Last time I have consider the sequence of x², x³,… x^n. We found that we can reduce their differential to a constant in n times. Could we apply the same method to 2^x, 3^x, 4^x?
Let’s start with 2:
1,2,4,8,16,32,64
Their differential:
1,2,4,8,16,32
The differential of this differential again yell:
1,2,4,8,16
Notice that since the differential is identical with the original sequence, therefore this method doesn’t work for 2^x..
Let’s tried with the sequence of 3^x:
1,3,9,27,81,243
Their differential:
2,6,18,54,162
The differential of their differential:
4,12,36,108
Although they may look different at the surface. Notice that:
2*(2,6,18,54,162)=(4,12,36,108,)
Thus, it is again a repeating sequence. We could thus said, the sequence of the n^x has the property that repeat itself in the differentiation process. Moreover, we also notice that for n^x, the differential of differential is a multiple of previous sequence by (n-1).
Let’s test this hypothesis:
Take 3,4,5 as example:
For 3, we have:
1,3,9,27,81,243;
2,6,18,54,162;
4,12,36,108;
8,24,72;
For 4, we have:
1,4,16,64,256,1024;
3,12,48,192,768;
9,36,144,576;
27,81,243;
For 5, we have:
1,5,25.125,625,3125;
4,20,100,500,2500;
16,80.400,2000;
64,320,1600.
We thus verify the hypothesis. One thing interesting is, if we look at the first item of each differential sequences, it formed a sequence identical to last sequence.
What can we make of this?
Donate n to the power x by k(x,n), we list them in order:
k(2,n)-k(1,n)=k(1,n-1)
k(3,n)-2*k(2,n)+k(1,n)=k(2,n-1)
k(4,n)-3*k(3,n)+3*k(2,n)-k(4,n)=k(3,n-1)
k(5,n)-4*k(4,n)+6*k(3,n)-4k(2,n)+k(1,n)=k(4,n-1)
There many equations we can form using them, I just list one type here:
(k(3,n)-2*k(2,n)+k(1,n))/(k(2,n)-k(1,n))=n
(k(4,n)-3*k(3,n)+3*k(2,n)-k(4,n))/(k(2,n)-k(1,n))=n^2
or (k(3,n)-2*k(2,n)+k(1,n))/(k(2,n)-k(1,n))= (k(4,n)-3*k(3,n)+3*k(2,n)-k(4,n))/(k(3,n)-2*k(2,n)+k(1,n))
we can form however complex relationship from this base.
