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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!
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 2 / 25
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.
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 3 / 25
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.
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 4 / 25
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}
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 5 / 25
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?
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 6 / 25
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 =∞
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 7 / 25
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|
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 8 / 25
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 ∣∣∣ .
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 9 / 25
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 ?
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 10 / 25
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)!
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 13 / 25
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.
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 14 / 25
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
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 15 / 25
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!
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 16 / 25
'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.
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 17 / 25
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.
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 18 / 25
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
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 19 / 25
Pardeigma
ex+y = exey .
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 20 / 25
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.
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 21 / 25
Pardeigma
d
dxex = ex .
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 22 / 25
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.
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 23 / 25
Pardeigma
∫exdx = ex + C .
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 24 / 25
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.
A. N. Giannakìpouloc (O.P.A) Logismoc II Earinì Exmhno 2017 25 / 25