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)!.

*******************************

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.

下文原貼於 2004-10-07 濟濟一堂學術討論區

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?

稍稍利用計數機,立即知 \ln8 \approx 2.079441542

若單純地設 \frac{1 + x}{1 - x} = 8,從而得出 x = \frac{7}{9}

但若把 x = \frac{7}{9} 代入上式 (*),你會發覺計算了很多項,也未能接近 2.079441542 這個數字(用 excel 試 ,要到第 22 項才可確定準確程度為小數點後 5 位的答案 2.07944),粗略地說,這是因為代入 x = \frac{7}{9},這個數字比較接近 1,那麼 (*) 中涉及 (\frac{7}{9})^n 的項,要相當大的 n,才可以使 (\frac{7}{9})^n 接近 0,換言之,代入 x = \frac{7}{9} 便會使 (*) 收斂得比較慢。

幸好,題目給了 \ln7 = 1.945910 這個資料,因為 \ln(\frac{8}{7}) = \ln8 - \ln7 所以,只要知道 \ln(\frac{8}{7}),便知 \ln8 了。

今設 \frac{1 + x}{1 - x} = \frac{8}{7},易得 x = \frac{1}{15}。這便會使 (\frac{1}{15})^n「快」一些接近零。即 (*) 可以快一些到達要求的準確程度。

試試看:

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)

於是 \ln8 \approx 2.07944(只要用 3 項,便有這個準確程度了,不似之前要 20 多項)。

因為題目沒有嚴格要求同學進行誤差評估,所以,處理這題,我們只要像中五時處理 method of bisection 般(注:現在已不在會考課程內),不停做 iteration, 當取小數點後某個位而答案不變時,則停止 iteration。

當然,我們可以進行誤差評估,步驟如下:

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

其中

計一計 f(x) 的導數,


觀眾可以注意到,取 x 愈接近 0,R 便愈「快」接近 0。

x = \frac{1}{15} 時,

要求小數點後 5 位的準確程度,我們只需設:

如何解出上式?trial and error 吧,但非常容易地試出 n = 2,換言之,當 x = \frac{1}{15},(*) 只要兩項,便達到所需的準確程度,所以,我們可以在式子 (2) 停下來,放心寫出 \ln8 \approx 2.07944 這個答案。

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