Answer to Question #265899 in Quantitative Methods for sagun poudel

Question #265899

Find the values of the first and second derivatives of y = 4x2+2x-1 for

x=1.25 from the following table using Lagrange’s interpolation

formula.

x : 1 1.5 2 2.5

y : 5 11 19 29


1
Expert's answer
2021-11-15T16:08:17-0500

We have 4 points, so the Lagrange polynomial will be of degree 3. The basic Lagrange polynomials are: 


L0(x)=(x1.5)(x2)(x2.5)(11.5)(12)(12.5)L_0(x)=\dfrac{(x-1.5)(x-2)(x-2.5)}{(1-1.5)(1-2)(1-2.5)}

L1(x)=(x1)(x2)(x2.5)(1.51)(1.52)(1.52.5)L_1(x)=\dfrac{(x-1)(x-2)(x-2.5)}{(1.5-1)(1.5-2)(1.5-2.5)}

L2(x)=(x1)(x1.5)(x2.5)(21)(21.5)(22.5)L_2(x)=\dfrac{(x-1)(x-1.5)(x-2.5)}{(2-1)(2-1.5)(2-2.5)}

L3(x)=(x1)(x1.5)(x2)(2.51)(2.51.5)(2.52)L_3(x)=\dfrac{(x-1)(x-1.5)(x-2)}{(2.5-1)(2.5-1.5)(2.5-2)}

P(x)=5L0(x)+11L1(x)+19L2(x)+29L3(x)P(x)=5L_0(x)+11L_1(x)+19L_2(x)+29L_3(x)

P(x)=203(x1.5)(x2)(x2.5)P(x)=-\dfrac{20}{3}(x-1.5)(x-2)(x-2.5)

+44(x1)(x2)(x2.5)+44(x-1)(x-2)(x-2.5)

76(x1)(x1.5)(x2.5)-76(x-1)(x-1.5)(x-2.5)

+1163(x1)(x1.5)(x2)+\dfrac{116}{3}(x-1)(x-1.5)(x-2)

=203x3+40x22353x+50=-\dfrac{20}{3}x^3+40x^2-\dfrac{235}{3}x+50

+44x3242x2+418x220+44x^3-242x^2+418x-220

76x3+380x2589x+282-76x^3+380x^2-589x+282

1163x340x2+174x116\dfrac{116}{3}x^3-40x^2+174x-116

The approximation for the pthp^{th} derivative of some function can be found by taking the pthp^{th} derivative of a polynomial approximation of the function

.

P(x)=276x2263P'(x)=276x-\dfrac{226}{3}

P(x)=276P''(x)=276


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