How do we prove that the McLearin series of $\ln(1+x)$ converges to $\ln(1+x)$ for every $x$ in the interval of convergence? Using Taylor's inequality, one can show this in the case $x > -1/2$. But, how do we show this in the case $x$ is between between $-1$ and $-1/2$?
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We can make use of the following theorem. If functions $f_n(x)$ are continuously differentiable on $[a,b]$ and the series $$S(x_0)=\sum_{n=1}^\infty f_n(x_0)$$ is convergent for some $a\le x_0\le b,$ while the series $$ T(x)=\sum_{n=1}^\infty f_n'(x)$$ is uniformly convergent on $[a,b],$ then the function $S(x)$ is differentiable and $S'(x)=R(x).$
We apply this theorem to $$f_k(x)= {(-1)^{k+1}x^k\over k},\quad |x|\le 1-\delta,\qquad x_0=0$$ Hence the function $$S(x)=\sum_{k=1}^\infty {(-1)^{k+1}x^k\over k},$$ satisfies $S(0)=0$ and $$S'(x)=\sum_{k=1}^\infty (-1)^{k+1}x^{k-1}={1\over 1+x},\quad |x|\le 1-\delta $$ Therefore $S(x)=\ln(1+x)$ for $|x|\le 1-\delta.$ As $\delta>0$ was arbitrary $S(x)=\ln(1+x)$ for $|x|<1.$
The same theorem can be used to prove that if a function $f(x)$is represented by a power series $f(x)=\sum a_nx^n,$ $|x|<1,$ then it is the MacLaurin series of $f.$
Remark No integral calculus is needed, nor estimates of the remainder in the MacLaurin formula.
Ryszard Szwarc
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