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If f(0) = 3, f(1) = 5, f(3) = 21, then the unique polynomial of degree 2 or less using Newton divided difference interpolation will be:
  • a)
    2x2 + 2x + 1
  • b)
    2x2 - 3x + 1
  • c)
    2x2 + 3
  • d)
    x2 + 3x - 2
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
If f(0) = 3, f(1) = 5, f(3) = 21, then the unique polynomial of degree...
Concept:
Newton’s divided difference polynomial method:
Second order polynomial interpolation using Newton’s divided difference polynomial method is as follows,
Given (x0,y0), (x1,y1), (x2,y2) be the data points and f(x) be the quadratic interpolant, then f(x) is given by
f(x) = b0 + b1(x – x0) + b2 (x – x0)(x – x1);
Where
b0 = f(x0);

Calculation:
Given f(0) = 3, f(1) = 5, f(3) = 21;
⇒ (0,3), (1,5), (3,21) are the data points;
The polynomial will be f(x) = b0 + b1(x) + b2 (x)(x – 1);
⇒ b0 = f(0) = 3;

Substituting the constant b0, b1, b2 in the quadratic interpolant,
⇒ f(x) = 3 + 2x + 2 (x)(x – 1) = 3 + 2x + 2x2 – 2x = 3 + 2x2;
The unique polynomial of degree 2 will be f(x) = 3 + 2x2;
Easy method:
To save time, simply substitute the data points in the polynomials given in options and find the polynomial that is satisfying all data points.
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If f(0) = 3, f(1) = 5, f(3) = 21, then the unique polynomial of degree 2 or less using Newton divided difference interpolation will be:a)2x2 + 2x + 1b)2x2 - 3x + 1c)2x2 + 3d)x2 + 3x - 2Correct answer is option 'C'. Can you explain this answer?
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