Quod Erat Demonstrandum

2008/11/18

Create an m.i. question

It is not difficult to create questions like:

Prove by mathematical induction that

\frac{3^3\times1}{4!} + \frac{3^4\times2}{5!} + \frac{3^5\times3}{6!} + \dots + \frac{3^{n+2}\times n}{(n+3)!} = \frac{9}{2} - \frac{3^{n+3}}{(n+3)!}

for any positive integer n.

It is all about the so-called telescoping principle.

What’s it? It’s just one of the methods of finding the sum (or product) of terms.

To evaluate the sum of

a_1 + a_2 + a_3 + \dots + a_n (say),

it will be easy when we can split each term into

a_k = b_k - b_{k+1}.

See

a_1 + a_2 + a_3 + \dots + a_n
= (b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) + \dots + (b_n - b_{n+1})
= b_1 - b_{n+1}

Thus, we may create an m.i. question in the form of

a_1 + a_2 + a_3 + \dots + a_n = b_1 - b_{n+1}"

To set up a common M.I. question, just take ‘reasonable’ b_k.

For example, we may take

b_k = \frac{a^k}{k!}

then

b_k - b_{k+1} = \frac{a^k}{k!} - \frac{a^{k+1}}{(k+1)!} = \frac{a^k(k + 1 - a)}{(k+1)!}

For the so-called beauty of the question, we may start from k = a, that is

\frac{a^a}{a!} - \frac{a^{a+1}}{(a+1)!} = \frac{a^a(1)}{(a+1)!}
\frac{a^{a+1}}{(a+1)!} - \frac{a^{a+2}}{(a+2)!} = \frac{a^{a+1}(2)}{(a+2)!}
\frac{a^{a+2}}{(a+2)!} - \frac{a^{a+3}}{(a+3)!} = \frac{a^{a+2}(3)}{(a+3)!}
\dots
\frac{a^{a+n-1}}{(a+n-1)!} - \frac{a^{a+n}}{(a+n)!} = \frac{a^{a+2}(n)}{(a+n)!}

Sum up the above n equations, we have

\frac{a^a}{a!} - \frac{a^{a+n}}{(a+n)!} = \frac{a^a(1)}{(a+1)!} + \frac{a^{a+1}(2)}{(a+2)!} + \frac{a^{a+2}(3)}{(a+3)!} + \dots +\frac{a^{a+2}(n)}{(a+n)!} – – – – – – (*)

By putting a = 1, 2, 3, 4, \dots into (*), we can create the following m.i. questions for students revision:

Prove the following are true for any positive integer n,

1. \frac{1}{2!} + \frac{2}{3!} + \frac{3}{4!} + \dots +\frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!}
2. \frac{2^2(1)}{3!} + \frac{2^3(2)}{4!} + \frac{2^4(3)}{5!} + \dots +\frac{2^{n+1}(n)}{(n+2)!} = 2 - \frac{2^{n+2}}{(n+2)!}
3. \frac{3^3(1)}{4!} + \frac{3^4(2)}{5!} + \frac{3^5(3)}{6!} + \dots +\frac{3^{n+2}(n)}{(n+3)!} = \frac{9}{2} - \frac{3^{n+3}}{(n+3)!} (The original question above)
4. \frac{4^4(1)}{5!} + \frac{4^5(2)}{6!} + \frac{4^6(3)}{7!} + \dots +\frac{4^{n+3}(n)}{(n+4)!} = \frac{32}{3} - \frac{4^{n+4}}{(n+4)!}
\dots

Furthermore, if we take

b_k = \sin(k), then

b_{k+1} - b_k = \sin(k+1) - \sin(k) = 2\cos\frac{2k+1}{2}\sin\frac{1}{2}

Hence, we have

\sin(2) - \sin(1) = 2\cos\frac{3}{2}\sin\frac{1}{2}
\sin(3) - \sin(2) = 2\cos\frac{5}{2}\sin\frac{1}{2}
\sin(4) - \sin(3) = 2\cos\frac{7}{2}\sin\frac{1}{2}
\sin(5) - \sin(4) = 2\cos\frac{9}{2}\sin\frac{1}{2}
\dots
\sin(n) - \sin(n - 1) = 2\cos\frac{2n-1}{2}\sin\frac{1}{2}

sum up the above, yields

\sin(n) - \sin(1) = 2\sin\frac{1}{2}(\cos\frac{3}{2} + \cos\frac{5}{2} + \cos\frac{7}{2} + \cos\frac{9}{2} + \dots + \cos\frac{2n-1}{2})

Thus, we create a question like:

prove by mathematical induction that

\cos\frac{3}{2} + \cos\frac{5}{2} + \cos\frac{7}{2} + \cos\frac{9}{2} + \dots + \cos\frac{2n-1}{2} = \frac{\sin(n) - \sin(1)}{2\sin\frac{1}{2}}

for all integers n \ge 2.

Students, try to let other b_k to create some boring m.i. questions at your command.

7 則迴響 »

  1. John Sir

    好耐之前係濟濟一堂… 你地 post 過一個由 David Morin 教授寫既一本 ebook, 內容係關於 mechanics and special relativity….
    因為個教授要出書… 順手已經 delete link,
    唔知你有無 save 低舊既 pdf file…
    有既可以 send 俾我嗎?

    thx

    迴響 由 Ricky — 2008/11/18 @ 11:18 下午 | 回覆

  2. AL時無留心上堂,或自己是無天份,總是學不懂telescoping principle…
    不過還是感謝阿sir的教導…

    迴響 由 lanven — 2008/11/19 @ 4:32 上午 | 回覆

  3. @Ricky

    對不起,我沒有存底。該網頁似乎有一些 lecture notes

    http://www.courses.fas.harvard.edu/~phys16/2004_lectures/

    不知可以用得著。

    @lanven

    節節相消原理本身很簡單,但有時應用起來可以是很難的,不一定是天份問題。謝謝大家留言!

    迴響 由 johnmayhk — 2008/11/19 @ 11:26 上午 | 回覆

  4. 在我的印象中, 所有課本的等比級數都是節節相消的變身.

    迴響 由 chin — 2012/05/14 @ 1:48 上午 | 回覆

  5. 在我的印象中, 所有課本的等比級數都是節節相消的變身.

    迴響 由 mm100100 — 2012/05/14 @ 1:52 上午 | 回覆

  6. Hi, the last term on the LHS of the trigo-related induction question should be cos {(2n – 1)/2} and that we prove for n >= 2. Thanks for sharing!

    迴響 由 wenshih — 2012/05/19 @ 12:23 下午 | 回覆

    • Corrected. Thank you Wenshih! Long time no see!

      迴響 由 johnmayhk — 2012/05/21 @ 6:30 上午 | 回覆


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