Question #153451

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)]


1
Expert's answer
2021-01-04T19:48:40-0500
x0=0,x1=d,x2=2d,x3=3d,x4=4d,x6=6d,x7=7d,x8=8d,x9=9d\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)=(xx2)(xx3)...(xx9)(x1x2)(x1x3)...(x1x9)×y1y=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

+(xx1)(xx3)...(xx9)(x2x1)(x2x3)...(x2x9)×y2+\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

+...+...

+(xx1)(xx2)...(xx8)(x9x1)(x9x2)...(x9x8)×y9+\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=5dx=5d

y5=3d(2d)(d)(d)(2d)(3d)(4d)d(2d)(3d)(5d)(6d)(7d)(8d)×y1y_5=\dfrac{3d(2d)(d)(-d)(-2d)(-3d)(-4d)}{-d(-2d)(-3d)(-5d)(-6d)(-7d)(-8d)}\times y_1

+4d(2d)(d)(d)(2d)(3d)(4d)d(d)(2d)(4d)(5d)(6d)(7d)×y2+\dfrac{4d(2d)(d)(-d)(-2d)(-3d)(-4d)}{d(-d)(-2d)(-4d)(-5d)(-6d)(-7d)}\times y_2

+4d(3d)(d)(d)(2d)(3d)(4d)2d(d)(d)(3d)(4d)(5d)(6d)×y3+\dfrac{4d(3d)(d)(-d)(-2d)(-3d)(-4d)}{2d(d)(-d)(-3d)(-4d)(-5d)(-6d)}\times y_3


+4d(3d)(2d)(d)(2d)(3d)(4d)3d(2d)(d)(2d)(3d)(4d)(5d)×y4+\dfrac{4d(3d)(2d)(-d)(-2d)(-3d)(-4d)}{3d(2d)(d)(-2d)(-3d)(-4d)(-5d)}\times y_4

+4d(3d)(2d)(d)(2d)(3d)(4d)5d(4d)(3d)(2d)(d)(2d)(3d)×y6+\dfrac{4d(3d)(2d)(d)(-2d)(-3d)(-4d)}{5d(4d)(3d)(2d)(-d)(-2d)(-3d)}\times y_6

+4d(3d)(2d)(d)(d)(3d)(4d)6d(5d)(4d)(3d)(d)(d)(2d)×y7+\dfrac{4d(3d)(2d)(d)(-d)(-3d)(-4d)}{6d(5d)(4d)(3d)(d)(-d)(-2d)}\times y_7

+4d(3d)(2d)(d)(d)(2d)(4d)7d(6d)(5d)(4d)(2d)(d)(d)×y8+\dfrac{4d(3d)(2d)(d)(-d)(-2d)(-4d)}{7d(6d)(5d)(4d)(2d)(d)(-d)}\times y_8

+4d(3d)(2d)(d)(d)(2d)(3d)8d(7d)(6d)(5d)(3d)(2d)(d)×y9+\dfrac{4d(3d)(2d)(d)(-d)(-2d)(-3d)}{8d(7d)(6d)(5d)(3d)(2d)(d)}\times y_9



=170y1+435y225y3+45y4=-\dfrac{1}{70}y_1+\dfrac{4}{35}y_2-\dfrac{2}{5}y_3+\dfrac{4}{5}y_4


+45y625y7+435y8170y9+\dfrac{4}{5}y_6-\dfrac{2}{5}y_7+\dfrac{4}{35}y_8-\dfrac{1}{70}y_9




y5=170(56(y4+y6)28(y3+y7)+8(y2+y8)(y1+y9))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)



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