Question

2. The solution to the boundary value problem y' + way=0, y(0) =0, y(1) - y'(1) = 0 is y(x) = an sin(Zral) T=1 where the an are Fourier coefficients and the Zn are zeros of tan(w) To compute the zeros we can solve the fixed point problem w= tan(w). (i) Draw a graph of y=w and y=tan(w) on the interval (-37, 37). (ii) How many zeros of f(w) =tan(w) - w do we expect for all w. (iii) As w → too, what do the zeros of f tend to? (This approximation can be used for large n). (iv) Use the intermediate value theorem to show there must be at least one root in the interval (/2, 37/2). (v) In some interval around the root w*, each iteration of Newton's method satisfies w* - Ww+1 < 1f"(w) Mw* - wil where M for all w in that interval. Solving iteratively, this implies 2 f'(w) w* - Wn+11 (M|W* - wo])21 where wo is the initial (seed) point. If the quantity M is very M large, then wo will have to be chosen very close to w* to obtain convergence. Assuming w =w* < 1.5 for the interval (1/2, 34/2), compute M and make a judgement about where the seed point can be relative tow*. = max 1

Answer 1

不少 OF 2. J=0 10) 一 21 三 13 y- tanw)

for all The number of zeros of tw)= tan (w)-W is the number of intersection of the curve y=tan(W) with the line y=w From the graph it is clear that the total number of zeros of f(w) is infinity. hii AS W IN the zeros of t tends to to. iv Intermediate value theorem lett be a continuous function defined on [a,b and let s be a number with f(a) cscf(b). Then there exists X E (a,b] such that f(x)=s. some on Here f(W) = tan (w)w is continuous (7 371) and we have tan (I) = too and tan (²1) =-0. Site o t (",). =) ( 1/2, 31 ) such that tanta f(w) .:3 WE ( 1 , 37) so that f(w) has atleast one zero.