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$\begingroup$

$(h-2.5i)^{1/2}$

I'm trying to isolate i, is it possible?

Cheers!

EDIT: $i$ is NOT $\sqrt{-1}$, it's just a variable and $h$ is a constant.

EDIT2: It's in sigma notation like so:

$\sum_{i=1}^{14}(h-2.5i)^{1/2}$

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    I've submitted an answer that I hope helps. If you can clarify on any of the points and explain how I could further help you, I'd be more than glad to expand my answer. Also, the index of summation is the part that changes with the sum. (That definition kinda sucks.) For example, $\sum_{j=1}^{n}h_j$. $j$ is the index. It indicates what changes as the sum is expanded.2012-01-29

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Here is the most constructive answer I can give:

The expression $\sum_{i=1}^{y}(h-ci)^{n}$ may have a closed form. Expansion would show: $\sum_{i=1}^{y}(h-ci)^{n}=(h-c)^n+(h-2c)^n+\dots+(h-(y-1)c)^n+(h-yc)^n$ There is a problem here: The basic binomial theorem only applies to the case $(a+b)^n$ where $n \in \mathbb{Z}^{+}$. (I do not know the conditions on $a$ and $b$.) Two different things are astray here: There is a coefficient greater than $1$ in every binomial but the first and in our specific case, $n$ is not a nonnegative integer.

These two issues are what makes this so hard to figure out. I can't seem to find any binomial series that works for the form $(h-ac)^{n}$ where $a \geq 1$ and $0. Even Newton's generalized binomial series does not suffice, which really shows just how cumbersome this particular scenario is.

Looking for a non-general way of helping you is even more cumbersome upon inspection: $\sum_{i=1}^{14}(h-2.5i)^{\frac{1}{2}}=\sqrt{(h-2.5)}+\sqrt{(h-5)}+\dots$ Do you see how this does not admit any sort of collecting of the terms in a more convenient fashion? You simply cannot add these terms and find a closed form. The only possible (slight) simplification of this involves turning the decimal into a fraction: $\sum_{i=1}^{14}\sqrt{h-\frac{5}{2}i}=\sum_{i=1}^{14}\sqrt{\frac{2h-5i}{2}}=\sum_{i=1}^{14}\frac{\sqrt{2h-5i}}{\sqrt{2}}=\frac{1}{\sqrt{2}}\sum_{i=1}^{14}\sqrt{2h-5i}$ You could then make life a little easier by rationalizing the denominator: $\frac{1}{\sqrt{2}}\sum_{i=1}^{14}\sqrt{2h-5i}=\frac{1}{2}\sum_{i=1}^{14}\sqrt{4h-10i}$ The above avoids as many radicals as possible and converts the problem into an integer form.

In conclusion: There isn't a lot you can do to make your life easier. Sorry. :/

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    Thanks so much! Yeah, it's so much more elegant? I don't know, I just don't like bringing other stuff into it.2012-01-29
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I'm trying to isolate i

Let's say you have: $y=(h-2.5i)^{1/2}$ then

$y^2=h-2.5i$, hence:

$i= (h-y^2)/2.5$

Edit: You could use binomial theorem formula below but notice the condition associated. (source: PDF-Bionomial Theorem:

enter image description here

Edit 2: A more generalized form is:

enter image description here

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    You are right. I suggest you post the full problem in a new thread with the diagram and all the relevant details without links, maybe someone would be able to help.2012-01-29
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Whenever $|h / (2.5 i)| \le 1|$ you can use the binomial theorem, where: $ {1/2 \choose k} = \frac{(-1)^{k - 1}}{2^{2k - 1} \cdot k} \cdot \binom{2k - 2}{k - 1} $ Don't know if this helps much.