If A is an n n matrix then e A n 0 A n n The significance of this is that x t e

If a is an n n matrix then e a n 0 a n n the

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If A is an n × n matrix, then e A = n =0 A n n ! . The significance of this is that x ( t ) = e At x 0 is the solution to the system of equations x 0 = Ax with initial condition x (0) = x 0 . For example, if A = 0 - 1 1 0 , then e At = cos t - sin t sin t cos t . If X ( t ) is a “fundamental matrix” (a matrix whose columns are linearly independent solutions to x 0 = Ax ), then e At = X ( t ) X (0) - 1 . (Matrix exponentials were not on the syllabus, but I explained this in class because I think it clarifies things a lot.) To solve an inhomogeneous system x 0 = Ax + f , choose a fundamental matrix X ( t ) and write x ( t ) = X ( t ) v ( t ). Then the equation to be solved is equivalent to v 0 ( t ) = X ( t ) - 1 f . This can be integrated to obtain v ( t ), and then multiplied by X ( t ) to give x ( t ). The answer will contain n integration constants, which correspond to adding a particular solution to x 0 = Ax + f to a general solution to the homogeneous equation x 0 = Ax . Differential equations chapter 10. The Fourier series of a periodic function f with period 2 π is given by f ( x ) = a 0 2 + X n =1 ( a n cos( nx ) + b n sin( nx )) where a n = 1 π Z π - π f ( x ) cos( nx ) dx and b n = 1 π Z π - π f ( x ) sin( nx ) dx. 5
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If f is an odd function then a n = 0 for all n ; and if f is an even function then b n = 0 for all n . If f and f 0 are piecewise continuous then the Fourier series for f ( x ) converges to the average of the left and right limits of f at x . The Fourier sine series of a function f on the interval [0 , π ] is f ( x ) X n =1 b n sin( nx ) where b n = 2 π Z π 0 f ( x ) sin( nx ) dx. The Fourier cosine series of f is f ( x ) a 0 2 + X n =1 a n cos( nx ) where a n = 2 π Z π 0 f ( x ) cos( nx ) dx. How to find some solutions to the heat equation by separation of vari- ables. The solution to the heat equation ∂u ∂t = β 2 u ∂x 2 for x [0 , π ] and t R with boundary conditions u (0 , t ) = u ( π, t ) = 0 and initial condition u ( x, 0) = f ( x ) is given by u ( x, t ) X n =1 b n e - βn 2 t sin( nx ) where the constants b n are the coefficients of the Fourier sine series of f . Here we ignore convergence issues. Similarly, if we change the boundary condition above to ∂u ∂x (0 , t ) = ∂u ∂x ( π, t ) = 0, then the solution is u ( x, t ) a 0 2 + X n =1 a n e - βn 2 t cos( nx ) where the constants a n are the coefficients of the Fourier cosine series of f . 6
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