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1. Solve 1 au 2 102 u 1 au + r ar ar u(1, 0) = sin 0. = 0 for r < 1,
1. Solve 1 au 2 102 u 1 au + r ar ar u(1, 0) = sin 0. = 0 for r < 1, 1. Solve 1 au 2 102 u 1 au + r ar ar u(1, 0) = sin 0. = 0 for r < 1, 1. Solve 1 au 2 102 u 1 au + r ar ar u(1, 0) = sin 0. = 0 for r < 1, 1. Solve 1 au 2 102 u 1 au + r ar ar u(1, 0) = sin 0. = 0 for r < 1,
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Given that hze Then 4 80 1 24 010 2 4 10 sin 0 f 0 116 3 2 n1 4 me 4 ...
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An Introduction to the Mathematics of financial Derivatives
Authors: Salih N. Neftci
2nd Edition
978-0125153928, 9780080478647, 125153929, 978-0123846822
