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STEP I 1997 question 7 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1997 question 7 solution

Fortunately, this question only asks us to find constants, it doesn't ask us to prove they are the only solution. If we guess that the answer will be in some sense symmetrical, this is about the shortest STEP question ever!

Consider what happens when P(x) = 1 (constant polynomial).

Then \int_{-1}^1 dt = a_1 + a_2 \implies a_1+a_2 = 2. Now I don't know about you, but at this point I'm thinking, surely we're going to have a_1 = a_2 = 1. So let's assume that and hope for the best.

Consider P(x) = x. Then \int_{-1}^1 t dt = a_1 u_1 + a_2 u_2 \implies 0 = a_1 u_1 + a_2 u_2. But we've assumed a_1 = a_2 = 1 and so we find u1 = - u2.

Consider P(x) = x^2. Then \int_{-1}^1 t^2 dt = a_1u_1^2 +a_2u_2^2 \implies 2/3 = a_1u_1^2 +a_2u_2^2

So if a_1 = a_2 = 1 and u_1 = -u_2, we get 2/3 = 2 u_1^2 and so we put u_1 = -1/\sqrt{3}, u_2 = 1/ \sqrt{3}.

So we've found our coefficients, and we know they work for P(x) = 1, x, x^2.

Consider P(x) = x^3. The integral is odd so = 0, and our formula gives us u1^3+u2^3 = 0 as u1 = - u2.

So our formula works for 1,x,x^2,x^3 and so works for all cubic polynomials by linearity.

Solution by DFranklin

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