Here is another finite difference problem

Hi,

Given $ u_1=1, u_1+u_3=5.41, u_4+u_5+u_6=18.47, u_7+u_8+u_9+u_{10}+u_{11}+u_{12}=90.36 $

Find the values of $ u_x $ for x=1,2,...,12.

Solution:-

I calculated $ S_3 =u_1+u_2+u_3=5.41+u_2, S_6=u_1+u_2+....+u_6=23.88+u_2, S_{12}=u_1+....+u_{12}=114.24+u_2 $

as 1,3,6,12 are at unequal intervals, I want to use Newton's divided difference interpolation formula.

for calculating $ u_2 $ shall I use $ (E-1)^2 u_x=0 $ when x=1

Problem solving hint is needed.

Re: Here is another finite difference problem

I am not sure what you are trying to do. What assumptions are you making about the distribution of the $u_i$'s?

Re: Here is another finite difference problem

How about

$S_0 = 0$

$S_3 = 5.41+u_2$

$S_6 = 23.88+u_2$

$S_9 = 23.88+u_2+k$

$S_{12} = 114.24+u_2$

Then you can get:

$S_x = \binom{x/3}{1}\left(5.41+u_2\right) + \binom{x/3}{2}(13.06-u_2) + \binom{x/3}{3}(k+u_2-31.53) + \binom{x/3}{4}(140.36-4k-u_2)$

The first thing you will want to do is solve for $k$. To do that, plug in values you know. $S_1 = 1$ and $S_2 = 1+u_2$. Plugging those in and solving the system of equations, you find $k = \dfrac{255761}{8625} \approx 29.65$ and $u_2 = \dfrac{7318}{2415} \approx 3.03$.

From here, you can solve for the rest of them.

The formula becomes:

$\begin{align*}S_x & \approx \binom{x/3}{1}(8.44) + \binom{x/3}{2}(10.03) + \binom{x/3}{3}(1.15) + \binom{x/3}{4}(18.73) \\ & = \dfrac{1}{194400}\left( 1873x^4-32334x^3+281331x^2-56646x \right) \end{align*}$

Then, $S_x - S_{x-1} = u_x$, so you can take the difference of the two formulas and find:

$u_x = \dfrac{1873x^3-27060x^2+166789x-93046}{48600}$

Note: It is slightly off, as the formula finds $u_1 \approx 0.999$ because I approximated the results for $u_2$ and $k$ above.