如何去解n次方的可分解不等式

我記得在中二時,數學老師叫我們依書本提供的方法去解2/3次方的不等式絕對是一件苦差,先嘗試把2/3次方的不等式因式分解成2/3個不等式的因子,然後再逐個個去試去解,一條數最快也要10分鐘才解完,做完數學功課,電視的兒童節目也做完了。於是我當時就想出了一個可大量節省時間的方法,看箸別人埋頭苦幹時自己可自由發夢,別有快感,不知現在愈來愈貴的教科書有沒有比以前進步,編者想到我想到的方法?(鄭重聲明,未經我同意教科書不可以拿來作謀利用途!)

這個方法其實用了邏緝作輔助,所以提高了效率。流程是這樣的,首先自然是把n次方的不等式化成n個因子,然後把所有不等式除以x的乘積,如三條不等式分別是:(2x-3)(3x-5)(4x-8)>0,則我們可分別列出(x-3/2)(x-5/3)(x-2)>0,再把所有因子用x的0次方的項數由小至大加以排列,即(x-3/2)(x-5/3)(x-2)>0,考慮到在任何自然數集下此關係一定為真(x-3/2)>(x-5 /3)>(x-2),現在再去思考一下,如果要三個因式的乘積為正,因為3是單數,所以一是3者同為正值,另外唯一可能性是首項為正,後兩項為負。在前一種情況,只要數值永遠為最少的一項即第三項為正值,其餘不可能變成負值,所以只要(x-2)>0或x>2則為此不等式的其中一解;在後一種情況,即首項為正值,後兩項為負值,首項為正即(x-3/2)>0或x>3/2及x-5/3<0或x<5/3(第三項可以不理),所以第二解為3/2換句話說,答案為3/22。

同理,要是我們要解(2x-3)(3x-5)(4x-8)<0,即(x-3/2)(x-5/3)(x-2)<0,我們已知在任何自然數集下此關係一定為真(x-3/2)>(x-5/3)>(x-2),再因為3是單數,要3個數的乘積為負數,只有兩個可能,一是三項因式同為負值,二是因式數值最細者為負,其餘因式為正。所以,要三項因式同為負值,只要首項為負值即可,所以即(x-3/2)<0或x<3/2;另外,要第三項為負及第二項為正,即(x-2)<0或x<2及(x-5/3)>0或x>5/3,所以解為2>x>5/3。
換句話說,答案為x<3/2及2>x>5/3。

用了此一方法去幫助,一條n次方的不等式,最多是考慮n+1次而不像傳統教科書要考慮2^n次,然後再用負數的單次次方一定為負數,負數的雙次次方一定為正數,正數的單雙數次次方一定為正數,尚可再減低要考慮的次數,例如要解一條7次方大於零的不等式,我們只要考慮分別是尾兩項、四項及六項為負的情況即可以。如此才算是教科書的增值,不是會叫會唱的容祖兒頭像在唱1+1=2!

Differential between x to the power n

This is first notice by me what I was in High School,

I discover that for the sequence of x²,
1,4,9,16,25.36

Their difference is respectively:
3,5,7,9,11

The difference of their difference is:
2,2,2,2

And the difference of difference of difference is all zero. For three identical steps we reduce the sequence to zero, and the power of x is 2. It also take 2 step to arrive at identical number.

Then I tried the sequence of x³:
1,8,27,64,125,216

Their difference is respectively:
7,19,37,61

The difference of their difference is:
12,18,24

The difference of their difference of their difference is:
6,6.

So now it take 3 iterations to reduce them to equal number, and it take four iterations to zero out all differentials. Notice the power of x is 3.

It doesn’t take long for me to realize the relationship between number of iterations to reduce the differential to equal number; and the number of iterations to zero out all differential is always one more than the power of x.

Therefore we can write the expression for the powers of x:
For x², (X(3)-X(2))-(X(2)-X(1))=(X(4)-X(3))-(X(3)-X(2))
or X(3)-2X(2)+X(1)= X(4)-2X(3)+X(2)
or X(4)-3X(3)+3X(2)-X(1)=0
Thus, X(4)=3X(3)-3X(2)+X(1)

For x³, ((X(4)-X(3))-(X(3)-X(2)))-((X(3)-X(2))-(X(2)-X(1)))=((X(5)-X(4))-(X(4)-X(3)))-((X(4)-X(3))-(X(3)-X(2)))
or X(4)-3X(3)+3X(2)-X(1)=X(5)-3X(4)+3X(3)-X(2)
or X(5)-4X(4)+6X(3)-4X(2)+X(1)=0
Thus, X(5)=4X(4)-6X(3)+4X(2)-X(1)

We could thus generalize the result for the sequence of X^n,
X(n+2)=(n+1)C(1)X(n)-(n+1)C(2)X(n-1)+(n+1)C(3)X(n-2)-(n+1)C(4)X(n-3)….+(-1)^(2+n)*(n+1)C(n+1)X(1)

Now for x^6, we have
X(8)=7X(7)-21X(6)+35X(5)-35X(4)+21X(3)-7X(2)+X(1)

Therefore, we have a more convenient way to calculated x to power n if we know the terms of the sequence from n+1 consecutive. Trust me, in my mind, that was better than the calculator to calculate the power of any number to any power at that time.

However, I notice that in the n iteration that the difference between x^n is equal to n!
i.e. For x², X(3)-2X(2)+X(1)=2, so we need only last 2 of the sequence in x²
For x³, X(4)-3X(3)+3X(2)-X(1)=6, so we need only last 3 of the sequence in x³

Therefore, for x^n, X(n+1)-nC(1)X(n)+nC(2)X(n-1)-nC(3)X(n-2)+nC(4)X(n-3)….+(-1)^(1+n)*nC(n+1)X(1)=n!

An alternative method to solve inequalities of Nth degree

I remember when I was in Middle School, it was a pain for other student to solve the quadratic and cubic inequalities. What the textbook taught is what I seen as a stupid and time-wasting method: Just like solving the quadratic and cubic inequalities, we first need to need all the terms in one side so we factorize it into their factors; then they list the factor and individually determine the signs of each factor. The stupidity lays in the need to determine the sign of each factor individually, I devise a faster method at that time.

I was thinking at that time, since all these factors are related, why must be they be deal with individually? Obviously, since all of them are referring to a single variable x, i.e. the signs of each factor are not independent, why we can’t take advantage of their inter-relationships? What is the implicit inter-relationship between the factors? If we transform all of the factor into comparable form, then we can list them in ascending or descending sequences like this: x+1>x-3>x-3/4… etc. Now, clever reader may already see the trick I play here: The central idea is we can save ourselves a few steps because we can’t have the case which a Greater factor being negative while a Smaller factor being positive. Thus we can list all feasible (and logical) case for the signs of the factors to be determined instead of blindly list ALL THE POSSIBLE case.

To save us more step, we know that we require an odd number of negative factor to resulted into negative, and we require an even number of negative factor to resulted into positive. And the result of multiplication of all positive factor must resulted into positive number.

Thus, we can draw the boundary somewhere in the factors. Assume one factor is being positive, then any factor which is greater than that factor would always be positive; and any factor smaller than that factor would always be negative. Thus, in the case of x+1>x-3>x-3/4, if we assume X+1>0, then we must have x-3<(x-3/4)<0,>-1 is the solution for (x+1)(x-3)(x-3/4)>0; now if we assume x-3>0, then we must have x+1>0 and x-3/4<0,>3 as the solution for (x+1)(x-3)(x-3/4)<0.>0, since then we must have x+1>x-3>x-3/4>0, therefore x>3/4 is the solution for (x+1)(x-3)(x-3/4)>0. Similarly, if we assume x+1<0,>(x-3)>(x-3/4), therefore x<-1 is the solution for (x+1)(x-3)(x-3/4)<0.

Using this methodology, an inequality of n-th degree only need to be evaluate n+1 times at maximum instead of evaluating 2^n times. Plus, using logical deduction with the axioms which only odd number of negative factor will result in negative number; and only even number of negative factor would guarantee a positive number, we can further reduce the cases which we need to evaluate. For instance, if we want to solve a seven factors inequality which is greater than zero, we can only cases of 2 negative, 4 negative, 6 negative factors and no negative factors.