User:Egm6341.s10.Team4.roni/HW5/5.4

= Problem 4: Comparing Richarson Extrap. Coeff. to Coeff. from Trap. Error Estimate =

Given
Richardson Extrapolation Coefficients on slide [[media:Egm6341.s10.mtg19.djvu|19-3]], which is:


 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle a_i=d_i*(f \overset{(2i-1)}{(b)} -f \overset{(2i-1)}{(a)})
 * style= |
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle d_i=- \frac{{B}_{2i}}{(2i)!}
 * style= |
 * }
 * }

$$\displaystyle $$ $$\displaystyle d_1=- \frac{1}{12} \ \ d_2= \frac{1}{720} \ \  d_3=- \frac{1}{30240} $$

Show
That these coefficients are equal to the following: equation (3) on slide [[media:Egm6341.s10.mtg27.djvu|27-1]]

$$\displaystyle $$ $$\displaystyle \bar{d}_{2r}= \frac{{P}_{2r} (t=1)}{(2^2r)!} $$

$$\displaystyle $$ $$\displaystyle d_1=\bar{d}_{2r}= \frac{{P}_{2r} (t=1)}{(2^2r)!} \ \ where  \ \  r=1 $$

$$\displaystyle $$ $$\displaystyle d_2=\bar{d}_{2r}= \frac{{P}_{2r} (t=1)}{(2^2r)!} \ \ where  \ \  r=2 $$

$$\displaystyle $$ $$\displaystyle d_3=\bar{d}_{2r}= \frac{{P}_{2r} (t=1)}{(2^2r)!} \ \ where  \ \  r=3 $$

Solution
Recalling:

equation (2) on slide [[media:Egm6341.s10.mtg21.djvu|21-2]]

P_1(t)=-t $$

equation (1) on slide [[media:Egm6341.s10.mtg21.djvu|21-3]]

P_2(t)=- \frac{t^2}{2}+\frac{1}{6} $$

equation (4) on slide [[media:Egm6341.s10.mtg21.djvu|21-3]]

P_3(t)=- -\frac{t^3}{6}+\frac{t}{6} $$

on slide [[media:Egm6341.s10.mtg21.djvu|26-3]]

P_4(t)=- \frac{t^4}{24}- \frac{t^2}{12}- \frac{72}{360} $$

In General, from the Paper by Patch Kessler we have:



P_{2k}(t)=c1 \frac{t^{2k}}{(2k)!} + c_2 \frac{t^{2k-2}}{(2k-2)!} + c_5 \frac{t^{2k-4}}{(2k-4)!} + c_{2k-3} \frac{t^{4}}{(4)!} + c_{2k-1} \frac{t^{2}}{(2)!} + c_{2k+1} \frac{t^{0}}{(0)!} $$

Giving (also from HW 5.2 and it's solution) [[media:Egm6341.s10.mtg27.djvu|27-1]]

P_6(t)= -\frac{t^{6}}{(720)}+\frac{t^{4}}{(144)}-\frac{7t^{2}}{(720)}+\frac{(31)}{(15120)} $$

We only need P1, P4 and P6

Giving,


 * {| style="width:20%" border="0"

$$\displaystyle d_1=\bar{d}_{2}= \frac{{P}_{2} (t=1)}{(4)} \ \ =-1/12 $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |


 * style= |
 * }
 * }


 * {| style="width:20%" border="0"

$$\displaystyle d_2=\bar{d}_{4}= \frac{{P}_{4} (t=1)}{(16)} \ \ = 1/16*(-1/241/12-7/360)=1/720 $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |


 * style= |
 * }
 * }


 * {| style="width:20%" border="0"

$$\displaystyle d_3=\bar{d}_{6}= \frac{{P}_{6} (t=1)}{(64)} \ \ = 1/64*(-1/720+1/144-7/720+31/15120)=-1/30240 $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |


 * style= |
 * }
 * }

Author
Solved and typed by - Reviewed by -