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Numerical Methods

by: Kiera Considine I

Numerical Methods EAD 210B

Kiera Considine I
GPA 3.9


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This 5 page Class Notes was uploaded by Kiera Considine I on Tuesday September 8, 2015. The Class Notes belongs to EAD 210B at University of California - Davis taught by Staff in Fall. Since its upload, it has received 51 views. For similar materials see /class/187494/ead-210b-university-of-california-davis in Engineering Applied Sci Davis at University of California - Davis.


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Date Created: 09/08/15
Numerical 39 quot and differentiation Difference between integration and differentiation Differentiation 0 First order forward f39xfxhfxh 0 First order backward f39xfxfxhh 0 Second order f39xfxhfxh2h 0 Second derivative Second order fquotxfxhfxh2fxh2 Interpolation o Interpolate lagrange using two points x1y1 xzyz fxxx1yzxzx1 xxzy1x1xz dfxdxyzy1xzx1 Interpolate lagrange using three points x1y1 XzyzX3y3 fXXX1XXzy3X3X1X3Xz xx1xX3yzxzx1xzx3 XX3XX2ylX139X3X139Xz o dfXdX2XX1Xzy3X3X1X3Xz 2XX1X3yzXzX1Xz X3 2XX3Xzy1X1X3X1Xz Integration We only discuss Riemann integrals o Lim Zi1quotfxiAxi 0 Open 7without the boundaries 0 Close with the boundaries Integration formulae EHquot wi fxi We can optimize either wi or the choice of xi or both Lagrange integrals 7 approximate function by lagrange polynomial and integrate the polynomials o fx2fxilix o Ifxdxl2fxilixdx2fxiI1ixdx o willixdx Method of undetermined coefficients 0 Equivalent request that integral is precise for all polynomials until degree n l 0 EHquot wi fxi Jfxdx for all polynomials up to degree nl 0 Solution equivalent to Lagrange integration 0 24quot w xik kadxxk1k1 bkllak1k1 Degree of integration 7 precision of integral If integral is precise for polynomial of degree nl then the integration is precise to degree n Newton quadratures fix xi select w o Midpoint rule 7 constant at center Mf2i1quot391fxixi12Axi o Trapezoid rule 7 linear in region Tf2i1quot391fxifxi12Axi o Simpsons rule 7 Second order interpolation o Sf2i1quot391fxi 4fxixi12fxi16Axi Estimate errors for quadratures o The error for the midpoint rule in a given interval h is fquotch324 where c is a point in the interval from taylor expansion in the middle of the section 0 The error for the trapezoid rule is approximately fquotch3 12 from the sum of talyor expansion on the two sides OOOO 0 Simpson error is fl4c2880h5 two order higher precision than both trapezoid and midpoint o Gauss quadratures O O O O O Selact both Xi and w to optimize order of integration In interpolary integration we had n free parameters we obtained a degree ofnl ifwe vary both Xi and wi we can obtain a 2nl degree Method of undermined coefficients for both Xi and wi results in 2n non linear equations Choose an orthogonal polynomial legender of degree 2nl that is orthogonal to all polynomials of lower degree The zeros of this polynomial are real and simple and in the integration interval The n point lagrange integration based on these zeros has a degree of 2n l Once we have determined the points in the interval ll we can change coordinates to bring any integration space to ll o Clenshaw Curtis quadrature 0 Almost as precise as gauss but instead of using legendre zeros use chebichev zeros easier to compute 0 Adaptive quadrature 0 Estimate error by comparing mid point and trapezoid or simpsons and keep dividing the section until the error is smaller than some threshold Extrapolation 0 Find an approximation for the data using fx 0 Linear axb nd a and b 0 Compute error EZ axib yi2 O S ZXiXi SK ZXi Sxy ZXiyi Sy Zyi dEdb2 Z axib yi0 nba 2 xi E yi dEda2 Z axib yixi0 b 239 xi a 2 xi2 Z xiyi SxanbSy o Sxxa5xbSxy o Parabola ax2bxc find ab and c Same method 0 dEda2 2 x3 axi2bxic fxixi0 O Sxxxxasxxxb Sxxc Sxxy o SxxxaSxxb chSxy o Sxxabe ncSy 0 General least square 0 fX 23139 gj X o E22ajgjxo yi2 00000 O dEdak0 gk xi2aj gjXi yi0 0 ZjajZi gk Xigjxi Zi gk Xiyi0 0 bjk2i gk XigjXi O ck2i gk XiYi o Bac 0 Continuous observations 7 same thing with integral 0 fX Zajgjx o 13 Eatgoo yzdx o dEdak0 2I gk x2ajgjx ydX0 o Ejajfgk max dx I gk 0de o bJH gk xgjxdx O Ckl gk Xydx o Bac o Legendre polynomials o If the polynomial are orthogonal with a weight function of 1 then B is diagonal and aicibii 0 Moreover each a does not depend on its predecessors 0 We can rescale the range of integration to 11 and the orthogonal polynomial will become the legendre polynomials o If the integration is over the range zw we replace X by 2Xw zwz 0 General theory 0 Given a set of n observations yi and a function with m components fX 2X1 gj X 0 We want ZngjXiyi 0 We denote aijgjxi and obtain Axy where A is nm a is a m vector and y is a n vector 0 Since A is over determined there will usually be no solution but we want to minimize the residual We want rAxy to be minimal and we use the Euclidean norm to nd the minima since it is a convex function it always exist 0 ler is minimal is equivalent to Her minimal 0 Hrl l 2rTrAxyTAxyxTATAxyTAxxTATeryTy 0 We want to nd the minima relative to a so that we want the gradient of the error to be minimal VE0 o ATAxATy0 0 Let is denote CATA and b ATy and we are back to Cxb where C is a square matrix If A has full rank then C is nonsingular 0 Now we will look for way to create a simple C with a low condition 0 Cholesky o If matrix M is positive de nite and symmetric we can fact prize MLLT where L is lower diagonal O 111 m1 1 2 0 Deal with every row of L or column of LT one after the other 0 Once we solved ATALLT we can solve LLTxATy 0 L2 ATy o LTxz o SVD 0 Any matrix Mmquot can be divided into MUnm2nmemT where U and V are orthogonal and Z is diagonal Gij0 when i j o This is denoted singular value decomposition SVD 0 We can divide 21 into 21 mm and Ommm Similarly we can divide Unm into Ulrmm and U2nnm o ATAxATb O Vmm2mnTUnnT Unn2nmemTX Vmm2mnTUnnTb O Vmquotm2mnT Z 4r1memTX Vmm2mnTUnnTb O Z Im mT Z 4r1memTX Z 4mrrTUnnTl o 21mmT 21mmvmme 21mmTU1mnTb o 21mvax U1me o xvmmzlmim39lU1minTb o QR o If Q is orthogonal then llell2QvTQvvTQTQvvTvllsz 0 Let us denote R as an upper triangular and Q as orthogonal while 0 is a zero matrix 0 if TA2QROT then lrll2AxbTAxblQROTxbll2 H ROTx Q bH 0 We denote QTb as c1czT HrHZHRxc1H2chH2 The minimum is obviously obtained when xicl o QR facrorization O O O O Housleholder matrix HI2vavTv Huu2vauvTv u2vTuvTvv If vux and 2vTuvTvl then Hux 0 Let us take x0ce1 o 2VTuVTV1 If we take 0c7signa1Hquy 2VTuVTV1 o If we take V0k7V2nk Hu u2V2Tu2V2TV2V and it is equal u for all indexes less than k 0 We can thus multiply a matrix and transform its rst column into zeros under the rst line We then continue to all other lines 0 End of approximation theory


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