Review : Taylor Series — Question 3

Question 3 – Solution

Strategy. Expose the first nonzero numerator term before removing the apparent singularity.

Step 1: Cancel and reindex. The coefficient of $x^{2k+1}$ in the numerator is $(-1)^k[1/(2k+1)!-1/(2k)!]$. It is zero when $k=0$. Setting $j=k-1$ for the surviving terms gives $$\boxed{H(x)=\sum_{j=0}^{\infty}\frac{(-1)^j\,2(j+1)}{(2j+3)!}x^{2j}.}$$ Sine and cosine converge absolutely for every real $x$; the new series also has infinite radius by the ratio test. It agrees with the quotient when $x\ne 0$.

Step 2: Fill the removable singularity. The new series defines an analytic extension with $$H(0)=\frac 13,\qquad H''(0)=2!\left(-\frac 1{30}\right)=-\frac 1{15}, \qquad H^{(4)}(0)=4!\frac 1{840}=\frac 1{35}.$$ Continuity forces the value $1/3$, so no other continuous extension is possible.

Step 3: Bound the truncated tail. The polynomial is $T_4(x)=1/3-x^2/30+x^4/840$. For $|x|\le 1$, successive term magnitudes have ratio $[(j+2)/(j+1)]x^2/[(2j+5)(2j+4)]\le 1/10$. They decrease to zero, so $$\boxed{|H(x)-T_4(x)|\le\frac{|x|^6}{45360}\le\frac 1{45360}.}$$

Step 4: Interpret symmetry and stability. Only even powers occur. Alternating bounds give $H(x)\ge 1/3-x^2/30\ge 3/10>0$ on $[-1,1]$. The numerator subtracts two quantities of order $x$ to obtain one of order $x^3$; the series evaluates the finite limiting value without that subtraction. This is a cancellation concern, not a failure of the exact quotient.