Inleiding tot de Cauchy-integraleformule
De video bouwt voort op het concept van contourintegralen in de complexe analyse, waarbij eerder is besproken dat de contourintegraal van een holomorfe functie binnen een eenvoudige gesloten kromme altijd nul is (de Understanding Cauchy’s Theorem and Complex Integrals Explained).
Formulering van de Cauchy-integraleformule
-
Stelling: Voor een holomorfe functie (f(z)) binnen een eenvoudige gesloten kromme (C) kan de waarde van (f(a)) voor elk complex getal (a) binnen (C) worden berekend via:
[ f(a) = \frac{1}{2\pi i} \oint_C \frac{f(z)}{z - a} , dz ]
-
Deze formule geldt alleen als (a) binnen de contour (C) ligt.
Bewijs van de formule
- Definitie van hulpfunctie: (g(z) = \frac{f(z)}{z - a}), die holomorf is binnen (C) behalve in (z = a).
- Constructie van een samengestelde contour: Bestaat uit (C) met een kleine cirkel rondom (a) (met een 'opening' zodat integralen kunnen worden onderverdeeld).
- Toepassing van de Cauchy-theorie: De integrale over de samengestelde contour is nul.
- Onaanboorderlijke limit: Door de openingen te laten sluiten, worden bepaalde integralen elkaar opheffen door integratie in tegengestelde richting.
- Reduceren naar kleine cirkelintegralen: Die makkelijk in poolcoördinaten kunnen worden uitgerekend.
- Het eindresultaat: Laat zien dat de integralen langs (C) en deze kleine cirkel gelijk zijn, leidend tot de formule.
Effecten en toepassingen
-
Waarde bepalen binnen de kromme: Men kan (f(a)) exact berekenen als men (f(z)) kent op (C).
-
Afgeleiden berekenen: Door (f(a)) te differentiëren naar (a) binnen de integraal kan men ook alle hogere-orde afgeleiden van (f) binnen dezelfde kromme bepalen:
[ f^{(n)}(a) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z - a)^{n+1}} , dz ]
-
Belangrijke inzichten: (a) wordt hier als variabele binnen de kromme gezien, wat differentiëren binnen de integraal rechtvaardigt. Dit sluit nauw aan bij het begrip van Introduction to Functions of Complex Variables and Holomorphicity.
Praktische tips
- Kies een kleine cirkel rondom (a) om berekeningen te vereenvoudigen.
- Vergeet niet dat de contour (C) gesloten en eenvoudig moet zijn.
- Om de begrenzing van contourintegralen beter te begrijpen, kan inzicht in de Understanding the ML Inequality: Bounding Contour Integrals in Complex Analysis nuttig zijn.
- De formule is fundamenteel voor complexanalyse, met toepassingen in wiskundige fysica, engineering en dynamische systemen.
Samenvatting
De video biedt een diepgaande uitleg en bewijs van de Cauchy-integraleformule, waarmee complexe functies binnen de contour volledig worden bepaald via hun randwaarden. Dit opent de deur naar krachtige analytische technieken zoals het vinden van functiewaardes en afgeleiden zonder expliciete kennis van de functie binnen het gebied.
If You Remember The Last video we Spoke about Contour integrals And how the Contour integral of a complex function
hom morphic Inside A Simple closed Curve c Is Always Zero this result Is What We Call c theorem but There is Another
Perhaps even More useful result that involves c and Contour integrals It's called the c integral Formula Here is
how it works If I have a differenti holic comple function f z Inside Simple Curve cve that aite Number of Corners
then I can find the value of that function at complex number a By The Following integral As Long As that
complex number a is located Inside The closed Curve c so My a is Inside The closed Curve c then I can find f a using
this integral Formula however If My a was outside the Curve c I'm integrating over then I can't use formul Why because
According to theor the cont inte Z once again I'm man up and Give You A Proof of this Formula begin by Letting a Be A
Point on the complexe Inside Simple closed Curve Let define A function G given by g z is f z over Z min a Where f
z is of course Our function That's holic on Inside c Let's use this diagram of what we have So Far in the we have a
point a enclosed Inside some this to make it More Clear add let draw Another Curve Circle of Radius R Around
The Point a but we Won't actually complete The Circle Leave a small gap at one end with the circumference Between
The Points B and e also make a gap in the Curve c and label The Two Points on either End of the gap a and D Now let go
back to my function G of that I Def Here Remember f is and f and G is that g z a term denin
What This Means is that G is also holic in the entire region Inside Z except for the point Where z equ a Where It's
undefined clearly The integral of G along this Path Which I'll Call w from B to a then from a to D Along The Curve c
then e And then finally from e back to B Along The Circle c Prime is zer by c's theorem Because The Point a is outside
the Area enclosed by this Whole Path we can split up this Contour integral in the chunks with each chunk representing
a single Path so Let's use Different Colors to make things easier since The integral of G along this Whole Path is
Zero The Sum of the integrals of G of Z from B to a from a to D from d to e then Back from e b is zer I Made this gap
earlier and I it so that later on I could eventually reduce that gap to Zero So What happens when I reduce this gap
Between The Line segments a and de to Zero so That a and de Now overlap well These two Contour integrals along These
segments become Contour integrals over virtually the same set of Points since The segments now overlap the only
difference is that the integrals occur over opposite directions it's easy to see that Contour integrals over the
exact Same set of Points in opposite directions would cancel out and That's What happens Here Now When The gap Here
vanes this Red integral Just Becomes an anti clockwise integral Along The Curve c and this Brown integral Becomes a
clockwise integral Along The Curve c Prime and So This is what we left with the closed integral anti clockwise along
c of G of Z DZ plus The closed integral clockwise along c Prime of G of Z DZ is zer Which means that the integrals of G
of Z along Both and Prime in the Anse are the Because The clockwise integral is nega of the an clockwise So
You Plug that in you get this equality Now That We proven that the integrals Along The Composite Curve and the much
simpler Circle are the same we can go ahead and work On The Circle and use that to make conclusions About The
integral More complicated Curve Because We working on Circle we can use Polar representation Of Numbers to write
Circle Just Z Which is function of Th equ a+ r e i theta Where theta varies from zer to 2 pi This a is there because
It symbol The Center Of The Circle and so That's Where You have to start your Z now because The Angle theta is the only
thing that varies along a fixed Circle Because of course Its Radius is constant It's Center is constant we can write The
differential DZ As I e d by differenti the equation of The Circle with respect to Th going back to
Our integral we can write g z as f over z a According to definition then we can use the equation of The Circle c Prime
Which We Just Found to rewrite z a and DZ in terms of Th in addition change the limits Of The integral From to 2 because
now we ch variables Circle some Common Us pi f z
i Now We Can choose Our Circle to be really really small so That We have a really really small Radius Doesn't
Matter How small The Radius is because It Ends up Getting cancelled Out Anyway so as we Make The Circle c Prime really
small f zesi value f as we radi circ small value of Fun f z on Circle f a
that approximation Becomes More valid The smaller The Radius so The closed in the Anse along c f z over Z D is
integral f a d that f CH on the one over the Curve c because
As you can see up Here The integral along c and c Prime are the same so I just did Two Steps in one doing some
rearrangement will Leave Us With The Final result and so we proven form but how do we
this this form means is that val of holic f on closed Curve c then I can find the value of f at any Inside The
closed Curve using this equation in other Words by Knowing What f Looks Like on a boundary the only thing I need to
perform the Contour integration over the boundary I can automatically DCE What f Looks Like Inside The boundary Now If I
Take This equation this c integral Formula I differentiate it with respect a n Times I can Event Ach with ispr for
value of f Any Inside CL Curve c and this makes Sense because I can use form to What My
function Looks Like I can also use it to find What n derivative Looks Like If I Take the derivative on the right and
Move It Inside The It Becomes partial deriv I don't know if you heard about this but called differentia Inside The
Anyway Take deriv respect A Can Use basic formul to n der of f can be taking integ of f z
over z a n+ m by n factorial over pi you Might be wondering How Come differentia With respect to a constant number is
that Kind of weird well the answer is that a Isn't really a fixed number It can represent Any Point Inside The
closed Curve c so in this context It's basically variable The formul We Just derived is like formul the
that exc the Any that do this video Next video very The res
De Cauchy-integraleformule is een stelling uit de complexe analyse die stelt dat de waarde van een holomorfe functie binnen een gesloten kromme berekend kan worden via een contourintegraal langs die kromme. Hiermee kan men functiewaardes en alle hogere-orde afgeleiden binnen dat gebied exact bepalen zonder de expliciete formule van de functie binnen het gebied te kennen.
Door de integrand ( \frac{f(z)}{z - a} ) te differentiëren naar (a) binnen de contourintegraal, ontstaat een formule voor de (n)-de afgeleide: ( f^{(n)}(a) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z - a)^{n+1}} , dz ). Hiermee kan elke hogere afgeleide berekend worden zolang (a) binnen de gesloten kromme ligt.
De contour (C) moet een eenvoudige, gesloten kromme zijn waarbinnen de functie (f(z)) holomorf is. Daarnaast moet het punt (a) waar de waarde of afgeleide wordt berekend strikt binnen deze contour liggen om de formule toe te kunnen passen.
Holomorfie betekent dat een complexe functie overal binnen het gebied complex differentieerbaar is. Dit is cruciaal omdat de formule berust op eigenschappen van analytische functies, zoals het kunnen definiëren van contourintegralen en garanderen dat deze integraalvorm geldig en convergent is.
Nee, voor de formule geldt dat het punt (a) binnen de gesloten kromme (C) moet liggen. Als (a) buiten (C) ligt, is de integraal gelijk aan nul en kan de formule de waarde of afgeleide van (f(a)) niet bepalen.
De formule maakt het mogelijk om functiewaarden en afgeleiden exact te bepalen aan de hand van randwaarden, wat nuttig is bij het oplossen van complexe differentiaalvergelijkingen, systeemanalyses en dynamische systemen in engineering en fysica. Het vereenvoudigt ook het berekenen van functies binnen gebieden waar directe evaluatie lastig is.
Kies een eenvoudige en goed gekozen gesloten kromme (C), bijvoorbeeld een kleine cirkel rondom (a), om de integraal te vereenvoudigen. Zorg dat (f(z)) holomorf is binnen en op (C), en benut de mogelijkheid om te differentiëren binnen de integraal voor afgeleiden. Begrip van begrenzingen zoals de ML-ongelijkheid helpt ook bij het inschatten van integraalwaarden.
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