# Quod Erat Demonstrandum

## 2008/10/10

### Say something about series in Applied Mathematics (II)

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 4:53 下午
Tags: ,

Here is just a typical question in AL Applied Mathematics (II).

For natural numbers $m, n$ ($m \ge n$).

Let $f(x) = x^ne^x$, evaluate $f^{(m)}(0)$.

The technique is all about series.

On expanding $f(x) = x^ne^x$, the coefficient of $x^m$ is $\frac{f^{(m)}(0)}{m!}$.

On the other hand,

$f(x) = x^ne^x = x^n(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots)$

Hence the coefficient of $x^m$ in the series above is $\frac{1}{(m - n)!}$

Yields,

$\frac{f^{(m)(0)}}{m!} = \frac{1}{(m - n)!}$
$f^{(m)}(0) = \frac{m!}{(m - n)!}$

Use exactly the same technique, I set up the following as a question in a quiz

Let $f(x) = (1 + x)^2\ln(1 + x)$, evaluate $f^{(n)}(0)$.

Using the fact that $\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$, I gave the following answer:

$n!(\frac{(-1)^{n - 1}}{n} + \frac{2(-1)^{n - 2}}{n - 1} + \frac{(-1)^{n - 3}}{n - 2})$.

But most of students gave a clean expression $2(-1)^{n - 1}(n - 3)!$ by usual practice: differentiation.

Hence, to a certain extent, my question was not well set.

Urm, well, at least I’d created an additional mathematics question:

Prove that, for natural number $n > 2$,

$n!(\frac{(-1)^{n - 1}}{n} + \frac{2(-1)^{n - 2}}{n - 1} + \frac{(-1)^{n - 3}}{n - 2}) = 2(-1)^{n - 1}(n - 3)!$.

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Just for reference of my students, here is a re-post of a discussion on textbook question about series.

(a) By expanding the integrand $\int_0^x \frac{dt}{1 + t}$ as a series of powers of $x$ and integrating term by term, find the series for $\ln(1 + x)$, assuming the method to be valid provided that $|x| < 1$.

(b) Write down the series for $\ln(1 - x)$ and hence obtain the series of $\ln\frac{1 + x}{1 - x}$.

(c) Hence calculate $\ln8$ with accuracy for 5 decimal places, given that $\ln7 = 1.945910$.

Part (a) Trivial.

Part (b) 答案是

$\ln\frac{1 + x}{1 - x} = 2(x + \frac{x^3}{3} + \frac{x^5}{5} + \dots)$ – – – – – – (*)

Part (c) 尋求 $\ln8$，並要求準確程度達小數點後 5 位。理論上，我們要先找出所謂 error term $R$，再設 $R \le 0.5 \times 10^{-5}$，來找出對應的 $n$，才知道要加多少項。但這題給了一個 Hint：$\ln7 = 1.945910$，為何用之？如何用之？

$2(\frac{1}{15}) + 1.945910 \approx 2.07924333$ – – – – – – (1)
$2(\frac{1}{15} + (\frac{1}{15})^3 + 1.945910 \approx 2.07944086$ – – – – – – (2)
$2(\frac{1}{15} + (\frac{1}{15})^3 + (\frac{1}{15})^5 + 1.945910 \approx 2.07944139$ – – – – – – (3)

Let $f(x) = \ln(\frac{1 + x}{1 - x})$

$x = \frac{1}{15}$ 時，