Infinite sum compression using the inverse of a formal power series
Infinite sum compression using the inverse of [formal power series
inverse element
$ A = \sum_{n=0}^\infty x^n
$ B = 1-x
At this time [$ A \times B = 1
$ A = 1 + x + x^2 + \ldots (1)
$ xA = x + x^2 + \ldots (2)
$ (1-x)A = 1 (1) - (2)
Common techniques for calculating sums of sequences of equal ratios
Generally when F is a formal power series without a constant term such that $ [x^0]F = 0 ,
$ (1-F) \times (\sum_{n=0}^\infty F^n) = 1
The proof requires defining a phase in the formal power ring and defining convergence
Count up the number of ways to express N as a sum of positive integers. However, distinguish between different orders of sums.
$ F = \sum_{n=1}^\infty x^n,[$ G = \sum_{n=0}^\infty F^n
$ [x^n]G is the answer
Replace the infinite sum of G with the inverse original
$ G = \sum_{n=0}^\infty F^n = \frac{1}{1 - F}
The infinite sum of F is also replaced by the inverse element
$ F = \sum_{n=1}^\infty x^n = \sum_{n=0}^\infty x^{n+1} = x\sum_{n=0}^\infty x^n = \frac{x}{1-x}
Substitute F for G
$ G = \frac{1}{1 - F} = \frac{1}{1-\frac{x}{1-x}} = (1-x) \frac{1}{1-2x}
where the $ 2x part is a formal power series, so we can return to infinite sums, which we will call J
$ J = \frac{1}{1-2x} = \sum_{n=0}^\infty (2x)^n
$ [x^n]J = 2^n
From here $ [x^n] G = [x^n](1-x)J = [x^n]J - [x^{n-1}]J = 2^n - 2^{n-1} = 2^{n-1}
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