bao cao chuyen de pth bpth

7/23/2019 Bao Cao Chuyen de Pth Bpth

1/22

P DNG DY S VO GII CC PHNGTRNH V BT PHNG TRNH HM

Minh Khoa

Hi Ton hc H Ni, email: [email protected]

V Quc B Cn

Archimedes Academy, email: [email protected]

Minh Khoa, V Quc B Cn ng dng dy s trong PTH, BPTH 17/10/2015 1 / 21
http://localhost/var/www/apps/conversion/tmp/scratch_5/[email protected]://localhost/var/www/apps/conversion/tmp/scratch_5/[email protected]://localhost/var/www/apps/conversion/tmp/scratch_5/[email protected]://localhost/var/www/apps/conversion/tmp/scratch_5/[email protected]://find/


2/22

Li dn

Cc bi ton phng trnh hm v bt phng trnh hm lun rt th vv li cun i vi ngi lm ton. T mt phng trnh no , ch bngmt vi php th n gin, ta c th tm c cc tnh cht c bit cahm s c cho hoc thm ch l cng thc tng qut ca hm. Tuynhin, khi i su vo vn ny th vic ch dng php th gii l cha

, c bit l vi cc bi ton bt phng trnh hm. Do vy, chng tacn c nhng cng c khc b tr tng cng thm tnh hiu qu caphng php th. Lc ny, vic s dng dy s v gii hn mt cch linhhot s gip con ng i tr nn sng sa v d dng hn rt nhiu.

Nhn thy vai tr rt ln ca dy s v gii hn trong cc bi tonphng trnh, bt phng trnh hm, chng ti quyt nh thc hin bivit ny chia s kinh nghim ca mnh trong cch s dng cng nh c hc hi v nhn c kin ng gp t qu ng nghip gn xa.

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3/22

Mt s k thut chn v kp thng s dng

Trong nhiu trng hp, ta cn tm cng thc tng qut ca hm s,khi mt trong cc hng i m ta c th ngh n l thit lp mtbt ng thc dng:

an f(x) bn, y (an), (bn)l hai dy c chn sao cho bt ng thc trnng vi mi n(ng vi mi xc nh). Lc ny, nu

lim an =lim bn =L(x)th bng cch chuyn sang gii hn, ta s tmc cng thc tng qut ca f(x)l L(x).

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4/22

Mt s k thut chn v kp thng s dng

Trong nhiu trng hp, ta cn tm cng thc tng qut ca hm s,khi mt trong cc hng i m ta c th ngh n l thit lp mtbt ng thc dng:

an f(x) bn, y (an), (bn)l hai dy c chn sao cho bt ng thc trnng vi mi n(ng vi mi xc nh). Lc ny, nu

lim an =lim bn =L(x)th bng cch chuyn sang gii hn, ta s tmc cng thc tng qut ca f(x)l L(x).Nu cn suy xt mt tnh cht no ca f(x),ta c th thit lpmt bt ng thc dng:

A(f) anB(f),trong A(f), B(f)l hai biu thc ca xv f(x),cn (an)l dyc chn sao cho bt ng thc trn ng vi mi n(ng vi mi xc nh). Lc ny, da trn s hi t ca an,ta c th a ra nhiu

kt lun cho A(f)v B(f),t suy ra tnhchtca f(x). Minh Khoa, V Quc B Cn ng dng dy s trong PTH, BPTH 17/10/2015 3 / 21
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5/22

Mt s v d c trng

V dTm tt c cc hm s f : R+ R+ tha mn

f(x)f(2y) =3f

x+yf(x)

(1)

vi mi cp s thc dng x, y.

Li gii.Trc ht, ta chng minh f(x) 2 vi mi x >0.Tht vy, gis tn ti x0 >0 sao cho f(x0) 0. Minh Khoa, V Quc B Cn ng dng dy s trong PTH, BPTH 17/10/2015 4 / 21
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6/22

Mt s v d c trng (tip theo)T , bng quy np, ta chng minh c

f(x) >an

vi mi n N, y (an)l dy s c xc nh bi

a1 =2

an+1 = 3anDo lim an =3 nn t bt ng thc trn, ta suy ra

f(x) lim an =3

vi mi x >0.Thay ybi yf(x) vo phng trnh (1), ta c

3f(x+y) =f(x)f

2yf(x)

3f(x) (2)

vi mi x, y >0.Vy f(x)l hm khng gim. Minh Khoa, V Quc B Cn ng dng dy s trong PTH, BPTH 17/10/2015 5 / 21
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7/22

Mt s v d c trng (tip theo)

Nu f(x) >3 vi mi x >0 th bt ng thc (2) khng xy ra dubng. Do trong trng hp ny f l hm tng ngt.

Thay ybi y2 vo (1), ta c

f(x)f(y) =3f

x+y

2f(x)

vi mi x, y >0.Do v tri ca ng thc trn i xng vi x, ynnv phi cng i xng. T ta c

f

x+y

2f(x)

= f

y+

x

2f(y)

.

Do ftng ngt nn ta c x+ y2 f(x) =y+ x2 f(y).Cho y=2,ta c

f(x) =

f(2)

2 1

x+2.

Chox 0

+,

ta suy ra mu thun. Minh Khoa, V Quc B Cn ng dng dy s trong PTH, BPTH 17/10/2015 6 / 21
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8/22


Do , tn ti c >0 sao cho f(c) =3.Thay x=cvo (1), ta c

f(3y+c) =f(2y).

Do 2y


9/22


V d (IMO Longlist 1979, APMO 1989)

Tm tt c cc song nh tng thc s f : R Rtho mnf(x) +g(x) =2x

vi mi x

R,trong g(x)l hm ngc ca f(x).

Li gii.t f0(x) =xv fn(x) =f

fn1(x)

,tng t cho gn(x).Thayxbi f(x)vo phng trnh bi, ta c f2(x) +x=2f(x).T ,bng quy np, ta chng minh c

fn+2(x) 2fn+1(x) +fn(x) =0.Gii phng trnh sai phn ny, ta tm c

fn(x) =x+n

f(x) x

vi mi s t nhin n. Minh Khoa, V Quc B Cn ng dng dy s trong PTH, BPTH 17/10/2015 8 / 21
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10/22

Mt s v d c trng (tip theo)Mt cch tng t, ta cng chng minh c

gn(x) =x+n

g(x) x= x nf(x) x.Do f(x)l song nh tng ngt nn fn(x)v gn(x)cng l cc hm tngthc s. Do , vi mi x >y,ta u c fn(x) >fn(y)v gn(x) >gn(y).Ni cch khc, ta c

x+ n

f(x)x >y+ nf(y)y nf(x)x f(y)y >yx,(3)

xnf(x)x >ynf(y)y nf(x)xf(y)y f(y)

yth bng cch cho n

+

trong (4), ta thuc iu mu thun. Cn nu f(x) x y.Ni cch khc, f(x) =x+cvimi x. Minh Khoa, V Quc B Cn ng dng dy s trong PTH, BPTH 17/10/2015 9 / 21
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11/22


V d

Tm tt c cc hm s f lin tc t Rvo Rtha mn iu kin:

f

x+f(y+z)

+f

y+f(z+x)

+f

z+f(x+y)

=0 (5)

vi mi b s thc x, y, z.

Li gii.D thy f(x) 0 l mt nghim ca phng trnh nn y tach cn xt f(x) 0.Thay x, y, zbi x2 vo (5), ta thu c

fx

2+f(x)=0

vi mi x R.Tip tc, thay xbi x2 +f(x)vo ng thc trn, ta c

f

x

22+

12

f(x)

=0

vi mi s thc x. Minh Khoa, V Quc B Cn ng dng dy s trong PTH, BPTH 17/10/2015 10 / 21
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12/22

Mt s v d c trng (tip theo)T y, bng quy np, ta chng minh c

f x

2n+1 + 12n f(x)

=0

vi mi x Rv n N.Trong ng thc trn, c nh xv chon +,sau s dng tnh lin tc ca f, ta suy ra f(0) =0.T y,bng cch thay x=yv z=

xvo (5), ta c

f

f(2x) x= 2f(x)vi mi x R.Do f(x) 0 nn tn ti x0sao cho f(x0) =0.Xt dy(an)c xc nh bi a0 =x0v an+1 =f(2an) an.Khi , d thy

f(an) = 2f(an1) = (2)2

f(an2) = = (2)n

f(x0)vi mi n N.Nu f(x0) >0 th ta c limn+ f(a2n) = +v

limn+

f(a2n1) = .

Cn nu f(x0)


13/22


Nhng trong c hai trng hp ny, ta u thy rng f(x)khng b chn,m flin tc nn n ton nh.

n y, bng cch thay y=z=0 vo (5), ta c

2f

f(x)

+f(x) =0

vi mi x R.Do fton nh nn ta c f(x) = x2 ,x R.Th li, tathy tha mn.

Vy phng trnh cho c hai nghim l f(x) =0 v f(x) = x2 .

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14/22


V d (IMO, 2013)

Cho hm s fxc nh trn tp Q+ v nhn gi tr trong tp Rtha mnng thi cc iu kin sau:

(1) f(x)f(y) f(xy)vi mi x, y Q+; (6)(2) f(x+y) f(x) +f(y)vi mi x, y Q+; (7)

(3) tn ti mt s hu t a >1 sao cho f(a) =a. (8)Chng minh rng f(x) =xvi mi x Q+.Li gii.Thay x=av y=1 vo (6), ta c a f(1) a,tc f(1) 1.T y, bng cch s dng quy np kt hp vi (7), ta chng minh c

f(n) nvi mi n N.Thay x= p

q vi p, q N, (p, q) =1 v y=qvo (6),

ta c

fpq f(q) f(p) p >0. Minh Khoa, V Quc B Cn ng dng dy s trong PTH, BPTH 17/10/2015 13 / 21
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15/22

Mt s v d c trng (tip theo)M f(q) q >0 nn t y ta c

f(x) >0

vi mi x Q+.Kt hp vi (7), ta suy ra f(x)l hm tng thc s trnQ+.By gi, t (6), s dng quy np kt hp vi f(a) =a,ta thu c

f(an) an

vi mi n N.Ta s chng minhf(x) x (9)

vi mi x 1, x Q+.Tht vy, gi s c x0 1 sao cho f(x0)


16/22

Mt s v d c trung (tip theo)Do , kt hp vi bt ng thc trn, ta thu c

xn

0 1 (x0 )n

vi mi n N.Nhng, mt iu r rng l bt ng thc ny s sai vi n ln. Mu thun ny cho ta iu va khng nh, tc (9) ng. Tiptheo, ta s chng minh

f(x) =x (10)vi mi x 1, x Q+.Chn n N sao cho an x+1.Khi , ta c

an f(an) =f

(an x) +x

f(an x) +f(x) (an x) +x=an.

Do vy, du ng thc trong cc nh gi phi xy ra v do (10) ng.Thay x= p

q vi p, q N, (p, q) =1 v y=qvo (6), ta suy ra

f(x) x

vi mi x Q+

. Minh Khoa, V Quc B Cn ng dng dy s trong PTH, BPTH 17/10/2015 15 / 21

M d ( i h )
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17/22


T (7) s dng phng php quy np, ta c

f(nx) nf(x)

vi mi x Q+ v n N.T , cho x= pq vi p, q N, (p, q) =1 vn=q,ta c

f(x) xvi mi x

Q+.Vy f(x) =xvi mi x

Q+.


M d ( i h )
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18/22


V d

Tm tt c cc hm s f : N

N

tha mn

2n+2016 ff(n) +f(n) 2n+2017.Li gii.Xt cc dy (an), (bn), (cn)v (dn)c xc nh nh sau

c1 =0, d1 =1

an = 2

cn+1

bn =2017 dn

cn+1cn+1 =

2an+1

=2(cn+1)

cn+3

dn+1 =2016 bn

an

+1 =

2016 2017dncn+1

2

cn+1 +1

=2016cn 1+dn

cn

+3 Minh Khoa, V Quc B Cn ng dng dy s trong PTH, BPTH 17/10/2015 17 / 21

Mt d t (ti th )
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19/22

Mt s v d c trng (tip theo)Ta chng minh c lim an =lim cn =1 v cn 23 vi mi n 2.T suy ra dn >0 vi mi n

1.Bng quy np theo n,ta s chng minh

cnx+dn f(x) anx+bn. (11)Do f(x) 1= c1x+d1 v 2017+2x f

f(x)

+f(x) 1+f(x)nn

f(x) 2x+2016= a1x+b1,khng nh trn ng vi n=1.Gi s khng nh ng vi n,khi ta c

2016+2x ff(x) +f(x) anf(x) +bn+f(x) = (an+1)f(x) +bn,suy ra

f(x) 2an+1x+2016 bnan+1 =cn+1x+dn+1.

S dng kt qu ny, ta c

2017+2x

ff(x)+f(x)

cn+1f(x)+dn+1+f(x) = (cn+1+1)f(x)+dn+1.


Mt d t (ti th )
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20/22

Mt s v d c trng (tip theo)Suy ra

f(x)

2

cn+1+1x+

2017 dn+1cn+1+1

=an+1x+bn+1.

Vy khng nh cng ng vi n+1.Ta c (11).

S dng kt qu (11) vi ch dn >0,ta c

f(x)>

cnxvi mi nnguyn dng. Cho n +,ta c f(x) xvi mi x >0.Tip theo, ta xt cc dy (en)v (fn)xc nh nh sau

e1 =2017

2fn =

2016 en2

en+1 =2017 fn

2 =

2017 2016en22

=2018+en

4


Mt d t (ti th )
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21/22


Ta chng minh c lim en = 20183 v lim fn = 20153 .Ngoi ra, bng cchquy np ging nh bc nh gi on trc, ta cng chng minh c

x+fn f(x) x+en

vi mi nnguyn dng. Cho n +,ta c

x+2015

3 f(x) x+2018

3 .

V f(x)ch nhn gi tr nguyn dng nn t y ta suy ra f(x) =x+ 672vi mi xnguyn dng. Th li, ta thy hm ny tha mn yu cu.


Li cm n
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22/22

Li cm n

Xin cm n qu thy c v cc bn hc sinh ch theo di bi bo cony. Chc mi ngi mt ngy lm vic vui v!

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