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AMATH 581 ( c○J. N. Kutz) 29O(∆t 2 )center-differenceschemesf ′ (t) =[f(t +∆t) − f(t − ∆t)]/2∆tf ′′ (t) =[f(t +∆t) − 2f(t)+f(t − ∆t)]/∆t 2f ′′′ (t) =[f(t +2∆t) − 2f(t +∆t)+2f(t − ∆t) − f(t − 2∆t)]/2∆t 3f ′′′′ (t) =[f(t +2∆t) − 4f(t +∆t)+6f(t) − 4f(t − ∆t)+f(t − 2∆t)]/∆t 4Table 2: Second-order accurate center-difference formulas.on t ∈ [a, b] withthegeneralboundaryconditionsdy(a)α 1 y(a)+β 1dtdy(b)α 2 y(b)+β 2dt= γ 1 (1.5.2a)= γ 2 . (1.5.2b)Thus the solution is defined over a specific interval and must satisfy the relations(1.5.2) at the end points of the interval.Before considering the general case, we simplify the method by consideringthe linear boundary value problemd 2 ydt 2on t ∈ [a, b] withthesimplifiedboundaryconditions= p(t)dy + q(t)y + r(t) (1.5.3)dty(a) =αy(b) =β.(1.5.4a)(1.5.4b)Taylor expanding the differential equation and boundary conditions will generatethe linear system of equations which solve the boundary value problem.To see how the Taylor expansions are useful, consider the following twoTaylor series:df (t)f(t +∆t) =f(t)+∆tdtf(t − ∆t) =f(t) − ∆tdf (t)dt+ ∆t22!+ ∆t22!d 2 f(t)dt 2d 2 f(t)dt 2where c i ∈ [a, b]. Subtracting these two expressions givesdf (t)f(t +∆t) − f(t − ∆t) =2∆t + ∆t3dt 3!+ ∆t33!− ∆t33!d 3 f(c 1 )dt 3d 3 f(c 2 )dt 3(1.5.5a)(1.5.5b)( d 3 f(c 1 )dt 3 + d3 f(c 2 )dt 3 ). (1.5.6)