Review of Gaussian Quadrature method


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1 Review of Gussin Qudrture method Nsser M. Asi Spring 006 compiled on Sundy Decemer 1, 017 t 09:1 PM 1 The prolem To find numericl vlue for the integrl of rel vlued function of rel vrile over specific rnge over the rel line. This mens to evlute I = f x) dx Geometriclly, this integrl represents the re under fx) from to Solution We cn lwys pproximte the re y dividing it in equl width strips nd then sum the res of ll the strips. In generl, there will lwys e n error in the estimte of the re using this method. The error will ecome smller the more strips we use which implies smller strip width). Hence we cn write N ) f x) dx = x f x i ) + E i=1 Where E is the error etween the ctul re nd the pproximted re using the ove method of numericl integrtion. N ove is the numer of strips or cn lso e refereed to s the numer of integrtion points. Insted of keep referring to the width of the strip ll the time, we will cll this quntity the weight w i tht we will multiply the vlue of the function with to otin the re. Hence the ove ecomes N ) f x) dx = w i f x i ) + E i=1 Using implied summtion on indices the ove ecomes f x) dx = w i f x i ) + E 1
2 y=fx) f x x Are under curve f x dx Figure 1: Integrting function f x f x 1 f x f x 4 y=fx) f x 1 4 x x 4 f x dx i 1 x f x i Figure : Numericl integrtion
3 In the ove we divided the rnge of the integrtion the distnce etween the upper nd lower limits of integrtion) into equl intervls. We cn decide to evlute fx i ) t the middle of the strip or t the strt of the strip or t the end of the strip. In the digrm ove we evluted the fx) t the left end of the strip. Our gol is to evlute this integrl such s the error E is minimum nd using the smllest numer of integrtion points. In sense this cn e considered n optimiztion prolem with constrints: minimize the error of integrtion using the smllest possile numer of points. To e le to do this minimiztion, we need to consider wht re the vriles involved. We see tht there re two degrees of freedom to this prolem. One is the width of the strip or w i. We do not hve to use fixed vlue of width, we cn use different width for different strips if the resulting integrl gives etter pproximtion. The second degree of freedom is the point t which to evlute fx i ) ssocited with ech strip. In the exmple ove we choose to evlute fx i ) t left end point of the strip. We cn choose to select different x i point if this will result in etter pproximtion. This is the min ide of Guss Qudrture numericl integrtion. It is to e le to choose specific vlues for these two degrees of freedom, the w i nd the x i. It turns out tht if the function fx) is polynomil, then there is n optiml solution. There is n optiml {w i, x i } for ech polynomil of order n. We cn determine these degrees of freedom such tht the error E is zero, nd with the lest possile numer of integrtion points. We re le to tulte these two degrees of freedom for ech polynomil of specific order. In other words, if the function fx) is polynomil of order n then we know efore the computtion strts wht these degrees of freedom should e. We know the loctions of x i nd we know weight w i tht we need to multiply fx i ) with to otin the re with minumum error. You might sk how cn this method of integrtion know the loctions of the integrtion points x i eforehnd without eing given the integrtion rnge of the function to integrte? It turns out tht we will mp f x) into new known nd specific rnge of integrtion from to +1) for the method tht we will now discuss..1 Guss Qudrture From now on we will ssume the function fx) to e integrted is polynomil in x of some order n. Guss qudrture is optiml when the function is polynomil The min strting point is to represent the function f x) s comintion of linerly independent sis. Insted of using strips of equl width, we ssume the width cn vry from one strip to the next. Let us cll the width of the strip w i. Insted of tking the height of the i th strip to e the vlue of the function t the left edge of the strip, let us lso e flexile on the loction of the x ssocited with strip w i nd cll the height of the strip w i s fx i ) where x i is to e determined. Hence the ove integrtion ecomes
4 N ) f x) dx = w i f x i ) + E i=1 N w i f x i ) i=1 So our gol is to determine w i nd x i such s the error E is minimized in the ove eqution. We would relly like to find w i nd x i such tht the error is zero with the smllest vlue for N. It seems s if this is very hrd prolem. We hve N unknown quntities to determine. N different widths, nd N ssocited x points to evlute the height of ech specific strip t. And we only hve s n input fx) nd the limits of integrtion, nd we need to determine these N quntities such tht the error in integrtion is zero. In other words, the prolem is to find w i, x i such tht I = f x) dx = w 1 f x 1 ) + w f x ) + w N f x N ) 1) One wy to mke some progress is to expnd fx) s series. We cn pproximte fx) s convergent power series for exmple. If fx) hppens to e polynomil insted, we cn represent it exctly using finite sequence of Legendre polynomils. It is in this second cse where this method mkes the most sense to use due to the dvntges we mke from the second representtion. We show oth methods elow. Expnding f x) s convergent power series over the rnge, gives Sustituting ) into 1) gives f x) = x + x + m x m + ) I = But x + x + m x m + ) dx = w 1 f x 1 ) + w f x ) + w N f x N ) ) x + x + m x m + ) dx = = 0 ) + 1 ) Sustituting the ove into ) results in 0 dx + + ) 1 x dx + x dx m m+1 m+1) m m x m dx + 4
5 ) ) m+1 m+1) 0 ) m + m + 1 = w 1 f x 1 ) + w f x ) + + w N f x N ) 4) But from ) we see tht f x 1 ) = x 1 + x 1 + m x m 1 + f x ) = x + x + m x m + Sustituting the ove into 4) gives f x N ) = x N + x N + m x m N + ) ) m+1 m+1) 0 ) m + = m + 1 w x 1 + x 1 + m x m 1 + ) +w x + x + m x m + ) Rerrnging gives +w N x N + x N + m x m N + ) ) ) m m ) 0 ) m + = m 0 w 1 + w + + w N ) + 1 w 1 x 1 + w x + + w N x N ) + w1 x 1 + w x + + w N x ) N + m w 1 x m 1 + w x m + + w N x m N) Equting coefficients on oth sides results in w 1 + w + + w N = 5) ) w 1 x 1 + w x + + w N x N = w 1 x 1 + w x + + w N x ) N = w 1 x m 1 + w x m + + w N x m N = m m ) m 5
6 Since we hve N unknown quntities to solve for N weights w i nd N points x i ) we need N equtions. In other words, we need to hve m = N. The ove set of simultneous N equtions cn now e solved for the unknown w i, x i nd this will give us the numericl integrtion we wnted. The ove ssumed tht the function fx) cn e represented exctly y the power series expnsion with m terms. We now consider the representtion of fx) s series of Legendre polynomils. We do this since when fx) itself is polynomil. We cn represent f x) exctly y finite numer of Legendre polynomils. But since Legendre polynomils P n x) re defined over [, 1] we strt y trnsforming f x) to this new rnge nd then we cn expnd the mpped f x) which we will cll g ζ)) in terms of the Legendre polynomils. y=fx) f x Liner trnsformtion g x An esy wy to find this mpping is to lign the rnges over ech others nd tke the rtio etween s the scle fctor. This digrm shows this for generl cse where we mp f x) defined over [, ] to new rnge defined over [c 1, c ] f x dx x g dx d c c 1 c 1 c d We see from the digrm tht ζ = c 1 + dζ But dζ dx is the sme rtio s Hence c c 1 dx dζ = c c 1 6) 6
7 The ove is clled the Jcoin of the trnsformtion. Now, From the digrm we see tht And dx = x Hence 6) ecomes dζ = ζ c 1 Hence we otin tht And x ζ c 1 = c c 1 ζ = x c c 1 ) + c 1 x = c c 1 ζ c 1 ) + For the specific cse when c 1 = nd c = +1 the ove expressions ecome Hence ζ = x ) 1 x ) = = x x = ζ + 1) + ) ζ + + ) = Before we proceed further, It will e interesting to see the effect of this trnsformtion on the shpe of some functions. Below I plotted some functions under this trnsformtion. The left plots re the originl functions plotted over some rnge, in this cse [4, 10] nd the right side plots show the new shpe the function g ζ)) over the new rnge [, 1] 7
8 We must rememer tht in the following nlysis, we re integrting now the function g ζ) over [, 1] nd not f x) over [, ]. Hence to otin the required integrl we need to trnsform ck the finl result. We will show how to do this elow. We cn pproximte ny function f x) s series expnsion in terms of weighted sums of set of sis functions. We do this for exmple when we use Fourier series expnsion. Hence we write f x) = i Φ i x) 7) i 8
9 We cn give n intuitive justifiction to the ove formultion s follows. If we think in terms of vectors. A vector V in n ndimensionl spce is written in terms of its components s follows V = 1 e 1 + e + N e N N = i e i i Where e i is the sis vectors in tht spce. If we now consider generl infinite dimensionl vector spce where ech point in tht spce is function, then we see tht we cn lso represent tht function s weighted series of sis functions just s we did for norml vector. There re mny sets of sis functions we cn choose to represent the function f x) with. The min requirements for the sis functions is tht they re complete This mens they spn the whole spce) nd there is defined n inner product on them. For our purposes here, we re interested in the clss of function f x) tht re polynomils in x. The sis we will select to use re the Legendre sis s explined ove. To do this, we trnsform f x) to g ζ) s shown ove nd then now our integrl ecomes f x) dx = ) ) f x ζ)) dζ This is ecuse we found tht dx = ) dζ from ove. If we cll f x ζ)) s g ζ) the integrl ecomes f x) dx = ) g ζ) dζ Since ) is the Jcoin of the trnsformtion, we write the integrl s f x) dx J g ζ) dζ Since the Jcoin is constnt in this cse, we will only worry out g ζ) dζ nd we just need to rememer to scle the result t the end y J. This is the integrl we will now numericlly integrte. Eqution 7) is now written s g ζ) = i P i ζ) i Where P i ζ) is the Legendre polynomil of order i nd g ζ) is polynomil in ζ or some order m. 9
10 Now we crry the sme nlysis we did when we expressed f x) s power series. The difference now is tht the limit of integrtion is symmetricl nd the sis re now the Legendre polynomils insted of the polynomils x n s the cse ws in the power series expnsion. So now eqution 1) ecomes I = g ζ) dζ = w 1 g ζ 1 ) + w g ζ ) + + w N g ζ N ) 8) And eqution ) ecomes g ζ) = 0 P 0 ζ) + 1 P 1 ζ) + P ζ) + m P m ζ) + 9) Sustituting 9) into 8) we get the equivlent of eqution ) I = 0 P 0 ζ) + 1 P 1 ζ) + P ζ) + m P m ζ) + ) dζ 10) = w 1 g ζ 1 ) + w g ζ ) + + w N g ζ N ) 1) 0 P P 1 + P + m P m + ) dζ = 0 P 0 dζ + 1 P 1 dζ + P dζ + + m P m dζ + = 0 ) = 0 The ove occurs since the integrl of ny Legendre polynomil of order greter thn zero will e zero over [, 1] Sustituting the ove into 10) we otin But from 9) we see tht I = 0 = w 1 g ζ 1 ) + w g ζ ) + + w N g ζ N ) 11) g ζ 1 ) = 0 P 0 ζ 1 ) + 1 P 1 ζ 1 ) + P ζ 1 ) + m P m ζ 1 ) + g ζ ) = 0 P 0 ζ ) + 1 P 1 ζ ) + P ζ ) + m P m ζ ) + g ζ N ) = 0 P 0 ζ N ) + 1 P 1 ζ N ) + P ζ N ) + m P m ζ N ) + Sustituting the ove in 11) gives 0 = w 1 0 P 0 ζ 1 ) + 1 P 1 ζ 1 ) + P ζ 1 ) + m P m ζ 1 ) + ) + + w 0 P 0 ζ ) + 1 P 1 ζ ) + P ζ ) + m P m ζ ) + ) + + w N 0 P 0 ζ N ) + 1 P 1 ζ N ) + P ζ N ) + m P m ζ N ) + ) 10
11 Rerrnging results in 0 = 0 w 1 P 0 ζ 1 ) + w P 0 ζ ) + + w N P 0 ζ N )) + 1 w 1 P 1 ζ 1 ) + w P 1 ζ ) + + w N P 1 ζ N )) + + m w 1 P m ζ 1 ) + w P m ζ ) + + w N P m ζ N )) Equting coefficients gives = w 1 P 0 ζ 1 ) + w P 0 ζ ) + + w N P 0 ζ N ) 0 = w 1 P 1 ζ 1 ) + w P 1 ζ ) + + w N P 1 ζ N ) 0 = w 1 P ζ 1 ) + w P ζ ) + + w N P ζ N ) 0 = w 1 P m ζ 1 ) + w P m ζ ) + + w N P m ζ N ) If we select the points ζ i to e the roots of P i we cn write the ove s = w 1 P 0 ζ 1 ) + w P 0 ζ ) + + w N P 0 ζ N ) 0 = w 1 P 1 ζ 1 ) + w P 1 ζ ) + + w N P 1 ζ N ) 0 = w 1 P ζ 1 ) + w P ζ ) + + w N P ζ N ) 0 = w 1 P m ζ 1 ) + w P m ζ ) + + w N P m ζ N ) References 1. Mthemtic Structurl Mechnics help pge. MIT course 16.0 lecture notes. MIT open course wesite.. Theory of elsticity y S. P. Timoshenko nd J. N. Goodier. chpter 10 11
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