Numerical Analysis/Order of RK methods/Quiz on RK order

<quiz display=simple> {What is the main importance of having a higher ODE method order |type="()"} -It always better to increase the accuracy of the solution +It requires fewer steps for the same accuracy restriction -We need the method to be better consistent -We need the method to be more stable -It provides better margin of stability

{The number of slope predictions (number of k's) in an RK method is related to the order of the method as follows. The order of the method is: |type="()"}

-exactly equal to the number slope predictions -equal to number of k's (slope evaluations) +1 +upper bounded by the number of k's -not related to the number of k's

{An n-th order of the method means that |type="()"} -the global truncation error is of the order $O(h^{n+1})$ -the local truncation error is of the order $O(h^n)$ +the global truncation error is of the order $O(h^n)$ -the maximum of the global and local error is n-th order

{The order of the local truncation error is related to the global truncation error in the following way |type="()"} -they are about equal -they are not related -the global is one higher order than the local +the local is one higher order than the global

{Is the following true? The global truncation error is always less then the local, since the errors partially cancel out over steps. |type="()"} -true +false

{What is the highest order that we can achieve with a Runge Kutta type method by using 5 k's? |type="()"} -We can achieve 5-th order -That method is not used, since it is not consistent. +We can achieve 4-th order -We can achieve even higher order, it is sixth order accurate.

{How is the order of single step methods related to consistency of the method? |type="()"} +At least first order of the method indicates that the method is consistent -To be consistent, a method must be second order or higher -not related at all, a single step method can be inconsistent and be of any order.

{What is the main "tool" to prove the order of a numerical method for solving ODE's of type $y^{\prime }=f(t,y(t))$? |type="()"} -Solving a specific "stiff" problem $y^{\prime }=f(t,y(t))=-Ky$ and showing that it converges for specific real positive K -Taylor series expansion for one variable +Taylor series expansion for one and two variables -Solving the ODE equation analytically and then numerically

{If we show that all terms in a numerical method for ODE recurrence equation match the terms in the Taylor series expansion up to the terms with $h^n$, but we do not analyze whether further cancellation of the terms exist, the following is true |type="()"} -The method is of the order n -The method is of the order n, since the global error is of the order $O(h^n)$ +The order of the method could be higher than n-1, but it is at least n-1

{If we use an explicit n-th (n>1) order accurate multistep method with s steps (s>2) to solve an ODE, but we use an n-1 order method to calculate the missing initial points that we need to start using the multistep method, what is the global order of accuracy of our calculation? |type="()"} -There is no such combined method, we cannot do that -The overall method is n-th order, since we just use the n-1 order method for few initial points +The overall method is of the order n-1, since the initial error from the lower accurate method remains in the cummulative error (global) -The order of the method could be either n or n-1 -The order of the method is something between n-1 and n, just like the RK 45 method

{What do we get as an order of error from the following expression

$\frac{1}{h}O(h^2)+10O(h^3)$?

|type="()"}

-Something between $O(h^2)$ and $O(h^3)$ -$O(h^2)$ -$O(h^3)$ +$O(h)$

{What do we get as an order of error from the following expression

$\sin (h^2)\frac{1}{h}O(h^2)$?

|type="()"}

-Something between $O(h^2)$ and $O(h)$ -$O(h^2)$ +$O(h^3)$ -$O(h)$

{What do we get as an order of error from the following expression

$e^h\cos (h)\frac{1}{h}O(h^2)$?

|type="()"}

-It must be of order $O(h^2)$ -$O(h^3)$ +$O(h)$

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