lets-f-0-1-R-lets-e-n-0-1-R-lets-a-n-0-1-R-such-that-f-x-i-1-3-a-i-x-e-i-x-0-1-e-i-x-e-j-x-dx-1-i-j-0-i-j-so-0-1-f-x-2-dx-

Question Number 891 by 123456 last updated on 17/Apr/15

lets f : [0, 1] \to R

lets e_{n} : [0, 1] \to R

lets a_{n} : [0, 1] \to R

such that

f (x) = \underset{i = 1}{\sum^{3}} a_{i} (x) e_{i} (x)

\underset{0}{\int^{1}} e_{i} (x) e_{j} (x) d x = {\begin{cases} 1 & i = j \\ 0 & i \neq j \end{cases}

so

\underset{0}{\int^{1}} {[f (x)]}^{2} d x = ?

Commented by prakash jain last updated on 16/Apr/15

{[f (x)]}^{2} = a_{1}^{2} e_{1}^{2} + a_{2}^{2} e_{2}^{2} + a_{3}^{2} e_{3}^{2} + 2 a_{1} e_{1} a_{2} e_{2} + 2 a_{1} e_{1} a_{3} e_{3}

+ 2 a_{2} e_{2} a_{3} e_{3}

I think the given input is insufficient .

Commented by 123456 last updated on 17/Apr/15

what if is given that

\frac{\partial f}{\partial x} = \underset{i = 1}{\sum^{3}} a_{i} \frac{\partial e_{i}}{\partial x}

Answered by prakash jain last updated on 17/Apr/15

Given \frac{\partial f}{\partial x} = \underset{i = 1}{\sum^{3}} a_{i} \frac{\partial e_{i}}{\partial x}

\int_{0}^{1} {[f (x)]}^{2} d x = \underset{i = 1}{\sum^{3}} a_{i}^{2} \int_{0}^{1} e_{1}^{2} d x

= \underset{i = 1}{\sum^{3}} a_{i}^{2}