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GENERALITZACIÓ DEL TEOREMA DE CAUCHY: TEOREMA DE GAS

Autor: Jordi Gas Riera

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Teorema de Gas (generalització del teorema de Cauchy per a n funcions, essent n>2) Siguin f1, f2, f3, …,fn n funcions reals de variable real definides en un interval [ ]b,a .

[ ]] [

] [ [ ] [ ] (c)f (a)f(b)f(c)f (a)f(b)f / ba, c

3n

ba,en derivables f......,,.........f,f,fba,en contínues f......,,.........f,f,f

i

n

ff , 1i1i1i1i

n

ff , 1iii

n321

n321

11n11n

ʹ′−=ʹ′−∈∃⇒

⇒⎪⎭

⎪⎬

⎫

≥

∑∑==

+++== ++

Demostració

Considerem la funció: G(x)= [ ] [ ][ ]∑==

+++

+

−−n

ff , 1ii1i1i1iii

11n

(x)f (a)f(b)f - (x)f (a)f(b)f

G(x) és una funció contínua en [ ]ba, per ser-ho f1, f2, f3, .........., fn i és derivable en ] [ba, per ser-ho f1, f2, f3, .........., fn . A més a més, anem a veure que G(a)=G(b) amb la qual cosa podrem aplicar el teorema de Rolle.

G(a)= [ ] [ ][ ]∑==

+++

+

−−n

ff , 1ii1i1i1iii

11n

(a)f (a)f(b)f - (a)f (a)f(b)f =

= [ ] [ ] (a)f (a)f(b)f - (a)f (a)f(b)f 122211 −− + + [ ] [ ] (a)f (a)f(b)f - (a)f (a)f(b)f 233322 −− + + [ ] [ ] (a)f (a)f(b)f - (a)f (a)f(b)f 344433 −− + + ............................................................... + + [ ] [ ] (a)f (a)f(b)f - (a)f (a)f(b)f 2-n1-n1-n1-n2-n2-n −− + + [ ] [ ] (a)f (a)f(b)f - (a)f (a)f(b)f 1-nnnn1-n1-n −− + + [ ] [ ] (a)f (a)f(b)f - (a)f (a)f(b)f n111nn −− = = (a)(a).ff(b)(a).ff(a)(a).ff(a)(b).ff 21212121 +−− + + (a)(a).ff(b)(a).ff(a)(a).ff(a)(b).ff 32323232 +−− + + (a)(a).ff(b)(a).ff(a)(a).ff(a)(b).ff 43434343 +−− + + ............................................................................... + + (a)(a).ff(b)(a).ff(a)(a).ff(a)(b).ff 1-n2-n1-n2-n1-n2-n1-n2-n +−− + + (a)(a).ff(b)(a).ff(a)(a).ff(a)(b).ff n1-nn1-nn1-nn1-n +−− + + (a)(a).ff(b)(a).ff(a)(a).ff(a)(b).ff 1n1n1n1n +−− =

= [ ]∑==

++

+

−n

ff , 1i1ii1ii

11n

(b)(a).ff(a)(b).ff

G(b)= [ ] [ ][ ]∑==

+++

+

−−n

ff , 1ii1i1i1iii

11n

(b)f (a)f(b)f - (b)f (a)f(b)f =

= [ ] [ ] (b)f (a)f(b)f - (b)f (a)f(b)f 122211 −− + + [ ] [ ] (b)f (a)f(b)f - (b)f (a)f(b)f 233322 −− +

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+ [ ] [ ] (b)f (a)f(b)f - (b)f (a)f(b)f 344433 −− + + ............................................................... + + [ ] [ ] (b)f (a)f(b)f - (b)f (a)f(b)f 2-n1-n1-n1-n2-n2-n −− + + [ ] [ ] (b)f (a)f(b)f - (b)f (a)f(b)f 1-nnnn1-n1-n −− + + [ ] [ ] (b)f (a)f(b)f - (b)f (a)f(b)f n111nn −− = = (a)(b).ff(b)(b).ff(b)(a).ff(b)(b).ff 21212121 +−− + + (a)(b).ff(b)(b).ff(b)(a).ff(b)(b).ff 32323232 +−− + + (a)(b).ff(b)(b).ff(b)(a).ff(b)(b).ff 43434343 +−− + + ............................................................................... + + (a)(b).ff(b)(b).ff(b)(a).ff(b)(b).ff 1-n2-n1-n2-n1-n2-n1-n2-n +−− + + (a)(b).ff(b)(b).ff(b)(a).ff(b)(b).ff n1-nn1-nn1-nn1-n +−− + + (a)(b).ff(b)(b).ff(b)(a).ff(b)(b).ff 1n1n1n1n +−− =

= [ ]∑==

++

+

−n

ff , 1i1ii1ii

11n

(b)(a).ff(a)(b).ff

Per tant, G(a)=G(b) i pel teorema de Rolle: ] [ 0(c)G / ba, c =ʹ′∈∃ . Ara bé:

[ ] [ ][ ]∑==

+++

+

ʹ′−ʹ′−=ʹ′n

ff , 1ii1i1i1iii

11n

(x)f.(a)f(b)f - (x)f.(a)f(b)f (x)G ⇒

⇒ [ ] [ ][ ]∑==

+++

+

ʹ′−ʹ′−=ʹ′n

ff , 1ii1i1i1iii

11n

(c)f.(a)f(b)f - (c)f.(a)f(b)f (c)G

⇒=ʹ′ 0(c)G [ ] [ ][ ] 0 (c)f.(a)f(b)f - (c)f.(a)f(b)f n

ff , 1ii1i1i1iii

11n

=ʹ′−ʹ′−∑==

+++

+

⇒

⇒ [ ] [ ] (c)f (a)f(b)f(c)f (a)f(b)f

i

n

ff , 1i1i1i1i

n

ff , 1iii

11n11n

ʹ′−=ʹ′− ∑∑==

+++== ++

Observació: Si apliquem aquest teorema per a n=3, prenent f3(x)=k (funció constant), resulta el teorema de Cauchy:

[ ]] [ ] [ [ ] [ ] (c)f (a)f(b)f (c)f (a)f(b)f / ba, c

ba,en derivables f,fba,en contínues f,f

12221121

21 ʹ′−=ʹ′−∈∃⇒⎭⎬⎫

Si en comptes de la funció constant, f3(x)=k, prenem com a tercera funció la suma de les altres dues, f3(x)=f1(x)+f2(x), també s´obté el teorema de Cauchy com fàcilment es pot comprovar. En fi, una altra manera d´obtenir el teorema de Cauchy és aplicant el teorema per a n=4 i prenent com a tercera i quarta funcions respectivament la funció suma i la funció diferència de les dues primeres: f1(x) , f2(x) , f3(x)=f1(x)+f2(x) , f4(x)=f1(x)-f2(x) o bé prenent f1(x) , f2(x) , f3(x)=f1(x)+f2(x) i f4(x)=f1(x)+f2(x)+f3(x)=2f1(x)+2f2(x).