Linear Least Squares Approximation(Calcus)?

Question

f(x) = 2*sqrt(x), 1<= x <= 4
this means using the polynomial basis set p1(x) = {1,x} with the standard Inner product for continuos functions C[a,b] to form an orthonormal basis. Then find g(x) = C0+c1 x so that it approximates the function f(x) over the interval [1,4]

Answer 1

First, we find an orthonormal basis of span {1, x}. First we find the norm of 1:

||1|| = √<1, 1> = √([1, 4]∫1*1 dx) = √3

Thus, the first vector in our orthonormal basis will be 1/√3. Next, we find <1/√3, x> = [1, 4]∫x/√3 dx = (16-1)/(2√3) = 15/(2√3). So the vector x-<1/√3, x>1/√3 = x-15/(2√3)*1/√3 = x-15/6 = x-5/2 is orthogonal to 1. Now we must find the norm of this vector:

||x-5/2|| = √ = √([1, 4]∫(x-5/2)² dx) = √((x-5/2)³/3 |[1, 4]) = √((27/8 + 27/8)/3) = √(9/4) = 3/2. So the unit vector orthogonal to 1 is (2x-5)/3

Thus our orthonormal basis is {1/√3, (2x-5)/3}. The projection onto the space spanned by these vectors, which is the least-squares approximation, is *1/√3 + *(2x-5)/3. We have:

= [1, 4]∫2√x/√3 dx = 4/(3√3)x^(3/2)|[1, 4] = 28/(3√3)
= [1, 4]∫2√x(2x-5)/3 dx = [1, 4]∫4x^(3/2)/3 - 10√x/3 dx = 8x^(5/2)/15 - 20x^(3/2)/9 |[1, 4] = (256/15 - 160/9) - (8/15 - 20/9) = 248/15 - 140/9 = 744/45 - 700/45 = 44/45

Thus our final function is:

28/(3√3) * 1/√3 + 44/45 * (2x-5)/3 = 28/9 + 88/135 x - 44/27 =

88/135 x + 40/27

Answer 2

Thus y=2sqrt(x) and y=ax+b, where a and b to be found; the technique implies:
S(a,b) = 1/(4-1)*integral{for x=1 intil 4} of (ax+b –2sqrt(x))^2 *dx, S(a,b) being min; thus dS/da = 0 and dS/db = 0;
Here(1): dS/da = (2/3)*int[(ax+b –2sqrt(x))*x*dx];
Here(2): dS/db = (2/3)*int[(ax+b –2sqrt(x))*dx];
I1 = int[axx+bx -2x^(3/2)]*dx = a*(x^3/3) +b*(xx/2) –(4/5)*x^(5/2)= {for x=1 until 4} = 21a +7.5b – 24.8 =0;
I1 = int[ax+b –2sqtr(x)]*dx = a*(xx/2) +b*x –(4/3)*x^(3/2)= {for x=1 until 4} = 7.5a +3b – 28/3 =0;
Thus system of 2 equations:
210a+75b=248
225a+90b=280
D0= 210*90-225*75 = 2025; Da = =248*90-75*280 = 1320; Db= 210*280-248*225 = 3000; a= Da/D0= 0.6519; b= Db/D0 =1.4815;
Yes “Pascal” is OK!

Answer 3

it will do you no good to cheat - it will catch up with you.if you don't understand your class, see your teacher.