# Error Formula

• Nov 10th 2011, 02:22 PM
veronicak5678
Error Formula
For the recursive algorithm p0=1, p1=1/3, pn=(10/3)(pn-1 - pn-2) for n >1 let the error be denoted by p'=pn+en for initial error e0, and assume the error has the form

p'0=p0+e0
p'1=p1+3e0
p'n=(10/3)(p'n-1 - p'n-2) for n>1

so that e1=3e0
Find an equation for en in terms of e0, n, and a constant.
• Nov 11th 2011, 05:57 AM
CaptainBlack
Re: Error Formula
Quote:

Originally Posted by veronicak5678
For the recursive algorithm p0=1, p1=1/3, pn=(10/3)(pn-1 - pn-2) for n >1 let the error be denoted by p'=pn+en for initial error e0, and assume the error has the form

p'0=p0+e0
p'1=p1+3e0
p'n=(10/3)(p'n-1 - p'n-2) for n>1

so that e1=3e0
Find an equation for en in terms of e0, n, and a constant.

$\displaystyle p'_n=(10/3)[p'_{n-1}-p'_{n-2}]$

is a linear difference equation, so suppose $\displaystyle p'_n=A^n$ for some real number $\displaystyle A$, then find the characteristic equation and solve for $\displaystyle A$. Form the general solution as a linear combination of the two solutions you will have found and fit to the initial conditions.

CB
• Nov 11th 2011, 11:35 AM
veronicak5678
Re: Error Formula
Could you please explain how to find the characteristic equation?
• Nov 11th 2011, 07:59 PM
CaptainBlack
Re: Error Formula
Quote:

Originally Posted by veronicak5678
Could you please explain how to find the characteristic equation?

You make the substitution $\displaystyle p'_n=A^n$ in the difference equarion to get:

$\displaystyle A^n=\frac{10}{3}[A^{n-1}-A^{n-2}}]$

Now divide through by $\displaystyle A^{n-2}$ to get:

$\displaystyle A^2=\frac{10}{3}[A-1]$

which is a quadratic in $\displaystyle A$ which rearranges to:

$\displaystyle A^2-(10/3)A+(10/3)=0$

which is the charateristic equation which you solve for $\displaystyle A$.

CB
• Nov 12th 2011, 11:31 AM
veronicak5678
Re: Error Formula
I see. Thanks!