24. Applications of Taylor Series

b. Derivatives

Recall the Taylor series formula: \[\begin{aligned} f\,(x) &=\sum_{n=0}^\infty \dfrac{f\,^{(n)}(a)}{n!}(x-a)^n \\ &=f\,(a)+f'(a)(x-a)+\dfrac{f''(a)}{2}(x-a)^2 \\[2 pt] &\quad+\dfrac{f\,^{(3)}(a)}{3!}(x-a)^3 +\cdots+\dfrac{f\,^{(n)}(a)}{n!}(x-a)^n+\cdots \end{aligned}\] In the chapter on Taylor series, we used this formula to find the power series about \(x=a\) for a given function \(f\,(x)\). However, if you already know the power series for a given function, then you can turn the Taylor series around and use it to compute \(f\,^{(n)}(a)\), the \(n^\text{th}\) derivative of \(f\,(x)\) evaluated at \(x=a\).

If \(f\,(x)=\ln(1-x^2)\), compute

1. \(f\,^{(14)}(0)\), the \(14^\text{th}\) derivative of \(f\,(x)\) evaluated at \(x=0\).
2. \(f\,^{(15)}(0)\), the \(15^\text{th}\) derivative of \(f\,(x)\) evaluated at \(x=0\).

Since we want derivatives evaluated at \(x=0\), we need a Taylor series for \(\ln(1-x^2)\) centered at \(x=0\). We substitute \(x\to-x^2\) into the known series \[ \ln(1+x)=\sum_{n=1}^\infty \dfrac{(-1)^{n-1}}{n}x^n =x-\dfrac{1}{2}x^2+\dfrac{1}{3}x^3-\dfrac{1}{4}x^4+\cdots \] to get \[\begin{aligned} \ln(1-x^2) &=\sum_{n=1}^\infty \dfrac{(-1)^{n-1}}{n}(-x^2)^n \\ &=(-x^2)-\dfrac{1}{2}(-x^2)^2 +\dfrac{1}{3}(-x^2)^3-\dfrac{1}{4}(-x^2)^4+\cdots \end{aligned}\] and simplify to get the specific series: \[ \ln(1-x^2)=\sum_{n=1}^\infty \dfrac{-1}{n}x^{2n}=-x^2-\dfrac{1}{2}x^4-\dfrac{1}{3}x^6-\dfrac{1}{4}x^8+\cdots \] We now compare this with the general Maclaurin series (Taylor series centered at \(x=0\)): \[\begin{aligned} f\,(x) &=\sum_{n=0}^\infty \dfrac{f\,^{(n)}(0)}{n!}x^n \\ &=f\,(0)+f'(0)x+\dfrac{f''(0)}{2}x^2 +\dfrac{f\,^{(3)}(0)}{3!}x^3+\cdots+\dfrac{f\,^{(n)}(0)}{n!}x^n+\cdots \end{aligned}\] Caution: Notice that \(n\) has a different meaning in the two series. In the specific series, the power of \(x\) is \(2n\). In the general series, the power of \(x\) is just \(n\).

1. To find \(f\,^{(14)}(0)\), we note that in the general series the coefficient of \(x^{14}\) is \(\dfrac{f\,^{(14)}(0)}{14!}\). In the specific series for \(\ln(1-x^2)\), the term with an \(x^{14}\) has \(n=7\) and its coefficient is \(-\dfrac{1}{7}\). We equate these terms and solve for \(f\,^{(14)}(0)\): \[ \dfrac{f\,^{(14)}(0)}{14!}x^{14}=-\dfrac{1}{7}x^{14} \qquad \Longrightarrow \] \[ f\,^{(14)}(0)=-\dfrac{14!}{7}=-2\cdot13! \]
2. To find \(f\,^{(15)}(0)\), we note that in the general series the coefficient of \(x^{15}\) is \(\dfrac{f\,^{(15)}(0)}{15!}\). In the specific series for \(\ln(1-x^2)\), there are no odd powers of \(x\). So the coefficient \(x^{15}\) is \(0\). We equate these terms and solve for \(f\,^{(15)}(0)\): \[ \dfrac{f\,^{(15)}(0)}{15!}x^{15}=0 \qquad \Longrightarrow \qquad f\,^{(15)}(0)=0 \]

Notice that computing the \(14^\text{th}\) derivative of \(\ln(1-x^2)\) directly would be a long, tedious process.