The solution to the boundary value problem y\" +w2y = 0, m0) = 0:. y(1) + y'(1)= 0 CC is at) = Z on sin[zn$) 11:] where the an are Fourier coefficients and the z\" are zeros of tan(w) + to. To compute the zeros we can solve the fixed point problem to = tan(w). {i} Draw a graph of y = w and y = tan{w) on the interval [Z'H, 21?]. (ii) How many zeros of re] = tan(w] + a; do we expect for all m E R. (iii) As a; > :too, what do the zeros tend to? Think about where the intersections of y = w and y = tan(w) occur relative to asymptotes. (This approximation can be used for large 11). (iv) Use the intermediate Value theorem to show there must be at least one root in the interval ([2,31TX2). We seek selected zeros of the of the function at) = (3-3le (xx-5(3). To compute the zeros we can solve the fixed point problem e'z!' 1" 2 (305(1). (i) Draw agraph of y = cznn and y = cor-3(3) on the interval ['\"{3, 9772]. (ii) How many zeros of f do we expect for .1: E R? (iii) As a: > 00, what do the zeros of f tend to? (if 2,; refers to the nth zero, this approximation can be used for large 11). {iv} Use the intermediate value theorem to show there must be at least one root in the interval (1r, 2:1}. 2 . (1) y=COS ( x ) 3tr (1) f (x ) = e 1 /10- cos( x ) we expect an infinite number of zeros of f for XER. The zeros will correspond to where eye = cos( x ) which will exist for x70 . (it) As x -00 we observe that exso. Hence we will be left with -cos( x ) for large x. Hence the zeros of + will concide with the zeros of - cos ( R ). That is, zn 7 ( 2 n+1 ) IT 2 for integer n; n positive and large ( iv) Since emio and cos ( 1 ) are continuous for all x, f is continuous for all x .* ( TT ) = e- 7/10 - cos ( TT) = evo + 1 . yo since e " f ( 271 ) = e - 21/10 - Cos (2TT ) -