
What Gets a value that will cost.
Visually, even if we all have a term of four terms as if we,
It takes eight terms, if we pass near the 16, roughly half term.
From left to right limit is a limit of zero function,
therefore, a divide in this case four, one half of that split the two gives.
PU split, a split in two is more than a right from the limit,
left from the limit of zero, plus zero, half of it,
It is giving it two in a divide and visually see it.
Analytically in our account x
a split or a split of the four we have put in place two
When this function, we calculated the value of the series.
Before we start with a little bit more because it is divided by two simple calculations
Let's say here: X instead we put one half time
cosine of pi is happening here because x denominator instead of one half of the two,
denominator that takes two cosine pi is going on, there's a minus sign in front.
Yet when we put a slash instead of two to three per x happens, three cosine of pi.
On the plus sign in front of her in that way we carry.
Here we put the two of them split a pita going five pi,
We are carrying with him a mark in already number in the denominator,
three, five intermediate double the number of zero
The numbers in the denominator due to expansion.
The number of k divided by a split pin comes even number k.
Now we come to the sinuses when we look at here,
x instead of a split, which we see happening here two times sine pi.
If you think trigonometric sine pi circle,
You start from scratch when you go to the west side pin angle is happening,
The E ratio means that the sine of zero is zero.
When you put a slash instead of x in the second term it is two pair of two sinus.
Sinus complete a full circle of the two pins
The time has come to eastern end you are trigonometric circle.
Sine there for all to all sinuses all these terms is zero zero
We see, all of these points in one half sine term is falling,
it is just that kosinüslü cost.
We hope to see this continue to do so in her sine terms.
Now let's put the value thereof.
Cosine of pi, minus one.
Because the western end of the trigonometric circle.
Beginning in value because it is a negative going 1.
Welcome to the western end of 3p still wandering circle minus 1 again.
It is because it is plus or minus sign.
This mark 5p, again minus 1.
Mark comes in changing that way.
See the divide where there was a 1 4 0 term this is coming from.
The others came from other than a and b k a k.
We download 1 over 4 terms here.
1 have pi over here.
We seemed to favor these numbers is now open and very interesting
See the situation we encounter an odd number of plus or minus plus
It is a collection of plus or minus minus.
The only numbers in the denominator can write very easily in this series.
We take the k-th power minus 1 to plus or minus to determine.
k 0 is the time when one is minus 1, plus or minus goes.
We put the denominator 2k plus one.
k is equal to 0 when the denominator in the 1 1 has the denominator.
This one produces.
When we first put k instead of minus 1 to 1.
A minus sign comes from the force.
K in the denominator 1 to 1 divided by 3.
Means that we have achieved in this way the series.
Perhaps not need much in the deal but did not want to deal with this issue
I'm here, I see a very beautiful mathematics.
Perhaps your most will love it.
You look like a lot of very higgledy-piggledy terms
s uncluttered layout structure emerges.
Instead of a regular structure emerges x 1 divided by 4
When we put this one had divided by 4 here.
See, the interesting thing is happening in kosinüsl term.
1 to 4 will be divided by 4 in the denominator when we put it,
cosine of pi over 2 stays here.
The PI divided by 2 again, the north pole of the trigonometric circle.
This arctic cosine 0.
When we get the next term is 3 divided by 2.
This spot at the south pole.
Here cosine 0.
It turns out that all the cosine of 0.
Because either the North Pole or the South Pole,
or it is happening in the North Pole or the South Pole.
1 over 2 he had fallen in the previous point for all sinus terms.
Here, too, are falling all kosinüsl terms.
It is just in terms of the sinus.
1 x instead of sine in terms of over 4 sheep,
sine pi divided by 4 pi over 2 stays.
Sadeleşip pi divided by 2 and 4 in the denominator denominator remains 2.
2. The denominator because the terms are 4 4 there,
sinus P remains.
Similarly pi divided by 2 range is going.
Yet there is an interesting situation.
Of course sinus p 0.
On the western tip of the trigonometric circle.
sinus p 0, sinus 6 p 0.
In contrast, the north pole sine pi divided by 2 plus 1.
Sinus 3 pi over 2 minus 1 in the South Pole.
As you can see here on the same terms as the previous one place it was 0.
It was negative, minus 1 divided by 3.
We still get a very different structure to the same series.
Therefore, this series will also do the same in this way.
Now we can just ask the question, "Do we know the limits of this series?" He said,
something that is very elegant.
Again, this is a very simple result emerges from chaos.
Some of you may remember the arc sine term,
Taylor arc tangent function of Pardon
if you found value in the first series Reveals open the same series.
Reverse odd number of plus or minus of plus or minus
Tangent is divided by the sum of which 1 is the angle at the point pi 4.
Because the sine and cosine is equal to pi divided by 4.
Tangent 1, arc tangent 1 pi divided by four.
So see how elegant structure that emerged.
This one had divided by 4, there are in both series.
Two series already arrived at the same place.
There have split a p 1 divided by pi.
Var p in the denominator.
We see in this series is pi divided by four.
Refer to cancel each other out when the battery in the second term, which simplify
time, initial pi over again sounds more as well as 4 divided by 1 divided by 4 1.
And 1 divided by the sum of the two is going 2.
The discontinuity that is already here now 2 for 1 split that visually
We have seen, we see it as analytical.
This accounts for some of you know is a teferuatl, some of you what it
I am excited about something so beautiful that I showed.
That image support, complement,
providing an analytical calculation happening.
The second issue is the existence conditions of the Fourier series.
That is, there is the Fourier series of each function or not?
In all of the examples we solve the Fourier series is now up to us
There were series and convergent.
We observed it visually, we may also find himself at the end of the series.
We are also here we take the general Fourier series,
just as the sine of kosinüsl, just as we take the exponential series,
This convergence will arise a problem.
This is so easy.
Now that we can make it through the general Fourier series.
This theorem shows that will prove right,
simple and elegant, and for that I want to do in a short proof of it
both series leads us to understand a little better.
He says this theorem, we immediately prove theorems,
Get this Calculate the integral f frame.
We opened this series between a and b.
Calculate these coefficients between a and a and b again,
Collect their square
and multiply the length of the B range.
There is a little difference in a 0 here.
1 divided by 2 does not come because you're doing it in front of one of the integral.
However, the integration of these are meant cosine and sine square frame,
1 divided by 2 is here.
In fact, this more than our simple geometry
expansion of the infinite space we know Pythagorean theorem.
What is in these respects those who prefer to call Pythagorean theorem of Pythagoras veyahut
He says: the square of the hypotenuse equals the sum of the squared edges.
If you interpret them the edge as if they were at work here, the coefficient of the rector
coefficients, if the Pythagoras theorem comments
as the sum of the square of the hypotenuse is equal to the carrying means.
This is important, more important than that regard, since if this is finite integral
will be equal to that infinite series, infinite series converges to say.
If this integration will be endless infinity, then the series from equal.
Thus, the series will be divergent.
Currently, he says, but keeping within the essence of this convergence.
Now let's move to prove it.
We wrote for x, we know FX here.
f x X. We have one more for shock.
Because the frames you need.
f x've written here, we wrote a fi x.
As we do, let's choose k indicator index in one collection,
We need to deselect k k is not because a certain number of the latter.
1 2 3 4 5 she goes forever changing.
Here k base say the same thing, but both give two series
k k with two of them have done well if we also have a large force,
We can not do that when the second collection.
You have to take it apart.
Now here are two terms here, there are two terms in the second f.
Here means that the first, and sinus of kosinüsl with kosinüsl customers
kosinüsl multiplied by the sine of the term sinus,
the product of the sine kosinüsl term future, including 4 terms.
4 of our term here, see cosine cosine,
kosinüs.biz the cosine and sine sinus sinus sinus
The need for the integral frame, so let's integration of the two sides.
When we take the integral of the left-hand frame work for the integral,
See if the integral of the right-hand side there's a product of cosine and cosine.
Cosine functions to each other just so short base
when it comes to equal k and the integral that occurs there cosine square 0
it would be minus a split integral becomes 2.
One exception is 0.
display zero frame and then k k k base because it is 0 to 1.
1 integration does not come from one split second.
Similarly, the product of a single term from sinus sinus also remains different from 0.
Because all of each other.
B squares comes from there.
Sine and kosinüsl terms, sinuses perpendicular to kosinüslü,
Whatever k and k base.
Here, too, the sine and cosine and
k and k whatever base they still have each other.
These terms and so integral to the self falls
we see that the sum of the squares of the coefficients.
Enteg one at the beginning, they have the length of the interval.
We can show said: This is the norm of the function f,
The length is a measure.
Here is the norm because of the cool sum of the squares of the coefficients,
i.e. the vector, the length of the hypotenuse edge of
theorem as a sum of the squares of equal length.
b minus a have the following: Because here's this for so DX frame
multiplied by a length.
The length thereof to which each of the repeat size in f
It needs to be multiplied.
Thus we come to a simple formula.
Just mad with this application, if these two integral
Or, these integrals are equal to the sum of two norms are equal,
bu integral sonluysa series of sonludur.
Thus, the series is convergent.
This series will be similar to the infinite this as an integral endless,
so it would be divergent.
Therefore Parseval's theorem says that by giving names
The convergence theorem we set
and it gives us the necessary and sufficient conditions for the divergence.
In practice of course this collection can not take forever.
Even if we can get an account on the computer doing large numbers.
We may have grown up in a place as we want.
But we can receive eternal.
So where marsupial pouch will occur if a fault.
So we would be a mistake if we take this series up to n from scratch.
It is this error.
Up to this collection from scratch n
If we take the difference with for the integral frame.
We also show a square frame of the norm because we are here,
For the length of the frames it.
It is that of the equivalent.
We were the last to cut it in the plus one
We neglect terms that means forever.
In this way, in a sense, we might find the review.
An example will clarify what we have seen by now.
We have seen before,
a function, box function, we examine the detail.
Why can function?
I have two reasons.
One can make a very simple calculation once.
Therefore boğulmaksız account of what we have made clear in the account details
as we can see.
In addition, digital signal communication marking
code in the box functions already in use.
Because if you calculate a function at various points,
You can not he calculated by the continuity of the computer,
You will calculate the discrete point, that he will give you the box functions.
Therefore, we examined the function box.
Accordingly, account is also simple.
In these pages we have seen in different ways,
We studied it in a reset interval.
Reset on a range of factors were at work in them,
now how much you need you to follow them?
Charter is not.
But here it is seen in the details.
The important thing to watch here thought.
We found these coefficients.
A from the coefficients of the cosine coefficient as seen from the table,
There was a small and were in sinus b are coefficients.
The numbers are the order of about 1.
These are short pie is divided longitudinally.
There is only one difference at zero.
These coefficients are going a divide dividing k.
See, there's a, there's zero, there's a, there's zero.
This quartet quartet revisited, cons and a cross hair as a cost.
b in a minus two, one, zero.
Again, this is repeated.
Here's a little more detail.
Here is a non-zero-zero, the first four non-zero-zero.
Non-zero-zero, non-zero-zero second quartet.
In this way it goes.
b counts the number of non-zero, non-zero number,
A number of non-zero-zero.
The first quartet.
Again, the first three is not zero, the fourth zero.
In this way it goes.
Now that we have our say white number,
see we have a number, we can calculate this series.
Reset integral function of f for a supremely simple because
divide among four zero, a value of between four to one half of a split.
A split in to two from zero, pardon me up a zero.
A divide that say only one that remains from the integration of a divided four to two
The integration between going.
It is also one half minus one divided by four, giving a split of the four.
This series also can be calculated.
be certain, a certain zero.
We found zero.
K are given here.
Therefore, these coefficients have collected by,
If you look a little more composed as a series.
This series,
This series can compare with the exact result.
Once the finite square integral of f out
here's the guarantee that the infinite series converges.
This one shows that one over four.
Now we find the square of f is the norm here for the integral frame.
What is the norm of f thus divided by the square root of it because we think a two.
If we multiply one half of a split with its own data four.
I mean, zero, five norms of these functions.
We yolks term, a term based on the number four terms,
eight terms, 16 terms, we take 32 terms, functions as you can see
not the norm in half this time approaching the series.
If you receive a term that comes naturally quite a difference,
20 percent off an error.
If we divide this f norm gives us the error rate.
If we take four terms error displeased, displeased we have eight terms.
The declining number of terms increases.
256 if you get one term,
512 if you get one term course these thousand will fall within a range.
See something very interesting, even if you take such a term honky subuk
a figure not quite expressive, you find a value close as possible.
This is significant in the following respects.
Usually the square of the fi function in practice
an energy or a contingency sum,
the probability, that things like data.
So these are numbers that mean anything.
Looking at you that you take even less remarkable terms,
you get a reasonable approximation.
This is a major strength of the Fourier series I can think of.
Series cosine and sine we've done here in general.
Others also did,
but it turns out that I've done because I did not show you the numbers are very similar.
Thus we are treated the convergence criteria.
for finite series is convergent if the integral of the square.
When can forever?
If this enteg, this function go to infinity in one place.
Mesale a split you take the integration of x squared,
This integration would also be zero to infinity forever.
Therefore, could the Fourier series converges.
Or it could go on forever somewhere in the middle, integration could be endless.
But take note of one over the square root of x weaker infinity in finite
will rise to a discontinuity for
and it can allow them to eternity Fourier series to converge.
This is the last closing note.
Thus we have completed our work on Fourier Series
Taylor Series will be but a very common one series.
A review of the differences and similarities with them
our lack of proof, I would put it.
Univariate Taylor series, you even see in multivariate function.
For example, one of the sine function, an exponential function more expansions
You need to have the issues you are experiencing.
We said in the Taylor series.
There is a function f.
Let its various derivatives, a x, calculate the zero point.
This k'yınc derivative is calculated when x is zero.
Now divide it with k factors.
This car get our k'yınc coefficient.
This we find in this way the coefficients c x minus x
We gather the force k'yınc hit zero zero indefinitely.
Here again, there is a convergence issue.
But it would be infinite in every series.
We also know that the convergence examination.
The functions of one variable that you look just the most mentioned
It is one of the topics.
When we compare it means that there is a similarity with the Taylor Series as follows:
Both have a base function.
Taylor series of x minus x is zero, zero many times Or, x
as possible to zero,
in the other, or exponential functions,
cosine or sine function is used as the base.
Taylor's team base in the function are not perpendicular to each other, while the other Fourier.
This is an important thing.
One of the most important difference,
We use derivative instruments to produce a series of Taylor series,
We use the integral of the Fourier series.
We account for derivative Taylor series at some point.
Here we are a range of accounts.
Therefore, Taylor series is a local series,
It has an opening near a point is a regional series of the Fourier series,
It is a series which opened in the EU range.
Taylor series, with discontinuities in the function,
It used in function to the infinite.
But Taylor series should be derived for each order.
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