Seja V=P3(R) e considere o produto interno dado por
lang<= f(x), g(x)rang<==integral0^1 f(x) g(x) d x
Calcule lang<= f(x), g(x)rang<=,||f(x)||,||g(x)|| e d(f(x), g(x)), em cada item abaixo.
f(x)=2 e g(x)=x^3+x+1

Passo 1

O enunciado define o espaço V=P3(R) com o produto interno
lang<= f,grang<==integral0^1f(x),g(x),dx
e pede para calcular
lang<= f,grang<=, |f|, |g| e d(f,g), onde f(x)=2 e g(x)=x^3+x+1.
Para isso usaremos as definições
|h|=raizlang<= h,hrang<= e d(f,g)=|f-g|

Passo 2

Calcular lang<= f,grang<= usando a definição do produto interno:
lang<= f,grang<= =integral0^12l(x^3+x+1r),dx =2integral0^1l(x^3+x+1r),dx
Agora
integral0^1x^3,dx=tfrac14, integral0^1x,dx=tfrac12, integral0^11,dx=1,
logo
lang<= f,grang<= =2l(tfrac14+tfrac12+1r) =2cdottfrac74 =tfrac72

Passo 3

Calcular |f|:
|f|=raizlang<= f,frang<= =raizintegral0^12cdot2,dx =raizintegral0^14,dx =raiz4cdot1 =2

Passo 4

Calcular |g|:
|g|=raizintegral0^1l(x^3+x+1r)^2dx
Expandindo
l(x^3+x+1r)^2=x^6+2x^4+2x^3+x^2+2x+1, temos
integral0^1x^6dx=tfrac17, integral0^12x^4dx=tfrac25, integral0^12x^3dx=tfrac12, integral0^1x^2dx=tfrac13, integral0^12x,dx=1, integral0^11,dx=1
Somando,
integral0^1(x^3+x+1)^2dx =tfrac17+tfrac25+tfrac12+tfrac13+1+1 =tfrac709210,
portanto
|g|=raiztfrac709210

Passo 5

Calcular d(f,g)=|f-g|. Note que
f(x)-g(x)=2-l(x^3+x+1r)=1-x-x^3
Então
|f-g|^2 =integral0^1(1-x-x^3)^2dx =integral0^1l(1-2x-2x^3+x^2+2x^4+x^6r)dx
Calculando cada termo:
integral0^11,dx=1,; integral0^1-2x,dx=-1,; integral0^1-2x^3,dx=-tfrac12,; integral0^1x^2,dx=tfrac13,; integral0^12x^4,dx=tfrac25,; integral0^1x^6,dx=tfrac17
Somando dá
|f-g|^2 =1-1-tfrac12+tfrac13+tfrac25+tfrac17 =tfrac79210,
logo
d(f,g)=|f-g|=raiztfrac79210

Resposta

lang<= f,grang<==tfrac72, |f|=2, |g|=raiztfrac709210, d(f,g)=raiztfrac79210

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