Comments by "MC116" (@angelmendez-rivera351) on "It's a Calculus Mistake" video.
-
0:52 - 0:55 Saying "I can keep adding 0s forever will continue giving me 0" is a mathematically incoherent statement, because of the non-concept of "adding forever." It is extremely easy to make mistakes in mathematics when using colloquial notions and intuition, when you should be using precise language instead. I think this here is the actual mistake in the "proof," rather than anything else. There is nothing that can be discussed at all about the validity in the proof until the terms in the proof are actually precisely defined, to begin with, so nothing else could possibly be meaningfully identified as the mistake. This is not very different from how, if I say "Gooblydegock is heavy and of the color martin," it amounts to nothing but literal gibberish, since the words "gooblydegock" and "martin" are undefined terms. There is no meaning in say the sentence is true or false: conceptually, it is not even really a sentence at all, to begin with. It is merely a nonsensical string of symbols. On that note, the idea being conveyed has multiple inequivalent ways of being formalized, but the most basic of those formalizations, and the one most commonly used, as well as the one most relevant to this video, would be to consider a sequence f(n) = 0. You can find the sequence of partial sums of f, and that gives you still s[f](n) = 0. This also means lim s[f](n) = 0. Yes, I do understand that colloquial language and intuition are important in the context of teaching. However, as far as proofs are concerned, those are completely inappropriate.
2:20 - 3:37 What is truly happening here is that we are talking about the partial sums of two different sequences. Earlier, we had f(n) = 0. Here, we have g(0) = π, g(n) = 0 otherwise. What the video is claiming is that lim s[g](n) = lim s[f](n). This is clearly false. The reason this is confusing, though, is that the notation used in the video (which is inappropriate when doing mathematics) makes it seem as though you are still talking about the sequence f and its partial sums, but in reality, it has been replaced by the sequence g with its partial sums, without the viewer noticing, because the notation being used is simply misleading and ill-defined. This goes back to what I said in my first paragraph. Since it is not even clear what the notation being used is even supposed to mean, it can be easily used to deceive people. This is not a matter of unnecessary pedantry, it lies at the very core of the problem.
2
-
@thebestof2341 No, it was not. The order of operations allows a + (b + c) = (a + b) + c. This is called associativity. In fact, the order of operations exists only because of associativity being true.
Also, I should point out that you are not actually obligated to follow the order of operations, since it is just a convention. Mathematicians could have chosen a different convention altogether, and it would have been just as valid. In fact, you can rewrite all of mathematics using Polish notation, where no order of operations is even needed, as the notation is unambiguous, unlike the infix notation most mathematicians use when publishing, which is the same notation we use. The ambiguity of infix notation is why the order of operations exists. However, with that being said, the order of operations was followed in the video. Anyone saying otherwise does not understand the order of operations.
2
-
1
-
1
-
1
-
1
-
In the first line only, 0 cannot be written as a sum of infinite zeroes.
Technically, there is no such a thing as an infinite sum. However, there is such a thing as the limit of a sequence. You can write 0 as the limit of the zero sequence. Consider this: z(m) = 0 everywhere. Let s[z](0) = 0, and s[z](m + 1) = z(m) + s[z](m) everywhere. Hence, s[z](m) = 0 everywhere. s[z] is the sequence of partial sums of z, so lim s[z] is the "value of the series" of z. lim s[z] = 0. This is completely valid.
As, if it is possible, then we can write the integral of a continuous function on [a, b], as the sum of the integral at each point.
No, that is actually not how that works. The integral of a function cannot be evaluated "at a single point." You can only evaluate the integral of a function over closed intervals, and finite unions or intersections of those closed intervals.
1
-
1
-
1
-
1
-
1
-
Since the cosine is not zero, you can cancel it out (divide by it) on both sides, and you get 0 = 2π.
This is not true. cos is nonzero, but cos could be a zero divisor, or, more simply, it may not be left-cancellable.
Divide both sides again, this time by 2, and you get 0 = π.
This is not justified. How have you discounted 2 = 0? Or 2 being not left-cancellable?
With 2 totally independent proofs, it has to be true, right?
2 is quite a small sample size, making your study statistically insignficant 😉
1
-
1
-
1