# Is there any way to expand this using the binomial theorem?

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

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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This is an ill-posed question. You need more specific context to the expression in order for it to make any sense. That is shown by the answer below, where Emmad demonstrates that it becomes very manageable when it is part of an equation. Also (while this may just be me), you can't isolate an index of a summation. . . There is a significant difference between a variable used in an expression and an index used in a summation. Sorry to sound so negative, but I mean all of this constructively. – 000 Jan 29 '12 at 3:55
I'm a little confused. What's an index of summation? It's not that I want to isolate the i, more just move the h (because it's constant) to the left of the sigma? – user1045696 Jan 29 '12 at 4:12
Actually, now that I think about it, I don't really need to remove h, I just need to be able to sum the series or to find where it converges. – user1045696 Jan 29 '12 at 4:15
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. – 000 Jan 29 '12 at 4:55

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 $0<n<1$. 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 a tonne. It's a shame I can't simplify it any further, I much prefer purely algebraic generalizations, but if it's not possibly, I suppose I could write a program to calculate widths, etc. Thanks again. – user1045696 Jan 29 '12 at 5:00
I, too, prefer purely algebraic generalizations! I felt the same disappointment as you do. I am looking at the question you linked to and I will see if I could possibly provide any help there. There may be another way of approaching this that circumvents the summation we're looking at. – 000 Jan 29 '12 at 5:01
Thanks so much! Yeah, it's so much more elegant? I don't know, I just don't like bringing other stuff into it. – user1045696 Jan 29 '12 at 5:04

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:

Edit 2: A more generalized form is:

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I should have mentioned, I'm using it with sigma notation, so it's: $\sum_{i=1}^{14}(h-2.5i)^{1/2}$ – user1045696 Jan 29 '12 at 3:44
Hm, interesting. I'll take a look, cheers! – user1045696 Jan 29 '12 at 3:46
It may just be my ignorance, but I don't see the edit as applicable to this problem. The expression on the left hand cannot be changed such that it is equivalent to the expression $(h-2.4i)^{\frac{1}{2}}$. You could make the substitution $x=-1+h-2.5i$, but I do not know if that is valid and I speculate that. EDIT: I speculate it because I think $x$ is confined to the reals and cannot be defined as an expression. I'm not sure if that is the case, however. – 000 Jan 29 '12 at 3:57
I made the edit before I read your first comment about the sum. I wanted to answer your original question about using the binomial expansion. I guess that reformatting the expression so that we could apply the binomial theorem won't make it an easier sum. Sorry that I can't find a good answer! – NoChance Jan 29 '12 at 4:04
It's all good, thanks for trying! I'll keep digging. I'm pretty sure I'll need a binomial series, and then I'll have to see what it converges to. It's all a bit over my head though, we covered binomials in class in a day, and never did fractional, etc exponents. – user1045696 Jan 29 '12 at 4:08

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.

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