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Unformatted text preview: Solutions to Homework Set 0 Differential calculus : The point of the exercise was to make sure that you know how to differentiate integrals with respect to their limits: d da integraldisplay b a f ( x ) dx = f ( a ) , d db integraldisplay b a f ( x ) dx = f ( b ) , and in general d dt integraldisplay b ( t ) a ( t ) f ( x, t ) dx = f ( b ( t )) b t f ( a ( t )) a t + integraldisplay b ( t ) a ( t ) t f ( x, t ) dx. Once you have this under control, the first problem is plug and chug: From y ( x ) = sin ( x L ) sin L integraldisplay x f ( t ) sin t dt + sin x sin L integraldisplay L x f ( t ) sin ( t L ) dt we get y prime ( x ) = cos ( x L ) sin L integraldisplay x f ( t ) sin t dt + cos x sin L integraldisplay L x f ( t ) sin ( t L ) dt. The two terms arising from the derivative of the integration limits have cancelled against each other. When we differentiate again, the two pieces arising from differentiating the factors outside the integral assemble to give 2 y ( x ). The two terms from differentiating the integrals are y primeprime ( x ) + 2 y ( x ) = f ( x ) cos ( x L ) sin x cos x sin ( x L ) sin L . The addition formula for sin( A + B ), now shows that this is equal to f ( x ) as required. For the second problem we have F prime ( x ) = K (0) f ( x ) + integraldisplay x x K ( x y ) f ( y ) dy = K (0) f ( x ) integraldisplay x f ( y ) y K ( x y ) dy = K (0) f ( x ) integraldisplay x y [ f ( y ) K ( x y )] dy + integraldisplay x f prime ( y ) K ( x y ) dy = K (0) f ( x ) K (0) f ( x ) + f (0) K ( x ) + integraldisplay x f prime ( y ) K ( x y ) dy = f (0) K ( x ) + integraldisplay x K...
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 Fall '07
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