Review : Taylor Series — Question 10

Question 10 – Solution

Strategy. A bound on derivatives throughout a segment controls Taylor’s remainder; center-only data control coefficients.

Step 1: Force the remainders to vanish. Taylor’s theorem uses some $\xi$ between $0$ and $x$. The hypothesis gives $$|f(x)-T_N(x)|\le\frac{C A^{N+1}(N+1)!}{(N+1)!}|x|^{N+1} =\boxed{C(A|x|)^{N+1}}.$$ This tends to zero whenever $|x|\le r$ and $A|x|<1$. On the specified smaller interval the uniform bound is $C(A\rho)^{N+1}\to 0$.

Step 2: Test the boundary honestly. At $A|x|=1$ the bound is only $C$, which does not tend to zero. The constant function $1$ satisfies the bound with $C=A=1$ and its Taylor series converges at $\pm 1$. For real $x$, $|x\pm i|\ge 1$, so the given identity yields $|g^{(n)}(x)|\le n!$. Yet the Maclaurin series of $g$ is $\sum_{k\ge 0}(-1)^kx^{2k}$; at $x=\pm 1$ its terms fail to tend to zero. Thus neither boundary outcome is forced.

Step 3: Rule out the bound for a flat function. If such constants existed on some $[-r,r]$, choose $0<|x|<\min(r,1/A)$. All Taylor polynomials would be zero, but Step 1 would force $f(x)=0$, contradicting positivity. Thus no finite positive $C,A$ can work on any such neighborhood.

Step 4: Separate coefficients from function values. A bound only at $0$ gives $|f^{(n)}(0)/n!|\le CA^n$. Geometric comparison proves absolute convergence of the Taylor series for $|x|<1/A$, but does not prove that its sum is $f(x)$. The smooth function $e^{-1/x^2}$ extended by zero at $0$ has every center derivative zero, so it satisfies every such center-only bound; its Taylor series is zero while the function is positive elsewhere.