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I have a simple series used to set trading quantities:

f(i) = a*b^i

We've been using it to make quantity steps, for example:

20*4^0 = 20
20*4^1 = 80
20*4^2 = 320
etc

a useful value for us is the sum of all terms for a specific exponent.

for example

f(0) = 20*4^0 = 20
f(1) = 20*4^0 + 20*4^1 = 100
f(2) = 20*4^0 + 20*4^1 + 20*4^2 = 420
etc

The exponent steps have always been integers, but I need now to make this more fine grained, so for example 1/4 intervals:

20*4^0.00 = 20
20*4^0.25 = 28
20*4^0.50 = 40
20*4^0.75 = 56
20*4^1.00 = 80
etc

however, I need to the sum of all the terms for a specific INTEGER exponent to be the same as above. for example, I had:

f(1) = 20*4^0 + 20*4^1 = 100

but now I have:

f(1) = 20*4^0.00 + 20*4^0.25 + 20*4^0.50 + 20*4^0.75 + 20*4^1.00 = 224

how can I rewrite the series so that for a given integer exponent and a given step (1/4 in my example), the sums match across all exponent values?

Thomas
  • 127

1 Answers1

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What you want is not possible.

For the first series, we have:

$$ S_1(n) = \sum_{i=0}^n a b^i = a \frac{b^{n+1}-1}{b-1} $$

The second series using powers $\frac{i}{k}$, and summing up to a term with integer power $n$, is:

$$ S_2(n) = \sum_{i=0}^{nk} c d^{i/k} = \sum_{i=0}^{nk} c (d^{1/k})^i = c \frac{(d^{1/k})^{nk+1}-1}{d^{1/k}-1} = c \frac{d^{n+1/k}-1}{d^{1/k}-1} $$

To eliminate $a$ and $c$ for now, look at the ratio of consecutive terms.

$$ \begin{align*} \frac{S_1(n+1)}{S_1(n)} &= \frac{b^{n+2}-1}{b^{n+1}-1} = b + \frac{b-1}{b^{n+1}-1} \\ \frac{S_2(n+1)}{S_2(n)} &= \frac{d^{n+1+1/k}-1}{d^{n+1/k}-1} = d + \frac{d-1}{d^{n+1/k}-1} \end{align*} $$

Since we want $S_1(n)=S_2(n)$ for every $n$, the limits of these ratios must also be equal.

$$ b = \lim_{n \to \infty} \frac{S_1(n+1)}{S_1(n)} = \lim_{n \to \infty} \frac{S_2(n+1)}{S_2(n)} = d $$

But then setting the ratios in terms of $n$ equal implies that for every $n$, $b^{n+1} = b^{n+1/k}$, which is not possible if $k>1$ and $b>1$.

aschepler
  • 9,449
  • That said, the right hand side of your first equation provides a natural algebraic expression that generalises to $n\in\mathbb R$ (if that could be helpful to OP) – Calvin Khor May 19 '20 at 12:05
  • Thanks for displaying the problem clearly. I understand the issue now! – Thomas May 19 '20 at 17:11