Question

The power series with nonzero radius of convergence is the

Answer 1

★ "Let aa,c0c0,c1c1,c2c2,... be in R, with at least one of c0c0,c1c1,c2c2,... nonzero. Let the power series

∑n=1∞cn(x−a)n∑n=1∞cn(x−a)n

have positive radius of convergence rr. Show that there exists a positive number ww < rr such that the sum of the series is nonzero for every real number xx such that 0 < |xx - aa| < ww."

EDIT: The attempted solution: A proof by contradiction showing that all coefficients cici = 0 for i = 0,1,2,3,....

assume such a delta does not exist. For all nn there exists xnxn such that 0 < |xnxn - aa| < 1/n1/n.

we note that ff is continuous in (a−r,a+ra−r,a+r) and the limit of xnxn as nn approaches infinity is equal to aa.

In the taylor expansion c0c0 = f(a)f(a) = lim f(xn)f(xn) as nnapproaches infinity which is equal to the lim 0 as nnapproaches infinity, which equals zero. So, c0c0 = 0.

Let us now look at c1c1.

f(x)=∑n=1∞cn(x−a)nf(x)=∑n=1∞cn(x−a)n

=(x−a)∑n=0∞cn+1(x−a)n=(x−a)∑n=0∞cn+1(x−a)n

, we let this be a new series and write l(x)=(x−a)∑∞n=0cn+1(x−a)nl(x)=(x−a)∑n=0∞cn+1(x−a)n. l(x)l(x) has the same radius of convergence and l(xn)l(xn) = 0 since f(xn)f(xn) = 0. Thus, the first term c1c1 = g(a)g(a) = 0.

Let us now use induction to show that cn+1cn+1 = 0.

=(x−a)n+1∑n=0∞cn+1(x−a)n=(x−a)n+1∑n=0∞cn+1(x−a)n

Let g(x)=(x−a)n+1∑∞n=0cn+1(x−a)ng(x)=(x−a)n+1∑n=0∞cn+1(x−a)n Then, cn+1cn+1 = g(a)g(a) = 0. Thus, all coefficients go to zero. Proof by contradiction.