Thursday, February 1, 2018

Interesting Log Relationships (2)

In a former entry, “Interesting Log Relationships 1” I showed how the infinite sums of reciprocals of the unique numbers associated with simple polynomial equations of the form (x – k)2 = 0, where k = 2, 3, 4, … are related to simple log type relationships.

So, again for example, when k = 2, the equation is (x – 2)2 = 0, i.e. x2 – 4x + 4 = 0.

Then by the same process by which the unique numbers forming the Fibonacci sequence is obtained from the equation x2 – x – 1 = 0, in like manner a unique number sequence can be obtained for x2 – 4x + 4 = 0, i.e. 1, 4, 12, 32, 80, … (listed as A001787 in the On-line Encyclopedia of Integer Sequences).

Then the infinite sum of reciprocals of this number sequence,

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

And in that piece, I provided a general formula to cover the infinite sum of reciprocals for all cases entailing unique number sequences of form (x – k)2= 0,

i.e. k{log k/(k – 1)}.

So for example when k = 6 we have (x – 6)2= x2 – 12x + 36 = 0.
And the unique sequence of numbers associated with this equation – again derived in a similar manner to the Fibonacci sequence – is 1, 12, 108, 864, (i.e. A053469 in the OEIS).

Therefore the infinite sum of reciprocals of this sequence

= 1 + 1/12 + 1/108 + 1/864 + …  = 6{log (6 – 1)/5} = 6 log 6/5 = 1.0939…


I also provided in that entry a general formula to cover all cases where (x + k)2 = 0, with k = 2, 3, 4, …

i.e. k{log (k + 1)/k}

In this case the same sequence of numbers is obtained as for (x – k)2 = 0, except that they alternate as between positive and negative values.

So the corresponding infinite sum of associated reciprocals

= 1 – 1/12 + 1/108 – 1/864 + …   = 6{log (6 + 1)/6} = 6 log 7/6 = .9249…

So interestingly the sum of the alternating sum of reciprocals can be directly obtained from the (positive) sum of reciprocals by increasing both numerator and denominator of the log expression in the latter formula by 1.


We can however proceed to obtain further log expressions for the infinite sums of reciprocals associated with the unique numbers associated with the polynomial equations of the form (x – k)3 = 0, where k = 1, 2, 3, 4, …    

For example, when k = 2, we have (x – 2)3 = x3 – 6x2  + 12x – 8 = 0.

And the unique sequence of numbers associated with this equation is 1, 8, 40, 160, 560, … (A001789 in OEIS).

So the corresponding sum of reciprocals of this sequence is given as

1 + 1/8 + 1/40 + 1/160 + 1/560 + …  = 4(1 – log 2).

Once again a general formula can be given to cover all such cases,

i.e. 2k{1 – (k – 1)[log k/(k – 1)]}

So for example when k = 3, we have (x – 3)3 = x3 – 9x2 + 27x – 27 = 0.

And the unique sequence of digits associated with this equation is 1, 9, 54, 270, 1215, …, (A027472  in OEIS).

Therefore the corresponding infinite sum of reciprocals

= 1 + 1/9 + 1/54 + 1/270 + 1/1215 + …

= 6{1 – [(3 – 1)log 3/(3 – 1)]} = 6{1 – 2 log (3/2)} = 1.1344 …

And as before, a related general formula can be given to cover the infinite sums of reciprocals of the unique number sequences associated with (x + k)3 = 0,

i.e. 2k{(k + 1)[log(k + 1/k)] – 1}.

So once again this provides the alternating case to the corresponding positive sum of reciprocals

Thus again for k = 3, the infinite sum of reciprocals of the unique number sequence associated with (x + 3)3

 = 1 – 1/9 + 1/54 – 1/270 + 1/1215 – …

Therefore from the formula the required sum = 6{[(3 + 1)log 3 + 1/(3)] – 1}

= 6{4 log (4/3) – 1} = .9043 …

A special case with respect to (x – k)3 = 0, occurs where k = 1 i.e. (x – 1)3
= x3 – 3x2 + 3x – 1 = 0.

And the unique sequence of numbers associated with this sequence is the set of triangular nos. i.e. 1, 3, 6, 10, 15, …

So the infinite sum of reciprocals of this sequence, i.e.

1 + 1/3 + 1/6 + 1/10 + 1/15 + … = 2{1 – (1 – 1)[log 1/(1 – 1)]} = 2{1 – 0[log (1/0)]}

= 2.

And the alternating sum of reciprocals, corresponding to the unique sequence of numbers associated with (x + 1)3 = 0, i.e.

1 – 1/3 + 1/6 – 1/10 + 1/15 – …


= 2{(1 + 1)[log(1 + 1/1)] – 1} = 2(2 log 2 – 1) = .772588 …

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