Question

(a) (b) Find the least squares approximation of f(x) = x2 + 3 over the interval [0, 1] by a function of the form y = ae? + bx, where a, b E R. You should write the coefficients a, b as decimal approximations, rounded to two decimal places. Let g(x) be the least squares approximation you found in the pre- vious problem. So g(x) = ae” + bx for some scalars a, b. Find the least squares approximation of g(x) over the interval [0, 1] by a function of the form cx? +d. You should write the coefficients c, d as decimal approximations, rounded to two decimal places. Let h(x) be the function you found in part (b), and g(x) the function you found in part (a). Without actually doing any integral computations, decide which of the following facts must be true: i. Sof(g(x) – (x2 + 3))?dx > Sof(g(x) – h(x))dx ii. Só (g(x) – (x2 + 3))?da 5 Só (g(x) – h(x))?dx And justify your assertion. (c)

Answer 1

0 Solution: - La scab) = (gen) - Fonya scaib) = 5 (oe++be--? J'a [ae'+bx -x3-5] (el) de x ²_3] da as да 0 as ab a [ ae*tb«-«?-3)(a) da OS as ab et as = 0 1 2 bez* dn 4bfeede- fects z 0 dx=0 (3.1945) a + b - 5.8731 0 da = 3.1945 de Saeede a enda=1 o

"de - 5.8731. as ob a face de 4b\'wd = face?:3) de Q + 0.33 3b = 1.15 - from ean 1 and ② 3.1945 5.8731 20:46) 0.333 1 1.75 Solving it 3.2058 b - 4. 3680 g (3.21) et -(4.7)x

g(x) - 3. let - 4.37x Now sccid) - Į (cerd-gend]² sccid) > S [catd - 3.21224 4.572) as 2 (C22+d - 3. 2)e4 +4:37 a] (x2)dx as Je (c2*d-30 (ca?+d – 3.2102 +4.372] Ci) da 0 OS -O ac 1 candien o faroles San a²( 3.21 e?+ U. 372) 0 da clŚ )+d() = 1.2132 0 20 C+0.333 d = 1.2182 as CH 2224 061) - Jeunes 4.37) da

0.3333 C d = 3.3307 → From and @ 0.20 0.3333 (1.2132 10.6 8.3307 0.3333 C = 1.16 d 2.94 lCa2+0 = 1.168² +9.94 09 (g(x) – (z?+3)]?dx = S'aca)henjave is true. Reason: Since gox) is a function Privolving expotential and bea) is a polynomial. :. approximalion of polynomial by expotential will obsiously be less accurate than ximation of polynomial by polynomial. appro-