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Unformatted text preview: E solution for which the numerical
method does not produce an exact solution. In the case of the leap frog method, this
would be any cubic polynomial, and we select y(t) = t3 . Substitution into the two
expressions for the local discretization error yields
t3 ; (tn ; 2h)3 ; 3(t ; h)2 = 6Ch2 :
The location of the point n is irrelevant for this exact solution and we nd C = 1=3.
= h y000( n)
n 2 (tn;2 tn ):
3 Problems 1. Using the method of undetermined coe cients, we can make some general observations regarding the coe cients of (5.1.2). Show, for example, that
i=0 4 i =0 (5.1.5a) if (5.1.2) is exact when y(t) = 1. Additionally, if (5.1.2) is exact when y(t) = tq ,
q = 1 2 : : : , show
i=1 iq i+q k
i=0 iq;1 i = 0 q = 1 2 ::: : (5.1.5b) These are order conditions for the linear multistep method. 5.2 Newton Divided-Di erence Polynomials
Speci c multistep formulas will be derived by approximating y(t) or f (t y) by interpolating polynomials and, respectively, di erentiating or integrating these polynomials. We've
already seen examples where one-step methods were constructed by using interpolating
polynomials with collocation. Thus, in this section, we'll review polynomial interpolation in an abstract setting that is removed from our from our primary task of solving
di erential equations.
The interpolation problem consists of nding a polynomial Pk (t) of degree k that
interpolates a function f (t) at k + 1 distinct points t0 t1 : : : tk , i.e., f (tj ) = Pk (tj ) j = 0 1 : : : k: (5.2.1) In practice, we express the polynomial in a convenient basis for the application. The
obvious basis of powers of monomial terms tk Pk (t) = a0 + a1 t + a2 t + : : : + ak =
i=0 leads to the linear algebraic system f (t0) = a0 + a1 t0 + a2 t2 + : : : + ak tk
f (t1) = a0 + a1 t1 + a2 t2 + : : : + ak tk
f (tk ) = a0 + a1tk + a2 t2 + : : : + ak tk
5 ai tk (5.2.2) and is rarely convenient. The Lagrange basis
Y t ; tj
Li (t) =
j =1 j 6=i ti ; tj (t ;...
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