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So I know this is a dumb question maybe, but it has been in my head recently. I once told someone, on the topic of introductory calculus, that using l'Hopital's rule to calculate $$\lim\limits_{x\rightarrow 0} \frac{\sin(x)}{x}$$ is circular reasoning, because the limit we are looking for is the definition of the derivative of $\sin(x)$ at $x=0$. They argued that it is not because you can work out the derivative of cosine via its series representation without ever recurring to the definition.

Now, there are two ways that I know of to obtain the series representation of cosine: the obvious one is Taylor expansion, which obviously requires knowledge of all the derivatives, the other one is to prove the following statement:

If there exists two functions $f$ and $g$ such that $f'=g$ and $g'=-f$, then they are unique.

Then you have to kind of get out of the hat the series representations and show that they respect the conditions listed above, which only involves deriving polynomials. Nonetheless, it seems to me that you still have to know the derivatives of sin and cos to then identify the two trigonometric functions with the series representations using their uniqueness.

My question is: is there really a way to obtain the series representation for sin and cos without knowing the derivatives, thus getting rid of the circularity of the reasoning?

This question is by no means practical and it's just a curiosity of mine, I'm not interested in other ways to calculate the limit, which is trivial.

EDIT:

I don't think this is a duplicate, as I said in my comment if you read the question that is linked, you will notice that what the OP means by "circular reasoning" is something on proving the limit using formulas about the area of circles, which has nothing to do with my question about series representations.

user438666
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  • This limit is not derivative of $cos(x)$ at $x=0$, but $cos(0)$ itself. – Przemek Jun 25 '17 at 17:49
  • @Przemek how embarrassing, I meant of $\sin(x)$, which is what you have to derive to apply the rule, editing now – user438666 Jun 25 '17 at 17:50
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    Chris, if you have a copy of Johnsonbaugh and Pfaffenberger (it's available in Dover) then you'll find a series definition of sine and cosine. In fact, all the usual features of sine and cosine can be derived from such analysis completely devoid of explicit geometric reasoning (as fills the alleged "duplicate"). See page 274-275 or so. Incidentally, I'm currently working on a paper which generalizes such observations to arbitrary associative algebras. So, divorcing sine and cosine from right triangle trig is dear to my heart. – James S. Cook Jun 25 '17 at 18:23
  • @JamesS.Cook thank you for your comments, I tried opening a meta post on this issue, see here, https://math.meta.stackexchange.com/questions/26553/linking-possible-duplicate-before-having-read-the-question. On your second comment, I don't have a copy of that book, I will look for this reference. I had not even thought about the fact that sin and cos are only defined in a geometrical way as far as I know! Can one take the statement I quoted as a definition? If one takes that as the definition, then sin and cos are merely two names for the series, and the circularity is gone, or is it not? – user438666 Jun 25 '17 at 18:30
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    Yes, the circularity is removed. Sine has derivative cosine by the term-by-term differentiation of series theorem. Hence, the usually circular Lhop argument is legit. Of course, for most users of calculus, the option to define sine and cosine via series comes a bit after they see Lhop rule. You can define $\pi$ in terms of series as well. – James S. Cook Jun 25 '17 at 18:35
  • Let's use an analogy. There are 3 ways to hold your left ear, first using your left hand, second using your right hand and finally the third taking your right hand at the back of head and then holding the left ear. The usage of L'Hospital's Rule to solve any limit of type $$\lim_{x\to a} \frac{f(x) - f(a)} {x-a} $$ is similar to the third way of holding your left ear explained above. Moreover this might not work always (try with $f(x) =x^{2}\sin(1/x),f(0)=0,a=0$). Use of L'Hospital's Rule in such cases is not circular but definitely very very roundabout. – Paramanand Singh Jun 26 '17 at 02:41
  • This was in the 70's but I think remember Moise defining $\arctan x = \int_0^x \dfrac{dt}{1+t^2}$ and proceeding from there. He even managed to define the "angle" of the point $(x,y)$ on the unit circle. – Steven Alexis Gregory Jun 26 '17 at 03:22
  • @stevengregory: I think the definition of arctan using integral is also very old. Hardy used it in his classic text A Course of Pure Mathematics at the beginning of 20th century. – Paramanand Singh Jun 26 '17 at 04:13
  • Apologies if this has come up before, but this question is quite related. – Arnaud D. Dec 03 '19 at 14:54

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We make the following definitions: $$ \sin(x) = x - \frac{1}{3!}x^3+ \frac{1}{5!}x^5+ \cdots = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{2k+1} $$ and $$ \cos(x) = 1 - \frac{1}{2!}x^2+ \frac{1}{4!}x^4+ \cdots = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} x^{2k} + \cdots$$ The series above converge for each $x \in \mathbb{R}$ by the ratio test. A well-known theorem of power series asserts that the derivative of a power series is indeed the power series of derivatives. This is not a trivial theorem, it requires some effort to prove. But, with the term-by-term differentiation theorem is obvious that \begin{align} \frac{d}{dx} \sin(x) &= 1+ \frac{2}{3!}x^2+ \cdots + \frac{2k+1}{(2k+1)!}x^{2k} + \cdots \\ &= 1+ \frac{1}{2!}x^2+ \cdots + \frac{1}{(2k)!}x^{2k} + \cdots \\ &= \cos(x) \end{align} Thus, applying L-Hopital's Rule: $$ \lim_{x \rightarrow 0} \frac{ \sin x}{x} = \lim_{ x \rightarrow 0} \frac{\cos(x)}{1} = \frac{\cos(0)}{1} = 1.$$ Notice, $\cos(0)=1$ and $\sin(0)=0$ are nearly manifest with the definitions I offer at the start of this answer.

Here's an interesting follow-up: how do we prove other aspects of the function theory of sine and cosine via series. For example, how to prove the $2\pi$-periodicity, or adding-angles formulas ? Much is known. I'll leave it at that for now.

James S. Cook
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  • incidentally, I usually say it's circular for students because I don't think they use the definitions offered here. You can also define sine and cosine as the unique solutions $y_1,y_2$ of $y''+y=0$ for which $y_1(0)=1$ and $y_1'(0)=0$ or $y_2(0)=0$ and $y_2'(0)=1$. Of course, $y_1$ is the cosine whereas $y_2$ is sine. – James S. Cook Jun 25 '17 at 19:07
  • Thank you. Just another curiosity, without exploiting the uniqueness of the solutions to the differential equations you mentioned, and the geometric representations, how can one be sure that the series actually are equivalent to the geometric definition? – user438666 Jun 25 '17 at 19:11
  • Well, that is an interesting question. Perhaps you should ask it. I'm not sure I've thought through that in its entirety. Certainly, the fact that $\sin^2 \theta+ \cos^2 \theta =1$ is a big part. So... how to define geometrically though? What does that entail? I'd start with right-triangle trigonometry then extend to other quadrants via sign considerations. Is reproducing polar coordinates enough? Is there more to the "geometry" of sine and cosine? I'm not sure. – James S. Cook Jun 25 '17 at 19:15
  • Related...this page includes (after other proofs) a geometric proof that sin(x)/x limits to 1: https://proofwiki.org/wiki/Limit_of_Sine_of_X_over_X – Bill Cook Jun 25 '17 at 20:12
  • Equipped with that fact, you can get the derivative of sine is cosine. – Bill Cook Jun 25 '17 at 20:13
  • I think this is too much of an overkill to evaluate the limit of $(\sin x) /x$ when it is obvious from your first equation (series definition of $\sin x$). Usage of L'Hospital's Rule is valid but very very inefficient and I never understood why everyone is charmed by L'Hospital's Rule. In fact the limit of $(\sin x) /x$ is always a trivial/immediate consequence of definition of $\sin x$ no matter what definition you choose. It is trivial even in case of geometrical definition contrary to what many believe. – Paramanand Singh Jun 26 '17 at 02:52
  • @ParamanandSingh you have some good company in your thinking. I think Apostol more or less says the same as you and refuses to bother proving it since it's "useless". I was mostly responding to the OP's question. That said, there are a lot of calculus students who have way way way more success with LHop. rule than series arguments. At least that has been my experience of late. I want to side with you. It should be easier to think in fancy-fied polynomials,but, it seems not for many folks (sadly). – James S. Cook Jun 26 '17 at 05:04
  • I like your phrase fancified polynomials for the series approach. +1 for the answer. – Paramanand Singh Jun 26 '17 at 05:23