테일러 다항식과 오차

Since $T_3(x)$ is the third order Taylor polynomial for $f(x)$ about $x=a$, we know the following things to be true:

• $f(a)=T_3(a)$
• $f'(a)=T'_3(a)$
• $f''(a)=T''_3(a)$
• $f'''(a)=T'''_3(a)$

But, some of these don't look super useful. For instance, if we try to use the first bullet, we get this equation: $$a^3\left[2\log a - \frac{11}{3}\right]=-\frac{2}{3}\sqrt{e^3}+3ea-6\sqrt{e}a^2+a^3$$ Solving this would be terrible. Instead, let's think about how the equations look when we move further down the list. Since $T_3(x)$ is a cubic equation, $T_3'''(x)$ is a constant (and so $T_3'''(a)$ does not depend on $a$). That sounds like it's probably the simplest option. Let's start differentiating. We'll need to know both $f'''(a)$ and $T_3'''(a)$.

$$\begin{aligned} f(x)&=x^3\left[2\log x - \frac{11}{3}\right]\\ f'(x)&=x^3\left[\frac{2}{x}\right]+3x^2\left[2\log x - \frac{11}{3}\right]=6x^2\log x -9x^2\\ f''(x)&=6x^2\cdot\frac{1}{x}+12x\log x - 18x=12x\log x - 12 x\\ f'''(x)&=12x\cdot\frac{1}{x}+12\log x - 12 = 12\log x\\ f'''(a)&=12\log a \\ \text{Now, let's move to the Taylor polynomial. Remember that $e$ is a constant.} \\ T_3(x)&=-\frac{2}{3}\sqrt{e^3}+3ex-6\sqrt{e}x^2+x^3\\ T_3'(x)&=3e-12\sqrt{e}x+3x^2\\ T_3''(x)&=-12\sqrt{e}+6x\\ T_3'''(x)&=6\\ T_3'''(a)&=6 \\ \text{The final bullet point gives us the equation:} \\ f'''(a)&=T_3'''(a)\\ 12\log a &= 6\\ \log a &= \frac{1}{2}\\ a&=e^{\tfrac{1}{2}} \end{aligned}$$

So, $a=\sqrt{e}$.