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Let $X=C[0,1]$ with the inner product $\langle x,y\rangle=\int_0^1 x(t)\overline y(t)\,dt$ $\forall$ $x(t),y(t)\in C[0,1]$

$X_0 =\{x(t) \in X :\int_0^1 t^2x(t)\,dt=0\}$and $X_0^\bot$ be the orthogonal complement of $X_0$.

let $y(t)=t^3$, $t\in [0,1]$ and $x_0\in X_0^\bot$ be the approximate of $y$. Then $x_0(t)$,t$\in$[0,1] is

(1)$\frac{4t^2}{5}$

(2)$\frac{5t^2}{6}$

(3)$\frac{6t^2}{7}$

(4)$\frac{7t^2}{8}$

answer i am obtaining is (5/6)$t^2$ but given ans is (4/5)$t^2$

i have done it as:

$X_0 =[span(t^2)]^\bot$

$X_0^\bot =[span(t^2)]^{\bot\bot}$ = span($t^2$)

now using othogonality,

<$t^3-at^2,t^2$>=0 which gives a=5/6

& therefore,$x_0(t)$=(5/6)$t^2$

is this correct or am i wrong somewhere

thanks in advance...

  • 0
    The best approximation with respect of which norm ?2017-01-24
  • 0
    @ Igor D. under $L^2$ norm2017-01-24

0 Answers 0