0

I was working with summations and I stumbled upon a sum that I could not reconcile: $$\sum_{i=0}^{\log_2(7)} 2^i$$

Using the formula $$\sum_{i = 0}^{n} x^i = \frac{x^{n+1}-1}{x-1}$$ found in a university textbook, with $x = 2$ and $n = \log_2(7),$ I got an incorrect answer: $$\frac{2^{\log_2(7)+1} - 1}{2-1} = 2^{\log_2(7)+1}-1 = (2^{\log_2(7)}*2)-1 = (7^{\log_2(2)}*2)-1 = 13.$$

I note that the above formula works correctly with natural numbers.

Blue
  • 75,673
Linnea117
  • 1
  • 2
    $\log_2(7) \notin \mathbb N$ so you cannot use that formula. (And it far from clear what that sum even means) – Henrik supports the community Jan 05 '22 at 10:58
  • $\log_2(7)\approx 2.81$. If you do $2^0+2^1+2^2$ you would get $7$ though this is really $\sum\limits_{i=0}^{\log_2(4)} 2^i = 4\times 2-1$ – Henry Jan 05 '22 at 11:23
  • 2
    Perhaps, the sum is over integers $i$ with $0 \leq i \leq log_2(7)$ in which case to have to repalce $log_2(7)$ by $[ log_2(7)]$ in your formula for the geometric sum. – Kavi Rama Murthy Jan 05 '22 at 11:24
  • @Linnea117 By what metric are you determining this is incorrect? It sounds as though you are trying to generalize the sigma notation form of summation to arbitrary real indices, in which case, you need a definition motivated by this type of generalization approach. But then, in that case, I fail to see how the answer you obtained could be incorrect. By what metric did you determine that $$\sum_{i=0}^{\log_2(7)}2^i\neq13?$$ – Angel Jan 05 '22 at 13:00

2 Answers2

2

$\displaystyle\sum_{i=0}^{\log_2(7)} 2^i$ is not mathematically correct.

When using the sigma notation, the indices are the elements of a finite set; not only natural numbers are allowed; you can sum on any finite set you want.

However, when we write $\displaystyle\sum_{i=0}^{p},$ we mean that $i\in\{0,1,2,\ldots,p\},$ where $p$ is a natural number.

ryang
  • 38,879
  • 14
  • 81
  • 179
Oussama Akar
  • 339
  • In order to define summation based on indices over a set, it not only has to be finite, it also needs a ordering (I don't remember the names of the relevant kind of ordering). Normally when summing over finite sets $S$, we write $\sum_{i\in S}$ (if addition in the relevant group is commutative) – Henrik supports the community Jan 05 '22 at 13:52
  • @Henriksupportsthecommunity No, that is incorrect. There is no requirement for the index set to have any ordering. What is required, for the sum to not be merely formal, is for the range of the function being summed to be a Hausdorff Abelian topological group, if not a finite Abelian group. – Angel Jan 05 '22 at 14:15
  • @Henriksupportsthecommunity I suggets you take a look at this: https://en.wikipedia.org/wiki/Series_(mathematics) and https://en.wikipedia.org/wiki/Index_set. – Angel Jan 05 '22 at 14:21
  • @Angel: I didn't say anything about the range (except for the comment that basically agrees with you). What I did say is that without an ordering on the (finite) set ${a, b, c}$, then $\sum_{i=a}^c f(i)$ doesn't make sense because you don't know if $f(b)$ is included, and the answer claims that it is as the indices are elements of a finite set. – Henrik supports the community Jan 05 '22 at 21:13
  • 1
    @Henriksupportsthecommunity I know you did not say anything about the range, I was imposing my own caveat. As for the rest of your reply, I think there is a misunderstanding here, regarding what the phrase "summation based on indeces over a set." I, and I assume the answer by Oussama as well, interpret the phrase to refer to a sum of the form $$\sum_{i\in{I}}f(i)$$ and not $$\sum_{m=a}^cf(m).$$ The latter is a form that mathematicians exclusively use when the indeces are integers. As far as I know, there is no generalization of that notation to any other kind of structure. – Angel Jan 06 '22 at 13:15
0

Suppose we have a function $f:\mathbb{N}\rightarrow\mathbb{M},$ where $\mathbb{M}$ is any commutative monoid with respect to $+,$: it could be the natural numbers, the rational numbers, the real numbers, the complex numbers, a ring of matrices, a vector space, etc. The way sigma notation works is that we say $$\sum_{m=0}^{n+1}f(m)=f(n+1)+\sum_{m=0}^nf(m).$$ As others have commented, $\log_2(7)\notin\mathbb{N},$ so as is, the symbol $$\sum_{m=0}^{\log_2(7)}2^m$$ is meaningless. However, there is the possibility of extending the meaning of the symbol. One thing you will realize is that if $\mathbb{M}^\mathbb{N}=\{\text{functions}\, f:\mathbb{N}\rightarrow\mathbb{M}\},$ then there exists a linear operator $S:\mathbb{M}^\mathbb{N}\rightarrow\mathbb{M}^\mathbb{N}$ such that $S[f](0)=f(0),$ $S[f](n+1)=S[f](n)+f(n+1).$ Now we have a recursion for $S[f]$ that does not rely on using sigma notation. We can find the extension of $f$ to $\mathbb{R},$ $*f:\mathbb{R}\rightarrow\mathbb{M}.$ by finding a solution to the equations $g_{*f}(x)=f(0),$ $g_{*f}(x+1)=*f(x+1)+g_{*f}(x).$

The issue is that, depending on what $\mathbb{M}$ is, there may be multiple solutions to the equation. For example, this is the case if $\mathbb{M}=\mathbb{R}.$ We have that if $f(n)=2^n,$ then the equation $g(x+1)=2^{x+1}+g(x)$ has multiple solutions satisfying $g(0)=1.$ In fact, there are uncountably many solutions (specifically, $2^{2^{\aleph_0}}$). One solution is given by $g(x)=2^{x+1}-1,$ but another solution is given by $g(x)=\cos(2\pi{x})(2^{x+1}-1).$

To uniquely determine a solution, $g$ needs to be uniquely specified in $[0,1),$ and with the condition $g(0)=1,$ you have yet to determine what $g$ is in $(0,1).$ One very natural requirement that will be helpful is if you require $g$ to be real-analytic, and also convex. This uniquely determines $g$ on $(0,1),$ and thus, uniquely determines $g$ everywhere. With this, it turns out that $g(x)=2^{x+1}-1$ is the unique solution. What this implies now is that $g[\log_2(7)]=2^{\log_2(7)}-1=13.$

Now, you said that $g[\log_2(7)]=2^{\log_2(7)}-1=13$ is the wrong answer. I have no idea of how you went about determining this. There are infinitely many solutions, without the analytic convexity requirement, that give that answer, and infinitely many that do not. So I do not know which of the extensions you are searching for. But if you are going with the most natural extension of them all, then $g[\log_2(7)]=2^{\log_2(7)}-1=13$ is definitely the correct answer.

Angel
  • 3,774
  • 12
  • 14