# EQsheet - Romberg R(n,0)= trap rule with 2^n subintervals...

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RombergR(n,0)= trap rule with 2^n subintervalsR(n,0)=1/2*R(n-1,0)+h*SUM(k=1,2^(n-1))of f[a+(2k-1)*h]h= (b-a)/2^nR(n,m) = R(n,m-1) +1/(4^m-1)*[R(n,m-1)-R(n-1,m-1)]Original L=L1Make ur n=n+1 so h becomes h/2 to get L2Multiply by constant that cancels ur (/2)^a on the first error term > > (2^a)L2Subtract ur original to cancel out the first error term (2^a-1)LSolve for L by dividing by (2^a-1)Re work eq to get new formula with smaller error termSimpson’s Scheme1)Integral of f from a to b = approx S(a,b)S(a,b)= ((b-a)/6)[f(a)+4f((a+b)/2)+f(b)]E= (-1/90)[.5(b-a)]^5*f^(4)(e)2) for h=(b-a)/2 doing the integral from a to a+2h= h/3[f(a)+4(a+h)+f(a+2h)]E=(-1/90)h^5*f^(4)(e)3) Composite 1/3 over n even subintervalsI(a,b)= h/3[f(a)+f(b)] +4h/3S(i=1,n/2)f[a+(2i-1)h] +2h/3S(i=1,(n-2)/2) f(a+2ih)h=(b-a)/nE= -1/180(b-a)h^4f^(4)(e)4) error test1/15abs[S(a,c)+S(c,b)-S(a,c)]<eGaussian Quadrature FormulasI(a,b)=approx S(i=0,n)Aif(xi)Ai=I(a,b)li(x)
li(x)=P(j=0 jneti,n)(x-xj)/(xi-xj)if q gives you I(a,b)x^kq(x)dx=0then xo,x1,…xn are zeros of qthese zeros are your nodesMake a table withf(x)