0
$\begingroup$

I'm trying to make the sum stop before the summation has a negative exponent.

For example, I would want the sum to stop at $2^0$ in

$2^3+2^2+2^1+2^0+2^{-1}$

The sum I'm dealing with is $$\sum P\left(\frac{a}{b}\right)^{\frac{1}{c}(d-fn)}$$ $$\text{Where}\; \{f,a,b,c,d,P\}\in \mathbb{N}$$

The first version I thought of was $$\sum\limits_{n=0}^{?}\frac{1}{2}P\left(\frac{a}{b}\right)^{\frac{1}{c}(d-fn)}(1+\text{sgn}(d-fn))$$ I don't know a good stopping point for this. Technically I could sum to infinity, but that seems unnecessary when the sum could stop after 3 terms. It also fails if $d=fn$ for the final sum as the last sum is divided by 2.

My next version was $$\sum\limits_{n=0}^{\alpha}P\left(\frac{a}{b}\right)^{\frac{1}{c}(d-fn)}$$ $$\alpha = \text{floor}\left(\frac{d}{f}\right)$$ But I don't know if placing $\alpha$ as the upper limit and stating $\alpha = \text{floor}\left(\frac{d}{f}\right)$ is acceptable notation.

Any input on this would be helpful.

$\scriptsize\text{(Hints for a possible closed form would also be appreciated)}$

$\endgroup$
  • $\begingroup$ If it's the correct terminating value, there's nothing wrong with $$\sum_{n=0}^{\lfloor d/f\rfloor}\cdots$$ $\endgroup$ – MPW Dec 30 '15 at 18:02
0
$\begingroup$

The most concise way to express it would be by a summation sign, with the subscripts under it being $n\in\Bbb N$ and $d−fn>0$.

However, there is indeed a closed form. Let $\alpha=d-fm$ be the smallest positive value of $d-fn$.

Then what you are looking for is, $\sum_{n=0}^m P(\frac{a}{b})^{\frac{1}{c}(d-fn)}$, which is the same thing as $$\sum_{n=0}^m P(\frac{a}{b})^{\frac{1}{c}(\alpha+fn)}=\left(P(\frac{a}{b})^{\frac{\alpha}{c}}\right)\sum_{n=0}^m\left(P(\frac{a}{b})^{\frac{f}{c}}\right)^n$$

Then, you would use the geometric series expansion to get the closed form for the above. Namely, if $\beta=P(\frac{a}{b})^{\frac{f}{c}}$, then the above would equal $\left(P(\frac{a}{b})^{\frac{\alpha}{c}}\right)*\frac{\beta^{m+1}-1}{\beta-1}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.