Quod Erat Demonstrandum

2009/04/17

可導性

Filed under: HKALE,Pure Mathematics — johnmayhk @ 5:23 下午
Tags: , ,

這是有關「可導性」(differentiablity)的討論,寫給那天沒有上復活假期補課班的中六同學。注:討論純粹以中學數學的觀點出發。

先請同學回答下面三道是非題:

1. Put x = 0 into the expression of f'(x) and it is undefined, then f(x) is not differentiable at x = 0. True?
2. If \lim_{x \rightarrow 0^-}f'(x) = \lim_{x \rightarrow 0^+}f'(x), then f(x) is differentiable at x = 0. True?
3. If \lim_{x \rightarrow 0^-}f'(x) is finite, then the value of \lim_{h \rightarrow 0^-}\frac{f(h) - f(0)}{h} is also finite. True?

=======================================

上面三題的答案皆「不一定是」。

=======================================

首先,要描述一個函數是否「可導」(differentiable),我們不是看 f'(x) 這個表達式是什麼,最安全的,都是返回「可導」的定義。

再看看上述的三題

Question 1

x = 0 入表達式 f'(x),而「計不到數」(undefined),並不一定代表”f(x) is not differentiable at x = 0”.

以下是一個經典的例子:

f(x) = x^2\sin\frac{1}{x} for x \ne 0 and f(0) = 0.

對於 x \ne 0,恆有:

f'(x) = -\cos\frac{1}{x} + 2x\sin\frac{1}{x} – – – – – – (*)

若把 x = 0 代入上式 (*),但因為當 x = 0\frac{1}{x} 是「計不到數」(沒有定義 undefined),同學往往以為 f(x)x = 0 處也是 undefined,故他們聲稱”f(x) is not differentiable”。

錯!

判別「可導」與否,一定要回歸「可導」的定義:

f(x) is differentiable at x = 0”的意思是「極限 \lim_{h \rightarrow 0}\frac{f(h) - f(0)}{h} 存在」。

一旦極限存在,我們才可寫 f'(0) = \lim_{h \rightarrow 0}\frac{f(h) - f(0)}{h},而並不是先用中四五的手法,找出 f'(x) 的表達式,再代入 x = 0,就得出 f'(0)

返回那經典例子:

f(x) = x^2\sin\frac{1}{x} for x \ne 0 and f(0) = 0.

當我們考慮極限

\lim_{h \rightarrow 0}\frac{f(h) - f(0)}{h}

根據定義,

f(h) = h^2\sin\frac{1}{h}, f(0) = 0,故

\lim_{h \rightarrow 0}\frac{f(h) - f(0)}{h}
= \lim_{h \rightarrow 0}\frac{1}{h}(h^2\sin\frac{1}{h} - 0)
= \lim_{h \rightarrow 0}h\sin\frac{1}{h}

\sin\frac{1}{h} 有界(bounded),且 \lim_{h \rightarrow 0}h = 0,故 \lim_{h \rightarrow 0}h\sin\frac{1}{h} = 0,即

\lim_{h \rightarrow 0}\frac{f(h) - f(0)}{h} = 0

亦即此極限存在(等於 0,即 f'(0) = 0)是故

f(x) is differentiable at x = 0

Question 2Question 3 可以一併考慮,隨便舉例

f(x) = x + 1 for x \ge 0 and
f(x) = \sin x - 1 for x < 0

對於 x \ne 0,我們易知

f'(x) = 1 for x > 0 and
f'(x) = \cos x for x < 0

易知

\lim_{x \rightarrow 0^-}f'(x) = \lim_{x \rightarrow 0^-}\cos x = \cos 0 = 1
\lim_{x \rightarrow 0^+}f'(x) = \lim_{x \rightarrow 0^+}1 = 1

於是

\lim_{x \rightarrow 0^-}f'(x) = \lim_{x \rightarrow 0^+}f'(x)

但,明顯地,f(x)x = 0 之處是不可導(not differentiable at x = 0)。

最簡單的原因是,f(x)x = 0 之處根本不連續(not continuous at x = 0),見下

\lim_{x \rightarrow 0^-}f(x) = \lim_{x \rightarrow 0^-}\sin x - 1 = \sin 0 - 1 = -1
\lim_{x \rightarrow 0^+}f(x) = \lim_{x \rightarrow 0^+}(x + 1) = 0 + 1 = 1

可見

\lim_{x \rightarrow 0^-}f(x) \ne \lim_{x \rightarrow 0^+}f(x)

是故 f(x)x = 0 之處不連續,從而不可導。

現在,具體地計算一下 f(x)x = 0 處的左導數(left-hand derivative),即

\lim_{h \rightarrow 0^-}\frac{f(h) - f(0)}{h}
= \lim_{h \rightarrow 0^-}\frac{(\sin h - 1) - 1}{h}
= \lim_{h \rightarrow 0^-}\frac{\sin h - 2}{h}
= +\infty

左導數不存在,導數也自然地不存在,順便回答了 Question 3

\lim_{x \rightarrow 0^-}f'(x) = 1  (finite)但
\lim_{h \rightarrow 0^-}\frac{f(h) - f(0)}{h} = +\infty (infinite)

重申一次,縱使存在以下關係:

\lim_{x \rightarrow 0^-}f'(x) = \lim_{x \rightarrow 0^+}f'(x)

我們也不能知道 f'(0) 存在與否,

亦即不能保證 f(x)x = 0 之處是可導。

習題

1. Let

f(x) = x + 1 for x \ge 0 and
f(x) = \sin x + a for x < 0

Suppose f(x) is differentiable at x = 0, evaluate a.

2. Let

f(x) = x + 1 for x \ge 0 and
f(x) = b\sin x for x < 0

Will f(x) be differentiable at x = 0 for some real number b ?

3. Let

f(x) = R(x) for x \ge 0 and
f(x) = L(x) for x < 0

Given that f(x) is differentiable at x = 0.

Is is true to write ”R'(0) = L'(0)”?

4 則迴響 »

  1. sir, here is my ans:

    1. a=+1
    2. differentiable when b=+1
    3. T

    but i actually don’t know what exactly what f'(x) is
    what is the different in meaning between
    lim f'(x)
    x->0-
    and
    lim (f(h)-f(0))/h
    h->o-
    and also
    lim f(x)
    x->0-

    i don’t know which one to consider when doing calculation

    迴響 由 harrison — 2009/04/17 @ 8:16 下午 | 回覆

  2. Harrison,

    The answer to Q.2 is FALSE.

    f(x) is NOT differentiable for ANY real number b.

    Just think about the graph of f(x), it is ALWAYS discontinuous at x = 0 (even for b = 1). (Or prove it by simple algebraic computation)

    What is f'(x)?

    It is a limit.

    By definition,

    If the limit \lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h} EXISTS, we define it as

    f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h}.

    What is \lim_{x \rightarrow 0}f'(x)?

    It is NOT f'(0) in general.

    By the definition above,

    f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h}.

    hence

    \lim_{x \rightarrow 0}f'(x)
    = \lim_{x \rightarrow 0}\lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h}

    You see, it is NOT f'(0) in general, because, by the definition, f'(0) is simply:

    f'(0) = \lim_{h \rightarrow 0}\frac{f(h) - f(0)}{h}

    SOMETIMES, f'(0) is really equal to \lim_{x \rightarrow 0}f'(x), as for example, when the function f'(x) is continuous at x = 0.

    Also, can you prove that the answer to Q.3 is TRUE?

    迴響 由 johnmayhk — 2009/04/18 @ 6:12 下午 | 回覆

  3. For Question 1, I used to teach my students in the same way as yours before at the beginning, but later I found that the answer of Q1 is true, and I also got a very simple proof.

    Let me first point out the problem of your counter-example.
    If f(x) = x^2 sin 1/x if x is not 0 and f(0) = 0,
    then what is f'(x)?
    Answer:
    f'(x) = (the expression you have) if x is not 0 and f'(0) = 0
    Therefore, when we put x = 0 , f'(0) = 0 which is defined.
    [ What you showed in your counter-example is in fact,
    lim {x -> 0} f'(x) . ]

    Back to the proof of the statement.
    Assume f(x) is differentiable at x = 0.
    Then f'(0) exists (defined), which leads to a contradiction.
    Therefore f'(x) is not differentiable at x = 0.

    迴響 由 Cheng Wing Kuen — 2009/04/21 @ 8:09 上午 | 回覆

  4. Thank you Cheng Wing Kuen.

    是我的表達出了問題。

    在 Question 1 中,

    Put x = 0 into the expression of f'(x)

    當中我希望說的,不是 f'(0)

    當中的 expression,其實指,把 f(x) 進行求導後得出的 expression。

    如上例

    f(x) = x^2\sin\frac{1}{x}

    利用求導法,得出的 expression 就是

    f'(x) = -\cos\frac{1}{x} + 2x\sin\frac{1}{x}

    Question 1 想說的,就是 put x = 0 into the expression above.

    它是 undefined 的。

    它不是 f'(0)

    我只是想指出,同學往往誤以為

    「把 f(x) D 完,再代個零入去,計唔到,就即係函數係 x = 0 個位 D 唔到」。

    是我的表達出問題,”expression of f'(x)”確實可以指

    f'(x) = -\cos\frac{1}{x} + 2x\sin\frac{1}{x} when x \ne 0 and f'(x) = 0 when x = 0

    那麼,當”Put x = 0 into the expression of f'(x)” 就被理解為 f'(0) 時,

    誠如 Cheng Wing Kuen 所言,

    Question 1 is trivially true.

    We can prove it simply by contrapositivity.

    f(x) is differentiable at x = 0 \Rightarrow f'(0) exists"

    is equivalent to

    f'(0) does not exist \Rightarrow f(x) is not differentiable at x = 0"

    P.S.

    Oh, Mr. cheng, are you the writer of “永權網頁"? I love the website, it is very useful.

    迴響 由 johnmayhk — 2009/04/21 @ 2:46 下午 | 回覆


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