# Quod Erat Demonstrandum

## 2007/10/05

### [AL][PM] Past Paper 1998 Paper II Q.11

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 5:01 下午

(a)

Let $f$ be a non-negative continuous function on [a , b]. Define $F(x) = \int_a^xf(t)dt$ for all $x\in [a,b]$.

Show that $F$ is an increasing function on [a , b]. Hence deduce that if $\int_a^bf(t)dt = 0$, then $f(x)\:=\:0$ for all $x\in [a,b]$.

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$F$ 的遞增性，得 $F(a)\:\le\:F(x)\:\le\:F(b)$，即 $\int_a^af(t)dt \le F(x) \le \int_a^bf(t)dt$，亦即 $0\le F(x)\le 0$，故 $F(x) = 0$ on [a , b]，意思是 $F(x)$ 在 [a , b] 區間內恆等於零，即 $F(x)$ 是 constant function 是也。那麼，對 $F(x)$ 求導，得 $F'(x) = f(x) = 0$，證畢。

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(b)

Let $g$ be a continuous function on [a , b]. If $\int_a^bg(x)u(x)dx = 0$ for any continuous function $u$ on [a , b], show that $g(x) = 0$ for all $x \in [a,b]$.

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Part (a) 的結論是 $f(x) = 0$，part (b) 的結論是 $g(x) = 0$，我們自然想到 $f(x)$$g(x)$ 有沒有什麼關聯？Part (a) 中，$f(x)$ 是一個非負函數；Part (b) 中，$g(x)$ 不一定是非負函數，所以我們斷不能直接 put $g(x) = f(x)$ 的。看看 $g$ 的條件：對任何連續函數 $u(x)$，恒有 $\int_a^bg(x)u(x)dx = 0$。Pure Mathematics 的題目常有這樣的技巧：既然對任意的 $u$ 也成立，那麼我們可否代入某個特別的 $u$，使我們做出一些結果？如果要運用 part (a)，我們要代入一個特別的 $u$，以產生一個非負函數。不妨設 $u(x) = g(x)$（這不是唯一選擇，$g^{2n-1}(x)$ 也可），則，由 $g(x)$ 的條件，有 $\int_a^bg(x)g(x)dx = 0$$\int_a^bg^2(x)dx = 0$，看，我們製作了一個非負函數 $g^2(x)$，我們便可運用 part (a)，設所謂的 $f(x)$$g^2(x)$，得 $g^2(x) = 0$，即 $g(x) = 0$.

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(c)

Let $h$ be a continuous function on [a , b]. Define $A = \frac{1}{b-a}\int_a^bh(t)dt$.

(i) If $v(x) = h(x) - A$ for all $x \in [a,b]$, show that $\int_a^bv(t)dt = 0$.

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– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

(ii) If $\int_a^bh(x)w(x)dx = 0$ for any continuous function $w$ on [a , b] satisfying $\int_a^bw(x)dx = 0$, show that $h(x) = A$ for all $x \in [a,b]$.

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1. $\int_a^bg(x)u(x)dx = 0$ for any continuous function $u$ on [a , b].

2. $\int_a^bh(x)w(x)dx = 0$ for any continuous function $w$ on [a , b] satisfying $\int_a^bw(x)dx = 0$.

$\int_a^bh(x)v(x)dx = 0$

$\int_a^b(A + v(x))v(x)dx = 0$

$\int_a^bAv(x)dx + \int_a^bv(x)v(x)dx = 0$

$\int_a^bv^2(x)dx = 0$ [因為 $\int_a^bv(x)dx = 0$]

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## 15 則迴響 »

1. […] please refer to the following post for the question I’d mentioned in the lesson today: https://johnmayhk.wordpress.com/2007/10/05/alpm-past-paper-1998-paper-ii-q11/ Leave a […]

通告 由 Solve DE by method of substitution « Quod Erat Demonstrandum — 2009/10/19 @ 5:35 下午 | 回應

2. To Matthew,

“why we sub u(x)=g(x), then isn’t that we only prove a specific example to make the statement correct, but not proving the statement generally right."

Prove

$99^2-1=9800$

How to prove it?

Suppose we know

$x^2-1\equiv (x-1)(x+1)$ … … (*)

Then we can substitute $x=99$ into (*), and obtain

$99^2=(99-1)(99+1)=9800$

OK?

Is $99^2-1=9800$ generally right?

Yes, of course.

Can we say $99^2-1=9800$ is true ONLY when $x=99$?

No.

$99^2-1=9800$ is generally right.

We can just say, if we use (*) to obtain the result $99^2-1=9800$, then substitute $x=99$ is the ONLY way.

Of course, there is other way to obtain $99^2-1=9800$, for example,

we can substitute $x=98$ into $(x+1)^2-1\equiv x(x+2)$ … … (**)

and yield

$99^2-1=9800$

If we put $x=99$ into (**), we CANNOT obtain $99^2-1=9800$ (we just take the wrong key to open a door) but we CANNOT say that “$99^2-1=9800$ is true ONLY when x equals to a PARTICULAR EXAMPLE $x=98$

Back to the original question, the condition that

$\int_{a}^{b}g(x)u(x)dx=0$ is true for ANY continuous function"

is very strong, it is true for ANY continuous function!

Just like that

$x^2-1 = (x-1)(x+1)$ … … (*)

is true for ANY real number.

Putting a particular case $u(x)=g(x)$ is just like putting a particular case $x=99$ into (*), it is a free choice and it is a right choice to obtain the truth $g(x)\equiv 0$, just like a right choice to obtain the truth $99^2-1=9800$. If not putting $u(x)=g(x)$, like say, putting $u(x)=\sin x$, then we CANNOT obtain the truth $g(x)\equiv 0$, but $g(x)\equiv 0$ is still the truth! Just like putting $x=98$ into (*), we CANNOT obtain $99^2-1=9800$, but it is still the truth!

Understand?

迴響 由 johnmayhk — 2011/04/03 @ 12:10 下午 | 回應

3. I know what I misunderstand, I do not pay attention to the condition “true for any contin. u(x) is true". Anyway, thx for ur clear explanation.

迴響 由 Matthew — 2011/04/03 @ 3:32 下午 | 回應

4. hi,

In part b, i think that the substitution using u=g is not general enough.

by putting u(x)=g(x), we only have

\int_a^b g(x)u(x)dx =0 for for any continuous function u on [a , b]
=> \int_a^b [g(x)]^2 dx =0
=> g(x)=0 for all x in [a,b]

In other words, we prove that A=>B=>C —-(*)

However, i think that in some cases, this doesnt make generality.
For example, we may show in other way that, AD=>(C or D) —–(**)

Comparing with (*), you may see that we may “lose some results" sometime.

In my method, i show the following things:
\int_a^b g(x)u(x)dx =0 for for any continuous function u on [a , b]
\int_a^b g(x){g(x)[u(x)]^2}dx =0 for for any continuous function u on [a , b]
=>\int_a^b {g(x)u(x)}^2dx =0
=>g(x)=0 for all x in [a,b] (by a and since u is arbitrary)

the problem I doubt is that
\int_a^b g(x)u(x)dx =0 for for any continuous function u on [a , b]
=> \int_a^b [g(x)]^2 dx =0

but

\int_a^b [g(x)]^2 dx =0 doesnt imply back to
\int_a^b g(x)u(x)dx =0 for for any continuous function u on [a , b]

Here is an example:
x^2=1
(x+1)(x-1)=0
x=1 or x=-1

however,
x^2=1
=>2ln(x)=0
=>x=1

There is “lost of results"

I implore you to reply me as soon as possible since I have to deal with the AL pure exam soon.

迴響 由 K.C. — 2011/04/05 @ 3:06 上午 | 回應

5. Oops, the comment cant show the imply and equivalent signs since they are in html coed…

Here is the corrected version:
(=＞ is the imply sign and ＜=＞ is the equivalent sign)

=======================================================================================================
hi,

In part b, i think that the substitution using u=g is not general enough.

by putting u(x)=g(x), we only have

\int_a^b g(x)u(x)dx =0 for for any continuous function u on [a , b]
=＞ \int_a^b [g(x)]^2 dx =0
=＞ g(x)=0 for all x in [a,b]

In other words, we prove that A=＞B=＞C —-(*)

However, i think that in some cases, this doesnt make generality.
For example, we may show in other way that, A＜=＞D=＞(C or D) —–(**)

Comparing with (*), you may see that we may “lose some results" sometime.

In my method, i show the following things:
\int_a^b g(x)u(x)dx =0 for for any continuous function u on [a , b]
＜=＞\int_a^b g(x){g(x)[u(x)]^2}dx =0 for for any continuous function u on [a , b]
=＞\int_a^b {g(x)u(x)}^2dx =0
=＞g(x)=0 for all x in [a,b] (by a and since u is arbitrary)

the problem I doubt is that
\int_a^b g(x)u(x)dx =0 for for any continuous function u on [a , b]
=＞ \int_a^b [g(x)]^2 dx =0

but

\int_a^b [g(x)]^2 dx =0 doesnt imply back to
\int_a^b g(x)u(x)dx =0 for for any continuous function u on [a , b]

Here is an example:
x^2=1
＜=＞(x+1)(x-1)=0
＜=＞x=1 or x=-1

however,
x^2=1
=＞2ln(x)=0
=＞x=1

There is “lost of results"

I implore you to reply me as soon as possible since I have to deal with the AL pure exam soon.

迴響 由 K.C. — 2011/04/05 @ 3:17 上午 | 回應

• To K.C.

The statement required to prove is

If $\int_{a}^{b}g(x)u(x)dx=0$ for any continuous function $u(x)$ on [$a,b$], then $g(x)=0$ on [$a,b$].

Put $u(x)=g(x)$ is perfectly okay to derive the fact that $g(x)=0$ on [$a,b$].

You write

$A\Leftrightarrow D \Rightarrow C$ or $D$

you mean

A:"$\int_{a}^{b}g(x)u(x)dx=0$ for any continuous function $u(x)$ on [$a,b$]"

D:"$\int_{a}^{b}g(x)(g(x)u^2(x))dx=0$ for any continuous function $u(x)$ on [$a,b$]"

C:"$g(x)=0$ on [$a,b$]"

right?

The required result is C.

And, the so-called missing result is D? Right?

Then I should say D is not the result, it is the condition.

Besides, you write

\int_a^b [g(x)]^2 dx =0 doesnt imply back to
\int_a^b g(x)u(x)dx =0 for for any continuous function u on [a , b]

$\int_{a}^{b}g^2(x)dx=0$
$\Rightarrow g(x)=0$ on [$a,b$] (by using (a))
$\Rightarrow \int_{a}^{b}g(x)u(x)dx=0$ for any continuous function $u(x)$ on [$a,b$]

? Accept?

Just a minor point, for considering real values,

$x^2=1$
$\Leftrightarrow \ln(x^2)=0$
$\Leftrightarrow \ln(|x|^2)=0$
$\Leftrightarrow 2\ln|x|=0$
$\Leftrightarrow x=\pm 1$

迴響 由 johnmayhk — 2011/04/05 @ 8:15 上午 | 回應

6. hi,

I have write something but i dont know how to post a photo here.

Here are the links of what I write:

P.1

P.2

for the last sentence, it is not clear. It should be u(x)=/=0.

迴響 由 K.C. — 2011/04/05 @ 3:28 下午 | 回應

• Just back home, reply you quickly.

If the condition is changed in this way:

$\int_{a}^{b}\frac{g(x)}{u(x)}dx=0$ for any continuous function $u(x)$ on [$a,b$],

of course, we CANNOT take $u(x)=\frac{1}{g(x)}$ as before, because of the fact that $g(x)$ is continuous on [$a,b$] DOES NOT IMPLY $\frac{1}{g(x)}$ is also continuous on [$a,b$]; but this is not a problem in the original part (b), taking $u(x)=g(x)$ is ‘legitimate’.

Next, apart from the continuity, there is another condition for $u(x)$, namely, $u(x)$ is non-zero on [$a,b$]. Thus, unlike the original (b), we need to have more careful choice on $u(x)$.

Your method is replacing $u(x)$ by $\frac{1}{g(x)+\frac{1}{u(x)}}$, but, before knowing the result that $g(x)=0$ on [$a,b$], how to ensure that $g(x)+\frac{1}{u(x)}$ is non-zero on [$a,b$]?

I’m still thinking.

迴響 由 johnmayhk — 2011/04/05 @ 6:56 下午 | 回應

• Because there may be some problems that K.C. cannot reply on the blog, the following are the dicussion through emails and I’d like to post here for further reference:

【K.C.】

In my method, all we need to do is NOT to ensure that g(x)+1/u(x) is non-zero but to ensure that 1/[g(x)+1/u(x)] is non-zero.

1/[g(x)+1/u(x)]
=u(x)/[g(x)*u(x)+1]
=/=0 since u(x)=/=0 form the initial condition.

Also, go back to the question part b, i think that substitution of u(x)=g(x) makes u(x) be particular but loss the arbitrary characteristics.

So, this particular case implies a particular result only, we cannot ensure that this result is complete enough.

However, replacing u(x)by u(x)+g(x), the new function u(x)+g(x) is still arbitrary.

So the result is complete.

This is my argument.

【johnmayhk】

Because taking u = 1/(g + 1/u), we need g + 1/u non-zero to ensure the u(x) is well-defined.

u is non-zero, but u/(gu + 1) may be not well-defined at some x in [a,b].

Sorry, I still cannot see the “loss of arbitrary characteristics" affect the proof in the original (b).

【K.C.】

For the problem considering u/(gu+1) may not be well-defined, i have a solution

Case 1 : g*u=/= -1 for all x in [a,b]

then the result follows my previous proof, ie g=0 for all x in [a,b]

Case2 : there exist some c in [a,b] such that g(c)*u(c)= -1 (ie. u/(gu+1) is not well-defined)

then g(c)(g(c)+1/u(c))=g(c)*[(g(c)*u(c)+1)/u(c)]=0

then we have

/int_a^b g(x)(g(x)+1/u(x) dx=0
/int_a^c g(x)(g(x)+1/u(x) dx + /int_c^b g(x)(g(x)+1/u(x) dx=0
then the result follows.

For how “loss of arbitrary characteristics" affects the proof in (b):
First of all, did you realize what my words “loss of arbitrary characteristics" mean? (not to query you but my ability to express myself is poor)
did you find that “loss of arbitrary characteristics" occurs when we sub u=g ?

Since i think that you only cant see how “loss of arbitrary characteristics" affects the proof in part b but not cant see there is a “loss of arbitrary characteristics", i want to ensure whether it is right or not.

迴響 由 johnmayhk — 2011/04/05 @ 10:02 下午

7. 老師你好，請問(x^3)/k (k是常數)　可以用quotient rule 微分嗎？

迴響 由 andy — 2011/04/05 @ 8:13 下午 | 回應

8. 哦。。謝謝，我只是想知道除了抽ｋ出來還有什麽方法,用quotient rule經常都會計錯數…

迴響 由 andy — 2011/04/05 @ 8:44 下午 | 回應

• $\frac{d}{dx}(\frac{f}{g})=\frac{d}{dx}(fg^{-1})$ 用 product rule 會好些嗎？

迴響 由 johnmayhk — 2011/04/05 @ 8:55 下午 | 回應

9. 對啊，所以我確定以後只用product rule

另外，請問你知道如何找出一個f(x)是連續還是不連續嗎？

迴響 由 andy — 2011/04/05 @ 9:43 下午 | 回應

• Check it by definition, or by experience…

迴響 由 johnmayhk — 2011/04/05 @ 9:47 下午 | 回應