Approximating an Integral
For all cubic polynomials f ( x ) , what positive value of k makes the following statement true?
∫ − 1 1 f ( x ) d x = f ( − k ) + f ( k )
Bonus question: What is this method of approximation known as?
The answer is 0.577.
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3 solutions
Let the polynomial be a x 3 + b x 2 + c x + d = f ( x ) . Now evaluating the value of the function f ( x ) at x = k , − k , we get: f ( k ) = a k 3 + b k 2 + c k + d f ( − k ) = − a k 3 + b k 2 − c k + d Now adding them gives I 1 = f ( k ) + f ( − k ) = 2 b k 2 + 2 d
Similarly, the integration equals to I 2 = ( 2 / 3 ) b + 2 d
But we know that I 1 = I 2 , therefore on comparing we have k 2 = 1 / 3 ⇒ k = 3 1 = 0 . 5 7 7 3 5 . □
Is anyone going to break the silence to tell us what this method of approximation is called?
This is from a technique called Gaussian Quadrature .
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Ah! Thank you. I haven't looked at that in many years. Obviously didn't recognise it.
it's called calculus dumb ass!
Because off their odd-ness, the cubic and linear terms do not contribute to either side of the equation. We will therefore prove the statement for a simple quadratic function.
∫ − 1 1 3 x 2 d x = x 3 ∣ ∣ − 1 1 = 2 ; 3 ( − k ) 2 + 3 k 2 = 6 k 2 = 2 ; k = 3 1 ≈ 0 . 5 7 7 3 5 0 2 6 9 .
(Adding a constant term is trivial: ∫ − 1 1 c d x = 2 c . )