91影视

To find the recurrence relation for the coefficients, we'll substitute the power series solution y(t) = 鈭戔倷鈧屸個 a鈧檛鈦� into Hermite's equation. Hermite's equation is y'' - 2ty' + 2渭y = 0. Take the derivatives: y'(t) = 鈭戔倷鈧屸個 na鈧檛^{n-1} y''(t) = 鈭戔倷鈧屸個 n(n-1)a鈧檛^{n-2} Now plug these into Hermite's equation: 鈭戔倷鈧屸個 n(n-1)a鈧檛^{n-2} - 2t 鈭戔倷鈧屸個 na鈧檛^{n-1} + 2渭鈭戔倷鈧屸個 a鈧檛鈦� = 0 Change the indices to match the powers of t: 鈭戔倷鈧屸倐 n(n-1)a鈧檛^{n-2} - 2鈭戔倷鈧屸倎 na鈧檛^{n} + 2渭鈭戔倷鈧屸個 a鈧檛鈦� = 0 Now we combine the series under the same summation: 鈭戔倷鈧屸個 [(n+2)(n+1)a_{n+2} - 2(n+1)a_{n+1} + 2渭a鈧橾t鈦� = 0 For the series to be equal to zero, the coefficients of the powers of t must be equal to zero, so we have: (n+2)(n+1)a_{n+2} - 2(n+1)a_{n+1} + 2渭a鈧� = 0 Rearrange the equation to find the recurrence relation: a_{n+2} = [2(n+1)a_{n+1} - 2渭a鈧橾 / [(n+2)(n+1)]

When 渭 = N, where N is a nonnegative integer, we can observe that since a_{N+2} depends on a鈧�, there will be a point where a鈧� becomes zero. This will cause all future coefficients in the series to be zero as well. Since no more terms are contributed after a certain term, this means the series converges to a polynomial of degree N. Thus, we can conclude that Hermite's equation will have a polynomial solution H_N(t) when 渭 = N is a nonnegative integer.

Verify that H鈧�(t) = 1 and H鈧�(t) = 2t satisfy Hermite's equation for 渭 = 0 and 渭 = 1 respectively by direct substitution. For H鈧�(t) = 1, we have: y'' = 0, y' = 0 (Substitute into Hermite's equation): 0 - 2t(0) + 2(0)(1) = 0, which is true. For H鈧�(t) = 2t, we have: y'' = 0, y' = 2 (Substitute into Hermite's equation): 0 - 2t(2) + 2(1)(2t) = 0, which is true. Thus, H鈧�(t) and H鈧�(t) do satisfy Hermite's equation for their respective values of 渭.

Use the recurrence relation and the requirement that H鈧�(t) = 2鈦縯鈦� + ... to determine H鈧�(t), H鈧�(t), H鈧�(t) and H鈧�(t). From the solution of part (a): a_{n+2} = [2(n+1)a_{n+1} - 2渭a鈧橾 / [(n+2)(n+1)] Start with H鈧�(t): 渭 = 2 a鈧� = (2(1)a鈧� - 2(2)a鈧�) / (2(3)) = (4 - 4) / 6 = 0 a鈧� = (2(2)a鈧� - 2(2)a鈧�) / (3(4)) = (-4) / 12 = -1/3 So, H鈧�(t) = 2鈦铰测伨t虏 - (1/3)(2)t鲁 = 4t虏 - (2/3)t鲁 Now find H鈧�(t): 渭 = 3 a鈧� = 0 (same as before) a鈧� = (6 - 2(3)a鈧�) / (3(4)) = (6 - 12) / 12 = -1/2 a鈧� = (12 - 2(3)(-1/2)) / (4(5)) = (12 + 6) / 20 = 9/10 So, H鈧�(t) = 2鲁t鲁 - (1/2)(2)t鈦� + (9/10)(2)t鈦� = 8t鲁 - 2t鈦� + (18/10)t鈦� Continue similarly for H鈧�(t): 渭 = 4 a鈧� = -1/2 (same as before) a鈧� = (4 - 4) / (20) = 0 a鈧� = (10 - 2(4)(-1/2)) / (5(6)) = (10 + 4) / 30 = 7/15 So, H鈧�(t) = 2鈦磘鈦� + (7/15)(2)t鈦� = 16t鈦� + (14/15)t鈦� Finally, find H鈧�(t): 渭 = 5 a鈧� = 0 (same as before) a鈧� = (5 - 2(5)(-1/2)) / (6(5)) = (5 + 5) / 30 = 1/3 a鈧� = (20 - 2(5)) / (6(7)) = (20 - 10) / 42 = 5/21 So, H鈧�(t) = 2鈦祎鈦� + (1/3)(2)t鈦� + (5/21)(2)t鈦� = 32t鈦� + (2/3)t鈦� + (10/21)t鈦�