My Math Forum Determine the error constant of a multistep method

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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 $\displaystyle \rho(e^z)-z\sigma(e^z)=\frac{1}{3}z^4+\mathcal{O}(z^{5})$

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