Polynomial Interpolation
|
|
- Cora Lawrence
- 5 years ago
- Views:
Transcription
1 Capter 4 Polynomial Interpolation In tis capter, we consider te important problem of approximating a function f(x, wose values at a set of distinct points x, x, x 2,,x n are known, by a polynomial P (x suc tat P (x i =f(x i, i =,, 2,,n Suc a polynomial is known as an interpolating polynomial An approximation of tis function is desirable if f(x is difficult to evaluate or manipulate like te error function erf(x = 2 x e t2 dt, π for example Note tat polynomials are easily evaluated, differentiated or integrated Anoter purpose of suc an approximation is tat a very long table of function values f(x i may be replaced by a sort table and a compact interpolation subroutine 4 Linear Interpolation Let f(x be given at two distinct points x i and x i+ Now, given f(x i =f i and f(x i+ =f i+, we wis to determine te interpolating polynomial P (x of degree (ie, a straigt line tat approximates f(x onteinterval[x i,x i+ ] Let P (x =αx + β From te conditions P (x i =f i and P (x i+ =f i+ we obtain te two equations ie, ( xi x i+ αx i + β = f i αx i+ + β = f i+, ( α β = ( fi f i+ Tis system is clearly nonsingular since x i x i+, and it as a unique solution: α = f i+ f i x i+ x i β = f ix i+ f i+ x i x i+ x i 68
2 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 69 Terefore, Example 4 wic gives ( fi+ f i P (x =f i + (x x i x i+ x i Using linear interpolation, approximate sin 65 from a table (a sin 6 =45; sin 7 =287 (b sin =;sin =765 Solution ( (a P (65 = (b P (65 = ( Te first answer is correct to 5 decimals wereas te second answer is correct only to 2 decimals! Conclusion: Linear interpolation is suitable only over small intervals 42 Polynomial Interpolation Since linear interpolation is not adequate unless te given points are closely spaced, we consider iger order interpolating polynomials Let f(x begiven at te selected sample of (n + points: x <x < <x n, ie, we ave (n +pairs(x i,f i, i =,, 2,,n Te objective now is to find te lowest degree polynomial tat passes troug tis selected sample of points f(x P n (x f(x x x x 2 x x n Figure 4: Interpolating te function f(x by a polynomial of degree n, P n (x Consider te nt degree polynomial P n (x =a + a x + a 2 x a n x n
3 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 7 We wis to determine te coefficients a j, j =,,,n, suc tat P n (x j =f(x j, j =,, 2,,n Tese (n + conditions yield te linear system a + a x + a 2 x a nx n = f a + a x + a 2 x a nx n = f a + a x 2 + a 2 x a n x n 2 = f 2 a + a x n + a 2 x 2 n + + a nx n n = f n or V a = f, were a T =(a,a,,a n, f T =(f,f,,f n, and V is an (n + (n +matrixgivenby x x 2 x n x x 2 x n V = x 2 x 2 2 x n 2 x n x 2 n x n n Te matrix V is known as te Vandermonde matrix nonzero determinant n det(v = (x i x j j= i>j It is nonsingular wit For n =, for example, det(v =(x x (x 2 x (x x (x 2 x (x x (x x 2 Hence a can be uniquely determined as a = V f Anoter proof of te uniqueness of te interpolating polynomial can be given as follows Teorem 4 Uniqueness of interpolating polynomial Given a set of points x <x < <x n, tere exists only one polynomial tat interpolates a function at tose points Proof Let P (x andq(x be two interpolating polynomials of degree at most n, for te same set of points x <x < <x n Also, let R(x =P (x Q(x Ten R(x is also a polynomial of degree at most n Since P (x i =Q(x i =f i, we ave, R(x i =fori =,,,n In oter words R(x as(n +roots From te Fundamental Teorem of Algebra owever, R(x cannot ave more tan n roots A contradiction! Tus, R(x, and P (x Q(x Example 42 Given te following table for te function f(x, obtain te lowest degree interpolating polynomial
4 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 7 x k 2 4 f k 4 4 Solution Since te number of nodes = = n +, terefore n =2 Let P 2 (x =a + a x + a 2 x 2, tat satisfies te following conditions at te points x =;x =2,x 2 =4: P ( = ; P (2 = 4; P (4 = a a = a 2 4 Using Gaussian elimination wit partial pivoting, we compute te solution a = a a = 4 2, a 2 Terefore, te interpolating polynomial is given by P 2 (x = 4 2x + x 2 We can use tis approximation to estimate te root of f(x tat lies in te interval [2, 4]: γ = =+ 5 2 An important remark is in order One, in general, sould not determine te interpolating polynomial by solving te Vandermonde linear system Tese systems are surprisingly ill-conditioned for n no larger tan For example, for <x <x < <x n = uniformly distributed in [,], large n yields a Vandermonde matrix wit almost linearly dependent columns, and te Vandermonde system becomes almost singular A more satisfactory form of te interpolating polynomial is due to Lagrange, P n (x =f(x L (x+f(x L (x+ + f(x n L n (x Here L j (x, for j n, are polynomials of degree n called fundamental polynomials or Lagrange polynomials From te(n + conditions P n (x i =f(x i, i n, we see tat L j (x i = { i j i = j
5 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 72 Te above property is certainly satisfied if we coose ie, L j (x = (x x (x x (x x j (x x j+ (x x n (x j x (x j x (x j x j (x j x j+ (x j x n L j (x = n i= i j (x x i (x j x i Te fact tat L j (x is unique follows from te uniqueness of te interpolation polynomial Recall tat tere is one and only one polynomial of degree n or less tat interpolates f(x atte(n +nodesx <x < <x n Tus,fortisgiven set of nodes, bot forms yield te same polynomial! Vandermonde Approac n P n (x = a i x i i= V a = f Operation count: It can be sown tat te Vandermonde system V a = f can be solved in O(n 2 aritmetic operations rater tan O(n operations because of te special form of te Vandermonde matrix Lagrange Approac P n (x = n f j L j (x = j= n j= g j n (x x i i= (x x j = n g j ν j j= were g j = f(x j n (x j x i i= i j Operation count: Te Lagrange form avoids solving te ill-conditioned Vandermonde linear, systems, and evaluates te polynomials ν j = (x x j n (x x i i=
6 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 7 for a given x Te cost of tese operations for eac j =,, 2,,n are: u(x = n (x x i (2n +ops i= ν j = u(x (x x j f j g j = n (x j x i i= i j (n +ops Now, te cost for evaluating te polynomial P n (x = 2n(n + ops (evaluated once only n g j ν j j= (2n +ops Terefore, te total number of operations are 5n +onceg j s are evaluated 4 Error in Polynomial Interpolation Let e n (x be te error in polynomial interpolation given by Since P n (x i =f(x i, we ave e n (x =f(x P n (x e n (x i = x <x < <x n In oter words e n (x is a function wit at least (n + roots Hence, we can f(x P n (x f(x x x x 2 x x n Figure 42: Computing te error in polynomial interpolation write e n (x =(x x (x x (x x n (x =π(x (x were (x is a function of x Let us define g(z =e n (z π(z(x
7 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 74 were x x j,for j n Tus, g(x =, g(x j =, j n, ie, g(z as(n + 2 distinct roots Assuming tat f(x as at least n + continuous derivatives, te (n + t derivative of g(z, ie, d (n+ dz g(z (n+ g(n+ (z as at least one root x Hence, g(z x x x 2 x x 4 g ( z g ( z g ( z g iv ( z x* Figure 4: Te (n + t derivative of a function g(x witn +2 roots(n = must ave at least one root g (n+ (z =e (n+ n (z (xπ (n+ (z, were π (n+ (z =(n +! Moreover, since e n (z =f(z P n (z, terefore, e (n+ n (z =f (n+ (z P (n+ (z in wic te last term P n (n+ (z is zero Using te above equations, we ave g (n+ (z =f (n+ (z (x(n +! We ad sown earlier tat g (n+ (z as at least one root x in te interval containing x and x j, j =,,n Tis implies tat g (n+ (x =, and terefore, (x = f (n+ (x (n +! n
8 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 75 Consequently, we ave te following representation of te error in polynomial interpolation e n (x =f(x P n (x =(x x (x x (x x n f (n+ (x (n +! were x is in te interval containing x and x j, j =,,n Example 4 Obtain te Lagrange interpolating polynomial from te data sin =, sin π 6 = 2, sin π = 2, sin π 2 =, and use it to evaluate sin π 2 and sin π 4 Solution Te interpolating polynomial is given by At x = π 2, Terefore, P (x = f L (x+f L (x+f 2 L 2 (x+f L (x = 2 L (x+ 2 L 2(x+L (x 2 L = 2 π 2 π ( 2 2 π 6 6 π 6 π = L 2 = 2 π 6 2 π ( 2 2 π π 6 π = L = 2 π 6 2 π ( 2 π 2 2 π 6 2 π = 6 P = Since sin π 2 =25882, we ave error =8 At x = π 4, L =5625, L 2 =5625, L = Terefore P = Since sin π 4 =77, we ave error =22 Let us estimate te error e n (x atx = π 4 : e = π 6 4 π 4 π f (4 (x 2 4! Since f(x =sinx; f (4 (x =sinx, and tus, f (4 (x
9 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 76 Terefore, ( π π e 4 4 π 2 π 2 π 4 4! =76 Similarly, e n (x atx = π 2 is bounded by ( π π e 2 2 π 2 π 4 5π 2 4! = Runge s Function and Equidistant Interpolation In tis section, we discuss a problem tat arises wen one constructs a global interpolation polynomial on a fairly large number of equally spaced nodes Consider te problem of approximating te following function on te interval [, ] f(x = +25x 2 using te interpolation polynomial P n (x on te equidistant nodes x j = + 2j n, j =,,,n Figure 44 sows ow te polynomials P 4 (x, P 8 (x, P 2 (x, and P 6 (x approximate f(x on[, ] We see tat as n increases, te interpolation error in te central portion of te interval diminises, wile increasing rapidly near te ends of te interval Suc beavior is typical in equidistant interpolation wit polynomials of ig degree Tis is known as Runge s penomenon, and te function f(x is known as Runge s function If te (n + interpolation nodes are not cosen equally spaced, but rater placed near te ends of te interval at te roots of te so called Cebysev polynomial of degree (n +, te problem wit Runge s function disappears (see Fig 45 Te roots of te Cebysev polynomial of degree (n + are given by ( k + 2 x k = cos n + π, k =,,,n 45 Te Newton Representation Te Lagrange form for te interpolating polynomial is not suitable for practical computations Te Newton form of te interpolating polynomial proves to be muc more convenient Let us obtain a different form of te lowest degree interpolating polynomial at te distinct interpolation nodes given by x <x <x 2 < <x n
10 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 77 n = 4 n = n = n = Figure 44: Equal spaced interpolation of Runge s function were f(x j =f j, j =,,,n Let te interpolating polynomial be Using te n conditions P n (x = c + c (x x +c 2 (x x (x x + c (x x (x x (x x c n (x x (x x (x x n P n (x j =f j, j =,,,n, we can construct a system of equations in te unknown c j For n =,for example, tis yields a system of linear equations of te form Lc = f, werel is given by, L = (x x (x 2 x (x 2 x (x 2 x (x x (x x (x x (x x (x x (x x 2 Note tat te evaluation of L for n nodes requires O(n 2 aritmetic operations Assuming tat te nodes are equidistant, ie, x i+ x i =, te system Lc = f
11 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 78 2 n = 4 2 n = n = n = Figure 45: Cebysev interpolation of Runge s function is given by Step c c c 2 c = f f f 2 f c = f f ( = (f f / f ( 2 = (f 2 f /2 f ( = (f f / Ten we ave te system order : Step 2 c c 2 = c c = f ( f ( f ( 2 f (
12 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 79 f (2 2 = (f ( 2 f ( / f (2 = (f ( f ( /2 Ten we get te system of order 2: ( ( c2 c = ( f (2 2 f (2 Step c 2 = f (2 2 f ( = (f (2 f (2 2 / Step 4 c = f ( MATLAB function c = InterpolateNewton(x,f % x: interpolating nodes % f: function values at x % c: coef of Newton s form of interpolating polynomial n = lengt(x; for k = :n- f(k+:n = (f(k+:n-f(k/(x(k+:n-x(k; end c = y; 46 Spline Interpolation Draftsmen used to draw smoot curves troug data points by using splines Tese are tin flexible strips of plastic or wood wic were laid on paper and eld wit weigts so as to pass troug te required data points (or nodes Te weigts are constructed in suc a way tat te spline is free to slip As a result, te flexible spline straigtens out as muc as it can subject to passing over tese points Teory of elasticity suggests tat tis mecanical spline is given by a cubic polynomial (degree polynomial in eac subinterval [x i,x i+ ], i =, 2,,n, wit adjacent cubics joint continuously wit continuous first and second derivatives Let s(x =a + a x + a 2 x 2 + a x, be te form of eac cubic spline Since we ave (n subintervals [x i,x i+ ], i n, ten we ave 4(n unknowns; (4 parameters for eac subinterval: x,a,a 2,anda Suppose s i (x denotestecubicsplineteit interval Te conditions to be satisfied by tese cubic splines are:
13 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 8 x x 2 x x 4 Figure 46: Spline interpolation Continuity at eac interior node: s i (x i =s i (x i, i =2,,,n 2 Continuity of first derivative at eac interior node: s i (x i=s i (x i, i =2,,,n Continuity of second derivative at eac interior node: s i (x i=s i (x i, i =2,,,n 4 Interpolation of function at eac node: s(x j =f(x j, j =, 2,,n Tese provide (n 2 + n =4n 6 conditions; owever, in order to completely specify te cubic splines, we still need 2 additional conditions Tese are conditions specified at te end points x and x n For te mecanical or natural spline we set s (x =s (x n = Let te nodes x j, j =, 2,,n, be equidistant, ie, x j+ x j = and let, ie, σ j = 6 s (x j s i (x i=s i (x i=6σ i
14 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 8 for all i satisfying condition Since we ave s(x =a + a x + a 2 x 2 + a x, s (x = a +2a 2 x +a x 2 s (x = 2a 2 +6a x, ie, s (x is a straigt line Tus, s i (x takes te following form (see Fig 47: i (x =x i+ x (6σ i + (x x i (6σ i+, i =, 2,,n s s'' i ( x 6σ i+ 6σ i x i x xi+ Figure 47: Computing spline witin an interval Integrating once, we ave s i (x = 6σ i (x i+ x 2 + 6σ i+ (x x i 2 + β 2 2 and integrating once more, we ave s i (x = σ i (x i+ x + σ i+ (x x i + β x + β 2 were β and β 2 are constants Clearly we can write s i (x as follows, s i (x = σ i (x i+ x + σ i+ (x x i + γ (x x i +γ 2 (x i+ x Using conditions 4 and, Terefore, s i (x i =f i f i = σ i 2 + γ 2 s i (x i+ =f i+ f i+ = σ i+ 2 + γ γ = f i+ σ i+, γ 2 = f i σ i,
15 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 82 and s i (x = σ i (x i+ x + σ i+ ( fi + σ i ( (x x i fi+ + (x i+ x s i (x = σ i (x i+ x 2 + σ i+ (x x i 2 + ( fi σ i σ i+ (x x i ( fi+ σ i+ Using condition 2, ie, s i (x i=s i (x i, we get ( ( fi+ f i fi f i σ i + (σ i+ σ i =σ i + (σ i σ i or σ i+ +4σ i + σ i = i i i =2,,,n, were i = f i+ f i Using te additional conditions σ = σ n =,weave 4σ 2 + σ = ( 2 / σ i +4σ i + σ i+ = ( i i / i =, 4,,n 2 σ n 2 +4σ n = ( n n 2 / Now, assigning g j = j j we ave te linear system = f j+ 2f j + f j 2, σ 2 σ σ n 2 σ n = g 2 g g n 2 g n Te tridiagonal system above, denoted by [, 4, ], may be factored as follows: µ λ µ 2 λ 2 µ λ n µ n 2
16 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 8 Te following conditions must be satisfied: µ =4, λ i µ i =, λ i + µ i+ =4 Te algoritm for computing L andu is given below Triangular decomposition for system [,4,] µ =4 for i =, 2,,n λ i = µ i µ i+ =4 λ i end; Once L andu are obtained, te above system is solved as sown in Capter
17 Cap 4 Polynomial Interpolation Numerical Metods Class Notes Appendix: Cubic Spline Interpolation In tis Appendix, we discuss ow to find out te coefficients in te cubic splines in an intuitive way (no integration involved Basically, it follows te 4 conditions to derive te coefficients but will result in a better way for uman beings to calculate Tecubicpolynomialoninterval[x i,x i+ ]is s i (x =a i + b i (x x i +c i (x x i 2 + d i (x x i, i =,,n ( Apply condition (iv, So, we ave all te values of a i s (2 Apply condition (i, Ten, s i (x i =a i = f(x i, i =,,n s i (x i+ =s i+ (x i+ =f(x i+, i =,,n 2 s i (x i+ =a i + b i (x i+ x i +c i (x i+ x i 2 + d i (x i+ x i = f(x i+, were i =,,n 2 Let i = x i+ x i Ten, a i + b i i + c i 2 i + d i i = f(x i+ =a i+, i =,,n 2 Since i is from to n 2, we migt define a n = f(x n Ten, a i+ = a i + b i i + c i 2 i + d i i, i =,,n (A ( Te first derivative of s i is Apply condition (ii, Ten, s i (x =b i +2c i (x x i +d i (x x i 2 s i(x i+ =s i+(x i+ =b i+, i =,,n 2 b i+ = s i(x i+ =b i +2c i i +d i 2 i, i =,,n 2 Since i is from to n 2, we migt define b n = s (x n Ten, b i+ = b i +2c i i +d i 2 i, i =,,n (B
18 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 85 (4 Te second derivative of s i is s i (x =2c i +6d i (x x i Apply condition (iii, s i (x i+ =s i+(x i+ =2c i+, i =,,n 2 Ten, 2c i+ = s i (x i+ =2c i +6d i i, i =,,n 2 Since i is from to n 2, we migt define c n = s (x n /2 Ten, c i+ = c i +d i i, i =,,n (C (5 We ave all a i s in (, for all i =, n From(C,weknow d i = c i+ c i, i =,,n (D i Introduce (D into (A, ten a i+ = a i + b i i + c i 2 i + d i i = a i + b i i + c i 2 i + c i+ c i i i = a i + b i i + 2 i (c i+ +2c i (E Introduce (D into (B, ten b i+ = b i +2c i i +d i 2 i = b i +2c i i +(c i+ c i i = b i + i (c i+ + c i (F (6 From (E, we get b i i = (a i+ a i 2 i (c i+ +2c i b i = (a i+ a i i i (c i+ +2c i (G Consider (F wit (G using indices i and i (ie (F is written as b i = b i + i (c i + c i, were i =2,,n Ten, (a i+ a i i i (c i+ +2c i = (a i a i i i (c i +2c i + i (c i + c i (a i+ a i (a i a i = i i i (c i+ +2c i i (c i +2c i + i (c i + c i (a i+ a i (a i a i i i = i (c i+ +2c i i (c i +2c i + i (c i + c i = i c i+ +2 i c i i c i 2 i c i + i c i + i c i = i c i+ +2 i c i +2 i c i + i c i = i c i +2c i ( i + i + i c i+,
19 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 86 were i =2,,n Ten, i c i +2c i ( i + i + i c i+ = (a i+ a i (a i a i, i i were i =2,,n (7 Now, use te natural conditions Since S (x n =, by te definition of c n we gave in (4, c n = On te oter and, because S (x =, s (x =2c +6d (x x = So,c = (8 Consider te results in (6 and (7 We can construct a linear system to solve for all c i s as below c 2( c 2 2 2( 2 + c n 2 2(n n 2 + n n c n c n = 2 (a a 2 (a 2 a (a 4 a 2 (a a 2 n (a n a n n 2 (a n a n 2 Let tis linear system be represented as Ac = b Matriax A is a strictly diagonally dominant matrix; terefore, is nonsingular So te above linear system as an unique solution for c (9 Up to now, we get all a i s and c i s We can obtain b i s and d i s by te equations we ad derived in (D and (G, respectively d i = i (c i+ c i b i = (a i+ a i i i (c i+ +2c i Example 44 Obtain te natural cubic spline interpolation for f(x wose values are given at four points: f( = 2, f( 2 =, f( =, f(4 = Solution Weuseatalbetoelpuscalculate
20 Cap 4 Polynomial Interpolation Numerical Metods Class Notes 87 = x =( a =2 b =( c = d = = x 2 =( 2 a 2 = b 2 =( c 2 = 4 29 d 2 =( 9 = x = a = b =( c =( d = x 4 =4 a 4 = b 4 =x c 4 = d 4 =x 78 We already ave te a i = f i for i =, 2,, 4 Ten, we need to find all c i s One can construct a linear system for c i s as c 8 c 2 2 c = 9 5 c 4 Ten, c = c 4 = Furtermore,c 2 = 4 29 and c =( After aving all c i s and a i s, we can decide te d i s and b i s as follow: d = d 2 = d = (c 2 c = 4 29 = 4 87 (c c 2 = 2 9 ( 9 87 =( 9 78 (c 4 c = = Te cubic splines are b = (a 2 a (c 2 +2c =( b 2 = (a a (c +2c 2 =( b = (a 4 a (c 4 +2c = s = 2+( s 2 = +( 92 4 (x ( (x ( + + (x ( (x ( 22 +( 9 78 s = + 76 (x + ( (x (x Te natural cubic spline interpolation S = s s 2 s (x ( 2
Polynomial Interpolation
Capter 4 Polynomial Interpolation In tis capter, we consider te important problem of approximatinga function fx, wose values at a set of distinct points x, x, x,, x n are known, by a polynomial P x suc
More informationLecture 15. Interpolation II. 2 Piecewise polynomial interpolation Hermite splines
Lecture 5 Interpolation II Introduction In te previous lecture we focused primarily on polynomial interpolation of a set of n points. A difficulty we observed is tat wen n is large, our polynomial as to
More informationNumerical Differentiation
Numerical Differentiation Finite Difference Formulas for te first derivative (Using Taylor Expansion tecnique) (section 8.3.) Suppose tat f() = g() is a function of te variable, and tat as 0 te function
More informationLecture XVII. Abstract We introduce the concept of directional derivative of a scalar function and discuss its relation with the gradient operator.
Lecture XVII Abstract We introduce te concept of directional derivative of a scalar function and discuss its relation wit te gradient operator. Directional derivative and gradient Te directional derivative
More informationHOMEWORK HELP 2 FOR MATH 151
HOMEWORK HELP 2 FOR MATH 151 Here we go; te second round of omework elp. If tere are oters you would like to see, let me know! 2.4, 43 and 44 At wat points are te functions f(x) and g(x) = xf(x)continuous,
More information5.1 introduction problem : Given a function f(x), find a polynomial approximation p n (x).
capter 5 : polynomial approximation and interpolation 5 introduction problem : Given a function f(x), find a polynomial approximation p n (x) Z b Z application : f(x)dx b p n(x)dx, a a one solution : Te
More informationCopyright c 2008 Kevin Long
Lecture 4 Numerical solution of initial value problems Te metods you ve learned so far ave obtained closed-form solutions to initial value problems. A closedform solution is an explicit algebriac formula
More informationChapter 4: Numerical Methods for Common Mathematical Problems
1 Capter 4: Numerical Metods for Common Matematical Problems Interpolation Problem: Suppose we ave data defined at a discrete set of points (x i, y i ), i = 0, 1,..., N. Often it is useful to ave a smoot
More informationConsider a function f we ll specify which assumptions we need to make about it in a minute. Let us reformulate the integral. 1 f(x) dx.
Capter 2 Integrals as sums and derivatives as differences We now switc to te simplest metods for integrating or differentiating a function from its function samples. A careful study of Taylor expansions
More informationOrder of Accuracy. ũ h u Ch p, (1)
Order of Accuracy 1 Terminology We consider a numerical approximation of an exact value u. Te approximation depends on a small parameter, wic can be for instance te grid size or time step in a numerical
More informationLECTURE 14 NUMERICAL INTEGRATION. Find
LECTURE 14 NUMERCAL NTEGRATON Find b a fxdx or b a vx ux fx ydy dx Often integration is required. However te form of fx may be suc tat analytical integration would be very difficult or impossible. Use
More information5 Ordinary Differential Equations: Finite Difference Methods for Boundary Problems
5 Ordinary Differential Equations: Finite Difference Metods for Boundary Problems Read sections 10.1, 10.2, 10.4 Review questions 10.1 10.4, 10.8 10.9, 10.13 5.1 Introduction In te previous capters we
More informationlecture 26: Richardson extrapolation
43 lecture 26: Ricardson extrapolation 35 Ricardson extrapolation, Romberg integration Trougout numerical analysis, one encounters procedures tat apply some simple approximation (eg, linear interpolation)
More information232 Calculus and Structures
3 Calculus and Structures CHAPTER 17 JUSTIFICATION OF THE AREA AND SLOPE METHODS FOR EVALUATING BEAMS Calculus and Structures 33 Copyrigt Capter 17 JUSTIFICATION OF THE AREA AND SLOPE METHODS 17.1 THE
More information(a) At what number x = a does f have a removable discontinuity? What value f(a) should be assigned to f at x = a in order to make f continuous at a?
Solutions to Test 1 Fall 016 1pt 1. Te grap of a function f(x) is sown at rigt below. Part I. State te value of eac limit. If a limit is infinite, state weter it is or. If a limit does not exist (but is
More informationExam 1 Review Solutions
Exam Review Solutions Please also review te old quizzes, and be sure tat you understand te omework problems. General notes: () Always give an algebraic reason for your answer (graps are not sufficient),
More informationNUMERICAL DIFFERENTIATION. James T. Smith San Francisco State University. In calculus classes, you compute derivatives algebraically: for example,
NUMERICAL DIFFERENTIATION James T Smit San Francisco State University In calculus classes, you compute derivatives algebraically: for example, f( x) = x + x f ( x) = x x Tis tecnique requires your knowing
More informationThe total error in numerical differentiation
AMS 147 Computational Metods and Applications Lecture 08 Copyrigt by Hongyun Wang, UCSC Recap: Loss of accuracy due to numerical cancellation A B 3, 3 ~10 16 In calculating te difference between A and
More informationHow to Find the Derivative of a Function: Calculus 1
Introduction How to Find te Derivative of a Function: Calculus 1 Calculus is not an easy matematics course Te fact tat you ave enrolled in suc a difficult subject indicates tat you are interested in te
More information2.8 The Derivative as a Function
.8 Te Derivative as a Function Typically, we can find te derivative of a function f at many points of its domain: Definition. Suppose tat f is a function wic is differentiable at every point of an open
More informationSection 3.1: Derivatives of Polynomials and Exponential Functions
Section 3.1: Derivatives of Polynomials and Exponential Functions In previous sections we developed te concept of te derivative and derivative function. Te only issue wit our definition owever is tat it
More informationThe Laplace equation, cylindrically or spherically symmetric case
Numerisce Metoden II, 7 4, und Übungen, 7 5 Course Notes, Summer Term 7 Some material and exercises Te Laplace equation, cylindrically or sperically symmetric case Electric and gravitational potential,
More informationPrecalculus Test 2 Practice Questions Page 1. Note: You can expect other types of questions on the test than the ones presented here!
Precalculus Test 2 Practice Questions Page Note: You can expect oter types of questions on te test tan te ones presented ere! Questions Example. Find te vertex of te quadratic f(x) = 4x 2 x. Example 2.
More information1 Calculus. 1.1 Gradients and the Derivative. Q f(x+h) f(x)
Calculus. Gradients and te Derivative Q f(x+) δy P T δx R f(x) 0 x x+ Let P (x, f(x)) and Q(x+, f(x+)) denote two points on te curve of te function y = f(x) and let R denote te point of intersection of
More informationPoisson Equation in Sobolev Spaces
Poisson Equation in Sobolev Spaces OcMountain Dayligt Time. 6, 011 Today we discuss te Poisson equation in Sobolev spaces. It s existence, uniqueness, and regularity. Weak Solution. u = f in, u = g on
More informationMath Spring 2013 Solutions to Assignment # 3 Completion Date: Wednesday May 15, (1/z) 2 (1/z 1) 2 = lim
Mat 311 - Spring 013 Solutions to Assignment # 3 Completion Date: Wednesday May 15, 013 Question 1. [p 56, #10 (a)] 4z Use te teorem of Sec. 17 to sow tat z (z 1) = 4. We ave z 4z (z 1) = z 0 4 (1/z) (1/z
More informationA h u h = f h. 4.1 The CoarseGrid SystemandtheResidual Equation
Capter Grid Transfer Remark. Contents of tis capter. Consider a grid wit grid size and te corresponding linear system of equations A u = f. Te summary given in Section 3. leads to te idea tat tere migt
More informationChapter 2 Limits and Continuity
4 Section. Capter Limits and Continuity Section. Rates of Cange and Limits (pp. 6) Quick Review.. f () ( ) () 4 0. f () 4( ) 4. f () sin sin 0 4. f (). 4 4 4 6. c c c 7. 8. c d d c d d c d c 9. 8 ( )(
More informationA = h w (1) Error Analysis Physics 141
Introduction In all brances of pysical science and engineering one deals constantly wit numbers wic results more or less directly from experimental observations. Experimental observations always ave inaccuracies.
More informationDifferentiation in higher dimensions
Capter 2 Differentiation in iger dimensions 2.1 Te Total Derivative Recall tat if f : R R is a 1-variable function, and a R, we say tat f is differentiable at x = a if and only if te ratio f(a+) f(a) tends
More informationMath 242: Principles of Analysis Fall 2016 Homework 7 Part B Solutions
Mat 22: Principles of Analysis Fall 206 Homework 7 Part B Solutions. Sow tat f(x) = x 2 is not uniformly continuous on R. Solution. Te equation is equivalent to f(x) = 0 were f(x) = x 2 sin(x) 3. Since
More informationTaylor Series and the Mean Value Theorem of Derivatives
1 - Taylor Series and te Mean Value Teorem o Derivatives Te numerical solution o engineering and scientiic problems described by matematical models oten requires solving dierential equations. Dierential
More informationBob Brown Math 251 Calculus 1 Chapter 3, Section 1 Completed 1 CCBC Dundalk
Bob Brown Mat 251 Calculus 1 Capter 3, Section 1 Completed 1 Te Tangent Line Problem Te idea of a tangent line first arises in geometry in te context of a circle. But before we jump into a discussion of
More informationContinuity and Differentiability of the Trigonometric Functions
[Te basis for te following work will be te definition of te trigonometric functions as ratios of te sides of a triangle inscribed in a circle; in particular, te sine of an angle will be defined to be te
More informationDifferential equations. Differential equations
Differential equations A differential equation (DE) describes ow a quantity canges (as a function of time, position, ) d - A ball dropped from a building: t gt () dt d S qx - Uniformly loaded beam: wx
More informationNew Fourth Order Quartic Spline Method for Solving Second Order Boundary Value Problems
MATEMATIKA, 2015, Volume 31, Number 2, 149 157 c UTM Centre for Industrial Applied Matematics New Fourt Order Quartic Spline Metod for Solving Second Order Boundary Value Problems 1 Osama Ala yed, 2 Te
More information4. The slope of the line 2x 7y = 8 is (a) 2/7 (b) 7/2 (c) 2 (d) 2/7 (e) None of these.
Mat 11. Test Form N Fall 016 Name. Instructions. Te first eleven problems are wort points eac. Te last six problems are wort 5 points eac. For te last six problems, you must use relevant metods of algebra
More information5.1 We will begin this section with the definition of a rational expression. We
Basic Properties and Reducing to Lowest Terms 5.1 We will begin tis section wit te definition of a rational epression. We will ten state te two basic properties associated wit rational epressions and go
More information1. Questions (a) through (e) refer to the graph of the function f given below. (A) 0 (B) 1 (C) 2 (D) 4 (E) does not exist
Mat 1120 Calculus Test 2. October 18, 2001 Your name Te multiple coice problems count 4 points eac. In te multiple coice section, circle te correct coice (or coices). You must sow your work on te oter
More informationClick here to see an animation of the derivative
Differentiation Massoud Malek Derivative Te concept of derivative is at te core of Calculus; It is a very powerful tool for understanding te beavior of matematical functions. It allows us to optimize functions,
More informationPolynomials 3: Powers of x 0 + h
near small binomial Capter 17 Polynomials 3: Powers of + Wile it is easy to compute wit powers of a counting-numerator, it is a lot more difficult to compute wit powers of a decimal-numerator. EXAMPLE
More informationFunctions of the Complex Variable z
Capter 2 Functions of te Complex Variable z Introduction We wis to examine te notion of a function of z were z is a complex variable. To be sure, a complex variable can be viewed as noting but a pair of
More informationMathematics 5 Worksheet 11 Geometry, Tangency, and the Derivative
Matematics 5 Workseet 11 Geometry, Tangency, and te Derivative Problem 1. Find te equation of a line wit slope m tat intersects te point (3, 9). Solution. Te equation for a line passing troug a point (x
More informationIEOR 165 Lecture 10 Distribution Estimation
IEOR 165 Lecture 10 Distribution Estimation 1 Motivating Problem Consider a situation were we ave iid data x i from some unknown distribution. One problem of interest is estimating te distribution tat
More information2.11 That s So Derivative
2.11 Tat s So Derivative Introduction to Differential Calculus Just as one defines instantaneous velocity in terms of average velocity, we now define te instantaneous rate of cange of a function at a point
More informationThe derivative function
Roberto s Notes on Differential Calculus Capter : Definition of derivative Section Te derivative function Wat you need to know already: f is at a point on its grap and ow to compute it. Wat te derivative
More informationChapter 5 FINITE DIFFERENCE METHOD (FDM)
MEE7 Computer Modeling Tecniques in Engineering Capter 5 FINITE DIFFERENCE METHOD (FDM) 5. Introduction to FDM Te finite difference tecniques are based upon approximations wic permit replacing differential
More informationDigital Filter Structures
Digital Filter Structures Te convolution sum description of an LTI discrete-time system can, in principle, be used to implement te system For an IIR finite-dimensional system tis approac is not practical
More information1 1. Rationalize the denominator and fully simplify the radical expression 3 3. Solution: = 1 = 3 3 = 2
MTH - Spring 04 Exam Review (Solutions) Exam : February 5t 6:00-7:0 Tis exam review contains questions similar to tose you sould expect to see on Exam. Te questions included in tis review, owever, are
More informationChapter 8. Numerical Solution of Ordinary Differential Equations. Module No. 2. Predictor-Corrector Methods
Numerical Analysis by Dr. Anita Pal Assistant Professor Department of Matematics National Institute of Tecnology Durgapur Durgapur-7109 email: anita.buie@gmail.com 1 . Capter 8 Numerical Solution of Ordinary
More information1. Consider the trigonometric function f(t) whose graph is shown below. Write down a possible formula for f(t).
. Consider te trigonometric function f(t) wose grap is sown below. Write down a possible formula for f(t). Tis function appears to be an odd, periodic function tat as been sifted upwards, so we will use
More informationMVT and Rolle s Theorem
AP Calculus CHAPTER 4 WORKSHEET APPLICATIONS OF DIFFERENTIATION MVT and Rolle s Teorem Name Seat # Date UNLESS INDICATED, DO NOT USE YOUR CALCULATOR FOR ANY OF THESE QUESTIONS In problems 1 and, state
More informationFlavius Guiaş. X(t + h) = X(t) + F (X(s)) ds.
Numerical solvers for large systems of ordinary differential equations based on te stocastic direct simulation metod improved by te and Runge Kutta principles Flavius Guiaş Abstract We present a numerical
More information1 The concept of limits (p.217 p.229, p.242 p.249, p.255 p.256) 1.1 Limits Consider the function determined by the formula 3. x since at this point
MA00 Capter 6 Calculus and Basic Linear Algebra I Limits, Continuity and Differentiability Te concept of its (p.7 p.9, p.4 p.49, p.55 p.56). Limits Consider te function determined by te formula f Note
More information2.1 THE DEFINITION OF DERIVATIVE
2.1 Te Derivative Contemporary Calculus 2.1 THE DEFINITION OF DERIVATIVE 1 Te grapical idea of a slope of a tangent line is very useful, but for some uses we need a more algebraic definition of te derivative
More informationERROR BOUNDS FOR THE METHODS OF GLIMM, GODUNOV AND LEVEQUE BRADLEY J. LUCIER*
EO BOUNDS FO THE METHODS OF GLIMM, GODUNOV AND LEVEQUE BADLEY J. LUCIE* Abstract. Te expected error in L ) attimet for Glimm s sceme wen applied to a scalar conservation law is bounded by + 2 ) ) /2 T
More informationMATHEMATICS FOR ENGINEERING DIFFERENTIATION TUTORIAL 1 - BASIC DIFFERENTIATION
MATHEMATICS FOR ENGINEERING DIFFERENTIATION TUTORIAL 1 - BASIC DIFFERENTIATION Tis tutorial is essential pre-requisite material for anyone stuing mecanical engineering. Tis tutorial uses te principle of
More informationCS522 - Partial Di erential Equations
CS5 - Partial Di erential Equations Tibor Jánosi April 5, 5 Numerical Di erentiation In principle, di erentiation is a simple operation. Indeed, given a function speci ed as a closed-form formula, its
More informationNotes on Multigrid Methods
Notes on Multigrid Metods Qingai Zang April, 17 Motivation of multigrids. Te convergence rates of classical iterative metod depend on te grid spacing, or problem size. In contrast, convergence rates of
More informationExercises for numerical differentiation. Øyvind Ryan
Exercises for numerical differentiation Øyvind Ryan February 25, 2013 1. Mark eac of te following statements as true or false. a. Wen we use te approximation f (a) (f (a +) f (a))/ on a computer, we can
More information3.1 Extreme Values of a Function
.1 Etreme Values of a Function Section.1 Notes Page 1 One application of te derivative is finding minimum and maimum values off a grap. In precalculus we were only able to do tis wit quadratics by find
More informationALGEBRA AND TRIGONOMETRY REVIEW by Dr TEBOU, FIU. A. Fundamental identities Throughout this section, a and b denotes arbitrary real numbers.
ALGEBRA AND TRIGONOMETRY REVIEW by Dr TEBOU, FIU A. Fundamental identities Trougout tis section, a and b denotes arbitrary real numbers. i) Square of a sum: (a+b) =a +ab+b ii) Square of a difference: (a-b)
More informationCombining functions: algebraic methods
Combining functions: algebraic metods Functions can be added, subtracted, multiplied, divided, and raised to a power, just like numbers or algebra expressions. If f(x) = x 2 and g(x) = x + 2, clearly f(x)
More informationDefinition of the Derivative
Te Limit Definition of te Derivative Tis Handout will: Define te limit grapically and algebraically Discuss, in detail, specific features of te definition of te derivative Provide a general strategy of
More informationSECTION 3.2: DERIVATIVE FUNCTIONS and DIFFERENTIABILITY
(Section 3.2: Derivative Functions and Differentiability) 3.2.1 SECTION 3.2: DERIVATIVE FUNCTIONS and DIFFERENTIABILITY LEARNING OBJECTIVES Know, understand, and apply te Limit Definition of te Derivative
More information2.3 Algebraic approach to limits
CHAPTER 2. LIMITS 32 2.3 Algebraic approac to its Now we start to learn ow to find its algebraically. Tis starts wit te simplest possible its, and ten builds tese up to more complicated examples. Fact.
More informationSolution for the Homework 4
Solution for te Homework 4 Problem 6.5: In tis section we computed te single-particle translational partition function, tr, by summing over all definite-energy wavefunctions. An alternative approac, owever,
More informationAnalytic Functions. Differentiable Functions of a Complex Variable
Analytic Functions Differentiable Functions of a Complex Variable In tis capter, we sall generalize te ideas for polynomials power series of a complex variable we developed in te previous capter to general
More informationFunction Composition and Chain Rules
Function Composition and s James K. Peterson Department of Biological Sciences and Department of Matematical Sciences Clemson University Marc 8, 2017 Outline 1 Function Composition and Continuity 2 Function
More informationA MONTE CARLO ANALYSIS OF THE EFFECTS OF COVARIANCE ON PROPAGATED UNCERTAINTIES
A MONTE CARLO ANALYSIS OF THE EFFECTS OF COVARIANCE ON PROPAGATED UNCERTAINTIES Ronald Ainswort Hart Scientific, American Fork UT, USA ABSTRACT Reports of calibration typically provide total combined uncertainties
More informationTHE IMPLICIT FUNCTION THEOREM
THE IMPLICIT FUNCTION THEOREM ALEXANDRU ALEMAN 1. Motivation and statement We want to understand a general situation wic occurs in almost any area wic uses matematics. Suppose we are given number of equations
More information160 Chapter 3: Differentiation
3. Differentiation Rules 159 3. Differentiation Rules Tis section introuces a few rules tat allow us to ifferentiate a great variety of functions. By proving tese rules ere, we can ifferentiate functions
More informationRules of Differentiation
LECTURE 2 Rules of Differentiation At te en of Capter 2, we finally arrive at te following efinition of te erivative of a function f f x + f x x := x 0 oing so only after an extene iscussion as wat te
More informationCubic Functions: Local Analysis
Cubic function cubing coefficient Capter 13 Cubic Functions: Local Analysis Input-Output Pairs, 378 Normalized Input-Output Rule, 380 Local I-O Rule Near, 382 Local Grap Near, 384 Types of Local Graps
More informationDerivatives. By: OpenStaxCollege
By: OpenStaxCollege Te average teen in te United States opens a refrigerator door an estimated 25 times per day. Supposedly, tis average is up from 10 years ago wen te average teenager opened a refrigerator
More informationChapters 19 & 20 Heat and the First Law of Thermodynamics
Capters 19 & 20 Heat and te First Law of Termodynamics Te Zerot Law of Termodynamics Te First Law of Termodynamics Termal Processes Te Second Law of Termodynamics Heat Engines and te Carnot Cycle Refrigerators,
More informationThe Verlet Algorithm for Molecular Dynamics Simulations
Cemistry 380.37 Fall 2015 Dr. Jean M. Standard November 9, 2015 Te Verlet Algoritm for Molecular Dynamics Simulations Equations of motion For a many-body system consisting of N particles, Newton's classical
More informationMA455 Manifolds Solutions 1 May 2008
MA455 Manifolds Solutions 1 May 2008 1. (i) Given real numbers a < b, find a diffeomorpism (a, b) R. Solution: For example first map (a, b) to (0, π/2) and ten map (0, π/2) diffeomorpically to R using
More informationVolume 29, Issue 3. Existence of competitive equilibrium in economies with multi-member households
Volume 29, Issue 3 Existence of competitive equilibrium in economies wit multi-member ouseolds Noriisa Sato Graduate Scool of Economics, Waseda University Abstract Tis paper focuses on te existence of
More informationSolutions to the Multivariable Calculus and Linear Algebra problems on the Comprehensive Examination of January 31, 2014
Solutions to te Multivariable Calculus and Linear Algebra problems on te Compreensive Examination of January 3, 24 Tere are 9 problems ( points eac, totaling 9 points) on tis portion of te examination.
More informationA CLASS OF EVEN DEGREE SPLINES OBTAINED THROUGH A MINIMUM CONDITION
STUDIA UNIV. BABEŞ BOLYAI, MATHEMATICA, Volume XLVIII, Number 3, September 2003 A CLASS OF EVEN DEGREE SPLINES OBTAINED THROUGH A MINIMUM CONDITION GH. MICULA, E. SANTI, AND M. G. CIMORONI Dedicated to
More informationSolution. Solution. f (x) = (cos x)2 cos(2x) 2 sin(2x) 2 cos x ( sin x) (cos x) 4. f (π/4) = ( 2/2) ( 2/2) ( 2/2) ( 2/2) 4.
December 09, 20 Calculus PracticeTest s Name: (4 points) Find te absolute extrema of f(x) = x 3 0 on te interval [0, 4] Te derivative of f(x) is f (x) = 3x 2, wic is zero only at x = 0 Tus we only need
More informationChapter 1. Density Estimation
Capter 1 Density Estimation Let X 1, X,..., X n be observations from a density f X x. Te aim is to use only tis data to obtain an estimate ˆf X x of f X x. Properties of f f X x x, Parametric metods f
More informationPOLYNOMIAL AND SPLINE ESTIMATORS OF THE DISTRIBUTION FUNCTION WITH PRESCRIBED ACCURACY
APPLICATIONES MATHEMATICAE 36, (29), pp. 2 Zbigniew Ciesielski (Sopot) Ryszard Zieliński (Warszawa) POLYNOMIAL AND SPLINE ESTIMATORS OF THE DISTRIBUTION FUNCTION WITH PRESCRIBED ACCURACY Abstract. Dvoretzky
More informationDerivatives and Rates of Change
Section.1 Derivatives and Rates of Cange 2016 Kiryl Tsiscanka Derivatives and Rates of Cange Measuring te Rate of Increase of Blood Alcool Concentration Biomedical scientists ave studied te cemical and
More informationGradient Descent etc.
1 Gradient Descent etc EE 13: Networked estimation and control Prof Kan) I DERIVATIVE Consider f : R R x fx) Te derivative is defined as d fx) = lim dx fx + ) fx) Te cain rule states tat if d d f gx) )
More informationLogarithmic functions
Roberto s Notes on Differential Calculus Capter 5: Derivatives of transcendental functions Section Derivatives of Logaritmic functions Wat ou need to know alread: Definition of derivative and all basic
More informationChapter 2. Limits and Continuity 16( ) 16( 9) = = 001. Section 2.1 Rates of Change and Limits (pp ) Quick Review 2.1
Capter Limits and Continuity Section. Rates of Cange and Limits (pp. 969) Quick Review..... f ( ) ( ) ( ) 0 ( ) f ( ) f ( ) sin π sin π 0 f ( ). < < < 6. < c c < < c 7. < < < < < 8. 9. 0. c < d d < c
More informationRecall from our discussion of continuity in lecture a function is continuous at a point x = a if and only if
Computational Aspects of its. Keeping te simple simple. Recall by elementary functions we mean :Polynomials (including linear and quadratic equations) Eponentials Logaritms Trig Functions Rational Functions
More information11.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
SECTION 11.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR 633 wit speed v o along te same line from te opposite direction toward te source, ten te frequenc of te sound eard b te observer is were c is
More informationNON STANDARD FITTED FINITE DIFFERENCE METHOD FOR SINGULAR PERTURBATION PROBLEMS USING CUBIC SPLINE
Global and Stocastic Analysis Vol. 4 No. 1, January 2017, 1-10 NON STANDARD FITTED FINITE DIFFERENCE METHOD FOR SINGULAR PERTURBATION PROBLEMS USING CUBIC SPLINE K. PHANEENDRA AND E. SIVA PRASAD Abstract.
More informationUniversity Mathematics 2
University Matematics 2 1 Differentiability In tis section, we discuss te differentiability of functions. Definition 1.1 Differentiable function). Let f) be a function. We say tat f is differentiable at
More information1 Lecture 13: The derivative as a function.
1 Lecture 13: Te erivative as a function. 1.1 Outline Definition of te erivative as a function. efinitions of ifferentiability. Power rule, erivative te exponential function Derivative of a sum an a multiple
More informationFractional Derivatives as Binomial Limits
Fractional Derivatives as Binomial Limits Researc Question: Can te limit form of te iger-order derivative be extended to fractional orders? (atematics) Word Count: 669 words Contents - IRODUCIO... Error!
More informationFinite Difference Method
Capter 8 Finite Difference Metod 81 2nd order linear pde in two variables General 2nd order linear pde in two variables is given in te following form: L[u] = Au xx +2Bu xy +Cu yy +Du x +Eu y +Fu = G According
More informationMath 1241 Calculus Test 1
February 4, 2004 Name Te first nine problems count 6 points eac and te final seven count as marked. Tere are 120 points available on tis test. Multiple coice section. Circle te correct coice(s). You do
More informationTHE IDEA OF DIFFERENTIABILITY FOR FUNCTIONS OF SEVERAL VARIABLES Math 225
THE IDEA OF DIFFERENTIABILITY FOR FUNCTIONS OF SEVERAL VARIABLES Mat 225 As we ave seen, te definition of derivative for a Mat 111 function g : R R and for acurveγ : R E n are te same, except for interpretation:
More informationSymmetry Labeling of Molecular Energies
Capter 7. Symmetry Labeling of Molecular Energies Notes: Most of te material presented in tis capter is taken from Bunker and Jensen 1998, Cap. 6, and Bunker and Jensen 2005, Cap. 7. 7.1 Hamiltonian Symmetry
More informationFinite Difference Methods Assignments
Finite Difference Metods Assignments Anders Söberg and Aay Saxena, Micael Tuné, and Maria Westermarck Revised: Jarmo Rantakokko June 6, 1999 Teknisk databeandling Assignment 1: A one-dimensional eat equation
More information158 Calculus and Structures
58 Calculus and Structures CHAPTER PROPERTIES OF DERIVATIVES AND DIFFERENTIATION BY THE EASY WAY. Calculus and Structures 59 Copyrigt Capter PROPERTIES OF DERIVATIVES. INTRODUCTION In te last capter you
More information