Interesting Log Relationships (1)

We have already considered the (infinite) reciprocal sequences of the unique numbers associated with (x – 1)ⁿ  = 0.

So once again, for example, where n = 3, the unique numbers associated with the polynomial equation (x – 1)³  = 0, i.e. x³ – 3x²  + 3x – 1 = 0 are,

1, 3, 6, 10, 15, 21, …

And the sum of the (infinite) reciprocal sequence of these numbers is

1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + …  = 2. 

We now consider (x – k)ⁿ = 0, with k an integer > 1, the sum of the (infinite) reciprocal sequences of the corresponding unique numbers associated with (x – k)ⁿ = 0, where n = 2.

For example when k = 2, we have (x – 2)² = 0, i.e. x²  – 4x + 4 = 0, and the unique numbers associated with this polynomial equation are, 

1, 4, 12, 32, 80, …

So the sum of the (infinite) reciprocal sequence of these numbers =

1 + 1/4 + 1/12 + 1/32 + 1/80 + …  = 1.3862…  = log 4 = 2(log 2/1) = 2(log 2).

Then, when k = 3, we have (x – 3)² = 0, i.e. x²  – 6x + 9 = 0, and the unique numbers associated with this polynomial equation are, 

1, 6, 27, 108, 405, 1458, …

Thus the sum of the (infinite) reciprocal sequence of these numbers,

= 1 + 1/6 + 1/27 + 1/108 + 1/405 + 1/1458 + … = 1.2163… = log 27/8 = 3(log 3/2).

In fact there is a general pattern at work here!

So log 4 and log 27/8 = log{k^k/(k – 1)^k} = k{log k/(k – 1)}, where k = 2 and k = 3 respectively.

This implies that the sum of the (infinite) reciprocal sequence of the unique numbers associated for example with (x – 4)² = 0,  = log {4⁴/3⁴} = log 256/81 = 4(log 4/3).

And (x – 4)² = 0 implies x²  – 8x + 16 = 0, and the unique numbers associated with this polynomial equation are 1, 8, 48, 256, 1280, 6144, …

Thus the sum of the (infinite) reciprocal sequence, 

= 1 + 1/8 + 1/48 + 1/256 + 1/1280 + 1/6144 + … = 1.1507… = 4(log 4/3).

Incidentally this pattern indicates clearly why the sum of the reciprocals of the natural numbers i.e. the harmonic series diverges to infinity.

The sum of this sequence is associated with (x – 1)² = 0, and the answer thereby according to the formula is log (1/0) i.e. log ∞ .  

There are also fascinating patterns associated with (x + k)² = 0, where k is now an integer ≥ 1.

These simple polynomial equations lead to the same unique number sequences associated with
(x – k)² = 0, except that the numbers keep alternating as between positive and negative signs.

Therefore in the case where k = 1, the unique number sequence associated with (x + 1)² = 0, 

i.e.  x²  + 2x + 1 = 0, are 1, – 2, 3, – 4, 5, – 6 …and the (infinite) sum of its reciprocals,

= 1 – 1/2 + 1/3 – 1/4 + 1/5 – 1/6 + …   = log 2.

Then when k = 2, the unique number sequence associated with (x + 2)² = 0, i.e.   

x²  + 4x + 4 = 0, is 1, – 4, 12, – 32, 80, – 192, …

Thus the corresponding (infinite) sum of its reciprocals

= 1 – 1/4 + 1/12 – 1/32 + 1/80 – 1/192 + …  = log 9/4 = 2(log 3/2)

Again a fascinating general pattern is at work here.

For example log 9/4 = log{(k + 1)^k/k^k} = k{log (k + 1)/k}, where k = 2  

This implies that when k = 3, the (infinite) sum of reciprocals associated with the unique number sequence corresponding to (x + 3)² = 0,  i.e. x²  + 6x + 9 = 0, = log{4³/3³}

= 3 log 4/3.

And the unique number sequence is 1, – 6, 27, – 108, 405, – 1458, …

Therefore the corresponding (infinite) sum of reciprocals

= 1 – 1/6 + 1/27 – 1/108 + 1/405 – 1/1458 + … = .8630 … = 3(log 4/3).

So the famous case of 1 – 1/2 + 1/3 – 1/4 + 1/5 – 1/6 + …   = log 2, represents just a special example corresponding to the (infinite) sum of reciprocals of the unique number sequence associated with the simple polynomial expression (x + k)² = 0, where k = 1.

And again the general solution for all integer values of k ≥ 1, is log{(k + 1)^k/k^k}

 = k{log (k + 1)/k}.