Answer to Question #153451 in Quantitative Methods for usman

Question

If y0, y1, ..., y9 are consecutive terms of a series, prove that

y5 = 1/70 [56(y4 + y6) – 28(y3 + y7) + 8(y2 + y8) – (y1 + y9)]

Answer 1

$$\begin{matrix} x_0=0, & x_1=d, & x_2=2d, \\ x_3=3d, & x_4=4d, & x_6=6d, \\ x_7=7d, &x_8=8d, & x_9=9d \end{matrix}$$

By Lagrange’s interpolation formula we have

$$y=f(x)=\dfrac{(x-x_2)(x-x_3)...(x-x_9)}{(x_1-x_2)(x_1-x_3)...(x_1-x_9)}\times y_1$$

$$+\dfrac{(x-x_1)(x-x_3)...(x-x_9)}{(x_2-x_1)(x_2-x_3)...(x_2-x_9)}\times y_2$$

$$+...$$

$$+\dfrac{(x-x_1)(x-x_2)...(x-x_8)}{(x_9-x_1)(x_9-x_2)...(x_9-x_8)}\times y_9$$

Put $x=5d$

$$y_5=\dfrac{3d(2d)(d)(-d)(-2d)(-3d)(-4d)}{-d(-2d)(-3d)(-5d)(-6d)(-7d)(-8d)}\times y_1$$

$$+\dfrac{4d(2d)(d)(-d)(-2d)(-3d)(-4d)}{d(-d)(-2d)(-4d)(-5d)(-6d)(-7d)}\times y_2$$

$$+\dfrac{4d(3d)(d)(-d)(-2d)(-3d)(-4d)}{2d(d)(-d)(-3d)(-4d)(-5d)(-6d)}\times y_3$$

$$+\dfrac{4d(3d)(2d)(-d)(-2d)(-3d)(-4d)}{3d(2d)(d)(-2d)(-3d)(-4d)(-5d)}\times y_4$$

$$+\dfrac{4d(3d)(2d)(d)(-2d)(-3d)(-4d)}{5d(4d)(3d)(2d)(-d)(-2d)(-3d)}\times y_6$$

$$+\dfrac{4d(3d)(2d)(d)(-d)(-3d)(-4d)}{6d(5d)(4d)(3d)(d)(-d)(-2d)}\times y_7$$

$$+\dfrac{4d(3d)(2d)(d)(-d)(-2d)(-4d)}{7d(6d)(5d)(4d)(2d)(d)(-d)}\times y_8$$

$$+\dfrac{4d(3d)(2d)(d)(-d)(-2d)(-3d)}{8d(7d)(6d)(5d)(3d)(2d)(d)}\times y_9$$

$$=-\dfrac{1}{70}y_1+\dfrac{4}{35}y_2-\dfrac{2}{5}y_3+\dfrac{4}{5}y_4$$

$$+\dfrac{4}{5}y_6-\dfrac{2}{5}y_7+\dfrac{4}{35}y_8-\dfrac{1}{70}y_9$$

$$y_5=\dfrac{1}{70}\big(56(y_4+y_6)-28(y_3+y_7)+8(y_2+y_8)-(y_1+y_9)\big)$$