since by using a computer we are limited to the use of elementary arithmetic operations


 Brett Norton
 1 years ago
 Views:
Transcription
1 > 4. Interpolation and Approximation Most functions cannot be evaluated exactly: x, e x, ln x, trigonometric functions since by using a computer we are limited to the use of elementary arithmetic operations +,,, With these operations we can only evaluate polynomials and rational functions (polynomial divided by polynomials).
2 > 4. Interpolation and Approximation Interpolation Given points and corresponding values x 0, x 1,..., x n find a function f(x) such that y 0, y 1,..., y n f(x i ) = y i, i = 0,..., n. The interpolation function f is usually taken from a restricted class of functions: polynomials.
3 > 4. Interpolation and Approximation > 4.1 Polynomial Interpolation Theory Interpolation of functions f(x) x 0, x 1,..., x n f(x 0 ), f(x 1 ),..., f(x n ) Find a polynomial (or other special function) such that What is the error f(x) = p(x)? p(x i ) = f(x i ), i = 0,..., n.
4 > 4. Interpolation and Approximation > Linear interpolation Linear interpolation Given two sets of points (x 0, y 0 ) and (x 1, y 1 ) with x 0 x 1, draw a line through them, i.e., the graph of the linear polynomial x 0 x 1 l(x) = x x 1 y 0 + x x 0 y 1 x y 0 y 0 x 1 x 1 x 0 1 l(x) = (x 1 x)y 0 + (x x 0 )y 1 x 1 x 0 (5.1) We say that l(x) interpolates the value y i at the point x i, i = 0, 1, or l(x i ) = y i, i = 0, 1 Figure: Linear interpolation
5 > 4. Interpolation and Approximation > Linear interpolation Example Let the data points be (1, 1) and (4,2). The polynomial P 1 (x) is given by P 1 (x) = (4 x) 1 + (x 1) 2 3 The graph y = P 1 (x) and y = x, from which the data points were taken. (5.2) Figure: y = x and its linear interpolating polynomial (5.2)
6 > 4. Interpolation and Approximation > Linear interpolation Example Obtain an estimate of e using the function values e 0.82 = , e 0.83 = Denote x 0 = 0.82, x 1 = The interpolating polynomial P 1 (x) interpolating e x at x 0 and x 1 is and P 1 (x) = while the true value s (0.83 x) (x 0.82) P 1 (0.826) = (5.3) e = to eight significant digits.
7 > 4. Interpolation and Approximation > Quadratic Interpolation Assume three data points (x 0, y 0 ), (x 1, y 1 ), (x 2, y 2 ), with x 0, x 1, x 2 distinct. We construct the quadratic polynomial passing through these points using Lagrange s folmula P 2 (x) = y 0 L 0 (x) + y 1 L 1 (x) + y 2 L 2 (x) (5.4) with Lagrange interpolation basis functions for quadratic interpolating polynomial L 0 (x) = L 1 (x) = L 2 (x) = (x x1)(x x2) (x 0 x 1)(x 0 x 2) (x x0)(x x2) (x 1 x 0)(x 1 x 2) (x x0)(x x1) (x 2 x 0)(x 2 x 1) Each L i (x) has degree 2 P 2 (x) has degree 2. Moreover L i (x j ) = 0, j i for 0 i, j 2 i.e., L L i (x i ) = 1 i (x j ) = δ i,j = the Kronecker delta function. P 2 (x) interpolates the data P 2 (x) = y i, i=0,1,2 { 1, i = j 0, i j (5.5)
8 > 4. Interpolation and Approximation > Quadratic Interpolation Example Construct P 2 (x) for the data points (0, 1), (1, 1), (2, 7). Then P 2 (x)= (x 1)(x 2) 2 ( 1)+ x(x 2) 1 ( 1)+ x(x 1) 2 7 (5.6) Figure: The quadratic interpolating polynomial (5.6)
9 > 4. Interpolation and Approximation > Quadratic Interpolation With linear interpolation: obvious that there is only one straight line passing through two given data points. With three data points: only one quadratic interpolating polynomial whose graph passes through the points. Indeed: assume Q 2 (x), deg(q 2 ) 2 passing through (x i, y i ), i = 0, 1, 2, then it is equal to P 2 (x). The polynomial has deg(r) 2 and R(x) = P 2 (x) Q 2 (x) R(x i ) = P 2 (x i ) Q 2 (x i ) = y i y i = 0, for i = 0, 1, 2 So R(x) is a polynomial of degree 2 with three roots R(x) 0
10 > 4. Interpolation and Approximation > Quadratic Interpolation Example Calculate a quadratic interpolate to e from the function values e 0.82 = e 0.83 = e 0.84 = With x 0 = e 0.82, x 1 = e 0.83, x 2 = e 0.84, we have P 2 (0.826) = to eight digits, while the true answer e = and P 1 (0.826) =
11 > 4. Interpolation and Approximation > Higherdegree interpolation Lagrange s Formula Given n + 1 data points (x 0, y 0 ), (x 1, y 1 ),..., (x n, y n ) with all x i s distinct, unique P n, deg(p n ) n such that P n (x i ) = y i, i = 0,..., n given by Lagrange s Formula P n (x) = n y i L i (x) = y 0 L 0 (x) + y 1 L 1 (x) + y n L n (x) (5.7) i=0 where L i (x)= n j=0,j i where L i (x j ) = δ ij x x j x i x j = (x x 0) (x x i 1 )(x x i+1 ) (x x n ) (x i x 0 ) (x i x i 1 )(x i x i ) (x i x n ),
12 > 4. Interpolation and Approximation > Divided differences Remark; The Lagrange s formula (5.7) is suited for theoretical uses, but is impractical for computing the value of an interpolating polynomial: knowing P 2 (x) does not lead to a less expensive way to compute P 3 (x). But for this we need some preliminaries, and we start with a discrete version of the derivative of a function f(x). Definition (Firstorder divided difference) Let x 0 x 1, we define the firstorder divided difference of f(x) by f[x 0, x 1 ] = f(x 1) f(x 2 ) x 1 x 2 (5.8) If f(x) is differentiable on an interval containing x 0 and x 1, then the mean value theorem gives f[x 0, x 1 ] = f (c), for c between x 0 and x 1. Also if x 0, x 1 are close together, then usually a very good approximation. f[x 0, x 1 ] f ( x0 + x 1 2 )
13 > 4. Interpolation and Approximation > Divided differences Example Let f(x) = cos(x), x 0 = 0.2, x 1 = 0.3. Then f[x 0, x 1 ] = cos(0.3) cos(0.2) = (5.9) while ( ) f x0 + x 1 = sin(0.25) = so f[x 0, x 1 ] is a very good approximation of this derivative.
14 > 4. Interpolation and Approximation > Divided differences Higherorder divided differences are defined recursively Let x 0, x 1, x 2 R distinct. The secondorder divided difference f[x 0, x 1, x 2 ] = f[x 1, x 2 ] f[x 0, x 1 ] x 2 x 0 (5.10) Let x 0, x 1, x 2, x 3 R distinct. The thirdorder divided difference f[x 0, x 1, x 2, x 3 ] = f[x 1, x 2, x 3 ] f[x 0, x 1, x 2 ] x 3 x 0 (5.11) In general, let x 0, x 1,..., x n R, n + 1 distinct numbers. The divided difference of order n f[x 0,..., x n ] = f[x 1,..., x n ] f[x 0,..., x n 1 ] x n x 0 (5.12) or the Newton divided difference.
15 > 4. Interpolation and Approximation > Divided differences Theorem Let n 1, f C n [α, β] and x 0, x 1,..., x n n + 1 distinct numbers in [α, β]. Then f[x 0, x 1,..., x n ] = 1 n! f (n) (c) (5.13) for some unknown point c between the maximum and the minimum of x 0,..., x n. Example Let f(x) = cos(x), x 0 = 0.2, x 1 = 0.3, x 2 = 0.4. The f[x 0, x 1 ] is given by (5.9), and f[x 1, x 2 ] = cos(0.4) cos(0.3) = hence from (5.11) f[x 0, x 1, x 2 ] = ( ) = (5.14) For n = 2, (5.13) becomes f[x 0, x 1, x 2 ] = 1 2 f (c) = 1 2 cos(c) 1 2 cos(0.3) = which is nearly equal to the result in (5.14).
16 > 4. Interpolation and Approximation > Properties of divided differences The divided differences (5.12) have special properties that help simplify work with them. (1) Let (i 0, i 1,..., i n ) be a permutation (rearrangement) of the integers (0, 1,..., n). It can be shown that f[x i0, x i1,..., x in ] = f[x 0, x 1,..., x n ] (5.15) The original definition (5.12) seems to imply that the order of x 0, x 1,..., x n is important, but (5.15) asserts that it is not true. For n = 1 f[x 0, x 1 ] = f(x 0) f(x 1 ) = f(x 1) f(x 0 ) = f[x 1, x 0 ] x 0 x 1 x 1 x 0 For n = 2 we can expand (5.11) to get f(x 0 ) f[x 0, x 1, x 2 ] = (x 0 x 1 )(x 0 x 2 ) + f(x 1 ) (x 1 x 0 )(x 1 x 2 ) + f(x 2 ) (x 2 x 0 )(x 2 x 1 ) (2) The definitions (5.8), (5.11) and (5.12) extend to the case where some or all of the x i coincide, provided that f(x) is sufficiently differentiable.
17 > 4. Interpolation and Approximation > Properties of divided differences For example, define f(x 0 ) f(x 1 ) f[x 0, x 0 ] = lim f[x 0, x 1 ] = lim x 1 x 0 x 1 x 0 x 1 x 0 f[x 0, x 0 ] = f (x 0 ) For an arbitrary n 1, let all x i x 0 ; this leads to the definition f[x 0,..., x 0 ] = 1 n! f (n) (x 0 ) (5.16) For the cases where only some of nodes coincide: using (5.15), (5.16) we can extend the definition of the divided difference. For example f[x 0, x 1, x 0 ] = f[x 0, x 0, x 1 ] = f[x 0, x 1 ] f[x 0, x 0 ] x 1 x 0 = f[x 0, x 1 ] f (x 0 ) x 1 x 0
18 > 4. Interpolation and Approximation > Properties of divided differences MATLAB program: evaluating divided differences Given a set of values f(x 0 ),..., f(x n ) we need to calculate the set of divided differences f[x 0, x 1 ], f[x 0, x 1, x 2 ],..., f[x 0, x 1,..., x n ] We can use the MATLAB function divdif using the function call divdif y = divdif(x nodes, y values) Note that MATLAB does not allow zero subscripts, hence x nodes and y values have to be redefined as vectors containing n + 1 components: x nodes = [x 0, x 1,..., x n ] x nodes(i) = x i 1, i = 1,..., n + 1 y values = [f(x 0 ), f(x 1 ),..., f(x n )] y values(i) = f(x i 1 ), i = 1,..., n + 1
19 > 4. Interpolation and Approximation > Properties of divided differences MATLAB program: evaluating divided differences function divdif y = divdif(x nodes,y values) % % This is a function % divdif y = divdif(x nodes,y values) % It calculates the divided differences of the function % values given in the vector y values, which are the values of % some function f(x) at the nodes given in x nodes. On exit, % divdif y(i) = f[x 1,...,x i], i=1,...,m % with m the length of x nodes. The input values x nodes and % y values are not changed by this program. % divdif y = y values; m = length(x nodes); for i=2:m for j=m:1:i divdif y(j) = (divdif y(j)divdif y(j1))... /(x nodes(j)x nodes(ji+1)); end end
20 > 4. Interpolation and Approximation > Properties of divided differences This program is illustrated in this table. i x i cos(x i ) D i E E E E E E E2 Table: Values and divided differences for cos(x)
21 > 4. Interpolation and Approximation > Newton s Divided Differences Newton interpolation formula Interpolation of f at x 0, x 1,..., x n Idea: use P n 1 (x) in the definition of P n (x): P n (x) = P n 1 (x) + C(x) } C(x) P n correction term = C(x i ) = 0 i = 0,..., n 1 = C(x) = a(x x 0 )(x x 1 )... (x x n 1 ) P n (x) = ax n +...
22 > 4. Interpolation and Approximation > Newton s Divided Differences Definition: The divided difference of f(x) at points x 0,..., x n is denoted by f[x 0, x 1,..., x n ] and is defined to be the coefficient of x n in P n (x; f). C(x) = f[x 0, x 1,..., x n ](x x 0 )... (x x n 1 ) p n (x; f) = p n 1 (x; f)+f[x 0, x 1,..., x n ](x x 0 )... (x x n 1 ) (5.17) p 1 (x; f) = f(x 0 ) + f[x 0, x 1 ](x x 0 ) (5.18) p 2 (x; f)=f(x 0 )+f[x 0, x 1 ](x x 0 )+f[x 0, x 1, x 2 ](x x 0 )(x x 1 ) (5.19). p n (x; f) = f(x 0 ) + f[x 0, x 1 ](x x 0 ) +... (5.20) +f[x 0,..., x n ](x x 0 )... (x x n 1 ) = Newton interpolation formula
23 > 4. Interpolation and Approximation > Newton s Divided Differences For (5.18), consider p 1 (x 0 ), p 1 (x 1 ). Easily, p 1 (x 0 ) = f(x 0 ), and [ ] f(x1 ) f(x 0 ) p 1 (x 1 )=f(x 0 )+(x 1 x 0 ) =f(x 0 )+[f(x 1 ) f(x 0 )]=f(x 1 ) x 1 x 0 So: deg(p 1 ) 1 and satisfies the interpolation conditions. Then the uniqueness of polynomial interpolation (5.18) is the linear interpolation polynomial to f(x) at x 0, x 1. For (5.19), note that p 2 (x) = p 1 (x) + (x x 0 )(x x 1 )f[x 0, x 1, x 2 ] It satisfies: deg(p 2 ) 2 and p 2 (x i ) = p 1 (x i ) + 0 = f(x i ), i = 0, 1 p 2 (x 2 ) = f(x 0 ) + (x 2 x 0 )f[x 0.x 1 ] + (x 2 x 0 )(x 2 x 1 )f[x 0, x 1, x 2 ] = f(x 0 ) + (x 2 x 0 )f[x 0.x 1 ] + (x 2 x 1 ){f[x 1, x 2 ] f[x 0, x 1 ]} = f(x 0 ) + (x 1 x 0 )f[x 0.x 1 ] + (x 2 x 1 )f[x 1, x 2 ] = f(x 0 ) + {f(x 1 ) f(x 0 )} + {f(x 2 ) f(x 1 )} = f(x 2 ) By the uniqueness of polynomial interpolation, this is the quadratic interpolating polynomial to f(x) at {x 0, x 1, x 2 }.
24 > 4. Interpolation and Approximation > Newton s Divided Differences Example Find p(x) P 2 such that p( 1) = 0, p(0) = 1, p(1) = 4. p(x) = p(x 0 ) + p[x 0, x 1 ](x x 0 ) + p[x 0, x 1, x 2 ](x x 0 )(x x 1 ) x i p(x i ) p[x i, x i+1 ] p[x i, x i+1, x i+1 ] p(x) = 0 + 1(x + 1) + 1(x + 1)(x 0) = x 2 + 2x + 1
25 > 4. Interpolation and Approximation > Newton s Divided Differences Example Let f(x) = cos(x). The previous table contains a set of nodes x i, the values f(x i ) and the divided differences computed with divdif D i = f[x 0,..., x i ] i 0 p n (0.1) p n (0.3) p n (0.5) True Table: Interpolation to cos(x) using (5.20) This table contains the values of p n (x) for various values of n, computed with interp, and the true values of f(x).
26 > 4. Interpolation and Approximation > Newton s Divided Differences In general, the interpolation node points x i need not to be evenly spaced, nor be arranged in any particular order to use the divided difference interpolation formula (5.20) To evaluate (5.20) efficiently we can use a nested multiplication algorithm P n (x) = D 0 + (x x 0 )D 1 + (x x 0 )(x x 1 )D (x x 0 ) (x x n 1 )D n with D 0 = f(x 0 ), D i = f[x 0,..., x i ] for i 1 P n (x) = D 0 + (x x 0 )[D 1 + (x x 1 )[D 2 + (5.21) + (x x n 2 )[D n 1 + (x x n 1 )D n ] ]] For example P 3 (x) = D 0 + (x x 0 )D 1 + (x x 1 )[D 2 + (x x 2 )D 3 ] (5.21) uses only n multiplications to evaluate P n (x) and is more convenient for a fixed degree n. To compute a sequence of interpolation polynomials of increasing degree is more efficient to se the original form (5.20).
27 > 4. Interpolation and Approximation > Newton s Divided Differences MATLAB: evaluating Newton divided difference for polynomial interpolation function p eval = interp(x nodes,divdif y,x eval) % % This is a function % p eval = interp(x nodes,divdif y,x eval) % It calculates the Newton divided difference form of % the interpolation polynomial of degree m1, where the % nodes are given in x nodes, m is the length of x nodes, % and the divided differences are given in divdif y. The % points at which the interpolation is to be carried out % are given in x eval; and on exit, p eval contains the % corresponding values of the interpolation polynomial. % m = length(x nodes); p eval = divdif y(m)*ones(size(x eval)); for i=m1:1:1 p eval = divdif y(i) + (x eval  x nodes(i)).*p eval; end
28 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation Formula for error E(t) = f(t) p n (t; f) n P n (x) = f(x j )L j (x) j=0 Theorem Let n 0, f C n+1 [a, b], x 0, x 1,..., x n distinct points in [a, b]. Then f(x) P n (x) = (x x 0 )... (x x n )f[x 0, x 1,..., x n, x] (5.22) = (x x 0)(x x 1 ) (x x n ) f (n+1) (c x ) (5.23) (n + 1)! for x [a, b], c x unknown between the minimum and maximum of x 0, x 1,..., x n.
29 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation Proof of Theorem Fix t R. Consider p n+1 (x; f) interpolates f at x 0,..., x n, t p n (x; f) interpolates f at x 0,..., x n From Newton p n+1 (x; f) = p n (x; f) + f[x 0,..., x n, t](x x 0 )... (x x n ) Let x = t. f(t) := p n+1 (t, f) = p n (t; f) + f[x 0,..., x n, t](t x 0 )... (t x) E(t) := f[x 0, x 1,..., x n, t](t x 0 )... (t x n )
30 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation Examples f(x) = x n, f[x 0,..., x n ] = n! n! = 1 or p n (x; f) = 1 x n f[x 0,..., x n ] = 1 f(x) P n 1, f[x 0, x 1,..., x n ] = 0 f(x) = x n+1, f[x 0,..., x n ] = x 0 + x x n R(x) = x n+1 p n (x; f) P n+1 R(x i ) = 0, i = 0,..., n R(x i ) = (x x 0 )... (x x n ) = x n+1 (x x n )x n +... }{{} P n = f[x 0,..., x n ] = x 0 + x x n. If f P m, f[x 0,..., x m, x] = { polynomial degree m n 1 if n m 1 0 if n > m 1
31 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation Example Take f(x) = e x, x [0, 1] and consider the error in linear interpolation to f(x) using x 0, x 1 satisfying 0 x 0 x 1 1. From (5.23) e x P 1 (x) = (x x 0)(x x 1 ) e cx 2 for some c x between the max and min of x 0, x 1 and x. Assuming x 0 < x < x 1, the error is negative and approximately a quadratic polynomial e x P 1 (x) = (x 1 x)(x x 0 ) e cx 2 Since x 0 c x x 1 (x 1 x)(x x 0 ) 2 e x 0 e x P 1 (x) = (x 1 x)(x x 0 ) e x 1 2
32 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation For a bound independent of x and e x 1 e on [0, 1] (x 1 x)(x x 0 ) max = h2 x 0 x x 1 2 8, h = x 1 x 0 e x P 1 (x) h2 8 e, 0 x 0 x x 1 1 independent of x, x 0, x 1. Recall that we estimated e = by e 0.82 and e 0.83, i.e., h = e x P 1 (x) h2 8 e = ex P 1 (x) = , 8 The actual error being , it satisfies this bound.
33 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation
34 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation Example Again let f(x) = e x on [0, 1], but consider the quadratic interpolation. e x P 2 (x) = (x x 0)(x x 1 )(x x 2 ) e cx 6 for some c x between the min and max of x 0, x 1, x 2 and x. Assuming evenly spaced points, h = x 1 x 0 = x 2 x 1, and 0 x 0 < x < x 2 1, we have as before e x P 2 (x) (x x 0 )(x x 1 )(x x 2 ) 6 e1 while (x x 0 )(x x 1 )(x x 2 ) max = h3 x 0 x x (5.24) hence e x P 2 (x) h3 e h3
35 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation For h = 0.01, 0 x 1 e x P 2 (x)
36 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation Let w 2 (x) = (x + h)x(x h) 6 = x3 xh 6 a translation along xaxis of polynomial in (5.24). Then x = ± h 3 satisfies 0 = w 2(x) = 3x2 h 2 6 and gives w 2 (± 3 h ) = h3 9 3 Figure: y = w 2 (x)
37 > 4. Interpolation and Approximation > Behaviour of the error When we consider the error formula (5.23) or (5.22), the polynomial ψ n (x) = (x x 0 ) (x x n ) is crucial in determining the behaviour of the error. Let assume that x 0,..., x n are evenly spaced and x 0 x x n. Figure: y = Ψ 6 (x)
38 > 4. Interpolation and Approximation > Behaviour of the error Remark that the interpolation error is relatively larger in [x 0, x 1 ] and [x 5, x 6 ], and is likely to be smaller near the middle of the node points In practical interpolation problems, highdegree polynomial interpolation with evenly spaced nodes is seldom used. But when the set of nodes is suitable chosed, highdegree polynomials can be very useful in obtaining polynomial approximations to functions. Example Let f(x) = cos(x), h = 0.2, n = 8 and then interpolate at x = 0.9. Case (i) x 0 = 0.8, x 8 = 2.4 x = 0.9 [x 0, x 1 ]. By direct calculation of P 8 (0.9), cos(0.9) P 8 (0.9) = Case (ii) x 0 = 0.2, x 8 = 1.8 x = 0.9 [x 3, x 4 ], where x 4 is the midpoint. cos(0.9) P 8 (0.9) = , a factor of 24 smaller than the first case.
39 > 4. Interpolation and Approximation > Behaviour of the error Example Le f(x) = 1 1+x 2, x [ 5, 5], n > 0 an even integer, h = 10 n x j = 5 + jh, j = 0, 1, 2,..., n It can be shown that for many points x in [ 5, 5], the sequence of {P n (x)} does not converge to f(x) as n. Figure: The interpolation to 1 1+x 2
40 > 4. Interpolation and Approximation > 4.3 Interpolation using spline functions x y The simplest method of interpolating data in a table: connecting the node points by straight lines: the curve y = l(x) is not very smooth. Figure: y = l(x): piecewise linear interpolation We want to construct a smooth curve that interpolates the given data point that follows the shape of y = l(x).
41 > 4. Interpolation and Approximation > 4.3 Interpolation using spline functions Next choice: use polynomial interpolation. With seven data point, we consider the interpolating polynomial P 6 (x) of degree 6. Figure: y = P 6 (x): polynomial interpolation Smooth, but quite different from y = l(x)!!!
42 > 4. Interpolation and Approximation > 4.3 Interpolation using spline functions A third choice: connect the data in the table using a succession of quadratic interpolating polynomials: on each [0, 2], [2, 3], [3, 4]. Figure: y = q(x): piecewise quadratic interpolation Is smoother than y = l(x), follows it more closely than y = P 6 (x), but at x = 2 and x = 3 the graph has corners, i.e., q (x) is discontinuous.
43 > 4. Interpolation and Approximation > Spline interpolation Cubic spline interpolation Suppose n data points (x i, y i ), i = 1,..., n are given and a = x 1 <... < x n = b We seek a function S(x) defined on [a, b] that interpolates the data S(x i ) = y i, i = 1,..., n Find S(x) C 2 [a, b], a natural cubic spline such that S(x)is a polynomial of degree 3on each subinterval [x j 1, x j ], j = 2, 3,...,n; S(x i ) = y i, i = 1,..., n, S (a) = S (b). On [x i 1, x i ]: 4 degrees of freedom (cubic spline) 4(n 1) DOF.
44 > 4. Interpolation and Approximation > Spline interpolation S(x i ) = y i : n constraints S (x i 0) = S (x i + 0) continuity 2 = 1,..., n 1 S (x i 0) = S (x i + 0) i = 2,..., n 1 S(x i 0) = S(x i + 0) i = 2,..., n 1 n + (3n 6) = 4n 6 constraints 3n 6 Two more constraints needed to be added, Boundary Conditions: S = 0 at a = x 1, x n = b, for a natural cubic spline. (Linear system, with symmetric, positive definite, diagonally dominant matrix.)
45 > 4. Interpolation and Approximation > Construction of the cubic spline Introduce the variables M 1,..., M n with M i S (x i ), i = 1, 2,..., n Since S(x) is cubic on each [x j 1, x j ] S (x) is linear, hence determined by its values at two points: S (x j 1 ) = M j 1, S (x j ) = M j S (x) = (x j x)m j 1 + (x x j 1 )M j x j x j 1, x j 1 x x j (5.25) Form the second antiderivative of S (x) on [x j 1, x j ] and apply the interpolating conditions we get S(x j 1 ) = y j 1, S(x j ) = y j S(x) = (x j x) 3 M j 1 + (x x j 1 ) 3 M j 6(x j x j 1 ) + (x j x)y j 1 + (x x j 1 )y j 6(x j x j 1 ) 1 6 (x j x j 1 ) [(x j x)m j 1 + (x x j 1 )M j ] (5.26) for x [x j 1, x j ], j = 1,..., n.
46 > 4. Interpolation and Approximation > Construction of the cubic spline Formula (5.26) implies that S(x) is continuous over all [a, b] and satisfies the interpolating conditions S(x i ) = y i. Similarly, formula (5.25) for S (x) implies that is continuous on [a, b]. To ensure the continuity of S (x) over [a, b]: we require S (x) on [x j 1, x j ] and [x j, x j+1 ] have to give the same value at their common pointx j, j = 2, 3,..., n 1, leading to the tridiagonal linear system x j x j 1 6 M j 1 + x j+1 x j 1 3 M j + x j+1 x j M j+1 (5.27) 6 = y j+1 y j x j+1 x j y j y j 1 x j x j 1, j = 2, 3,..., n 1 These n 2 equations together with the assumption S (a) = S (b) = 0: M 1 = M n = 0 (5.28) leads to the values M 1,..., M n, hence the function S(x).
47 > 4. Interpolation and Approximation > Construction of the cubic spline Example Calculate the natural cubic spline interpolating the data {(1, 1), (2, 12 ), (3, 13 ), (4, 14 ) } n = 4, and all x j 1 x j = 1. The system (5.27) becomes and with (5.28) this yields which by (5.26) gives S(x) = Are S (x) and S (x) continuous? 1 6 M M M 3 = M M M 4 = 1 12 M 2 = 1 2, M 3 = x3 1 4 x2 1 3 x + 3 2, 1 x x x2 7 3 x , 2 x x , 3 x 4
48 > 4. Interpolation and Approximation > Construction of the cubic spline Example Calculate the natural cubic spline interpolating the data x y n = 7, system (5.27) has 5 equations Figure: Natural cubic spline interpolation y = S(x) Compared to the graph of linear and quadratic interpolation y = l(x) and y = q(x), the cubic spline S(x) no longer contains corners.
49 > 4. Interpolation and Approximation > Other interpolation spline functions So far we only interpolated data points, wanting a smooth curve. When we seek a spline to interpolate a known function, we are interested also in the accuracy. Let f(x) be given on [a, b], that we want to interpolate on evenly spaced values of x. For n > 1, let h = b a n 1, x j = a + (j 1)h, j = 1, 2,..., n and S n (x) be the natural cubic spline interpolating f(x) at x 1,..., x n. It can be shown that max f(x) S n(x) ch 2 (5.29) a x b where c depends on f (a), f (b) and max a x b f (4) (x). The reason why S n (x) doesn t converge more rapidly (have an error bound with a higher power of h) is that f (a) 0 f (b), while by definition S n(a) = S n(b) = 0. For functions f(x) with f (a) = f (b) = 0, the RHS of (5.29) can be replaced with ch 4.
50 > 4. Interpolation and Approximation > Other interpolation spline functions To improve on S n (x), we look for other interpolating functions S(x) that interpolate f(x) on a = x 1 < x 2 <... < x n = b Recall the definition of natural cubic spline: 1 S(x) cubic on each subinterval [x j 1, x j ]; 2 S(x), S (x) and S (x) are continuous on [a, b]; 3 S (x 1 ) = S (x n ) = 0. We say that S(x) is a cubic spline on [a, b] if 1 S(x) cubic on each subinterval [x j 1, x j ]; 2 S(x), S (x) and S (x) are continuous on [a, b]; With the interpolating conditions S(x i ) = y i, i = 1,..., n the representation formula (5.26) and the tridiagonal system (5.27) are still valid.
51 > 4. Interpolation and Approximation > Other interpolation spline functions This system has n 2 equations and n unknowns: M 1,..., M n. By replacing the end conditions (5.28) S (x 1 ) = S (x n ) = 0, we can obtain other interpolating cubic splines. If the data (x i, y i ) is obtained by evaluating a function f(x) y i = f(x i ), i = 1,..., n then we choose endpoints (boundary conditions) for S(x) that would yield a better approximation to f(x). We require S (x 1 ) = f (x 1 ), S (x n ) = f (x n ) or S (x 1 ) = f (x 1 ), S (x n ) = f (x n ) When combined with( ), either of these conditions leads to a unique interpolating spline S(x), dependent on which of these conditions is used. In both cases, the RHS of (5.29) can be replaced by ch 4, where c depends on max x [a,b] f (4) (x).
52 > 4. Interpolation and Approximation > Other interpolation spline functions If the derivatives of f(x) are not known, then extra interpolating conditions can be used to ensure that the error bond of (5.29) is proportional to h 4. In particular, suppose that x 1 < z 1 < x 2, x n 1 < z 2 < x n and f(z 1 ), f(z 2 ) are known. Then use the formula for S(x) in (5.26) and s(z 1 ) = f(z 1 ), s(z 2 ) = f(z 2 ) (5.30) This adds two new equations to the system (5.27), one for M 1 and M 2, and another equation for M n 1, M n. This form is preferable to the interpolating natural cubic spline, and is almost equally easy to produce. This is the default form of spline interpolation that is implemented in MATLAB. The form of spline formed in this way is said to satisfy the notaknot interpolation boundary conditions.
53 > 4. Interpolation and Approximation > Other interpolation spline functions Interpolating cubic spline functions are a popular way to represent data analytically because 1 are relatively smooth: C 2, 2 do not have the rapid oscillation that sometime occurs with the highdegree polynomial interpolation, 3 are relatively easy to work with on a computer. They do not replace polynomials, but are a very useful extension of them.
54 > 4. Interpolation and Approximation > The MATLAB program spline The standard package MATLAB contains the function rm spline. The standard calling sequence is y = spline(x nodes, y nodes, x) which produces the cubic spline function S(x) whose graph passes through the points {(ξ i, η i ) : i = 1,..., n} with (ξ i, η i ) = (x nodes(i), y nodes (i)) and n the length of x nodes (and y nodes). The notaknot interpolation conditions of (5.30) are used. The point (ξ 2, η 2 ) is the point (z 1, f(z 1 )) of (5.30) and (ξ n 1, η n 1 ) is the point (z 2, f(z 2 )).
55 > 4. Interpolation and Approximation > The MATLAB program spline Example Approximate the function f(x) = e x on the interval [a, b] = [0, 1]. For n > 0, define h = 1/n and interpolating nodes x 1 = 0, x 2 = h, x 3 = 2h,..., x n+1 = nh = 1 Using spline, we produce the cubic interpolating spline interpolant S n,1 to f(x). With the notaknot interpolation conditions, the nodes x 2 and x n are the points z 1 and z 2 in (5.30). For a general smooth function f(x), it turns out that te magnitude of the error f(x) S n,1 (x) is largest around the endpoints of the interval of approximation.
56 > 4. Interpolation and Approximation > The MATLAB program spline Example Two interpolating nodes are inserted, the midpoints of the subintervals [0, h] and [1 h, 1]: x 1 =0, x 2 = 1 2 h, x 3 =h, x 4 =2h,..., x n+1 =(n 1)h, x n+2 =1 1 2 h, x n+3 =1 Using spline results in a cubic spline function S n,2 (x); with the notaknot interpolation conditions conditions, the nodes x 2 and x n+2 are the points z 1 and z 2 of (5.30). Generally, S n,2 is a more accurate approximation than is S n,1 (x). The cubic polynomials produced for S n,2 (x) by spline for the intervals [x 1, x 2 ] and [x 2, x 3 ] are the same, thus we can use the polynomial for [0, 1 2 h] for the entire interval [0, h]. Similarly for [1 h, h]. n E n (1) Ratio E n (2) Ratio E E E E E E E E Table: Cubic spline approximation to f(x) = e x
INTERPOLATION. Interpolation is a process of finding a formula (often a polynomial) whose graph will pass through a given set of points (x, y).
INTERPOLATION Interpolation is a process of finding a formula (often a polynomial) whose graph will pass through a given set of points (x, y). As an example, consider defining and x 0 =0, x 1 = π 4, x
More information4.3 Lagrange Approximation
206 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION Lagrange Polynomial Approximation 4.3 Lagrange Approximation Interpolation means to estimate a missing function value by taking a weighted average
More informationPiecewise Cubic Splines
280 CHAP. 5 CURVE FITTING Piecewise Cubic Splines The fitting of a polynomial curve to a set of data points has applications in CAD (computerassisted design), CAM (computerassisted manufacturing), and
More informationSome Notes on Taylor Polynomials and Taylor Series
Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited
More informationLimit processes are the basis of calculus. For example, the derivative. f f (x + h) f (x)
SEC. 4.1 TAYLOR SERIES AND CALCULATION OF FUNCTIONS 187 Taylor Series 4.1 Taylor Series and Calculation of Functions Limit processes are the basis of calculus. For example, the derivative f f (x + h) f
More information1 if 1 x 0 1 if 0 x 1
Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or
More informationNovember 16, 2015. Interpolation, Extrapolation & Polynomial Approximation
Interpolation, Extrapolation & Polynomial Approximation November 16, 2015 Introduction In many cases we know the values of a function f (x) at a set of points x 1, x 2,..., x N, but we don t have the analytic
More information4.5 Chebyshev Polynomials
230 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION 4.5 Chebyshev Polynomials We now turn our attention to polynomial interpolation for f (x) over [ 1, 1] based on the nodes 1 x 0 < x 1 < < x N 1. Both
More informationTOPIC 3: CONTINUITY OF FUNCTIONS
TOPIC 3: CONTINUITY OF FUNCTIONS. Absolute value We work in the field of real numbers, R. For the study of the properties of functions we need the concept of absolute value of a number. Definition.. Let
More informationHOMEWORK 4 SOLUTIONS. All questions are from Vector Calculus, by Marsden and Tromba
HOMEWORK SOLUTIONS All questions are from Vector Calculus, by Marsden and Tromba Question :..6 Let w = f(x, y) be a function of two variables, and let x = u + v, y = u v. Show that Solution. By the chain
More informationHomework set 1: posted on website
Homework set 1: posted on website 1 Last time: Interpolation, fitting and smoothing Interpolated curve passes through all data points Fitted curve passes closest (by some criterion) to all data points
More informationFunctions and Equations
Centre for Education in Mathematics and Computing Euclid eworkshop # Functions and Equations c 014 UNIVERSITY OF WATERLOO Euclid eworkshop # TOOLKIT Parabolas The quadratic f(x) = ax + bx + c (with a,b,c
More information1 Review of Newton Polynomials
cs: introduction to numerical analysis 0/0/0 Lecture 8: Polynomial Interpolation: Using Newton Polynomials and Error Analysis Instructor: Professor Amos Ron Scribes: Giordano Fusco, Mark Cowlishaw, Nathanael
More informationEECS 556 Image Processing W 09. Interpolation. Interpolation techniques B splines
EECS 556 Image Processing W 09 Interpolation Interpolation techniques B splines What is image processing? Image processing is the application of 2D signal processing methods to images Image representation
More information5 Numerical Differentiation
D. Levy 5 Numerical Differentiation 5. Basic Concepts This chapter deals with numerical approximations of derivatives. The first questions that comes up to mind is: why do we need to approximate derivatives
More informationEigenvalues and eigenvectors of a matrix
Eigenvalues and eigenvectors of a matrix Definition: If A is an n n matrix and there exists a real number λ and a nonzero column vector V such that AV = λv then λ is called an eigenvalue of A and V is
More informationPUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.
PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include
More information1 Review of Least Squares Solutions to Overdetermined Systems
cs4: introduction to numerical analysis /9/0 Lecture 7: Rectangular Systems and Numerical Integration Instructor: Professor Amos Ron Scribes: Mark Cowlishaw, Nathanael Fillmore Review of Least Squares
More informationRolle s Theorem. q( x) = 1
Lecture 1 :The Mean Value Theorem We know that constant functions have derivative zero. Is it possible for a more complicated function to have derivative zero? In this section we will answer this question
More informationQuotient Rings and Field Extensions
Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.
More informationHøgskolen i Narvik Sivilingeniørutdanningen STE6237 ELEMENTMETODER. Oppgaver
Høgskolen i Narvik Sivilingeniørutdanningen STE637 ELEMENTMETODER Oppgaver Klasse: 4.ID, 4.IT Ekstern Professor: Gregory A. Chechkin email: chechkin@mech.math.msu.su Narvik 6 PART I Task. Consider twopoint
More information106 Chapter 5 Curve Sketching. If f(x) has a local extremum at x = a and. THEOREM 5.1.1 Fermat s Theorem f is differentiable at a, then f (a) = 0.
5 Curve Sketching Whether we are interested in a function as a purely mathematical object or in connection with some application to the real world, it is often useful to know what the graph of the function
More informationSOLVING POLYNOMIAL EQUATIONS
C SOLVING POLYNOMIAL EQUATIONS We will assume in this appendix that you know how to divide polynomials using long division and synthetic division. If you need to review those techniques, refer to an algebra
More informationTOPIC 4: DERIVATIVES
TOPIC 4: DERIVATIVES 1. The derivative of a function. Differentiation rules 1.1. The slope of a curve. The slope of a curve at a point P is a measure of the steepness of the curve. If Q is a point on the
More informationNatural cubic splines
Natural cubic splines Arne Morten Kvarving Department of Mathematical Sciences Norwegian University of Science and Technology October 21 2008 Motivation We are given a large dataset, i.e. a function sampled
More informationROOTFINDING : A PRINCIPLE
ROOTFINDING : A PRINCIPLE We want to find the root α of a given function f(x). Thus we want to find the point x at which the graph of y = f(x) intersects the xaxis. One of the principles of numerical
More informationExamination paper for Solutions to Matematikk 4M and 4N
Department of Mathematical Sciences Examination paper for Solutions to Matematikk 4M and 4N Academic contact during examination: Trygve K. Karper Phone: 99 63 9 5 Examination date:. mai 04 Examination
More information3.8 Finding Antiderivatives; Divergence and Curl of a Vector Field
3.8 Finding Antiderivatives; Divergence and Curl of a Vector Field 77 3.8 Finding Antiderivatives; Divergence and Curl of a Vector Field Overview: The antiderivative in one variable calculus is an important
More informationUNIT  I LESSON  1 The Solution of Numerical Algebraic and Transcendental Equations
UNIT  I LESSON  1 The Solution of Numerical Algebraic and Transcendental Equations Contents: 1.0 Aims and Objectives 1.1 Introduction 1.2 Bisection Method 1.2.1 Definition 1.2.2 Computation of real root
More information2 Complex Functions and the CauchyRiemann Equations
2 Complex Functions and the CauchyRiemann Equations 2.1 Complex functions In onevariable calculus, we study functions f(x) of a real variable x. Likewise, in complex analysis, we study functions f(z)
More informationAPPLICATIONS OF DIFFERENTIATION
4 APPLICATIONS OF DIFFERENTIATION APPLICATIONS OF DIFFERENTIATION So far, we have been concerned with some particular aspects of curve sketching: Domain, range, and symmetry (Chapter 1) Limits, continuity,
More informationSolutions to SelfTest for Chapter 4 c4sts  p1
Solutions to SelfTest for Chapter 4 c4sts  p1 1. Graph a polynomial function. Label all intercepts and describe the end behavior. a. P(x) = x 4 2x 3 15x 2. (1) Domain = R, of course (since this is a
More informationFIXED POINT ITERATION
FIXED POINT ITERATION We begin with a computational example. solving the two equations Consider E1: x =1+.5sinx E2: x =3+2sinx Graphs of these two equations are shown on accompanying graphs, with the solutions
More informationTaylor and Maclaurin Series
Taylor and Maclaurin Series In the preceding section we were able to find power series representations for a certain restricted class of functions. Here we investigate more general problems: Which functions
More informationThe Method of Partial Fractions Math 121 Calculus II Spring 2015
Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method
More informationApplication. Outline. 31 Polynomial Functions 32 Finding Rational Zeros of. Polynomial. 33 Approximating Real Zeros of.
Polynomial and Rational Functions Outline 31 Polynomial Functions 32 Finding Rational Zeros of Polynomials 33 Approximating Real Zeros of Polynomials 34 Rational Functions Chapter 3 Group Activity:
More information10.3 POWER METHOD FOR APPROXIMATING EIGENVALUES
58 CHAPTER NUMERICAL METHODS. POWER METHOD FOR APPROXIMATING EIGENVALUES In Chapter 7 you saw that the eigenvalues of an n n matrix A are obtained by solving its characteristic equation n c nn c nn...
More informationLearning Objectives for Math 165
Learning Objectives for Math 165 Chapter 2 Limits Section 2.1: Average Rate of Change. State the definition of average rate of change Describe what the rate of change does and does not tell us in a given
More informationSection 3.2 Polynomial Functions and Their Graphs
Section 3.2 Polynomial Functions and Their Graphs EXAMPLES: P(x) = 3, Q(x) = 4x 7, R(x) = x 2 +x, S(x) = 2x 3 6x 2 10 QUESTION: Which of the following are polynomial functions? (a) f(x) = x 3 +2x+4 (b)
More informationJUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson
JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials
More informationWe can display an object on a monitor screen in three different computermodel forms: Wireframe model Surface Model Solid model
CHAPTER 4 CURVES 4.1 Introduction In order to understand the significance of curves, we should look into the types of model representations that are used in geometric modeling. Curves play a very significant
More informationPolynomials and Vieta s Formulas
Polynomials and Vieta s Formulas Misha Lavrov ARML Practice 2/9/2014 Review problems 1 If a 0 = 0 and a n = 3a n 1 + 2, find a 100. 2 If b 0 = 0 and b n = n 2 b n 1, find b 100. Review problems 1 If a
More informationInner Product Spaces
Math 571 Inner Product Spaces 1. Preliminaries An inner product space is a vector space V along with a function, called an inner product which associates each pair of vectors u, v with a scalar u, v, and
More information8 Divisibility and prime numbers
8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express
More informationMA107 Precalculus Algebra Exam 2 Review Solutions
MA107 Precalculus Algebra Exam 2 Review Solutions February 24, 2008 1. The following demand equation models the number of units sold, x, of a product as a function of price, p. x = 4p + 200 a. Please write
More information6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives
6 EXTENDING ALGEBRA Chapter 6 Extending Algebra Objectives After studying this chapter you should understand techniques whereby equations of cubic degree and higher can be solved; be able to factorise
More informationMATH 304 Linear Algebra Lecture 20: Inner product spaces. Orthogonal sets.
MATH 304 Linear Algebra Lecture 20: Inner product spaces. Orthogonal sets. Norm The notion of norm generalizes the notion of length of a vector in R n. Definition. Let V be a vector space. A function α
More information7.6 Approximation Errors and Simpson's Rule
WileyPLUS: Home Help Contact us Logout HughesHallett, Calculus: Single and Multivariable, 4/e Calculus I, II, and Vector Calculus Reading content Integration 7.1. Integration by Substitution 7.2. Integration
More informationTaylor Polynomials and Taylor Series Math 126
Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will
More informationHomework # 3 Solutions
Homework # 3 Solutions February, 200 Solution (2.3.5). Noting that and ( + 3 x) x 8 = + 3 x) by Equation (2.3.) x 8 x 8 = + 3 8 by Equations (2.3.7) and (2.3.0) =3 x 8 6x2 + x 3 ) = 2 + 6x 2 + x 3 x 8
More informationDIFFERENTIABILITY OF COMPLEX FUNCTIONS. Contents
DIFFERENTIABILITY OF COMPLEX FUNCTIONS Contents 1. Limit definition of a derivative 1 2. Holomorphic functions, the CauchyRiemann equations 3 3. Differentiability of real functions 5 4. A sufficient condition
More informationIntroduction to Calculus for Business and Economics. by Stephen J. Silver Department of Business Administration The Citadel
Introduction to Calculus for Business and Economics by Stephen J. Silver Department of Business Administration The Citadel I. Functions Introduction to Calculus for Business and Economics y = f(x) is a
More informationIntegrals of Rational Functions
Integrals of Rational Functions Scott R. Fulton Overview A rational function has the form where p and q are polynomials. For example, r(x) = p(x) q(x) f(x) = x2 3 x 4 + 3, g(t) = t6 + 4t 2 3, 7t 5 + 3t
More informationDifferentiation and Integration
This material is a supplement to Appendix G of Stewart. You should read the appendix, except the last section on complex exponentials, before this material. Differentiation and Integration Suppose we have
More informationLimits and Continuity
Math 20C Multivariable Calculus Lecture Limits and Continuity Slide Review of Limit. Side limits and squeeze theorem. Continuous functions of 2,3 variables. Review: Limits Slide 2 Definition Given a function
More informationCHAPTER 1 Splines and Bsplines an Introduction
CHAPTER 1 Splines and Bsplines an Introduction In this first chapter, we consider the following fundamental problem: Given a set of points in the plane, determine a smooth curve that approximates the
More informationReview of Fundamental Mathematics
Review of Fundamental Mathematics As explained in the Preface and in Chapter 1 of your textbook, managerial economics applies microeconomic theory to business decision making. The decisionmaking tools
More informationPractice with Proofs
Practice with Proofs October 6, 2014 Recall the following Definition 0.1. A function f is increasing if for every x, y in the domain of f, x < y = f(x) < f(y) 1. Prove that h(x) = x 3 is increasing, using
More informationWinter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com
Polynomials Alexander Remorov alexanderrem@gmail.com Warmup Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)).
More information5.3 Improper Integrals Involving Rational and Exponential Functions
Section 5.3 Improper Integrals Involving Rational and Exponential Functions 99.. 3. 4. dθ +a cos θ =, < a
More information3.4 Complex Zeros and the Fundamental Theorem of Algebra
86 Polynomial Functions.4 Complex Zeros and the Fundamental Theorem of Algebra In Section., we were focused on finding the real zeros of a polynomial function. In this section, we expand our horizons and
More informationLinear and quadratic Taylor polynomials for functions of several variables.
ams/econ 11b supplementary notes ucsc Linear quadratic Taylor polynomials for functions of several variables. c 010, Yonatan Katznelson Finding the extreme (minimum or maximum) values of a function, is
More informationContents. Introduction and Notes pages 23 (These are important and it s only 2 pages ~ please take the time to read them!)
Page Contents Introduction and Notes pages 23 (These are important and it s only 2 pages ~ please take the time to read them!) Systematic Search for a Change of Sign (Decimal Search) Method Explanation
More information> 2. Error and Computer Arithmetic
> 2. Error and Computer Arithmetic Numerical analysis is concerned with how to solve a problem numerically, i.e., how to develop a sequence of numerical calculations to get a satisfactory answer. Part
More informationDERIVATIVES AS MATRICES; CHAIN RULE
DERIVATIVES AS MATRICES; CHAIN RULE 1. Derivatives of Realvalued Functions Let s first consider functions f : R 2 R. Recall that if the partial derivatives of f exist at the point (x 0, y 0 ), then we
More informationcorrectchoice plot f(x) and draw an approximate tangent line at x = a and use geometry to estimate its slope comment The choices were:
Topic 1 2.1 mode MultipleSelection text How can we approximate the slope of the tangent line to f(x) at a point x = a? This is a Multiple selection question, so you need to check all of the answers that
More informationMATH 2300 review problems for Exam 3 ANSWERS
MATH 300 review problems for Exam 3 ANSWERS. Check whether the following series converge or diverge. In each case, justify your answer by either computing the sum or by by showing which convergence test
More informationIn this lesson you will learn to find zeros of polynomial functions that are not factorable.
2.6. Rational zeros of polynomial functions. In this lesson you will learn to find zeros of polynomial functions that are not factorable. REVIEW OF PREREQUISITE CONCEPTS: A polynomial of n th degree has
More informationFinite Elements for 2 D Problems
Finite Elements for 2 D Problems General Formula for the Stiffness Matrix Displacements (u, v) in a plane element are interpolated from nodal displacements (ui, vi) using shape functions Ni as follows,
More informationMA4001 Engineering Mathematics 1 Lecture 10 Limits and Continuity
MA4001 Engineering Mathematics 1 Lecture 10 Limits and Dr. Sarah Mitchell Autumn 2014 Infinite limits If f(x) grows arbitrarily large as x a we say that f(x) has an infinite limit. Example: f(x) = 1 x
More informationRoots of Polynomials
Roots of Polynomials (Com S 477/577 Notes) YanBin Jia Sep 24, 2015 A direct corollary of the fundamental theorem of algebra is that p(x) can be factorized over the complex domain into a product a n (x
More informationUnderstanding Basic Calculus
Understanding Basic Calculus S.K. Chung Dedicated to all the people who have helped me in my life. i Preface This book is a revised and expanded version of the lecture notes for Basic Calculus and other
More informationMicroeconomic Theory: Basic Math Concepts
Microeconomic Theory: Basic Math Concepts Matt Van Essen University of Alabama Van Essen (U of A) Basic Math Concepts 1 / 66 Basic Math Concepts In this lecture we will review some basic mathematical concepts
More information15. Symmetric polynomials
15. Symmetric polynomials 15.1 The theorem 15.2 First examples 15.3 A variant: discriminants 1. The theorem Let S n be the group of permutations of {1,, n}, also called the symmetric group on n things.
More informationNumerical Methods for Solving Systems of Nonlinear Equations
Numerical Methods for Solving Systems of Nonlinear Equations by Courtney Remani A project submitted to the Department of Mathematical Sciences in conformity with the requirements for Math 4301 Honour s
More information3. Interpolation. Closing the Gaps of Discretization... Beyond Polynomials
3. Interpolation Closing the Gaps of Discretization... Beyond Polynomials Closing the Gaps of Discretization... Beyond Polynomials, December 19, 2012 1 3.3. Polynomial Splines Idea of Polynomial Splines
More information88 CHAPTER 2. VECTOR FUNCTIONS. . First, we need to compute T (s). a By definition, r (s) T (s) = 1 a sin s a. sin s a, cos s a
88 CHAPTER. VECTOR FUNCTIONS.4 Curvature.4.1 Definitions and Examples The notion of curvature measures how sharply a curve bends. We would expect the curvature to be 0 for a straight line, to be very small
More informationThe Steepest Descent Algorithm for Unconstrained Optimization and a Bisection Linesearch Method
The Steepest Descent Algorithm for Unconstrained Optimization and a Bisection Linesearch Method Robert M. Freund February, 004 004 Massachusetts Institute of Technology. 1 1 The Algorithm The problem
More informationAn important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate.
Chapter 10 Series and Approximations An important theme in this book is to give constructive definitions of mathematical objects. Thus, for instance, if you needed to evaluate 1 0 e x2 dx, you could set
More informationx if x 0, x if x < 0.
Chapter 3 Sequences In this chapter, we discuss sequences. We say what it means for a sequence to converge, and define the limit of a convergent sequence. We begin with some preliminary results about the
More informationMathematical Procedures
CHAPTER 6 Mathematical Procedures 168 CHAPTER 6 Mathematical Procedures The multidisciplinary approach to medicine has incorporated a wide variety of mathematical procedures from the fields of physics,
More informationTHE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS
THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear
More informationContinued Fractions and the Euclidean Algorithm
Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction
More informationMATRIX ALGEBRA AND SYSTEMS OF EQUATIONS. + + x 2. x n. a 11 a 12 a 1n b 1 a 21 a 22 a 2n b 2 a 31 a 32 a 3n b 3. a m1 a m2 a mn b m
MATRIX ALGEBRA AND SYSTEMS OF EQUATIONS 1. SYSTEMS OF EQUATIONS AND MATRICES 1.1. Representation of a linear system. The general system of m equations in n unknowns can be written a 11 x 1 + a 12 x 2 +
More informationContinuous Functions, Smooth Functions and the Derivative
UCSC AMS/ECON 11A Supplemental Notes # 4 Continuous Functions, Smooth Functions and the Derivative c 2004 Yonatan Katznelson 1. Continuous functions One of the things that economists like to do with mathematical
More informationSlant asymptote. This means a diagonal line y = mx + b which is approached by a graph y = f(x). For example, consider the function: 2x 2 8x.
Math 32 Curve Sketching Stewart 3.5 Man vs machine. In this section, we learn methods of drawing graphs by hand. The computer can do this much better simply by plotting many points, so why bother with
More informationSection 3.7 Rational Functions
Section 3.7 Rational Functions A rational function is a function of the form where P and Q are polynomials. r(x) = P(x) Q(x) Rational Functions and Asymptotes The domain of a rational function consists
More informationZeros of a Polynomial Function
Zeros of a Polynomial Function An important consequence of the Factor Theorem is that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we
More informationMath 120 Final Exam Practice Problems, Form: A
Math 120 Final Exam Practice Problems, Form: A Name: While every attempt was made to be complete in the types of problems given below, we make no guarantees about the completeness of the problems. Specifically,
More informationNumerical Solution of Differential
Chapter 13 Numerical Solution of Differential Equations We have considered numerical solution procedures for two kinds of equations: In chapter 10 the unknown was a real number; in chapter 6 the unknown
More information3 Polynomial Interpolation
3 Polynomial Interpolation Read sections 7., 7., 7.3. 7.3.3 (up to p. 39), 7.3.5. Review questions 7. 7.4, 7.8 7.0, 7. 7.4, 7.7, 7.8. All methods for solving ordinary differential equations which we considered
More informationRoots and Coefficients of a Quadratic Equation Summary
Roots and Coefficients of a Quadratic Equation Summary For a quadratic equation with roots α and β: Sum of roots = α + β = and Product of roots = αβ = Symmetrical functions of α and β include: x = and
More informationSection 3.7. Rolle s Theorem and the Mean Value Theorem. Difference Equations to Differential Equations
Difference Equations to Differential Equations Section.7 Rolle s Theorem and the Mean Value Theorem The two theorems which are at the heart of this section draw connections between the instantaneous rate
More information0.8 Rational Expressions and Equations
96 Prerequisites 0.8 Rational Expressions and Equations We now turn our attention to rational expressions  that is, algebraic fractions  and equations which contain them. The reader is encouraged to
More informationReal Roots of Univariate Polynomials with Real Coefficients
Real Roots of Univariate Polynomials with Real Coefficients mostly written by Christina Hewitt March 22, 2012 1 Introduction Polynomial equations are used throughout mathematics. When solving polynomials
More informationUNCORRECTED PAGE PROOFS
number and and algebra TopIC 17 Polynomials 17.1 Overview Why learn this? Just as number is learned in stages, so too are graphs. You have been building your knowledge of graphs and functions over time.
More informationparametric spline curves
parametric spline curves 1 curves used in many contexts fonts (2D) animation paths (3D) shape modeling (3D) different representation implicit curves parametric curves (mostly used) 2D and 3D curves are
More informationComputational Geometry Lab: FEM BASIS FUNCTIONS FOR A TETRAHEDRON
Computational Geometry Lab: FEM BASIS FUNCTIONS FOR A TETRAHEDRON John Burkardt Information Technology Department Virginia Tech http://people.sc.fsu.edu/ jburkardt/presentations/cg lab fem basis tetrahedron.pdf
More informationA Brief Review of Elementary Ordinary Differential Equations
1 A Brief Review of Elementary Ordinary Differential Equations At various points in the material we will be covering, we will need to recall and use material normally covered in an elementary course on
More informationSeparable First Order Differential Equations
Separable First Order Differential Equations Form of Separable Equations which take the form = gx hy or These are differential equations = gxĥy, where gx is a continuous function of x and hy is a continuously
More information