Lecture 6: =1=Truncation Error, Consistency, and Convergence

Lecture 6:Truncation Error, Consistency, andConvergenceJ.K. Ryan@tudelft.nlWI3097TUDelft Institute of Applied MathematicsDelft University of Technology8 October 2012() Truncation Error, Consistency, and Convergence 8 October 2012 1 / 54

Outline1 ReviewSingle Step MethodsStabilityIVP StabilityAmplification factor2 Local Truncation Error & Consistency3 Global truncation error & Convergence() Truncation Error, Consistency, and Convergence 8 October 2012 2 / 54

Single-step methodsSummaryWe have 4 schemes to solvey ′ (t) = f (t, y(t))y(t 0 ) = y 0 .Forward Euler: w n+1 = w n + hf (t n , w n )Backward Euler: w n+1 = w n + hf (t n+1 , w n+1 )Trapezoidal method:Modified Euler:w n+1 = w n + h 2 (f (t n, w n ) + f (t n+1 , w n+1 ))¯w n+1 = w n + hf (t n , w n )w n+1 = w n + h 2 (f (t n, w n ) + f (t n+1 , ¯w n+1 ))() Truncation Error, Consistency, and Convergence 8 October 2012 3 / 54

StabilityLax Equivalence Theorem: For a linear scheme,Stability + Consistency = ConvergenceStability is how much small perturbations affect the solution.For this we have to look atStability of the initial value problem (IVP)Stability of the numerical method() Truncation Error, Consistency, and Convergence 8 October 2012 4 / 54

StabilityIVP StabilityFor the stability of the initial value problem (IVP), we considerɛ(t) =There are three possibilities:ỹ(t) −y(t)}{{}perturbed problem1 It is unstable if lim t→∞ |ɛ(t)| → ∞2 It is stable if lim t→∞ |ɛ(t)| < ∞ (finite)3 It is unconditionally stable if lim t→∞ |ɛ(t)| → 0() Truncation Error, Consistency, and Convergence 8 October 2012 5 / 54

StabilityIVP StabilityFor the test equationy ′ = λyy(0) = y 0 , (1)lim t→∞ |ɛ(t)| → 0 equates to λ < 0.() Truncation Error, Consistency, and Convergence 8 October 2012 6 / 54

StabilityNumerical StabilityFor numerical stability, we must consider the amplificationfactor, Q(hλ) inThe possiblities are:w n+1 = Q(hλ)w n = (Q(hλ)) n+1 y 0 .1 The method is unstable if lim h→0 |Q(hλ)| > 1.2 The method is stable if lim h→0 |Q(hλ)| ≤ 1.We use these conditions to choose an appropriate bound on h.() Truncation Error, Consistency, and Convergence 8 October 2012 7 / 54

StabilityAmplification factorw n+1 = Q(hλ)w nThe amplification factor for our methods is given by:Forward Euler: Q(hλ) = 1 + hλ, stable for 0 < h < 2|λ| .Backward Euler: Q(hλ) = 11−hλ, stable for 0 < h.Trapezoidal method: Q(hλ) = 1+ h 2 λ1− h λ,stable for 0 < h.2modified Euler: Q(hλ) = 1 + hλ + 1 2 (hλ)2 , stable for0 < h < 2|λ| .() Truncation Error, Consistency, and Convergence 8 October 2012 8 / 54

Local Truncation ErrorLast time we asked: How good are these methods?For this, we need convergence.But, convergence is difficult to prove directly.The Lax equivalence theorem says that we can instead provestability and consistency.() Truncation Error, Consistency, and Convergence 8 October 2012 9 / 54

StabilityLast time, we covered stability.Stability is given by the amplification factor,w n+1 = Q(hλ)w n .For this, we choose h so that |Q(hλ)| ≤ 1.() Truncation Error, Consistency, and Convergence 8 October 2012 10 / 54

Local Truncation ErrorFor consistency, we will need the local truncation error. Butfirst we need a few definitions:y n+1 = exact solution at time t n+1 .z n+1 = Q(hλ)y n = approximation applied to the exactsolution at time t n .w n+1 = (Q(hλ)) n+1 y 0 = approximation applied n + 1times.() Truncation Error, Consistency, and Convergence 8 October 2012 11 / 54

Local Truncation ErrorDefinition ( Local truncation error)The local truncation error at time step n + 1 is given byτ n+1 (h) = y n+1 − z n+1.hNotice, it is the difference between the exact solution and theapproximation applied to the exact solution at time t n .() Truncation Error, Consistency, and Convergence 8 October 2012 12 / 54

ConsistencyDefinition (Consistent)A numerical method is consistent iflim τ n+1(h) = 0.h→0Consistent ↔We can replicate the results.Solution of the numerical method converges to the exactsolution of the IVP.y n+1 − z n+1 → 0 faster than h → 0.By definition, this means that we need to examine thetruncation error.() Truncation Error, Consistency, and Convergence 8 October 2012 13 / 54

Local truncation errorConsider again the test equation:y ′ = λy,y(0) = y 0which has solution y(t) = y 0 e λt .The solution at t n+1 isy n+1 = y(t n+1 ) = y 0 e λt n+1= e λ(n+1)h y 0 = e λh y n = y n+1 .() Truncation Error, Consistency, and Convergence 8 October 2012 14 / 54

Local truncation errorThe numerical method is given byw n+1 = Q(hλ)w n .Applied to the exact solution it isz n+1 = Q(hλ)y n .The local truncation error is thenτ n+1 (h) = y n+1 − z n+1h= eλh y n − Q(hλ)y nh= (eλh − Q(hλ))y n.h() Truncation Error, Consistency, and Convergence 8 October 2012 15 / 54

Local truncation errorThis means that the truncation error is determined bye λh − Q(λh).For this, we need two things:1 The Taylor series for e λhe λh = 1+hλ+ 1 2! (hλ)2 + 1 3! (hλ)3 +· · · =r−1 ∑i=02 Q(λh), which is given by the numerical method.1i! (hλ)i +O(h r )() Truncation Error, Consistency, and Convergence 8 October 2012 16 / 54

Local truncation errorForward EulerThe amplification factor for Forward Euler is given byQ(hλ) = 1 + hλ.Therefore, the local truncation error isτ n+1 (h) = (eλh − Q(λh))y n⎡h= 1 (h ⎢ 1 + hλ + 1 )⎣ 2! (hλ)2 + O(h 3 ) − (1 + hλ)} {{ }} {{ }Taylor series for e λh= 1 ( )1h 2! (hλ)2 + O(h 3 ) y n( )1=2! hλ2 + O(h 2 ) y n ⇒ τ n+1 (h) = O(h)⎤⎥ y n⎦Amplification factor() Truncation Error, Consistency, and Convergence 8 October 2012 17 / 54

Forward EulerEuler forward,w n+1 = w n + hF (t n , w n ),Is an explicit method.Has amplification factor Q(hλ) = 1 + hλ.Is stable for 0 < h < 2|λ| .Has local truncation error of order one.Stability + Consistency = Convergence. It is a first ordermethod.() Truncation Error, Consistency, and Convergence 8 October 2012 18 / 54

ConsistencyLet us do the same for the other methods.Convergence is difficult to prove.Lax theorem says that we can instead prove stability andconsistency.Stability is given by ensuring the amplification factor is lessthan one.|Q(hλ)| ≤ 1.For consistency, we need the local truncation error tovanish as the step size becomes smaller.lim τ 1n+1(h) = limh→0 h→0 h (y n+1 − z n+1 ).() Truncation Error, Consistency, and Convergence 8 October 2012 19 / 54

Local truncation errorModified EulerFor the modified Euler, the amplification factor is given byQ(hλ) = 1 + hλ + (hλ)2 .2It is stable for 0 < h < 2|λ| .For consistency, we examine the local truncation error:τ n+1 (h) = (eλh − Q(λh))y nh= 1 [(1 + hλ + 1 h2! (hλ)2 + 1 )3! (hλ)3 + O(h 4 ) ..()].. − 1 + hλ + (hλ)2 y n2= 1 ( )1h 3! (hλ)3 + O(h 4 )=( 1 3! h2 λ 3 + O(h 3 ))y n ⇒ τ n+1 (h) = O(h 2 )() Truncation Error, Consistency, and Convergence 8 October 2012 20 / 54y n

Modified EulerModified Euler is given by¯w n+1 = w n + hF (t n , w n ),w n+1 = w n + h 2 (F (t n, w n ) + F (t n+1 , ¯w n+1 )) (2)Is an explicit method.Has amplification factor Q(hλ) = 1 + hλ + (hλ)22.Is stable for 0 < h < 2|λ| .Has local truncation error of order two.Stability + Consistency = Convergence. It is a secondorder method.() Truncation Error, Consistency, and Convergence 8 October 2012 21 / 54

Local truncation errorImplicit methodsFor implicit methods we generally have to work a bit harder.Amplification factor is not in polynomial form.Have to put in polynomial form.⇒ Requires Taylor expansion.() Truncation Error, Consistency, and Convergence 8 October 2012 22 / 54

Local truncation errorBackward EulerFor Backward Euler, the amplification factor is given byQ(hλ) = 11 − hλ .Before we plug this into the formula for the local truncationerror, we must perform a Taylor expansion:when |hλ| < 1.11 − hλ = 1 + hλ + (hλ)2 + (hλ) 3 + O(h 4 ),() Truncation Error, Consistency, and Convergence 8 October 2012 23 / 54

Local truncation errorBackward EulerPlugging in11−hλ = 1 + hλ + (hλ)2 + (hλ) 3 + O(h 4 ) into theformula for local truncation error givesτ n+1 (h) = (eλh − Q(λh))y nh= 1 [(1 + hλ + 1 )h2! (hλ)2 + O(h 3 ) ..].. − (1 + hλ + (hλ) 2 + O(h 3 )) y n= 1 (− 1 )h 2 (hλ)2 + O(h 3 ) y n(= − 1 )2 hλ2 + O(h 2 ) y n ⇒ τ n+1 (h) = O(h)() Truncation Error, Consistency, and Convergence 8 October 2012 24 / 54

Backward EulerEuler backward,w n+1 = w n + hF (t n+1 , w n+1 ),Is an implicit method.Has amplification factor Q(hλ) = 11−hλ .Is stable for h > 0.Has local truncation error of order one.Stability + Consistency = Convergence. It is a first ordermethod.() Truncation Error, Consistency, and Convergence 8 October 2012 25 / 54

Local truncation errorTrapezoidal methodFor the Trapezoidal method, the amplification factor,Q(hλ) = 1 + hλ 21 − hλ 2also has to be expanded into a Taylor series:when1 + hλ 21 − hλ 2=∣ hλ 2 ∣ < 1.(1 + hλ 2)(1 + hλ 2 + (hλ 2 )2 + ( hλ )2 )3 + O(h 4 )= 1 + hλ + 1 2 (hλ)2 + 1 4 (hλ)3 + O(h 4 )() Truncation Error, Consistency, and Convergence 8 October 2012 26 / 54

Local truncation errorTrapezoidal methodPlugging in 1+ hλ 2= 1 + hλ + 1 1− hλ 2 (hλ)2 + 1 4 (hλ)3 + O(h 4 ) into the2formula for local truncation error givesτ n+1 (h) = (eλh − Q(λh))y nh= 1 [(1 + hλ + 1 h2! (hλ)2 + + 1 )3! (hλ)3 + O(h 4 ) .... − (1 + hλ + 1 2 (hλ)2 + 1 ]4 (hλ)3 + O(h 4 )) y n= 1 (− 1 )h 6 (hλ)3 + O(h 4 ) y n(= − 1 )6 h2 λ 3 + O(h 3 ) y n ⇒ τ n+1 (h) = O(h 2 )() Truncation Error, Consistency, and Convergence 8 October 2012 27 / 54

Trapezoidal methodTrapezoidal method,w n+1 = w n + h 2 (F (t n, w n ) + F (t n+1 , w n+1 )),Uses the average of Euler forward and Euler backward.Is an implicit method.Has amplification factor Q(hλ) = 1+ hλ 21− hλ 2Is stable for h > 0.Has local truncation error of order two.Stability + Consistency = Convergence. It is a secondorder method..() Truncation Error, Consistency, and Convergence 8 October 2012 28 / 54

Local truncation errorTo find the local truncation errorwhere z n+1 = Q(hλ)y n .τ n+1 (h) = 1 h (y n+1 − z n+1 )1 Write the exact solution in terms of y n : y n+1 = y n e hλ .τ n+1 (h) = y nh (ehλ − Q(hλ))2 Expand e hλ in a Taylor series.3 Expand the amplification factor, Q(hλ) in a Taylor series(if necessary).4 Simplify the quantity e hλ − Q(hλ)5 Multiply by ynh .() Truncation Error, Consistency, and Convergence 8 October 2012 29 / 54

Local truncation errorObserve: All methods can be written asw n+1 = Q(hλ)w n .Explicit methods can be written asz n+1 = y n + hΦ(z n , t n , h).This gives a truncation error of:τ n+1 (h) = 1 h [y n+1 − z n+1 ] = 1 h [y n+1 − (y n + hΦ(z n , t n , h))]= 1 h (y n+1 − y n ) − Φ(z n , t n , h)} {{ }} {{ }Depends upon methodAll explicit methods() Truncation Error, Consistency, and Convergence 8 October 2012 30 / 54

Local truncation errorFor explicit methods, the local truncation error can be writtenasτ n+1 (h) = 1 h (y n+1 − y n ) − Φ(z n , t n , h).We can simplify this further. Expand y n+1 around y n :y n+1 = y n + h (y n ) ′ + h2(y n ) ′′ +O(h 3 )} {{ } 2 } {{ }=f (t n,y n) =(f (t n,y n)) ′ =(f t+ff y ) n= y n + hf (t n , y n ) + h22 (f t + ff y ) n + O(h 3 )() Truncation Error, Consistency, and Convergence 8 October 2012 31 / 54

Local truncation errorThenτ n+1 (h) = 1 h (y n+1 − y n ) − Φ(z n , t n , h)= 1 ()(y n + hf (t n , y n ) + h2h2 (f t + ff y ) n + O(h 3 )) − y n− Φ(z n , t n , h)= f (t n , y n ) + h 2 (f t + ff y ) n + O(h 2 ) − Φ(z n , t n , h)Euler Forward: Φ(z n , t n , h) = f (t n , y n )Modified Euler: Φ(z n , t n , h) = f (t n , y n ) + h(f t + ff y )() Truncation Error, Consistency, and Convergence 8 October 2012 32 / 54

Local truncation errorFinding Φ(z n , t n , h) for modified Euler:¯z n+1 = y n + hf (t n , y n )z n+1 = y n + h 2 (f (t n, y n ) + f (t n+1 , ¯z n+1 ))We need to expand f (t n+1 , ¯z n+1 ) in a Taylor series around(t n , y n ).f (t n+1 , ¯z n+1 ) =f (t n , y n ) + f t (t n , y n ) (t n+1 − t n )} {{ }=h(+ f tt (t n , y n )(t n+1 − t n ) 2 ..+f y (t n , y n ) (¯z n+1 − y n )} {{ }=hf (t n,y n).. + 2f ty (t n+1 − t n )(¯z n+1 − y n ) + f yy (¯z n+1 − y n ) 2)+ O(h 3 )=f (t n , y n ) + h(f t (t n , y n ) + f (t n , y n )f y (t n , y n )) + O(h 2 )() Truncation Error, Consistency, and Convergence 8 October 2012 33 / 54

Global truncation errorThe local truncation error,τ n+1 (h) = 1 h (y n+1 − z n+1 ),says how well the method performs over one time-step. It tellsus whether the method is consistent.What is the cumulative effect?() Truncation Error, Consistency, and Convergence 8 October 2012 34 / 54

Global truncation errorFor that we need:Definition (Global truncation error)The global truncation error of a scheme is defined ase n+1 = y n+1 − w n+1 ,where y n+1 = y(t n+1 ) is the exact solution at t n+1 andw n+1 = (Q(hλ)) n+1 y 0 is the approximation applied to the initialcondition n + 1 times.This is the cumulative effect of the errors from theapproximation over the entire interval.() Truncation Error, Consistency, and Convergence 8 October 2012 35 / 54

Global truncation errorDefinition (Convergence)A numerical method is convergent iflim e n+1 = 0.h→0The global truncation error goes to zero as we increase thenumber of subintervals.The local & global truncation errors are of the same order.() Truncation Error, Consistency, and Convergence 8 October 2012 36 / 54

ConvergenceWe now have the three elements in the Lax EquivalenceTheorem: For a linear scheme,Stability + Consistency = ConvergenceProof is in the book...() Truncation Error, Consistency, and Convergence 8 October 2012 37 / 54

Higher Order MethodsThe methods we have discussed are all 1 st or 2 nd ordermethods. Sometimes, we would like a higher order method.When we require high accuracy.When we must solve the equation over long periods intime.() Truncation Error, Consistency, and Convergence 8 October 2012 38 / 54

Higher Order Methods4 th −order Runge-Kutta (RK4)One higher-order method is the family of Runge-Kutta methods.The 4 th −order Runge-Kutta scheme is:k 1 = hf (t n , w n )k 2 = hf (t n + h 2 , w n + k 12 )k 3 = hf (t n + h 2 , w n + k 22 )k 4 = hf (t n + h, w n + k 3 )RK4 is the average of these slopesw n+1 = w n + 1 6 (k 1 + 2(k 2 + k 3 ) + k 4 )() Truncation Error, Consistency, and Convergence 8 October 2012 39 / 54

Higher Order Methods4 th −order Runge-Kutta (RK4)RK4:Uses an average of different slopes.This allow us to avoid taking higher derivativesThis is a 4 th −order method.The amplification factor isQ(hλ) = 1 + hλ + 1 2 (hλ)2 + 1 3! (hλ)3 + 1 4! (hλ)4The stability region is 0 < h < 2.8|λ| .() Truncation Error, Consistency, and Convergence 8 October 2012 40 / 54

ExampleExampleConsider the numerical scheme given byu ⋆ = u n + βhf (t n , u n )u n+1 = u ⋆ + (1 − β)hf (t n + βh, u ⋆ )1 Show the truncation error is O(h) for each value of β.Show that there does not exist a β for which the scheme isO(h 2 ).2 What is the amplification factor?3 Consider the nonlinear equation y ′ = 2y − 4y 2 and takeβ = 1 2. Determine the maximal step size such that themethod is stable in the neighbourhood of y = 1 2 .() Truncation Error, Consistency, and Convergence 8 October 2012 41 / 54

ExampleTruncation errorThe numerical method applied to the exact solution, y n , isgiven byz ⋆ = y n + βhf (t n , y n )z n+1 = z ⋆ + (1 − β)hf (t n + βh, z ⋆ )We first need to expand f (t n + βh, z ⋆ ) in a Taylor series around(t n , y n ) :f (t n + βh, z ⋆ ) =f (t n , y n ) + (f t ) n (t n + βh − t n ) + (f y ) n (z ⋆ − y n )+ 1 2 β2 h 2 (f tt ) n (t n + βh − t n ) 2 + 1 2 (f yy) n (z ⋆ − y n ) 2+ βh(f ty ) n (t n + βh − t n )(z ⋆ − y n ) + O(h 3 )() Truncation Error, Consistency, and Convergence 8 October 2012 42 / 54

ExampleTruncation errorSubstituting z ⋆ = y n + βhf (t n , y n ) into the expansion, we havef (t n + βh, z ⋆ ) = f (t n , y n ) + βh((f t ) n + (f f y ) n ) + O(h 2 ).Now, we can consider z n+1 :z n+1 =z ⋆ + (1 − β)hf (t n + βh, z ⋆ )=y n + βhf (t n , y n )+ (1 − β)h(f (t n , y n ) + βh((f t ) n + (f f y ) n ) + O(h 2 ))=y n + βhf (t n , y n )+ h(f (t n , y n ) + βh(f t + f f y ) n + O(h 2 ))−βh(f (t n , y n ) + βh(f t + f f y ) n + O(h 2 ))() Truncation Error, Consistency, and Convergence 8 October 2012 43 / 54

ExampleTruncation errorTherefore we have thatz n+1 = y n +hf (t n , y n )+βh 2 (f t +f f y ) n +(βh) 2 (f t +f f y ) n +O(h 3 )We also needy n+1 = y n + h (y n ) ′ + 1 } {{ } 2 h2 (y n ) ′ ′ + O(h 3 )} {{ }=f (t n,y n) =(f t+ff y ) n= y n + hf (t n , y n ) + 1 2 h2 (f t + ff y ) n + O(h 3 ) = y n+1() Truncation Error, Consistency, and Convergence 8 October 2012 44 / 54

ExampleTruncation errorWhich gives the local truncation error asτ n+1 (h) = 1 h (y n+1 − z n+1 )= 1 ((y n + hf (t n , y n ) + 1 h2 h2 (f t + ff y ) n + O(h 3 )) ..).. − (y n + hf (t n , y n ) + βh 2 (f t + f f y ) n + . . .)= 1 2 h(f t + ff y ) n − βh(f t + f f y ) n + . . .=O(h)The local truncation error is O(h) for any value of β.() Truncation Error, Consistency, and Convergence 8 October 2012 45 / 54

ExampleAmplification factorTo determine the amplification factor, we apply the numericalmethod to the test equation,This givesy ′ = λy, y(0) = y 0 , λ < 0.u ⋆ = u n + βhλu nu n+1 = u ⋆ + (1 − β)hλu ⋆() Truncation Error, Consistency, and Convergence 8 October 2012 46 / 54

ExampleAmplification factorSimplifying,u n+1 = u ⋆ + (1 − β)hλu ⋆= (u n + βhλu n ) + (1 − β)hλ(u n + βhλu n )= (1 + ✟✟✯ ✟ βhλ + hλ + β(hλ) 2 − ✟✟✯ ✟ βhλ − β 2 (hλ) 2 )u n= (1 + hλ + β(1 − β)(hλ) 2 )} {{ }Q(hλ)The Amplification factor is given byu nQ(hλ) = (1 + hλ + β(1 − β)(hλ) 2 )() Truncation Error, Consistency, and Convergence 8 October 2012 47 / 54

ExampleNonlinear equationThe nonlinear equation isy ′ = f (y) = 2y − 4y 2 .Notice, this is zero at y = 0 and y = 1 2 . If β = 1 2, we want todetermine the maximum step-size, h, for which the scheme isstable. We proceed as followsFirst, we linearize f (y) around y = 1 2 .() Truncation Error, Consistency, and Convergence 8 October 2012 48 / 54

ExampleNonlinear equationThe linearization is given byL(y) = f ( 1 2 ) + f ′ ( 1 (2 ) y − 1 ).2This givesL(y) = 1 − 2y.Therefore, to consider the stablity of the nonlinear equationy ′ = 2y − 4y 2 .it is enough to consider the stability of the linear equationy ′ = −2yThis is just our test equation with λ = −2.() Truncation Error, Consistency, and Convergence 8 October 2012 49 / 54

ExampleNonlinear equationFor stability, we consider the amplification factor with β = 1 2and λ = −2 :The method is stable ifQ(hλ) =1 + hλ + β(1 − β)(hλ) 2(1 − 1 2=1−2h + 1 2=1 − 2h + h 2)(−2h) 2|Q(hλ)| = |1 − 2h + h 2 | ≤ 1.() Truncation Error, Consistency, and Convergence 8 October 2012 50 / 54

ExampleNonlinear equation|Q(hλ)| = |1 − 2h + h 2 | ≤ 1This means−1 ≤ 1 − 2h + h 2 ≤ 1−1 ≤ (h − 1) 2 ≤ 1−2 ≤ h(h − 2) ≤ 0Recall: h > 0. Thus:0 < h ≤ 2 for stability.() Truncation Error, Consistency, and Convergence 8 October 2012 51 / 54

Summary tableFE BE ME TM1+ hλ 21+ hλ 2Amplification factor 1 + hλ 1 − hλ 1 + hλ + (hλ)22Type method Explicit Implicit Explicit ImplicitStability range 0 < h < 2|λ|h > 0 0 < h < 2|λ|h > 0Method order One One Two TwoΦ(z n , t n , h) f n f n + h(f t + ff y ) nFE = Forward Euler; BE = Backward Euler; ME ModifiedEuler; TM = Trapezoidal method() Truncation Error, Consistency, and Convergence 8 October 2012 52 / 54

SummaryLax Equivalence Theorem: For a linear scheme,Stability + Consistency = ConvergenceStability: given by amplification factor, choose h such that|Q(hλ)| < 1.Consistency: given by the local truncation error:τ n+1 = 1 h (y n+1 − z n+1 ).Convergence: given by the global truncation error:e n+1 = y n+1 − w n+1 .() Truncation Error, Consistency, and Convergence 8 October 2012 53 / 54

Material addressed1 ReviewSingle Step MethodsStabilityIVP StabilityAmplification factor2 Local Truncation Error &Consistency3 Global truncation error &ConvergenceMaterial in book:Chapter 6, Sections 1-7Useful exercises: 1-5() Truncation Error, Consistency, and Convergence 8 October 2012 54 / 54