Math Operations And Reasoning: Fill & Download for Free

Because half-open intervals [math]a \le x < b[/math] are easier to compose and reason with. You never have to think about fencepost problems when writing down the nice propertieslen(range(m)) == mlen(range(a, b)) == b - arange(a, a) == []range(a, b) + range(b, c) == range(a, c)n % m in range(m)range(a, b) == [x for c in range(a, b, k) for x in range(c, c + k)](for any k dividing b - a)randrange(m) in range(k) with probability k/mwhich, in an alternate universe where Python had instead chosen to promote closed intervals [math]a \le x \le b[/math], would all have to be carefully sprinkled with + 1s and - 1s.The same reasoning applies to the slice operator, and I find that one of the most common string processing operations—removing a known prefix or suffix of a string—crucially depends on the nice properties of half-open intervals:if str.startswith('prefix')  str = str[len('prefix'):] if str.endswith('suffix')  str = str[:-len('suffix')] It is an unfortunate quirk of human language that half-open intervals are somewhat more cumbersome to describe in words, which perhaps makes them feel less intuitive to beginners (like array indices that start at 0). But a moderate amount of experience will convince you that they are far more pleasant to program with.See also Indices point between elements, for a nice explanation of why this is intuitive.

Ok, now that we’ve gotten that out of the way…There are tons of questions on Quora about this matter, and by now they have mostly been merged into these two:What's the intuition behind the equation [math]1+2+3+\cdots = -\tfrac{1}{12}[/math]?Ramanujan proved that a sum of all positive integers is -1/12. How is it possible?I couldn’t possibly read all the hundreds of answers to those questions, and I don’t expect I can contribute anything that wasn’t already said in one of those answers, several of which I’ve upvoted. so I’ll use this personal question to offer my personal perspective on the question and the cultural mini-phenomenon around it.So first off, in case it wasn’t clear earlier: no, the “sum of all the positive integers” isn’t [math]-1/12[/math], and Ramanujan didn’t prove that it is.Next, “believing” is the wrong word here. I don’t “believe” in mathematical results I understand, I know them. I understand quite well the concept of infinite series, and I don’t need to believe or disbelieve anything to do with them.The (in)famous Numberphile video about this, which I will refrain from linking to, has generated great publicity of a very wrong kind for mathematics. It confuses more than it teaches, in my mind, and it entrenches common beliefs that math is arcane, weird and meaningless and mathematicians are unreasonable people who revel in confusing honest folks.Mathematical truths are truths. They are the most universally true, clear, unambiguous and unassailable facts available to the human mind.To communicate a mathematical truth or fact we use language, symbols, notation. This works the way all communication works: if you understand the symbols I’m using the way I intended, then you know what I’m saying. If you don’t know what they mean, or if interpret them in some other way, then you’re not accessing the truth I’m trying to convey. You’re seeing something else, which may be true, or false, or neither. Either way it’s not what I had meant.This isn’t deep, mysterious or interesting, and it has nothing to do with mathematics. It’s communication. There are clear, useful and meaningful mathematical truths associated with the misleading notation [math]1+2+3+4+\ldots\overset{??}{=}-1/12[/math]. There really are. But when presented in this confusing way, they – surprise! – confuse people. This isn’t a failure of math, or of your intuition. Your intuition is fine. This is exactly what the Captain said: What we have here is failure to communicate.So, what are the facts?Number one: algebraic operations don’t apply to infinitely many objects.It is reasonable to assume that we all agree and understand what it means to add two rational numbers, so when I write [math]\frac{2}{3}+\frac{3}{4}=\frac{17}{12}[/math], we all understand the same simple fact. The rational number represented by the fraction [math]\frac{2}{3}[/math], when added to the number represented by [math]\frac{3}{4}[/math], yields the number [math]\frac{17}{12}[/math], which is also the number [math]1\frac{5}{12}[/math].This mathematical operation, [math]+[/math], is at this point applicable only to two numbers. Asking “what is the sum of the set [math]\{1,2,3\}[/math]" or “what is [math]1+2+3[/math]" is a category error. It’s not defined and it doesn’t mean anything.However, thankfully, this [math]+[/math] operation has two neat features: it is commutative, and it is associative. This just means that [math]a+b[/math] is the same thing as [math]b+a[/math], whatever numbers [math]a[/math] and [math]b[/math] are. Also, [math](a+b)+c[/math] is the same thing as [math]a+(b+c)[/math], where the parentheses clarify the order in which are applying the addition operation.Thanks to those wonderful properties, we can define the sum of three numbers [math]a,b,c[/math] to be [math](a+b)+c[/math], and we allow the notation [math]a+b+c[/math] to denote it. We know it doesn’t matter if we’re adding [math]1,2,3[/math] or [math]3,2,1[/math], it’s the same thing. We can similarly define the sum of four, five, a hundred or a trillion numbers.Great.But.Do we, right now, know what it means to add an infinite collection of numbers?WE DO NOT.There’s no such thing at this point. We have absolutely no idea what to do with infinitely many numbers.Algebraic operations don’t apply to infinitely many objects.Number two: there is a standard notion of “sum” of an infinite series, but it doesn’t always exist.For many good and valid reasons we’ve decided to introduce new machinery, various new functions, that do accept infinitely many inputs and do, sometimes, yield a result. One of those machineries is usually referred to as a “sum”, or more precisely “a sum of an infinite series”. The other such machineries are rarely if ever called that.Incredibly, those methods turned out to be applicable to real-world problems in physics and technology, so they’re not just an abstract game with numbers. This is a simple empirical observation: math doesn’t know about the real world. We, the people, are just happy (and somewhat shocked) that the real world behaves so nicely and complies with our abstract mathematical definitions.The most standard way by far to define the “sum” of infinitely many numbers is this:[math]\displaystyle \cal{S}(a_1,a_2,a_3,\ldots) = \lim_{n \to \infty} (a_1+a_2+\ldots+a_n)[/math]This means the following: to determine [math]\cal{S}[/math] for an infinite sequence [math]a_1,a_2,a_3,\ldots[/math], we first calculate the partial sums[math]S_1 = a_1[/math][math]S_2 = a_1+a_2[/math][math]S_3 = a_1+a_2+a_3[/math][math]S_4 = a_1+a_2+a_3+a_4[/math]and so on. We now have a new infinite sequence of numbers, [math]S_1, S_2, S_3, S_4,\ldots[/math], and an infinite sequence of numbers sometimes converges to a limiting value [math]S[/math]. If it does, and only when it does, we declare [math]\cal{S}[/math] of our original sequence [math]a_1,a_2,a_3,\ldots[/math] to be this limiting value [math]S[/math].(You may have not seen the idea of convergence before, and that’s ok. The important thing to understand is that some sequences of numbers get closer and closer to some limiting value, which is said to be their “limit”, and some other sequences don’t. If it so happens that [math]S_1,S_2,S_3,\ldots[/math] is kind enough to get closer and closer to some number [math]S[/math], then we say that [math]S[/math] is the [math]\cal{S}[/math]-value of the infinite sequence [math]a_1,a_2,a_3,\ldots[/math].)For example:[math]{\cal{S}}(1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\ldots)=2[/math][math]{\cal{S}}(0.9,0.09,0.009,0.0009,\ldots)=1[/math][math]{\cal{S}}(1,-1,1,-1,1,-1,\ldots)[/math] doesn’t exist.[math]{\cal{S}}(1,2,3,4,\ldots)[/math] doesn’t exist. Repeat, doesn’t exist. The sequence of partial sums is [math]S_1=1[/math], [math]S_2=3[/math], [math]S_3=6[/math], [math]S_4=10[/math], and so on, with [math]S_n=\frac{n(n+1)}{2}[/math]. This sequence does not converge.You may notice that I’m calling this thing “[math]\cal{S}[/math]”, not “the sum” or “the value of the infinite series”. It’s a new, shiny, thing, and we should treat it with respect.The common notations for this [math]\cal{S}[/math] thingie is[math]\displaystyle {\cal{S}}(a_1,a_2,a_3,\ldots) = \sum_{n=1}^\infty a_n = a_1+a_2+a_3+\ldots[/math]We are overusing the symbols “[math]\sum[/math]” and the [math]+[/math] symbol here. This is not the usual sum of numbers and it’s not the usual [math]+[/math] operation. The usual [math]+[/math] operation plays a part in the definition, as we’ve seen, but the definition is a lot more elaborate than merely “adding things up”. If this feels confusing, just ignore this notation and keep using [math]{\cal{S}}(a_1,a_2,a_3,\ldots)[/math]. It’s clearly reminding us that this is a new, special, sophisticated machine (or “operator” in math-speak) that takes in infinitely many values and sometimes returns a value.Number three: There are other ways of assigning a value to an infinite series. There’s no reason to call any of them “the sum” of the series.Another well-known way of doing this is called Cesàro summation. It works like this:[math]\displaystyle {\cal{C}}(a_1,a_2,a_3,\ldots) = \lim_{n \to \infty} \frac{1}{n}(S_1+S_2+\ldots+S_n)[/math]where the [math]S_i[/math] are the partial sums [math]S_i=a_1+a_2+\ldots+a_i[/math], as before.We now find:[math]{\cal{C}}(1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\ldots)=2[/math][math]{\cal{C}}(0.9,0.09,0.009,0.0009,\ldots)=1[/math][math]{\cal{C}}(1,-1,1,-1,1,-1,\ldots)=\frac{1}{2}[/math].[math]{\cal{C}}(1,2,3,4,\ldots)[/math] still doesn’t exist. Repeat, it still doesn’t exist.A few interesting things emerge: [math]\cal{C}[/math] seems to be the same as [math]\cal{S}[/math] whenever [math]\cal{S}[/math] exists. This is, indeed, always true. [math]\cal{C}[/math] extends the standard summation operator by mimicking it whenever possible, and filling in a bunch of new results on top.For example, [math]\cal{C}[/math] manages to assign a value of [math]\frac{1}{2}[/math] to the input sequence [math]1,-1,1,-1,1\ldots[/math], whose partial sums are [math]1,0,1,0,1,0,\ldots[/math]. This makes nice intuitive sense, doesn’t it? Some people choose to write[math]\displaystyle 1-1+1-1+1-1+\ldots = \frac{1}{2}[/math]but you can now see how misleading that is. Under the usual [math]\cal{S}[/math] operator, [math]1-1+1-1+1-1+\ldots[/math] wasn’t anything at all.Finally, [math]\cal{C}[/math] still doesn’t manage to assign any reasonable value to [math]1,2,3,4,\ldots[/math]. The “sum of all positive integers” is still “nothing” with this improved summation operator.There are many other “summability methods[1]” which attach values to infinite sequences. There’s Nørlund summation, Euler summation, Abel summation, Ramanujan summation (kind of), zeta-function regularization and more. Some of them agree with [math]\cal{S}[/math] whenever it exists; some don’t. Some enjoy other reasonable properties of “sum”, such as being linear, or being stable to shifts. No method has all nice properties together. And none of them is called, or deserves to be called, “the sum of the infinite series”. They are simply operators that receive infinitely many numbers as input, and produce an output. They can be useful for various purposes, but they are not, for crying out loud, sums.Ramanujan’s method, incidentally, isn’t even a summability method. To work it needs more than an infinite series of numbers: it needs a function of a real variable, so that its derivatives can be examined. It is a very cool and very neat operator, but it’s not an operator of infinite series. When applied to the function [math]f(x)=x[/math], it indeed yields [math]-\frac{1}{12}[/math].Zeta-function regularization is another summability method, which also yields [math]-\frac{1}{12}[/math] to the series [math]1,2,3,4,\ldots[/math]. It is linear, but not stable, so using the “+” notation for it is truly, deeply misleading. It means that [math]a+b+c+\ldots[/math] isn’t necessarily the same as [math]0+a+b+c+\ldots[/math]. Not quite what you expect from “sum”.But here is the thing: if someone, such as Numberphile, wanted to popularize zeta-function regularization, and they would call it something like [math]Z(1,2,3,\ldots)[/math], and they would demonstrate that [math]Z(1,2,3,\ldots)=-\frac{1}{12}[/math], nobody would care. Right? It would be an obscure result in an obscure part of mathematics.However, rewriting this as [math]1+2+3+4+\ldots=-\frac{1}{12}[/math] is suddenly seen as cool. It’s like, wow, adding all the whole numbers is negative! And a fraction! That’s impossible! Well, guess what, it’s not just impossible, it’s not true. Adding together all the whole numbers doesn’t yield anything.As is common with many summability methods, some of them can be seen to match various manipulations done with the infinite series, like shifting it and adding it to itself and so on. The thing is, such manipulations can just as easily yield different answers, or no answers at all, or answers that don’t fit the summability method we are wondering about.Showing how the “sum” of [math]-\frac{1}{12}[/math] comes out of such manipulations is not just manipulative, it is plain wrong. Zeta-function regularization, for instance, is not stable, so shifting the series (or adding a [math]0[/math] in the beginning) changes its value. For example, per the same method,[math]0+1+2+3+4+\ldots = \frac{5}{12}[/math]while[math]2+3+4+5+\ldots = -\frac{7}{12}[/math]which is not what you’d expect (it’s not [math]1[/math] less than [math]-\frac{1}{12}[/math]).These observations simply highlight the fact that [math]Z(a_1,a_2,a_3,\ldots)[/math] is just an operator with various nice properties, nothing more. It is not “the sum [math]a_1+a_2+a_3+\ldots[/math]", and Hardy and Littlewood didn’t think it was.Really, that’s all there is to it. I wish that popular piece of wrong math didn’t confuse so many people.Footnotes[1] Divergent series - Wikipedia

Every infinite non-abelian group is an infinite non-abelian group “under [math]+[/math]”. You may choose to write the group operation any way you like, and if you want to use the symbol “[math]+[/math]”, you can.This is very rarely done, since we are so used to [math]+[/math] denoting a commutative operation, but it’s important to realize that there’s absolutely no difference between “infinite non-abelian groups under [math]+[/math]” and “infinite non-abelian groups under [math]\times[/math]” or “infinite non-abelian groups under [math]\circ[/math]”. The symbol of the operation is immaterial.Perhaps you meant to ask: what would be an example of an algebraic operation which can reasonably be symbolized as “[math]+[/math]”, but which nevertheless represents an infinite non-abelian group?Nothing to do with numbers is reasonably denoted with [math]+[/math] while being non-commutative. However, if I had to pick an example, I would probably go with string concatenation.In many programming languages, “hello“+”world” is “helloworld” . This gives you a very natural [math]+[/math] operation which is clearly non-commutative: “world”+”hello“ is “worldhello“. So we have an infinite set (strings of letters in English) with a non-commuative binary operation, and there’s even an identity element: the empty string “” can be added to anything without changing it.The only problem is that this isn’t a group, because we don’t have inverses. There’s nothing you can concatenate to “Desserts” to make it “”. To remedy this, we introduce the following idea: uppercase and lowercase letters are decreed to be inverses, meaning “Xx” == “xX” == ””, for any letter x whose uppercase version is X. Nothing special happens to bE or By. Only lowercase/uppercase pairs of the same letter cancel each other out.So now, for example,“Quora” + "ARte” == “QuoraARte" == "QuorRte" = "Quote".And indeed,“Desserts” + “STRESSEd” == “”.We now have a group. Every string (“word”) has an inverse, obtained by reversing its letters and flipping their cases. Checking that the operation is still associative takes some work, but it can be done, and this is indeed a tremendously important group: it’s called the free group, in this case the free group on [math]26[/math] generators (or “letters”).Group theorists and other mathematicians would universally denote the group operation in a free group either no symbol at all, merely juxtaposing the words next to each other, or using a symbol like *. But computer programmers are quite used to having + denote concatenation, so there you go.