Quod Erat Demonstrandum


Trivial reflection of mathematics teaching

Filed under: HKALE,Pure Mathematics — johnmayhk @ 6:04 下午

Pure Mathematics is a subject of mathematical “techniques". It provides students with certain tools and procedures to solve certain mathematical problems.

However, for me, Pure Mathematics is a subject of “art".

“Techniques" can be taught, while “art" cannot be taught easily.

In symbols,

P \Rightarrow x“, it’s easy to determine x; while for
x \Rightarrow Q“, it may not be easy to determine x, sometimes.

Let’s take an example from textbook. Evaluate

\lim_{x \rightarrow \infty}x((1 + \frac{1}{x})^x - e).

[I think the question requires students to consider x \rightarrow +\infty only.]

Well, I believe that most of the students could apply l’Hôpital’s rule.

\lim_{x \rightarrow \infty}x((1 + \frac{1}{x})^x - e).
= \lim_{x \rightarrow \infty}\frac{(1 + \frac{1}{x})^x - e}{\frac{1}{x}}
= \lim_{x \rightarrow \infty}\frac{(1 + \frac{1}{x})^x(\ln(1 + \frac{1}{x}) - \frac{1}{1 + x})}{\frac{-1}{x^2}}

Okay, what will be the next?

l’Hôpital’s rule again?

Urm, I believe that, some students may be struggling. “Should I differentiate the stuff further?" But, a “monster" is predictable.

Well, some students may have a better way. Once we know

\lim_{x \rightarrow \infty}(1 + \frac{1}{x})^x = e

We may just consider

\lim_{x \rightarrow \infty}\frac{\ln(1 + \frac{1}{x}) - \frac{1}{1 + x}}{\frac{-1}{x^2}}

And if the limit above exists and equal to L (say), then the required limit is eL.

Okay, even we consider the above, what is next? Differentiate directly? Or make some changes before we go on?

Well, let’s do something first, some students may arrive at

\lim_{x \rightarrow \infty}x^2(\frac{1}{1 + x} - \ln(1 + \frac{1}{x}))
= \lim_{x \rightarrow \infty}\frac{x^2(1 - (1 + x)\ln(1 + \frac{1}{x}))}{1 + x}

Next? l’Hôpital’s rule again? More horrible monster will come into existence, as expected. Then? Differentiate it again? Okay, I do it, resulting in

\lim_{x \rightarrow \infty}[x^2(-(1+x)(\frac{1}{1+1/x})(-\frac{1}{x^2}) - \ln(1+\frac{1}{x})) + 2x(1 - (1+x)\ln(1+\frac{1}{x}))]

Okay, here is another choice of life, continue or not to continue?

For me, I quit.

Looking back to the station

\lim_{x \rightarrow \infty}\frac{\ln(1 + \frac{1}{x}) - \frac{1}{1 + x}}{\frac{-1}{x^2}}

If I choose to differentiate both the numerator and the denominator of the fraction directly, then we will obtain

\lim_{x \rightarrow \infty}\frac{\frac{1}{1 + 1/x}(-\frac{1}{x^2}) + \frac{1}{(1 + x)^2}}{2/x^3}

Oh, no stuff of natural logarithm something and there is only a rational function…seems to see the sun light from the cloud…

The limit
= \lim_{x \rightarrow \infty}\frac{x^3}{2}[\frac{-1}{x(1 + x)} + \frac{1}{(1 + x)^2}]
= \lim_{x \rightarrow \infty}\frac{x^3}{2}\times\frac{-1}{x(1 + x)^2}
= \lim_{x \rightarrow \infty}-\frac{1}{2}\times\frac{1}{(1 + 1/x)^2}
= -\frac{1}{2}

Sucess! That is, \lim_{x \rightarrow \infty}x((1 + \frac{1}{x})^x - e) = -\frac{e}{2}.

You see, what I want to say is, the above “microscopic" procedure is not just technique, I think, it seems to be certain kind of “art".

Students’ FAQ (frequently asked question) is “HOW to think it out?" or “WHY you are so clever to consider that stuff/ procedure?"

Well, you may try another question in the same exercise:

\lim_{x \rightarrow \infty}x[(1 + \frac{1}{x})^x - e\ln(1 + \frac{1}{x})^x] = ?

to feel the sense of “art". (A bit “body catching fire and entered by demons"…)

Apart from telling students that “do more", “learn from experience" etc., it is always a challenge to lead students to the correct track of thinking, so as to be in line with the solution path, especially, in a microscopic way. Sometimes, I could just say that “it’s a kind of feeling".

Similar situation appears when we are doing curve sketching problems. In 2005 AL Pure Mathematics Paper 2 Q.7, it required students to sketch the graph of

y = f(x) = -x + |x|\sqrt{\frac{x}{x + 2}}

Well, it is likely that the setter of this question had chosen simple coefficents involved to avoid the so-called “double punishment", however, for me, it is still not “easy" to determine the expressions of f'(x) and f''(x) correctly.

For a student being under the real examination pressure, it may also be time-consuming to work out correct expressions.

The marking scheme tells us that, students could only be given 1 mark each for perfectly correct expressions of f'(x) and f''(x). All the “microscopic" procedures in differentiation are in the “black box", never be shown in the marking scheme.

When students are performing badly in similar questions as mentioned above, should they be blamed that they are weak in differentiation? Should they be blamed that they don’t know l’Hôpital’s rule? How to give comment in the marker’s report? “Students should be more careful in doing differentiation"? Well, if a student reads the statement in the report, will it be helpful to him/her? (Just like when seeing the hint:"by above" in a structural question, what is the use?)

May be I don’t like complicated algebraic computation from day 1, and hence I do not emphasize it in my daily teaching of pure mathematics, am I right?

There may be many “mother-is-female" type articles (and even (action) research) in education, sorry to give another one.


3 則迴響 »

  1. Here is how I would do the problem: (I will give two ways)

    Let y = 1/x, then the limit is equivalent to:

    lim{y to 0}[(1 + y)^{1/y} – e]/y = lim{y to 0} [f(y) – f(0)]/y = f'(0), where f(x) = (1 + x)^{1/x}. (note that f(0) = e).

    We then have f’/f = [x/(1 + x) – log(1 + x)]/x^2.
    Then L’hopital’s rule yields: lim(x to 0) f’/f = [1/(1 + x)^2 – 1/(1 + x)]/2x = lim -1/2(1 + x)^2 = -1/2.
    Thus, f'(0) = f(0)(-1/2) = -e/2.

    Another way: Make the same substitution, (since I like limit going to 0 better)
    Write (1 + y)^{1/y} = exp((1/y)log(1 + y)).
    Now, use the Taylor series for log(1 + y) to write log(1 + y) = y – y^2/2 + y^3/3 – ….
    Thus (1 + y)^{1/y} = exp[1 – y/2 + y^2/3 – …] = (e^1)e^{-y/2}e^{y^2/3}…
    =e[1 – y/2 + (-y/2)^2/2! + …][1 + y^2/3 + (y^2/3)^2/2! + …]…
    =e[1 – y/2 + higher powers of y]
    Therefore, [(1 + y)^{1/y} – e]/y = -e/2 + y*(some power series in y)
    Hence, the limit is -e/2.
    However, I guess my second method is out of syllabus.

    迴響 由 koopakoo — 2009/04/22 @ 4:26 下午 | 回應

  2. Very nice method! Thank you Koopa!!

    迴響 由 johnmayhk — 2009/04/22 @ 6:08 下午 | 回應

  3. Hi,

    I agree with you. It’s also the art of looking at a problem deeply, creatively and imaginatively.

    Thanks for sharing!

    Wen Shih

    迴響 由 Wen Shih — 2009/04/25 @ 7:24 上午 | 回應

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