04-Divided Difference
计算方法-插值b
…………
2
f [ x , x 0 , x1 ]
x x1
f [ x , x 0 , ... , x n 1 ] f [ x 0 , ... , x n ] ( x x n ) f [ x , x 0 , ... , x n ]
n1
1
←
←
3
←……
←
n1
f ( x ) f ( x 0 ) f [ x 0 , x 1 ]( x x 0 ) f [ x 0 , x 1 , x 2 ]( x x 0 )( x x 1 ) ...
(k+1)阶差商：
f [ x0 , x1 , ... , xk ] f [ x1 , ... , xk , xk 1 ] f [ x0 , ... , xk 1 ] x 0 x k 1 f [ x0 , ... , xk 1 , xk ] f [ x0 , ... , xk 1 , xk 1 ] x k x k 1
( )
实际计算过程为
x0 x1 x2 …
xn1 xn
k !
, ( x min , x max )
f (x0) f (x1) f (x2) … f (xn1) f (xn)
xn+1 f (xn+1)
f [x0, x1] f [x1, x2] …… …… f [xn1, xn] f [xn, xn+1]
f [x0, x1 , x2] …… …… f [xn2, xn1, xn] f [xn1, xn, xn+1]
f [x0, …, xn] f [x1, …, xn+1]
f [x0, …, xn+1]
计算方法牛顿插值
称为在xi,xj,xk处的2阶差商
k阶差商：
f [ x0 , x1 ,, xk 1 ] f [ x1 , x2 ,, xk ] f [ x0 , x1 ,, xk ] x0 xk
利用插值条件和差商,可求出Nn(x)的系数 Ai :
A0 f ( x0 ) f [ x0 ] A1 f [ x0 , x1 ]
差商具有如下性质
性质1 (差商与函数值的关系)
f ( xi ) f x0 , x1 ,..., xn i 0 ' ( xi )
性质2 （对称性）：差商的值与结点排列顺序无关
f x0 , , xi , , x j , , xn f x0 ,, x j ,, xi ,, xn
f ( xi ) f ( xi h) f ( xi )
f ( xi ) f ( xi ) f ( xi h)
一阶中心差分 f ( x ) f ( x h ) f ( x h ) i i i /* centered 2 2 difference */
一般地,称k阶差分的差分为k+1阶差分,如二阶 向前和向后差分分别为
f [ x, x0 ] f [ x0 , x1 ] ( x x1 ) f [ x, x0 , x1 ]
1
2
n1
…………
f [ x, x0 , ... , xn1 ] f [ x0 , ... , xn ] ( x xn ) f [ x, x0 , ... , xn ]
1 + (x x0) 2 + … … + (x x0)…(x xn1)
性质5(差分与导数的关系）
k fi k !h k f [ xi , xi 1 ,, xi k ] hk f ( k ) ( ), ( xi xi k )
4-插值法
Nn ( x) A0 A1 ( x x0 ) A2 ( x x0 )( x x1) ... An ( x x0 )...( x xn1)
利用插值条件Nn(xj)=f(xj), j=0,1，…，n代入 上式，得关于Ak (k=0,1，…，n)的线性代数方 程组
y0
y1
x1
yn
0 a x0
xn b
X
2015-6-11
9
定理 n次插值问题的解是存在而且唯一的。
证明： 设 Pn (x)=a0 + a1x + a2x2 + ……+ anxn 是 y=f(x)在[a,b]上的n+1个互异节点x0,x1,…,xn的 插值多项式，则求Pn (x)问题归结为求系数 a0,a1,…,an。 由插值条件： Pn (xk)= yk (k = 0,1,…,n)
1 lk ( x j ) 0
2015-6-11
j k (j,k=0,1,…,n) jk
19
则称这n+1个n次多项式l0(x), l1(x),…, ln(x)为节点 x0 ,x1,…,xn上的n次插值基函数。
由n=1,2时的讨论可得
( x x0 ) ( x xk 1 )(x xk 1 ) ( x xn ) lk ( x ) ( xk x0 ) ( xk xk 1 )(xk xk 1 ) ( xk xn )
2015-6-11 11
2、Lagrange插值
2.1 线性插值与抛物插值 1.线性插值：n=1情形 给定插值节点 x0,x1， y0=f(x0)，y1=f(x1). 求线性插值多项式L1 (x)=a0+ a1x，使满足: L1(x0)=y0 , L1(x1)=y1. y= L1 (x)的几何意义就是过点(x0, y0)，(x1, y1)的直线。
数学中常用英文
代数部分1、基础add，plus加subtract减difference差multiply times乘product积divide除divisible可被整除得divided evenly被整除dividend被除数divisor因子，除数quotient商remainder余数factorial阶乘power乘方radical sign, root sign根号round to四舍五入to the nearest四舍五入2、有关集合union并集proper subset真子集solution set解集3、有关代数式、方程与不等式algebraic term代数项like terms, similar terms同类项numerical coefficient数字系数literal coefficient字母系数inequality不等式triangle inequality三角不等式range值域original equation原方程equivalent equation同解方程等价方程linear equation线性方程(e、g、5x+6=22) 4、有关分数与小数proper fraction真分数improper fraction假分数mixed number带分数vulgar fraction，common fraction普通分数simple fraction简分数complex fraction繁分数numerator分子denominator分母(least)common denominator(最小)公分母quarter四分之一decimal fraction纯小数infinite decimal无穷小数recurring decimal循环小数tenths unit 十分位5、基本数学概念arithmetic mean算术平均值weighted average加权平均值geometric mean几何平均数exponent指数，幂base乘幂得底数,底边cube立方数，立方体square root平方根cube root立方根common logarithm常用对数digit数字constant常数variable变量inverse function反函数complementary function余函数linear一次得，线性得factorization因式分解absolute value绝对值，e、g、｜-32｜=32 round off四舍五入。
参考资料-牛顿差分
S ince f(>i)g, the functio has Y 1 p,{z;), fo rR»u e's T heor er n ïm Pplt ;yät a numb er } in (o.in [O, b j. The G e ner alïZ6dg‹•)(})0, SO0 f'n' f J•"(*)•pol ynomi al of de$fe•riwhose l eaöin g cœffic ie nt is f 1x0!>i!- •pin) ) = ri! f lx‹,• l , ... , x n1›J O,¡31vd t es of x.ASäC O fl S q l e 'l G-e. .z,J,Newton’s interpolat o fÿn!d iv id ed-dif f erence form ula can be ex pressed in * *Pform when <o. <i. -p, are arranged co nsec uöv ely with equ¥spacing• J»*****-•* inœoduc e the notaüon h = =‹+—x¡, f or each im 0, 1, . • • › n 1 and let x = >o +-* Then the diff erenc e x- x, can be wrineng, p q, (s —i)ù. S o @. (3.10) b ecomesP,(x) ——P,(/ -i- ›) ,J[ x0] + sh f r i o! ul +S * )2* 0' * ^2’-j-- --j- s{s Uf(s —* + )h‘/$^ ' ^ ' ' ‘•A=OUsing bin O mia l-Cœ Ncient notaöon,s s{s - 1) $ — k -I- 1)k k!wc can express P>‹x) c ompactly asSn x} m P,(x$ -J- sh} f{x9} 3- k)*!*This formula is called the Newton forseard divid ed-difference fomu+•fom, called the Newtonthe forward difference notation A introduced in Aitken’s A2 me@od.fp g :l *h - f(*o)0›*i — <o12h 2/t2‘2*(* )'Aj°(x ) —Ö f{y )tt Since Pn{ z)ïs a3 2 Divided £//,t/gFg/jCand, in general,127Then, Eq. (3.11) has the following formula.Newton For war d-Dif ference F ormula£f (<o)(3.12)If the i nterpolating nodes are reordered as z… z,—i, ... , = . a for mu la sil nildf tO Eq. (3.10) results:+ - j- f[xz, ... , x o] • - xz){x —x.—i) ‹x — xi).H the nodes are equally spaced wilx xz + sh and x xi -i- ( S 3- n —i)h, thenPz{x) ——P,{x, -i- sh)f{x n + SP f[xz, xz-i1 -1- s(s -¥1)h2J f<n. <.—i. *n—2] +-J- s{s -1- 1)- - (S + Zt I)h‘ [ n• -• . , x o]-This form is called the Newton backward divided-dNerence formula. It is used to derive a more com monly applied formula known as the Newton backward-dNerenceformul a . T O di SC uss thi s f O rmula, we need the following d efinition.Definition $.T Given the se quence {p•! n=o• define the backward difference 4 pz (read nabla p,) by"Up. ——p. —p.-i. for n ¿; 1.Higher powers are define d recwsively bypn '!!!- °'p»), for k 2i 2.De finit ion 3.7 i mplies fatf xz, x,-ifand, i n ge ner&,2h2Con sequently,f[xz, x z-;, ... •='k!h*k f••)pp g) f g n ] 3- s4 f{xz) +1'( n)•n 1)2 n!2) • '('-c H A P T L R 3 • intet polaiion anö PoJyn p NJa ï Appro ximat loH ’’’I f we e X t e fl d the b1Il Or»ial co effic ient no tatio n to includ e all real values o › by l ( -J- 1-)- - (J + k - 1 "- • — — 3— i (——* k + 1)— _ •)’ . ).thenP,[x} == j"[gq] —b(— 1)1—sQjf(y q )-ÿ(—1)2 2)-ÿ = •-f(—1)‘whieh gives the following result.Newton Back ward- DiHere nce For mul aP n {x } —— j"[x,](— 1) kkm IXAMPLE 2The divided-difference Table 3.9 eorresponds to the data in Example 1.bbte 3.91.00.76519771.30.6200860First divided differences—0.4837057Second Third ,F divided divided di differencesdiserencesdiff—0.1087339 —0.5489460 0.0658784 . 1.60.4554022—0.0494433 0.&l—0.57861200.0680685 1.90.28181860.0118183 **” ”***”*—0.5715210****””””*”Only one in t er pol ati n g poly nom ial of de gree £tt m O s t 4 u S we will organize the data QOlfltS tO O btai n t hebest eS thèsefive data po 1, 2, and 3. T his will give us a sense the given value of x. i 'l t°'P l £t u O n £f pproximation s O I of aceuracy O f t he f t 3 ll rth-deg ree approxim äIf a n a p r O xiIR t i O n t f( . ) is req uir e d, the re il s o nab le e hoiew ^o = 1 0, x i = 1.3, a:2 = 1.6, 3 = 1.9, ande for the nodes possible use of the data poiEtS cl os est to xp. 1, and alsO m ake s dif ference. This impli es that ù 0.3 and s o the Newuse of the fourtb form ll l il l S u S e d with the d ivide d dif fe re nc es fhàt ÏÎt C n f orward divided-diä…,, ave ä SO lid un derscore in T able 3•4 ( -1) = P 4 (1.0 -j- (0.3))= 0.7651997 -p (p)(—0.4837057)"* Z ( 2) (0.3)°( —0. 1087332J-s=i.3.2 f/ivJded Difi erencœ+ 31+ 30.7196480.(0.3)3(0.0658784)(0.3)4(0.0018251)To app r oz¡i » £f te a value when x is close to the end of the tabulate d V alues, s £f y, x = 2.0, we would l g • llk t o ma ke ie earliest use of ie data p O i fl tS closest to z. This requires using Öe N ewt O n backwa rd d ivided-diä erence formula w 1Î S = 2 and the divided d if feren ces in Tabl e 3.9 that have a dashed underscore: 4(2.0)4 2.2 — 2(0.3))= 0.1103623 — 2 (0.3)(—0.5715210) — 2) (0.3)2(0.0118183)3 3J)4) (0.3)3(0.0680685) — 3 J) 4-) J) (0.3)4(0.0018251)0.2238754.The Newton formulas are not appropriate for approximating f(x) when z lies near the center of ie table since employing either ie backward or forward method in such a way that the highest-order difference is involved will not allow <o to be close to x. A number of divided-diYerence formulas are available for this case, each of which has situations when it can be used to maximum advantage. These methods are Aown as centered-difference formulas. There are a number of such melods, but we will present only one, Stirling’s melod, and again refer ie interested reader to [Hild] for a more complete presentation.For the centered-difference f or mulas, we choose =0 near the point being approxi- mated and label the nodes direcXy below =o as x… x 2. . and those directly above asx _… x _2, ....With thi s co n vention, S tirling’s formu l il i s given by s/i p x) P;w p (x) f{x o] if[x -i. •o1 + J [x o. <il) + s 2h 2(3.14) nl+ 2g21)J 3.fIx —2. x-;. <o.=il + Cf=-i.=o. xz])21*—i, *o, *i5-p ... 3- s 2 (s21)(s 2 — 4) • • • ls - [ttt - l )2)h f $x ,z, ... , x,z)J(i 2 — 1-)- - (s 2 — 2)h**+'** 2(/t=— -i. . , xp! -t- f ïx q, ... , <q+,]),«» = 2« -)- 1 is o dd. If ü- 2m is even, we use the same formula but delete the last line. The entri es used for this f ormu la are underûned in Table 3.10 on page 130. c o fl s i d erl e tabl e O f d£t W °* given in the previous examples. To use Stirling’s formula to ap proxi fna te J ’(1.5) wi1*o = 1.6, we use the underlined entries in the difference Table3.11.c H A P T E R 3- lnte‹polai ion an0 P ol y nomial Appro xima tionTable 3.10First Secon ddivided divide dd if ferenc es dis erenc esThirddi videddifferen ces diffepggf{x-i, x-i. zoJfi«-i. x0l f'•- 2°-' °0’"x0 /l›0l/[ 0› lJ f lx- . x0.•ff[x-i,x0. x; , >2!x2 /lx213bble 3.JJ1.0 0.76519770.62008601.6 0.45540221.9 0,28181862.2 0.1103623 —0.5786120-0.57152100.01181830.0680685Th e f O rmul il , with A = 0.3, Up = 1.6, and s =j"(1.5) 4 1.6 -j- —} ) (0.3))3'beco mes= 0.4554022 -F0.3((—0.5489460) -j- (—0.5786120)) .(0.3)2(—0.04 94433)(0.3)3(0.0658784 + 0.0680685)= 0.51l820g. (0.3)4(0.0018251)divided diYerencesSeconddivideddifferencesThirddivideddifferences—0.48370J7—0.1087339—0.5489 460 0.0658784—0.0494433。
带重节点的牛顿插值法
带重节点的牛顿插值法牛顿插值法是一种在给定数据点的情况下通过插值函数来逼近真实函数的方法。
它属于插值法中的一种，是一种非常有用和广泛使用的数值计算方法。
牛顿插值法使用一个多项式函数来逼近真实函数，该函数使用给定的数据点来确定多项式系数。
带重节点的牛顿插值法是一种牛顿插值法的扩展形式，可以在数据点重复的情况下使用。
在牛顿插值法中，我们首先将给定的数据点按照节点值从小到大排序。
然后我们需要计算每一个节点上的差商，通过差商可以得到一个多项式函数。
差商的定义如下：$f[x_0,x_1]=\frac {f(x_1)-f(x_0)}{x_1-x_0}$以此类推，差商的递归计算可以用以下公式表示：$f[x_i,x_{i+1},\dots,x_{i+j}]=\frac{f[x_{i+1},x_{i+2},\dots,x_{i+j}]-f[x_i,x_{i+1},\dots,x_{i+j-1}]}{x_{i+j}-x_i} $使用差商的定义和递归公式，我们可以得到一个多项式函数：$P_n(x)=f[x_0]+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1)+\dots+f[x_0,x_1, \dots,x_n](x-x_0)(x-x_1)\dots(x-x_{n-1})$其中，$f[x_0,x_1,\dots,x_n]$表示一个$n$阶的差商。
但是，如果在给定的数据点中有重复的节点，那么上述算法将不再适用。
为了使用带重节点的牛顿插值法，我们需要使用多重差商（divided difference）来计算插值函数。
在插值点为$x_0,x_1,\dots,x_n$且有$m+1$个插值点重复的情况下，$m+1$重差商（divided difference）定义为：对于$m+1$个插值点重复的情况，用多重差商计算插值函数的公式如下：其中，$f(x_0,x_0,\dots,x_0,x_i)$是$i+1$重差商，$\prod_{j=0}^{i-1}(x-x_j)$是拉格朗日基函数。
代数英语
(0,2) 插值||(0,2) interpolation0#||zero-sharp; 读作零井或零开。
0+||zero-dagger; 读作零正。
1-因子||1-factor3-流形||3-manifold; 又称“三维流形”。
AIC准则||AIC criterion, Akaike information criterionAp 权||Ap-weightA稳定性||A-stability, absolute stabilityA最优设计||A-optimal designBCH 码||BCH code, Bose-Chaudhuri-Hocquenghem codeBIC准则||BIC criterion, Bayesian modification of the AICBMOA函数||analytic function of bounded mean oscillation; 全称“有界平均振动解析函数”。
BMO鞅||BMO martingaleBSD猜想||Birch and Swinnerton-Dyer conjecture; 全称“伯奇与斯温纳顿－戴尔猜想”。
B样条||B-splineC*代数||C*-algebra; 读作“C星代数”。
C0 类函数||function of class C0; 又称“连续函数类”。
CA T准则||CAT criterion, criterion for autoregressiveCM域||CM fieldCN 群||CN-groupCW 复形的同调||homology of CW complexCW复形||CW complexCW复形的同伦群||homotopy group of CW complexesCW剖分||CW decompositionCn 类函数||function of class Cn; 又称“n次连续可微函数类”。
Cp统计量||Cp-statisticC。
《城市轨道交通供电系统工程施工质量验收规范》标准全文
量安全监督总站负责技术内容的解释。执行过程中，请各有关单位结合实际，不断总结经验，
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本规范由浙江省住房和城乡建设厅负责管理和对强制条文的解释，由杭州市建设工程质
浙江省工程建设标准
城市轨道交通供电系统施工质量验收规范
Code for acceptance of construction quality of power supply system engineering of urban rail transit
DB33/T ××/××××-201×
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1
2
3
Page 189
4
Reading
Page 190 – Example 4.1
5
Reading
Pages 192 & 193 – Examples 4.2 & 4.3
Home Assignment
Home Assignment
Page 217 – Exercises 1 and 2
17
Interpolation
Newton’s Divided Difference Polynomial Method
What is Interpolation ?
Given (x0,y0), (x1,y1), …… (xn,yn), find the value of ‘y’ at a value of ‘x’ that is not given.
which passes through (x0,y0), (x1,y1), (x2,y2), and (x3,y3).
15
The general form of Lagrange interpolating polynomial (p208)
16
Reading
Page 207 – Example 4.6 Page 210 – Example 4.7
21
Example
The path of a rapid laser is given by these specifications. If the laser is traversing from x = 2 to x = 4.25 in a linear path, find the value of y at x = 4 using the Newton’s Divided difference polynomial method.
27
x2 5.25, y( x2 ) 6.0
Quadratic Interpolation (contd)
b0 y( x0 )
7 .2 y( x1 ) y( x0 ) 7 .1 7 . 2 b1 4.25 2.00 x1 x0 0.04444 y ( x2 ) y ( x1 ) y ( x1 ) y ( x0 ) 6.0 7.1 7.1 7.2 x2 x1 x1 x0 b2 5.25 4.25 4.25 2.00 x 2 x0 5.25 2.00 1.1 0.04444 3.25
0.04444
23
7.1
7.1 7.08 x s 10
0
5
Hale Waihona Puke 05 x s range x desired
10 x s 10
1
Linear Interpolation (contd)
7.2 7.2 7.18 7.16 ys f ( range ) f x desired 7.14
a 7.2735 7.1111 100 7.2735
2.2327 %
29
General Form
f 2 ( x) b0 b1 ( x x0 ) b2 ( x x0 )(x x1 )
where
b0 f [ x0 ] f ( x0 ) f ( x1 ) f ( x 0 ) b1 f [ x1 , x0 ] x1 x0 f ( x 2 ) f ( x1 ) f ( x1 ) f ( x0 ) f [ x 2 , x1 ] f [ x1 , x0 ] x 2 x1 x1 x0 b2 f [ x 2 , x1 , x0 ] x 2 x0 x 2 x0
6.5
6
6
2 2
2.5
3
3.5
4
4.5
5
5.5 5.25
x s range x desired
y( x) b0 b1 ( x x0 ) b2 ( x x0 )(x x1 ) x0 2.00, y( x0 ) 7.2
x1 4.25, y( x1 ) 7.1
Rewriting
f 2 ( x) f [ x0 ] f [ x1 , x0 ](x x0 ) f [ x2 , x1 , x0 ](x x0 )(x x1 )
30
General Form
Given (n 1) data points, x0 , y0 , x1 , y1 ,......,xn1 , yn1 , xn , yn as
x (m) y (m)
Path of a robot 8 7 6 5 4 3 2 1 0 0 5 X 10 15
2 4.25 5.25 7.81 9.2 10.6
26
7.2 7.1 6 5 3.5 5
Y
Quadratic Interpolation (contd)
7.56258 8 7.5 ys f ( range ) f x desired 7
Linear interpolation: Given ( x0 , y0 ), ( x1 , y1 ), pass a linear interpolant through the data
f1 ( x) b0 b1 ( x x0 )
where
b0 f ( x0 )
b1 f ( x1 ) f ( x0 ) x1 x0
Page 195 – Exercises 1(b) & 2(b)
6
Polynomial Approximation
(p200)
7
8
9
10
Reading
Page 200 – Example 4.4
Home Assignment
Page 205 – Exercise 2
11
Different from previous method (using system of linear equation to calculate the polynomial coefficients), we could directly give the coefficients without derivation:
31
General form
The third order polynomial, given ( x0 , y0 ), ( x1 , y1 ), ( x2 , y 2 ), and ( x3 , y3 ), is
f 3 ( x) f [ x0 ] f [ x1 , x0 ](x x0 ) f [ x2 , x1 , x0 ](x x0 )(x x1 ) f [ x3 , x2 , x1 , x0 ](x x0 )(x x1 )(x x2 )
19
Interpolants
Polynomials are the most common choice of interpolants because they are easy to: Evaluate Differentiate, and Integrate.
20
Newton’s Divided Difference Method
x (m) y (m)
Path of a robot 8 7 6 5 4 3 2 1 0 0 5 X 10 15
2 4.25 5.25 7.81 9.2 10.6
22
7.2 7.1 6 5 3.5 5
Y
Linear Interpolation
y( x) b0 b1 ( x x0 ) x0 2.00, y( x0 ) 7.2
Lagrange linear polynomial
13
Lagrange quadratic interpolating polynomial
Page 210
which passes through (x0,y0), (x1,y1), and (x2,y2).
14
Lagrange cubic interpolating polynomial
2.00 x 5.25
At x 4,
y(4.00) 7.2 0.04444 (4.00 2.00) 0.32479 (4.00 2.00)(4.00 4.25) 7.2735 in.
The absolute relative approximate error a obtained between the results from the first and second order polynomial is
f n ( x) b0 b1 ( x x0 ) .... bn ( x x0 )(x x1 )...(x xn1 )
where
b0 f [ x0 ]
b1 f [ x1 , x0 ]
b2 f [ x2 , x1 , x0 ]
bn1 f [ xn1 , xn2 ,....,x0 ] bn f [ xn , xn1 ,....,x0 ]
-- Lagrange Polynomial -- Newton Polynomial
12
Lagrange Approximation
The French mathematician Lagrange noticed
which passes through (x0,y0) and (x1,y1).
24
7.1111in.
Quadratic Interpolation
Given ( x0 , y0 ), ( x1 , y1 ), and ( x2 , y 2 ), fit a quadratic interpolant through the data.