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let me start by saying that my formatting may be way off, but it's the best I can do, and has little to do with the question, and I will make sure I am as clear as humanly possible, including showing all my work

So I was playing with squares and stuff, I found that I could calculate out every successive perfect square with: $$\sum_{n=0}^{x-1} (2n+1) = x^2$$ where $$ 0 \le n\le(x-1)|_{||}| 1\le(n+1)\le x$$

For example: $5^2=\sum_{n=0}^4(2n+1)=1+3+5+7+9=25$

and then found through trial and error that I could also calculate out successive cubes with: $$ x^3=\sum_{n=0}^{x-1}(3n^2+3n+1) $$ with the same rules for n.

For example: $5^3=\sum_{n=0}^4 (3n^2+3n+1)=1+7+19+37+61=125$

Then I discovered that the cubic formula can be factored so that: $$x^3=\biggr(\sum_{n=0}^{x-1}\bigr((2n+1)*(n+1)\big)\biggr) $$

I noticed each even power can be represented by: $x^{m_{even}}=\biggr(\sum_{n=0}^{x-1}(2n+1)\biggr)^{\frac m 2}$

While odd powers can be represented with: $x^{m_{odd}}=\biggr(\sum_{n=0}^{x-1}(2n+1)\biggr)^{\frac{m-1} 2}*(n+1)$

where $(n+1)=x$ , but only the final term is replaced with $x$.

Which eventually lead me to an interesting relation (at least to me): $$x^m=\bigg(\sum_{n=0}^{x-1}(2n+1)\bigg)^{\frac {m-3} 2}*\bigg(\sum_{n=0}^{x-1}(3n^2+3n+1)\bigg)$$

I get that I'm essentially just multiplying squares and cubes together and essentially saying that: $$x^m=\big((x^2)^{\frac {m-3} 2}\big)*x^3$$ and all the math works out to give you: $$x^{m-3}=x^{m-3}$$ However, I want to know if this is something unique, or are there many other ways to represent $x^m$ than what I've found? Is there anything useful somebody could do with something like this, or is it only good for looking cool?

p.s. Sorry if what I tried to convey didn't make any sense, I'm not familiar with how I'm supposed to phrase things in math.

Some loony with a calculator
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1 Answers1

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You are implicitly using telescoping series with your sums giving the integer squares and cubes. For the first case, you have

$$\begin{equation}\begin{aligned} & \sum_{n=0}^{x-1}(2n+1) \\ & = \sum_{n=0}^{x-1}((n^2 + 2n+1) - n^2) \\ & = \sum_{n=0}^{x-1}((n+1)^2 - n^2) \\ & = (1^2 - 0^2) + (2^2 - 1^2) + \ldots + ((x-1)^2 - (x-2)^2) + (x^2 - (x-1)^2) \\ & = x^2 - 0^2 \\ & = x^2 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

The final result occurs because all of the terms cancel each other out, except for the $x^2$ and $0$ terms. Similarly, you also have

$$\begin{equation}\begin{aligned} & \sum_{n=0}^{x-1}(3n^2+3n+1) \\ & = \sum_{n=0}^{x-1}((n^3 + 3n^2+3n+1) - n^3) \\ & = \sum_{n=0}^{x-1}((n+1)^3 - n^3) \\ & = (1^3 - 0^3) + (2^3 - 1^3) + \ldots + ((x-1)^3 - (x-2)^3) + (x^3 - (x-1)^3) \\ & = x^3 - 0^3 \\ & = x^3 \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Regarding your statement of

$$x^m=\bigg(\sum_{n=0}^{x-1}(2n+1)\bigg)^{\frac {m-3} 2}*\bigg(\sum_{n=0}^{x-1}(3n^2+3n+1)\bigg) \tag{3}\label{eq3A}$$

It's true, as you determined by substituting and simplifying. As for any particular use for this, I don't know of any offhand. Also, regarding other ways to express $x^m$, there are many possibilities, with the particular choice often depending on what you're checking on or trying to determine.

John Omielan
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  • weird question, but would there be a way to make this work with non-integers and/or arbitrary arguments for $m$ ? – Some loony with a calculator Apr 20 '20 at 02:36
  • @Someloonywithacalculator What you're asking is not, in my opinion, a "weird" question. For positive integers $x$ (note: the following doesn't necessarily apply if you use negative $x$), your formula, i.e., the one I've shown in ($3$), will work for all real $m$. This is because the second factor is always $x^3$, and $(x^p)^q = x^{pq}$ for all real $p$ & $q$, so the first factor is $(x^2)^{\frac{m-3}{2}} = x^{m-3}$. Finally, since $(x^p)(x^q) = x^{p+q}$ for all real $p$ & $q$, the product of the $2$ would be $(x^{m-3})(x^3) = x^{(m-3)+3} = x^{m}$. – John Omielan Apr 20 '20 at 02:42
  • Would that also mean that if either $p$ or $q$ (not both) were imaginary arguments, along with $x$ that this might work as well? It seems to work with negative values as if $x$ is negative, the second term will always be, and if m is even, the first term will be as well. So as long as I take an imaginary (not complex) to another imaginary, I should have a real number right? Or is that a bit presumptive? – Some loony with a calculator Apr 20 '20 at 03:22
  • @Someloonywithacalculator If I understand what you're stating correctly then, yes, you can certainly extend it in a way basically as you describe. However, note that when you start dealing with complex numbers, the situation can become more "complex" (pun intended). For example, certain rules with exponents don't always hold, logarithms are no longer single-valued, etc. You need to be careful with what your assumptions are and what specific manipulations you are trying to do. However, discussing this in the comments is not an appropriate place to go into a lot of details. – John Omielan Apr 20 '20 at 03:27
  • @Someloonywithacalculator FYI, very coincidentally, a question regarding issues with powers of complex numbers & their manipulations was just asked at Problem using existing index laws with complex numbers?. – John Omielan Apr 20 '20 at 03:37
  • @Someloonywithacalculator The formula you stated of $x^m=(x^2)^{\frac {m-3} 2}*(x^3)$ implicitly uses $(x^2)^{\frac{m-3}{2}} = x^{\frac{2(m-3)}{2}} = x^{m-3}$. However, the accepted answer to the other question states $x^{\frac{a}{b}} = (x^a)^{\frac{1}{b}}$ doesn't work in general. My comment above perhaps was not clear, but you need to be careful when dealing with situations like this when dealing with complex numbers. You yourself seem to acknowledge this when you state $(-1^{\frac 1 2})^3)=-i$ & $(-1^3)^{\frac 1 2}=i$. Just please be very careful when dealing with powers & complex #'s. – John Omielan Apr 20 '20 at 05:07