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 July 29th, 2015, 10:36 AM #1 Newbie   Joined: Jul 2015 From: Edinburgh Posts: 1 Thanks: 0 Determine the error constant of a multistep method I have the following problem but I cannot solve part B in the way suggested by my professor in this past exam paper. I can solve it in a different way, but not in the specific way he's suggesting. Thanks a lot in advance! QUESTION: A family of multistep methods, dependent on a parameter $\displaystyle \theta\in\left[0,1/2\right]$, is given by $\displaystyle y_{n+2}-4\theta y_{n+1}-(1-4\theta)y_n=h\left[(1-\theta)y_{n+2}'+(1-3\theta)y_n'\right]$ A) Write out the characteristic polynomials $\rho$ and $\sigma$ for this method. B) Determine the error constant for this method (it will depend on $\displaystyle \theta$) by computing the leading nonzero coefficient in the expansion of $\displaystyle \rho(e^z)-z\sigma(e^z)$ in powers of $\displaystyle z$. ANSWERS: A) We have - $\displaystyle \alpha_0=4\theta-1$, $\alpha_1=-4\theta$, and $\alpha_2=1$ - $\displaystyle \beta_0=1-3\theta$, $\beta_1=0$, and $\beta_2=1-\theta$ This gives the characteristic polynomials: - $\displaystyle \rho(\zeta)=\zeta^2-4\theta\zeta+4\theta-1$ - $\displaystyle \sigma(\zeta)=(1-\theta)\zeta^2+1-3\theta$ B) So I substitute and get the following: $\displaystyle e^{2z}-4\theta e^{z}+4\theta-1-z e^{2z}+z \theta e^{2z}-z+3z\theta=e^{2z}(1+z(\theta-1))-4\theta e^z+4\theta-1-z+3z\theta$ But I know I have done something wrong because it makes no sense for the next part of the exercise. I also don't get the same result when I calculate the error constant (it is $\displaystyle 5/12$) of the Adams-Bashforth method using the formula given in this exercise (instead of another using another method). The Adams-Bashforth is given below. $\displaystyle y_{n+2}-y_{n+1}=h\left[\frac{3}{2}y_{n+1}'-\frac{1}{2}y_n'\right]$ Another example, for $\displaystyle y_{n+3}+y_{n+2}-y_{n+1}-y_{n}=h\left(\frac{8}{3}y_{n+2}'+\frac{2}{3}y_{n+1 }'+\frac{2}{3}y_{n}'\right)$ Apparently, the example says that substituting $\displaystyle \rho(\zeta)=\zeta^3+\zeta^2-\zeta-1$ $\displaystyle \sigma(\zeta)=\frac{8}{3}\zeta^2+\frac{2}{3}\zeta+ \frac{2}{3}$ into $\displaystyle \rho(e^z)-z\sigma(e^z)=\mathcal{O}(z^{p+1})$ gives $\displaystyle \rho(e^z)-z\sigma(e^z)=\frac{1}{3}z^4+\mathcal{O}(z^{5})$ Tags constant, determine, differential equation, error, method, multistep, ode Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post WWRtelescoping Algebra 7 March 2nd, 2014 11:00 AM Epsilone Real Analysis 3 September 1st, 2013 10:35 AM moulion Applied Math 1 May 13th, 2013 12:13 PM lu5t Applied Math 0 June 10th, 2012 11:54 AM Epsilone Applied Math 1 December 31st, 1969 04:00 PM

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