1 polynomial interpolation ( 多項式內插法 ). interpolation vs. extrapolation interpolation (...

1

Polynomial Interpolation Polynomial Interpolation ((多項式內插法多項式內插法 ))

Interpolation vs. Extrapolation

• Interpolation ( 內插法 )• Data to be found are within the range of observed data• Used to estimate values between data points• 與迴歸分析的差異

• Goes through data points, no error at data points

• The most common method is the polynomial interpolation

• Extrapolation (外插法 )• Data to be found are beyond the range of observation data (not

reliable)2

Same data points, different curve fitting

Polynomial Interpolation

Given n data points, fit an (n-1)th-order polynomial through them

Use polynomial interpolation to determine ai’s

For consistency with MATLAB, use

3

1nn

2321 xaxaxaaxf ...

n1n2n

21n

1 pxpxpxpxf )(

Interpolating Polynomials

4

First-order Second-order Third-order

Coefficients of an Interpolating Polynomial

Newton and Lagrange polynomials are well-suited for determining values between points However, they do not provide a convenient polynomial of

conventional form

Use n data points to determine n coefficients

5

nn1n2n

n21n

n1n

n21n2n

221n

212

n11n2n

121n

111

pxpxpxpxf

pxpxpxpxf

pxpxpxpxf

)(

)(

)(

n1n2n

21n

1 pxpxpxpxf )(

Coefficients of an Interpolating Polynomial

Can be solved with any matrix method, but inefficient

There are more efficient methods to find p’s The above equations are notoriously ill-conditioned (病態條

件 ), especially for large n Limit yourself to lower-order polynomials

6

)(

)(

)(

)(

n

3

2

1

n

3

2

1

n2n

n1n

n

32n

31n

3

22n

21n

2

12n

11n

1

xf

xf

xf

xf

p

p

p

p

1xxx

1xxx

1xxx

1xxx

Polynomial Coefficients

Example – use last four points of Table 14.1

7

432

23

141

432

23

131

432

23

121

432

23

111

432

23

1

)500()500()500(457.0)( ; 500

)400()400()400(525.0)( ; 400

)300()300()300(616.0)( ; 300

)250()250()250(675.0)( ; 250

)(

ppppxfx

ppppxfx

ppppxfx

ppppxfx

pxpxpxpxf

4570

5250

6160

6750

p

p

p

p

1500500500

1400400400

1300300300

1250250250

4

3

2

1

23

23

23

23

.

.

.

.

)()(

)()(

)()(

)()(

Vandermonde Matrices ( 凡德芒矩陣 )

Vandermonde Matrices

8

>> A=[250^3 250^2 250 1; 300^3 300^2 300 1; 400^3 400^2 400 1; 500^3 500^2 500 1]

A =

15625000 62500 250 1

27000000 90000 300 1

64000000 160000 400 1

125000000 250000 500 1

>> b=[0.675; 0.616; 0.525; 0.457]

b =

0.6750

0.6160

0.5250

0.4570

>> format long

>> p = A\b

p =

-0.00000000260000

0.00000427000000

-0.00293700000000

1.18300000000000

>> cond(A)

ans =

9.306535523991324e+009ill-conditioned matrix

18300000001x00293700000

x00000427000x00000000260xf 23

..

..)(

Newton Interpolation

Use Newton’s “divided differences” of the functional values

The lower order coefficients bi do not change when the order of interpolation is increased

Easy to add more data points and try a higher order polynomial

9

)())((

))(()()(

1n21n

2131211n

xxxxxxb

xxxxbxxbbxf

Newton Linear Interpolation

Start with linear interpolation: f1(x) = b1 + b2 (x - x1) Similar triangles

10

)()()(

)()(

)()()()(

112

1211

12

12

1

11

xxxx

xfxfxfxf

xx

xfxf

xx

xfxf


Newton linear interpolation formula

Example 1: interpolate e2 using e1 and e5

Example 2: Interpolate e2 using e1.5 and e2.5

11

112

1211 xx

xx

xfxfxfxf

4233614

71832411487183212

15

eee2f

151

1 ...

.

33218501

4817418251248174512

5152

eee2f

515251

1 ....

....

...

3891.7Exact 2 e


12

Small interval x provides a better estimate

Linear estimates of ln(2)

Logarithmic function

Accuracy of Interpolation

Quadratic Interpolation ( 二次內插 )

Quadratic interpolation – need three points Use the parabola ( 拋物線 )

This is the same as

14

2131212 xxxxbxxbbxf

33

231322

2131211

23212

ba

xbxbba

xxbxbba

xaxaaxf )(

where

Quadratic Interpolation

To get coefficients b’s 1) Set x = x1, get b1 = f(x1)

2) Use b1 and set x = x2 to get b2

15

1121113112112 xfbxxxxbxxbbxf

12

122

222123122122

xx

xfxfb

xfxxxxbxxbxfxf

2131212 xxxxbxxbbxf


3) Use b1 and b2, and set x = x3 to get b3

16

13

12

12

23

23

3

3231331312

12132

xx

xx

xfxf

xx

xfxf

b

xfxxxxbxxxx

xfxfxfxf

b1 is a constant (0th order)

b2 gives slope (finite difference)

b3 gives curvature (difference of finite differences)

Hand Calculation Example

例 : interpolate e2 using e1, e3, and e5

17

41148xf 5x

08620xf 3x

71832xf 1x

33

22

11

.

.

.

8701315

13

ee

35

ee

b ;6836813

eeb ;71832b

1335

3

13

21 ...

468123212870131268368718322f2 .*.*..

38917e Exact 2 .

Hand Calculation Example — Different xi

例 : interpolate e2 using e1, e1.5, and e2.5

182512xf 52x

48174xf 51x

71832xf 1x

33

22

11

..

..

.

78272152

151

ee

5152

ee

b ;52683151

eeb ;71832b

1515152

3

151

21 ..

.....

.

...

.

6365751212782721252683718322f2 ..*.*..

38917e Exact 2 .

Order of Interpolation

Linear, quadratic and cubic estimates of ln(2)

21Higher-order interpolation improves the estimate

Logarithmic function

Newton’s Interpolating Polynomials

General form for Newton’s interpolating polynomials

Bracketed functions are finite differences

22

xxxxxxxxb

xxxxbxxbbxf

1n321n

2131211n

121nnn

1233

122

11

xxxxfb

xxxfb

xxfb

xfb

,,,,

,,

,

Newton’s Divided Differences

First finite difference ( 一次有限分割差分 )

Second finite difference (二次有限分割差分 )

The difference of two finite differences

23

ji

jiji xx

xfxfxxf

,

ki

kjjikji xx

xxfxxfxxxf

,,

,,

Newton’s Divided Differences

The nth finite difference

24

1n

122n1n231nn121nn xx

xxxxfxxxxfxxxxf

,,,,,,,,,,,,

An iterative procedure

1. Evaluate all first-order finite differences; save f (x1) for b1

2. Evaluate second-order from firsts; save f [x2, x1] for b2

3. Continue to nth-order, saving needed ones


No need to solve a system of simultaneous equations Spacing may be non-uniform and xi may be in arbitrary

order 25

15

12342345123455

14

12323412344

13

12231233

12

12122

111

1n1n2131211n

xx

xx xxfxx xxfxx xxxfb

xx

xx xfxx xfxx xxfb

xx

x xfx xfx xxfb

xx

xfxfx xfb

xfxfb

xxxxbxxxxbxxbbxf

,,,,,,,,,,

,,,,,,,

,,,,

,

)(

)()())(()()(

)(

,)(

,,,)(

,,,,,,)(

,,,,,,,,,,)(

,,,,,,,,,,)(

,,,,,,,)(

66

5655

4564544

34563453433

2345623452342322

1234512341231211

i4ii3ii1i2ii1iiii

xfx6

xxfxfx5

xxxfxxfxfx4

xxxxfxxxfxxfxfx3

xxxxxfxxxxfxxxfxxfxfx2

xxxxxfxxxxfxxxfxxfxfx1

xxfxxfxxxfxxfxfyxi

1n

122n1n131nn121nnn

1n1n2131211n

xx

x xxxfx xxxfx xxxfb

xxxxbxxxxbxxbbxf

,,,,,,,,,,

)()())(()()(

,,

Use the top element of each column (= bn) to evaluate the interpolated

functional value f(x)

First Second Third Fourth


Relative error decreases with increasing order of the interpolating polynomial

Error is also sensitive to the position and sequence of the original data (x1 , x2 , x3 , x4 ,…, xn)

Percentage Relative Error

)3,4,1,0()x,x,x,(x using 2xat ef(x) Estimate 4321x

93618754x1xx57197011xx8937523x718282112f

22407121xx8937523x718282112f

3

2

.))((.)(..)(

.)(..)(

085542034

43x

5126134598155443

13x 14x

60966282932917718282212

03x 04x 01x

571970189375237182821000000101

xxxxfxxxfxxfxfxi i1i2i3ii1i2ii1iii

.

..

...

....

,,,,,,)(

Hand Calculations: Newton Interpolation

Error Estimate

Error for Newton’s interpolating polynomial Estimate from

The leading term in the truncated polynomial Similar to the truncation error in a Taylor series

29

n2111nn1nn xxxxxxxxxxfR ,,,,

30

function yint = Newtint(x,y,xx)% Newton interpolation. Uses an (n - 1)-order Newton% interpolating polynomial based on n data points (x, y)% to determine a value of the dependent variable (yint)% at a given value of the independent variable, xx.% input:% x = independent variable% y = dependent variable% xx = value of independent variable at which% interpolation is calculated% output:% yint = interpolated value of dependent variable % compute the finite divided differences in the form of a% difference tablen = length(x);if length(y)~=n, error('x and y must be same length'); endb = zeros(n,n);% assign dependent variables to the first column of b.b(:,1) = y(:); % the (:) ensures that y is a column vector.for j = 2:n for i = 1:n-j+1 b(i,j) = (b(i+1,j-1)-b(i,j-1))/(x(i+j-1)-x(i)); endend% use the finite divided differences to interpolate xt = 1;yint = b(1,1);for j = 1:n-1 xt = xt*(xx-x(j)); yint = yint+b(1,j+1)*xt;end

Modified M-File: yint may be evaluated at multiple pointsfunction [b,yint] = Newtint2(x,y,xx)% yint = Newtint(x,y,xx):% Newton interpolation. Uses an (n - 1)-order Newton% interpolating polynomial based on n data points (x, y)% to determine a value of the dependent variable (yint)% at a given value of the independent variable, xx. % compute the finite divided differences in the form of a% difference tablen = length(x);if length(y)~=n, error('x and y must be same length'); endb = zeros(n,n);% assign dependent variables to the first column of b.b(:,1) = y(:); % the (:) ensures that y is a column vector.for j = 2:n for i = 1:n-j+1 b(i,j) = (b(i+1,j-1)-b(i,j-1))/(x(i+j-1)-x(i)); endend% use the finite divided differences to interpolate for k = 1:length(xx) xt = 1; yint(k) = b(1,1); for j = 1:n-1 xt = xt*(xx(k)-x(j)); yint(k) = yint(k)+b(1,j+1)*xt; endend

>> x = [0 1 4 3 1.5 2.5]x = 0 1.0000 4.0000 3.0000 1.5000 2.5000>> y = exp(x)y = 1.0000 2.7183 54.5982 20.0855 4.4817 12.1825>> xx = 0:0.1:4; yy=exp(xx);>> [b, yint] = Newtint2(x, y, xx)b = 1.0000 1.7183 3.8938 1.5720 0.3312 0.0697 2.7183 17.2933 8.6097 2.0687 0.5055 0 54.5982 34.5126 9.6440 2.8270 0 0 20.0855 10.4026 5.4035 0 0 0 4.4817 7.7008 0 0 0 0 12.1825 0 0 0 0 0yint = Columns 1 through 8 1.0000 1.1355 1.2670 1.4000 1.5394 1.6893 1.8536 2.0356 Columns 9 through 16 2.2384 2.4651 2.7183 3.0008 3.3153 3.6649 4.0525 4.4817 Columns 17 through 24 4.9561 5.4801 6.0584 6.6964 7.4003 8.1770 9.0343 9.9809 Columns 25 through 32 11.0266 12.1825 13.4606 14.8744 16.4387 18.1698 20.0855 22.2054 Columns 33 through 40 24.5506 27.1441 30.0107 33.1774 36.6729 40.5284 44.7770 49.4543 Column 41 54.5982>> H = plot(x,y,'mo',xx,yy,'r',xx,yint,'bx'); set(H,'LineWidth',3);

Newton Interpolating Polynomial

Table

Coefficients of Newton interpolating

polynomial

f(x) = ex, Interpolation at [0 1 4 3 1.5 2.5]

Newton Interpolating Polynomial

Lagrange Interpolating Polynomials

Give the same result as the Newton’s polynomials, but different approach

35

)())(())((

)())(())((

)(

)()(

)()()()()()()()()(

1121

1121

1

122111

niiiiiii

nii

ii

in

ijj ji

ji

i

n

iinnn

xxxxxxxxxx

xxxxxxxxxx

xP

xP

xx

xxxL

xfxLxfxLxfxLxfxLxf

ijji

ji

ii

iiii

xL

0xL ij

1xP

xPxL i ;j

Note

)(

)(;

)(

)()(

:

Lagrange Interpolation

1st-order Lagrange polynomial

Second-order Lagrange polynomial

Third-order Lagrange polynomial

36

)()()()()()( 212

11

21

222111 xf

xx

xxxf

xx

xxxfxLxfLxf

)())((

))(()(

))((

))(()(

))((

))(()( 3

2313

212

3212

311

3121

322 xf

xxxx

xxxxxf

xxxx

xxxxxf

xxxx

xxxxxf

)())()((

))()(()(

))()((

))()((

)())()((

))()(()(

))()((

))()(()(

4342414

3213

432313

421

2423212

4311

413121

4324

xfxxxxxx

xxxxxxxf

xxxxxx

xxxxxx

xfxxxxxx

xxxxxxxf

xxxxxx

xxxxxxxf

Linear Lagrange Interpolation

Both L1(x) and L2(x) are straight lines37

)()()()()()( 212

11

21

222111 xf

xx

xxxf

xx

xxxfxLxfLxf

L2(x)f(x2)

L3(x)f(x3)L1(x)f(x1)

x1 x2 x3

Quadratic Lagrange Interpolation

First-Order Lagrange Interpolation

39

f(x) = ex

Interpolation at [0 4]

Second-Order Lagrange Interpolation

40

f(x) = ex, Interpolation at [0 1 4]

Third-Order Lagrange Interpolation

41

f(x) = ex, Interpolation at [0 1 4 3]

例： Lagrange Interpolation

Estimate f(x) = exp(x) at x = 2 using (x0, x1, x2, x3 ) = (0,1,4,3) Second-order: (x1, x2, x3 ) = (0,1,4)

Third order: (x1, x2, x3, x4 ) = (0,1,4,3)

42

224067125982541404

1202718281

4101

420201

4010

42122f

4f1404

1x0x1f

4101

4x0x0f

4010

4x1xxf

2

2

.).())((

))(().(

))((

))(().(

))((

))(()(

)())((

))(()(

))((

))(()(

))((

))(()(

936187508554206

4598254

12

2718282

6

401

12

22f

3f431303

4x1x0x4f

341404

3x1x0x

1f314101

3x4x0x0f

304010

3x4x1xxf

3

3

.).().().().()(

)())()((

))()(()(

))()((

))()((

)())()((

))()(()(

))()((

))()(()(

43

function yint = Lagrange(x,y,xx)% yint = Lagrange(x,y,xx):% Lagrange interpolation. Uses an (n - 1)-order Lagrange% interpolating polynomial based on n data points (x, y)% to determine a value of the dependent variable (yint)% at a given value of the independent variable, xx.% input:% x = independent variable% y = dependent variable% xx = value of independent variable at which the% interpolation is calculated% output:% yint = interpolated value of dependent variable n = length(x);if length(y)~=n, error('x and y must be same length'); ends = 0;for i = 1:n product = y(i); for j = 1:n if i ~= j product = product*(xx-x(j))/(x(i)-x(j)); end end s = s+product;endyint = s;

M-file 執行例

給定不同溫度 t之空氣密度 d如下 t1 = -40,d(t1) = 1.52

t2 = 0, d(t2) = 1.29t3 = 20, d(t3) = 1.2t4 = 50, d(t4) = 1.09

求 15℃時的空氣密度

44

Coefficients of Lagrange Polynomial

Save the coefficients of Lagrange polynomial Evaluate the interpolated values at multiple locations

45

Evaluation of Interpolated Values

46

» x=[0 4]x = 0 4» y=exp(x)y = 1.0000 54.5982» c=Lagrange_coef(x,y)c = -0.2500 13.6495» t=2; p=Lagrange_eval(t,x,c)p = 27.7991

» x=[0 1 4 3]x = 0 1 4 3» y=exp(x)y = 1.0000 2.7183 54.5982 20.0855» c=Lagrange_coef(x,y)c = -0.0833 0.4530 4.5498 -3.3476» t=2; p=Lagrange_eval(t,x,c)p = 5.9362

» x=[0 1 4]x = 0 1 4» y=exp(x)y = 1.0000 2.7183 54.5982» c=Lagrange_coef(x,y)c = 0.2500 -0.9061 4.5498» t=2; p=Lagrange_eval(t,x,c)p = 12.2241

First-order Second-order

Third-order

Exact solution

e2 = 7.389056

Lagrange Interpolation

Very convenient for the same abscissas ( 橫座標 ) but different yi (e.g., measurements always taken using the same independent variables xi)

Lk(x) needs to be computed only once

However, less convenient when additional data may be added

48

1n1nnn2211

nn2211

nn2211

y,x, y,x,, y,x, y,x :2new

z,x,, z,x, z,x :1new

y,x,, y,x, y,x:original

Inverse Interpolation (反內插 )

Given x’s and f(x)’s – interpolation enables us to obtain new f(x) from new x

What about new x from new f(x)? Example: f(x) = 1/x

49

Switch x and f(x) and do new interpolation. However, non-uniform spacing in [x vs. f(x)] often leads to oscillations (振盪、擺盪 ) in the resulting interpolating polynomial (橫軸上數值分佈間隔不均常使得多項式內插造成振盪 )

Fit an nth-order polynomial to the original data [f(x) vs. x], then use root-finding techniques to find x. ( 求根 )

反內插求解例

給定若干數據點 (1,1) 、 (2,0.5) 、 (3,0.3333) 、(4,0.25) 、 (5,0.2) 、 (6,0.1667) 、 (7,0.1429) ，求f(x)=0.3 的 x值配適二次多項式到三個點 (2,0.5) 、 (3,0.3333) 、 (4,0.25) ，求

f(x)=0.3 的 x值 f2(x) = 0.041667x2-0.375x+1.08333 (why?)

0.3 = 0.041667x2-0.375x+1.08333 ，求根求得根為 5.704158 或 3.295842 哪一個為所求？

50

Extrapolation ( 外插 )

估計在已知基準點 x1, x2,…, xn範圍之外的 f(x)值的程序外差的開放特性代

表將進入未知領域，因此將曲線拓展到超過已知的區域

可能與真實的曲線偏離，和預測不同

使用時應特別小心

51

此例說明外插可能引發的偏離情形；此例之外插是基於前三個已知點所擬合的拋物線配適

此例說明外插可能引發的偏離情形；此例之外插是基於前三個已知點所擬合的拋物線配適

例：外插的危險性

美國 1920—2000 年的人口數

以七階的多項式來配適前 8個點 (1920—1990) ，使用所得結果外插計算 2000 年的人口數

Reasonable to use interpolation, but not the extrapolation52

Oscillations ( 振盪 )

Higher-order polynomials tend to be ill-conditioned Tend to be highly sensitive to round-off error

例：倫基函數 (Runger’s functions)

54

2251

1)(

xxf

Runge’s function

Difficulties with polynomial interpolation: Humped or flat data

Runge’s Function

55

» x2=-1:0.2:1x2 = Columns 1 through 7 -1.0000 -0.8000 -0.6000 -0.4000 -0.2000 0 0.2000 Columns 8 through 11 0.4000 0.6000 0.8000 1.0000» y2=1./(1+25*x2.^2)y2 = Columns 1 through 7 0.0385 0.0588 0.1000 0.2000 0.5000 1.0000 0.5000 Columns 8 through 11 0.2000 0.1000 0.0588 0.0385» c=Lagrange_coef(x2,y2)c = Columns 1 through 7 0.1035 -1.5830 12.1102 -64.5875 282.5702 -678.1684 282.5702 Columns 8 through 11 -64.5875 12.1102 -1.5830 0.1035» x=-1:0.02:1; y=1./(1+25*x.^2);» t=x; p=Lagrange_eval(t,x2,c);» H=plot(x,y,'r',t,p,'b-',x2,y2,'mo');» set(H,'LineWidth',2.5)» print -djpeg075 poly5.jpg

2251

1)(

xxf

Oscillations

56

MATLAB Functions: polyfit and polyval

>> x = linspace(-1,1,5); y = 1./(1+25*x.^2)>> xx = linspace(-1,1); p = polyfit(x,y,4)p = 3.3156 0.0000 -4.2772 0.0000 1.0000>> y4 = polyval(p,xx);>> yr = 1./(1+25*xx.^2);>> H=plot(x,y,'o',xx,y4,xx,yr,'--')>> x=linspace(-1,1,11); y = 1./(1+25*x.^2)y = Columns 1 through 8 0.0385 0.0588 0.1000 0.2000 0.5000 1.0000 0.5000 0.2000 Columns 9 through 11 0.1000 0.0588 0.0385>> p = polyfit(x,y,10)p = Columns 1 through 8 -220.9417 0.0000 494.9095 -0.0000 -381.4338 0.0000 123.3597 -0.0000 Columns 9 through 11 -16.8552 0.0000 1.0000>> y10 = polyval(p,xx);>> H = plot(x,y,'o',xx,y10,xx,yr,'--')>> set(H,'LineWidth',3,'MarkerSize',12)>> print -djpeg Fig14_13.jpg

4th-order polynomial

10th-order polynomial

2251

1)(

xxf

Runge’s Function with polynomial fit

58

4th-order 10th-order

» [x,y]=examplex = Columns 1 through 7 0 0.2000 0.8000 1.0000 1.2000 1.9000 2.0000 Columns 8 through 10 2.1000 2.9500 3.0000y = Columns 1 through 7 0.0100 0.2200 0.7600 1.0300 1.1800 1.9400 2.0100 Columns 8 through 10 2.0800 2.9000 2.9500» c=Lagrange_coef(x,y)c = Columns 1 through 7 -0.0007 0.0512 -2.4384 8.3366 -7.7423 37.5192 -61.2208 Columns 8 through 10 26.4741 -1.1493 0.8958» t=0:0.05:3; p=Lagrange_eval(t,x,c);» H=plot(x,y,'ro',t,p,'b',x,x,'m');» set(H,'LineWidth',2.5); print -djpeg075 poly4.jpg

function [x, y] = examplex = [0.00 0.20 0.80 1.00 1.20 1.90 2.00 2.10 2.95 3.00];y = [0.01 0.22 0.76 1.03 1.18 1.94 2.01 2.08 2.90 2.95];

Noisy straight line

Noisy straight line

1 polynomial interpolation ( 多項式內插法 ). interpolation vs. extrapolation interpolation (...

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