Logismoc II, Dunamoseirècyanath/teaching/calculus2/powerseries_slides.pdf0 +R den gnwrÐzw. P¸c ja...

Logismoc II, Dunamoseirèc

A. N. Giannakìpouloc

Tm ma Statistik c

O.P.A

Earinì Exmhno 2017

A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 1 / 25

Dunamoseir

Orismìc

'Estw {an} mÐa pragmatik akoloujÐa kai x0 ∈ R dedomèno.MÐa dunamoseir eÐnai mia sullog apo seirèc thc morf c

a0 + a1(x − x0) + · · ·+ an(x − x0)n + · · · =∞∑n=0

an(x − x0)n,

ìpou to x jewroÔme oti metablletai sto R.

Gia kje x ∈ R paÐrnoume kai mia diaforetik arijmhtik seir!


Pardeigma

H seir

∞∑n=0

1

n!xn = 1 + x +

1

2x2 + · · ·+ 1

n!xn + · · · ,

eÐnai mÐa dunamoseir me kèntro to x0 = 0.

An p.q. x = 1 paÐrnoume thn seir

∞∑n=0

1

n!= 1 + 1 +

1

2+ · · ·+ 1

n!+ · · · = e.

An p.q. x = 2 paÐrnoume thn seir

∞∑n=0

1

n!2n = 1 + 2 +

1

222 + · · ·+ 1

n!2n + · · · = e2.


Pardeigma

H seir

∞∑n=0

xn = 1 + x + x2 + · · ·+ xn + · · · ,

eÐnai mÐa dunamoseir me kèntro to x0 = 0.

An p.q. x = 1/2 paÐrnoume thn seir

∞∑n=0

1

2

n

= 1 +1

2+

(1

2

)2+ · · ·+

(1

2

)n+ · · · = 1

1− 12= 2.

An p.q. x = 1/3 paÐrnoume thn seir

∞∑n=0

1

3

n

= 1 +1

3+

(1

3

)2+ · · ·+

(1

3

)n+ · · · = 1

1− 13=

3

2.


Er¸thma

MÐa dunamoseir einai mia peirh sullog apo arijmhtikec seirèc -

gia kje pijan tim tou x ∈ R kai apo mÐa diaforetikh!

EÐnai dunatìn gia orismenec epilogèc tou x ∈ R h antÐstoiqh seir nasugklÐnei kai gia llec epilogèc tou x ∈ R h antÐstoiqh seir na mhnsugklÐnei?

An to parapnw ìntwc sumbaÐnei ja mporoÔsa na brw to sÔnolo twn

tim¸n tou x gia tic opoÐec h dunamoseir sugklÐnei?

Ja jela loipìn na qarakthrÐsw to sunolo

C := {x ∈ R : h dunamoseir∞∑n=0

an(x − x0)n, sugklÐnei}


To sÔnolo C eÐnai èna uposÔnolo tou R.

Ti morf ja èqei?

EÐnai èna sÔnolo apo diakrit shmeÐa?

EÐnai èna eujÔgrammo tm ma sthn eujeÐa twn pragmatikwn?

EÐnai mia ènwsh apo eujÔgramma tm mata?


Je¸rhma

Uprqei kpoioc pragmatikìc arijmìc R > 0 (h aktÐna sugklishc thcseirc) tètoioc ¸ste

gia kje x ∈ (x0 − R, x0 + R) h dunamoseir sugklÐneigia kje x ∈ (−∞, x0 − R) ∪ (x0 + R,∞) h dunamoseir apoklÐneigia x = x0 − R kai x = x0 + R den gnwrÐzw.

P¸c ja mporoÔsa na kajorÐsw to R ?

To R mporeÐ na eÐnai

Peperasmèno −M < R < M gia kpoio M,Mhdenikì R = 0,

'Apeiro R =∞


Upologismìc tou R

Ac prw èna opoiod pote x ∈ R, opìte h dunamoseir antistoiqeÐ sthnarijmhtik seir ∑

n=0

bn, bn = an(x − x0)n.

Ac efarmìsw èna apo ta krit ria sugklÐshc pou xèrw gia thn

arijmhtik aut seir, p.q. to krit rio tou lìgou.

∣∣∣∣|bn+1bn∣∣∣∣ = ∣∣∣∣an+1an

∣∣∣∣ |x − x0|,ra

limn→∞

∣∣∣∣bn+1bn∣∣∣∣ = |x − x0| limn→∞

∣∣∣∣an+1an∣∣∣∣ =: r |x − x0|


SÔmfwna me to krit rio tou lìgou h seir∑∞

n=0 bn ja sugklÐnei an

limn→∞

∣∣∣bn+1bn ∣∣∣ < 1.Apo ton parapnw upologismì autì ja sumbaÐnei an

r |x − x0| < 1 =⇒ |x − x0| <1

r=

1

limn→∞

∣∣∣an+1an ∣∣∣ .SÔmfwna me to krit rio tou lìgou h seir

∑∞n=0 bn ja apoklÐnei an

limn→∞

∣∣∣bn+1bn ∣∣∣ > 1.Apo ton parapnw upologismì autì ja sumbaÐnei an

r |x − x0| > 1 =⇒ |x − x0| >1

r=

1

limn→∞

∣∣∣an+1an ∣∣∣ .


Er¸thma

Ti ja gÐnontan an eÐqa qrhsimopoi sei èna dioforetikì krit rio gia

thn sÔgklish thc antÐstoiqhc arijmhtik c seirc?

Ja ebriska thn Ðdia diaforetikh tim gia thn aktÐna sugklishc R ?


Pardeigma

H dunamoseir∞∑n=0

1

n!xn

èqei aktÐna sÔgklishc R =∞, dhladh sugklÐnei gia kaje x ∈ R.

Autì shmaÐnei oti gia opoiod pote x ∈ R mpor¸ na upologÐsw thnarijmhtik seir

∑∞n=0

1n!x

n kai met na knw thn antistoiqÐa

x 7→∞∑n=0

1

n!xn

kai bsei autoÔ na orÐsw mia sunrthsh f : R→ R me eikìna

f (x) =∞∑n=0

1

n!xn.

H sunrthsh aut eÐnai h ekjetikh sunarthsh f (x) = ex .A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 11 / 25

Pardeigma

H dunamoseir∞∑n=0

xn

èqei aktÐna sÔgklishc R = 1, dhladh sugklÐnei gia kaje x ∈ (−1, 1).

Autì shmaÐnei oti gia opoiod pote x ∈ (−1, 1) mpor¸ na upologÐswthn arijmhtik seir

∑∞n=0 x

n kai met na knw thn antistoiqÐa

x 7→∞∑n=0

xn

kai bsei autoÔ na orÐsw mia sunrthsh f : R→ R me eikìna

f (x) =∞∑n=0

xn.

H sunrthsh aut eÐnai h sunarthsh f (x) = 11−x .A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 12 / 25

Pardeigma

H dunamoseir

∞∑n=0

n!xn

èqei aktÐna sÔgklishc R = 0, dhladh sugklÐnei mìno gia x = 0.

Sthn perÐptwsh aut den mpor¸ na orÐsw mia sunrthsh me ton

parapnw trìpo (para mìno sto shmeÐo x = 0)!


GiatÐ na endiafèromai gia tic dunamoseirèc?

GiatÐ me thn bo jeia touc mpor¸ na orÐsw sunart seic pou eÐnai pio

perÐplokec apo ta polu¸numa!

EpÐshc ja doÔme ìti ìqi mìno mpor¸ na tic orÐsw all kai na knw

logismì me autèc, dhlad na tic paragwgÐsw kai na tic oloklhr¸sw!

Pra pollèc apo tic gnwstèc mac sunart seic den eÐnai par

suntomografÐec gia sugklÐnousec dunamoseirèc.


1

1− x=

∞∑n=0

xn, x ∈ (−1, 1),

ex =∞∑n=0

1

n!xn, x ∈ R,

ln(1 + x) =∞∑n=0

(−1)n+1

nxn, x ∈ (−1, 1),

sin(x) =∞∑n=0

(−1)n

(2n + 1)!x2n+1, x ∈ R,

cos(x) =∞∑n=0

(−1)n

(2n)!x2n, x ∈ R


Prxeic metaxÔ dunamoseir¸n

Sto disthma sÔgklishc touc mpor¸ na knw prxeic me tic

dunamoseirec san na einai peperasmèna ajroÐsmata!

Oi prxeic autèc mporeÐ na eÐnai prxeic algebrikèc akìma kai

prxeic logismoÔ!


'Ajroisma dunamoseir¸n

An

f (x) =∞∑n=0

an(x − x0)n,

g(x) =∞∑n=0

bn(x − x0)n,

tìte

(f + g)(x) = f (x) + g(x) =∞∑n=0

(an + bn)(x − x0)n.


Grammikìc sundiasmoc dunamoseir¸n

An

f (x) =∞∑n=0

an(x − x0)n,

g(x) =∞∑n=0

bn(x − x0)n,

tìte

(λ1f + λ2g)(x) = λ1f (x) + λ2g(x) =∞∑n=0

(λ1an + λ2bn)(x − x0)n.


Ginìmeno dunamoseir¸n

An

f (x) =∞∑n=0

an(x − x0)n,

g(x) =∞∑n=0

bn(x − x0)n,

tìte

(f g)(x) = f (x)g(x) =∞∑n=0

cn(x − x0)n,

cn =n∑

i=0

aibn−i


Pardeigma

ex+y = exey .


Parag¸gish dunamoseir¸n

An

f (x) =∞∑n=0

an(x − x0)n,

tìte

df

dx(x) =

∞∑n=0

cn(x − x0)n,

cn = (n + 1)an+1,

dhlad mporoÔme na paragwgÐsoume mia dunamoseir

paragwgÐzontac ìro proc ìro.


Pardeigma

d

dxex = ex .


Olokl rwsh dunamoseir¸n

An

f (x) =∞∑n=0

an(x − x0)n,

tìte ∫f (x)dx = C +

∞∑n=0

cn(x − x0)n,

cn =an

n + 1,

dhlad mporoÔme na oloklhr¸soume mia dunamoseir paragwgÐzontac

ìro proc ìro.


Pardeigma

∫exdx = ex + C .


Efarmogèc stic pijanìthtec: Ropogenn tria

An X mia diakrit tuqaia metablht pou paÐrnei timèc sto N mepn = P(X = n), mporoÔme na orisoume thn dunamoseir,

∞∑n=0

pnsn.

H dunamoseir aut orÐzei mia sunrthsh h opoÐa onomazetai

ropogenn tria thc X ,

φX (s) =∞∑n=0

pnsn = E[sX ], |s| ≤ 1.


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