Find the least squares line through the points: (-3. 3). (-1. 2). (0-1). (1. -l) and (3.4) In class: Find the Least Squares line through the population data: (it may simplifyr things to consider 1950 as time zero). Population (in billions) Using the same data use an exponential model for the population versus time: P : Poe\" in the spirit of least squares lines to nd P and r. For both the line and exponential. compute the value the method gives for the population in 2000. The true value was 6.08. Set up the matrix equation needed to approximate the population using a least squares quadratic P 2 at: + In +c Session 10 Polynomial Interpolation; Lagrange Polynomials and Error Please view the videos and work the example problems on the Polynomial Interpolation and Error page Read the first page of the Newton Divided Difference Proof file. Do now: Question: Use a straightforward - brute force approach to find the lowest degree polynomial through the points (-1, -3) (0, -4) (2, 0) and (3, 5). Set up the equations. Use the Lagrange Polynomial method and the Newton Divided Difference Method for these points. If Lo (x), L (x), L, (x), and L, (x) are cubic Lagrange Polynomials with the following properties L (0) =1;2 (0) =0; 12 (0) =0 and Z, (0) =0 Lo (1) = 0; 4 (1) =1;12 (1) =0 and Z, (1) =0 Lo (2) = 0; 1 (2) =0;12 (2) = 1 and I, (2) =0 Lo (3) = 0:1 (3) =0;12 (3) =0 and I, (3) =1 Find Lo (x) + 21 (x)+312 (x) +413 (x). Explain. Taylor series - match a function's value and that of its derivatives at a single point. Lagrange Polynomial Interpolation - match a function's values at a bunch of points. (there are in between methods of matching function values and some derivatives at points - we won't discuss this Our goal in this section is to fit polynomials - either a single polynomial or piecewise polynomial functions - through the set of points. Three ways to connect two points with a line - of course it's the same line. . Point-slope form y-y =(x-x ) (y minus one of the y values equals the slope * - No times x minus the corresponding x value) . Slope-intercept form y = mx + b where m is the slope and b is the y intercept . Lagrange-polynomial form P (x) =- (x -X ) (x - X ) (Xo - X]We can prove that 2 points determine a line, 3 points a quadratics, etc. and that there is only one line (or polynomial of lower degree, i.e., a constant) through two points, only one quadratic (or lower degree polynomial) through 3 points and so forth. To prove, first show there is at least one such polynomial - write out the nxn system. Columns are independent. Next, suppose 2 quadratics p(x) and q(x) pass through same 3 points. Subtract to get a quadratic, r(x) with, rx)=(x)=r(x2)=0 Sor(x)=(x-x)(x-x])(x-x2)v(x) which isn't be a quadratic so unique n+1 points determine an nith degree polynomial Hard, messy, sensitive to small errors as add more points The question is: how do we find the equation of the polynomial? Show - Using a straightforward - brute force approach to find the lowest degree polynomial through the points (-1, -3) (0, -4) (2, 0) and (3, 5). Set up the equations. Show the Lagrange Polynomial method and the Newton Divided Difference Method. Demonstrate for these points. Derive the error term for Lagrange Polynomials. Derive Taylor Series error if time: Taylor series matches up function and the derivatives at x = a so it is easy to see that: P (x) = Co+G(x-a)+...+C, (x-a)" must satisfy f (a) = co f' ( a) = q f" (a) = 2c2... f() (a) =nic, Consider: , (x) = P. (x)+K(x-a)" so , (x) satisfies matching all the derivatives through the n th at a. Let K be chosen so that . (x) also agrees with the original function fat b so that 4. (b) = f(b) Then f (b) = 0 (b) = P. (b)+K(b-a)" = K_J(6)-P (b) (b-a) " Now define F(x) = f (x)-. (x) Since F(a) =F(b)=0 by Rolle's Theorem there is at least one point on [a, b] call it c, where F' (C ) =0 Now F'(a) = f'(a)-. '(a) =0 and F'(c ) =0so there's a c, on [a,c ] where F" (C ) = 0Similarly, we can do this through n derivatives and F"(a) =0 and F(") (C, ) = 0 so there's a co- on [a. ca] where F() (Cq=])=0. Now differentiating F (x) = f (x) -. (x) =f(x)-P (x)-K(x-a) n+1 times and plugging in x = Co. yields F(#" (Cq]) = f(" (Can1 ) - 0-(n+1)!k 0 =f (Cm]) -(n+1)!K K =J f (Cz-1 ) (m +1)! Finally, f (b) = P.(b)-J () (b-a)" (n+1)! END) First let's derive the error formula for writing an interpolation polynomial through the points Xo= X1=-.-: X, as P( x) = C+G(x-X)+c, (x-x) (xx-x)+...+6x (xx-x) (xx-x)...(xx-X_) Define: D ( 2 ) = (2-x) ..(2-X,) G(z) =f(z)-P(z)-R(x)@(z) where R(x) is defined so that G(z) =0 at a certain value z = x Thus G(z) =0 whenever z=x, and xos X1:... x, since f=P and @ =0 at x0= X1:---, X Since G=0 at n+2 points then the first derivative must be 0 at n+ 1 points (Rolle's Theorem) and so on and the n+ 1st derivative must be 0 at least at one point on the interval so that 0 = G(* (g) = f(**"($)-0-R(x)(n+1)! since P has degree n and thus its n+ 1st derivative is zero. So R(x) = and f (x) = P(x) + f(5(x)) (n+1)! (n+1)! -(x - x0 ) ... (x -Xx ) Summary: f (x) - R(x)S- max (x - Xo) (x- x) 2! max f"(x)| Note that the value of max (x-xo) (x - ) = 4 h where h is the spacing between adjacent x- values. This value occurs at the mid-point of the interval.2 For quadratic Lagrange Polynomials, we have max (x-X, ) (x-X ) (X-X ) = h . This value 313 h occurs at x* = I V3 In video: Example: If f (x) =et on 0, 1] then the line segment connecting the endpoints has error at any given point of no more than: f ( x) - P, (x)/S max (x -X ) (x -x 2! max f (xx )