M. Jimbo, T. Miwa's Algebraic Analysis of Solvable Lattice Models PDF

By M. Jimbo, T. Miwa

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In the present problem we don’t care what is happening to the left of 0. Plugging in the value 1/2 as auxiliary point between 0 and 1, we get 21 ≥ ( 21 )2 , so we see that in [0, 1] the curve y = x is the higher. To the right of 1 we plug in the auxiliary point 2, obtaining 22 ≥ 2, so the curve y = x2 is higher there. Therefore, the area between the two curves has to be broken into two parts: 1 2 (x − x2 ) dx + area = 0 (x2 − x) dx 1 since we must always be integrating in the form right higher - lower dx left In some cases the ‘side’ boundaries are redundant or only implied.

This idea occurs in all basic physical science and engineering. In particular, for a projectile near the earth’s surface travelling straight up and down, ignoring air resistance, acted upon by no other forces but gravity, we have acceleration due to gravity = −32 feet/sec 2 Thus, letting s(t) be position at time t, we have s¨(t) = −32 We take this (approximate) physical fact as our starting point. From s¨ = −32 we integrate (or anti-differentiate) once to undo one of the derivatives, getting back to velocity: v(t) = s˙ = s(t) ˙ = −32t + vo where we are calling the constant of integration ‘vo ’.

We would obtain 4x + 1 x3 + 4x2 − x + 1 =x+5+ x(x − 1) x(x − 1) since the quotient is x + 5 and the remainder is 4x + 1. Thus, in this situation x3 + 4x2 − x + 1 dx = x(x − 1) x+5+ 4x + 1 dx x(x − 1) Now we are ready to continue with the first algebra trick. In this case, the first trick is applied to 4x + 1 x(x − 1) We want constants A, B so that 4x + 1 A B = + x(x − 1) x x−1 As above, multiply through by x(x − 1) to get 4x + 1 = A(x − 1) + Bx and plug in the two values 0, 1 to get 4 · 0 + 1 = −A 4·1+1=B That is, A = −1 and B = 5.

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Algebraic Analysis of Solvable Lattice Models by M. Jimbo, T. Miwa


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