## Sunday, December 9, 2018

### Fun with the Infinite Series 1 + x + x^2 + x^3 + x^4 + x^5 + ...

Fun with the Infinite Series 1 + x + x^2 + x^3 + x^4 + x^5 + ...

The Series and Its Derivatives

Let F be the infinite series:

F = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ... = ∑ x^k from k = 0 to ∞.

Working with derivatives:

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First Derivative of F:  (F' = dF/dx)

F' = 1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5 + 7*x^6 + ...
= ∑ (k+1)*x^k from k = 0 to ∞

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Second Derivative of F:  (F'' = d^2F/dx^2)

F'' = 2 + 6*x + 12*x^2 + 20*x^3 + 30*x^4 + 42*x^5 + 56*x^6 + ....

Factor out a 2:
= 2 * (1 + 3*x + 6*x^2 + 10*x^3 + 15*x^4 + 21*x^5 + 28*x^6 + .... )

Note the sequence 1, 3, 6, 10, 15, 21, 28...  These are triangle numbers, denoted as T_n.

T_1 = 1
T_2 = 1 + 2 = 3
T_3 = 1 + 2 + 3 = 6
T_4 = 1 + 2 + 3 + 4 = 10
and so on.

Using summation notation,  T_n = ∑ k from k = 1 to n

Going back to the series:

F'' = 2 + 6*x + 12*x^2 + 20*x^3 + 30*x^4 + 42*x^5 + 56*x^6 + ....
= 2 * (1 + 3*x + 6*x^2 + 10*x^3 + 15*x^4 + 21*x^5 + 28*x^6 + .... )
= 2 * (∑ x^k * T_k+1 from k = 0 to ∞)

In nested summation notation:

= 2 * (∑ x^k * (∑ m from m = 1 to k+1) from k = 0 to ∞)

Addition with F, F', and F"

F + F' = 2 + 3*x + 4*x^2 + 5*x^3 + 6*x^4 + 7*x^5 + 8*x^6 + ...
= ∑ ((k + 2) * x^k from k = 0 to ∞)

----

F + F' + F'' = 4 + 9*x + 16*x^2 + 25*x^3 + 36*x^4 + 49*x^5 + 64*x^6 + ...
= ∑ ((k + 2)^2 * x^k from k = 0 to ∞)

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F' + F" = 3 + 8*x + 15*x^2 + 24*x^3 + 35*x^4 + 48*x^5 + 63*x^6 + ...

Note the sequence 3, 8, 15, 24, 35, 48, 63... where
3 = 4 - 1 = 2^2 -1
8 = 9 - 1 = 3^2 - 1
15 = 16 - 1 = 4^2 - 1
24 = 25 - 1 = 5^2 - 1
35 = 36 - 1 = 6^2 - 1
48 = 49 - 1 = 7^2 - 1
63 = 64 - 1 = 8^2 - 1
and so on...

This can be summarized as ∑( (k + 2)^2 - 1 from k = 0 to ∞)

Hence:
F' + F" = 3 + 8*x + 15*x^2 + 24*x^3 + 35*x^4 + 48*x^5 + 63*x^6 + ...
= ∑  ((k + 2)^2 - 1) * x^k from k = 0 to ∞)

Multiplying F and F' by x and x^2

F = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ... = ∑ x^k from k = 0 to ∞.
x * F = x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + ... = ∑ x^(k+1) from k = 0 to ∞.
x^2 * F =  x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + ... = ∑ x^(k+2) from k = 0 to ∞.

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F + x * F = 1 + 2*x + 2*x^2 + 2*x^3 + 2*x^4 + 2*x^5 + 2*x^6 + ...
= 2 - 1 + 2*x + 2*x^2 + 2*x^3 + 2*x^4 + 2*x^5 + 2*x^6 + ...
= 2 + 2*x + 2*x^2 + 2*x^3 + 2*x^4 + 2*x^5 + 2*x^6 + ... - 1
= 2 * F - 1

This is one way to dervie the formula for the Infinite Geometric Series (for |x| < 1), to solve for F:

F + x * F  = 2 * F - 1
F + x * F - 2 * F = -1
F * (1 + x - 2) = -1
F * (x - 1) = -1
F = -1 / (x - 1)
F = 1 / (1 - x)   (keep this mind, this is true only when |x| < 1)

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F - x * F = (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ... ) - (x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + ... )
= F * (1 - x)

For |x| < 1,

F * (1 - x) = 1 / (1 - x) * (1 - x) = 1

In general:
F - x * F = (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + ... ) - (x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + ... )
= 1 + (x - x) + (x^2 - x^2) + (x^3 - x^3) + (x^4 - x^4) + (x^5 - x^5) + (x^6 - x^6) + ...
= 1

F - x * F = 1

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F' = 1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5 + 7*x^6 + ...
= ∑ ((k + 1) * x^k from k = 0 to ∞)

x * F' = x + 2*x^2 + 3*x^3 + 4*x^4 + 5*x^5 + 6*x^6 + 7*x^7 + ...
= ∑ ((k + 1) * x^(k + 1) from k = 0 to ∞)

x^2 * F' = x^2 + 2*x^3+ 3*x^4 + 4*x^5 + 5*x^6 + 6*x^7 + 7*x^8 + ...
= ∑ ((k + 1) * x^(k + 1) from k = 0 to ∞)

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F' + x * F' = 1 + 3*x + 5*x^2 + 7*x^3 + 9*x^4  + 11*x^5 + 13*x^6 + ...

The sequence of 1, 3, 5, 7, 9, 11, 13, ... is the sequence of odd numbers which can be summarized as:

∑(2 * k + 1 from k = 0 to ∞)

Then:

F' + x * F'  =  F' * (1 + x) = ∑( (2*k + 1) * x^k from k = 0 to ∞)

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F' + x * F' + x^2 * F' = 1 + 3*x + 6*x^2 + 9*x^3 + 12*x^4 + 15*x^5 + 18*x^6 + ...
= 1 + ( 3*x + 6*x^2 + 9*x^3 + 12*x^4 + 15*x^5 + 18*x^6 + ... )
= 1 + ∑(3 * k * x^k from k = 0 to ∞)

Eddie

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