Problem 4. Let w be a positive continuous function and let n be a nonnegative integer....

or equal to n. we that is LED au where problem G Pn (IR) is the space of an polynomials of degree lesser know {Vi} a basis for Pn (2) V = 3 xi for osisn & Gram-schonidt orthogonalisation process get orthogonal rectors (polynomials forms of Pn (IB) the orthogonal Now we by Can Still Call basis which basis {wx3 ko Let

J-0 ug V. and uk = UK proj.

(OK) for all kxl. Now obserserve that proju; (UK) (uk, u;) we claim that, (ej u;) deg K for all K=OC) n. uo = Vo = x°= 1 → deg (no)=o. uj -k For k=o,

Now let assume NOW let Then uno Vino uk Now as uk uk degree same nool But has we so, that deg (uk) =k for all ock <no. for some no <n.

K=no. no-1 E Projur (un) K=0 proj (und) has since ocksnoole, E proju (ono) degree s no-l: And already have deg (Uno) = no. degree (uno) = no. * Thus, by the principle of mathematical induction, we conclude that deg (ux) = k for Now take, || «K 11 And thus, thus, {Pk3 k2 becomes orthonormal basis of Pn CIR) such that deg (P, () = deg(uk)=k for KE {O,1,.-:,n}. 6 we we also also know, Gram - Schmidt process assures of the orthogonalisation of the {:} the happens in to that span {4, U2, U,...

Up} = span{use, ..., Up} for all o.fr sn. all ocksn. ak Pk for all K. an all set vectors set {u;} .0 so such way

> Wk-1 = xk-1} - , PK 3. (Notations has been kept unaltered) Now for any K, P ! has degree (k-1).

» Pk & span {n9=1, span {uo, uk-3. span {po, of Pi). (. Every multiple of => PK = E ciPi Pi) (Pk, P:) = 0 (By orthogonality Hence proved) ui is a Kt i-D ki => (Pk, P4) (Pk) E ſ=0 i=0 of pis).