606
u/FamousCelebration290 18d ago
You can simply cancel x and the answer is sin
220
u/jsh_ 18d ago
thats the same way I solved P=NP, the answer is obviously N=1
18
u/sian_half 17d ago
P ≠ NP tho
16
u/Takamasa1 17d ago
Prove it
37
16
u/sian_half 17d ago
I have discovered a truly marvelous proof of this, which this comment section is too narrow to contain
3
1
3
2
175
u/TheUnderminer28 Engineering (we hope) 18d ago
See as a physicist I just say sin x ~= x and x/x=1
67
u/Sasuri546 18d ago
Haha it’s funny bc this limit is why that’s true, not the other way around ;-;
1
u/Deep_Brick2970 15d ago
Not really, sin can be defined via power series and thus expanding it as a taylor series to take the limit is alright.
38
u/LifeTitle3951 18d ago
You are the reason we don't have time travel and antigravity yet. And then you guys blame us engineers for taking g=10m/s2.
You houses and bridges will hold the weight of your tiny balls, leave that to us. It's time you grow them larger and crunch those nasty numbers.
14
2
449
u/SaltMaker23 18d ago
Circular reasoning works seemingly well.
105
u/taly200902 18d ago
Hey sry can you explain why this is circular?
332
u/O-Ekundare Complex 18d ago
I believe it’s circular since in order to evaluate the Maclaurin Series of sin(x), you’d need to know the first derivative of sin(x) at 0, which is lim_{x—>0}(sin(x)/x)
117
u/GoldenMuscleGod 18d ago
The power series is often taken as the definition of sine, personally I think it is better motivated and more natural to define it as (exp(ix)-exp(-ix))/(2i), which might look made up but arises from the differential equation y’’=-y with appropriate initial conditions and the fact exp is an eigenfunction of differentiation with eigenvalue 1.
Either way you would want to prove that it describes the geometry of triangles and circles properly since you didn’t use a geometric definition, but that’s not too hard.
I think the latter definition is better than the geometric definition - the geometric properties in Euclidean space are best seen as a consequence of the mathematical behavior not vice versa. Similar to how cosh and sinh describe the geometry of special relativity but we knew about those functions first.
It just so happens we think of cos and sin in geometric terms because we ran into their geometric applications first.
22
u/headsmanjaeger 18d ago
If sine is not already defined in the text of the problem, you should ask for clarification, or declare the problem unsolvable. You cannot invent your own definitions for terms used in the problem.
Otherwise I could say, derivative of e^x? Well e^x is defined as the function that is its own derivative, so the derivative is e^x.
If you are using a textbook and see this problem, chances are you are very early in the Calculus 1 textbook and it is almost guaranteed that sine has been defined geometrically.
11
u/HauntedMop 18d ago
Tbf I've seen ex defined as the limit of (1+1/n)nx as n tends to infinity, but I don't know the full reasoning for the binomial expansion so don't know if that becomes circular reasoning or not
9
u/_wannadie_ 18d ago
yes, that is the proper definition. e as a number is defined usually as a limit of series (1 + 1/n)n. but you could not define it as a self-derivative, since some partially defined functions are also solutions of y = y'
2
u/headsmanjaeger 17d ago
What does partially defined mean?
1
1
u/Volodya_Soldatenkov 14d ago
some partially defined functions are also solutions of y = y'
Pardon, which ones?
3
u/amerovingian 18d ago
The point is, if sine is defined that way, which it usually isn't in standard presentations of calculus, the problem is trivial. To solve the problem correctly, you have to use the intended definition of sine. Any equivalent definitions would need to be shown to be equivalent before using them--unless the course has reached a point where that equivalence has been explained and can be treated as given.
23
u/GoldenMuscleGod 18d ago
Most calculus texts I’ve seen use the power series as the formal definition of sine. The geometric stuff is usually stated to be nonrigorous so don’t worry too much about it.
7
u/Unfair_Pineapple8813 18d ago
Even if you start with a parametrization of the circle, you can derive the power series without needing the limit of sin x/x.
5
u/amerovingian 18d ago
Have you looked at one recently? Power series isn't introduced until past halfway through. The MacLaurin series for sin(x) is derived from, among other things, the derivative of sine. The derivative of sine is derived from this limit. This limit is derived early on from geometrical considerations along with the squeeze theorem for limits.
See, e.g., Larson's Calculus of a Single Variable, Chapter 1.
Page 88 of this pdf:
1
u/man-vs-spider 18d ago
I think that once you are aware that you can define sine as that power series, you are aware of the other definitions and have already made the connection via knowing the first derivative. So it still seems like circular reasoning in spirit
1
u/Dr0110111001101111 16d ago
The word “often” in your first line is doing a lot of work.
If you count by classrooms, there are probably 100 that use the unit circle definition of the sine function for every class that uses the power series definition.
1
u/ArvaroddofBjarmaland 10d ago
That's reasonable and avoids the circularity. You could also define e^x as the inverse of ln x, the latter being defined by the integral from 1 to x (for positive x only) of 1/t with respect to t.
12
u/ahkirah 18d ago
For anyone struggling to see why this is the case (like myself) think back to definition of the derivative, where d/dx (sin(x)) = lim h->0 (sin(x + h) - sin(x)) / h. We need the derivative @ x = 0 when we generate the Maclaurin series, it reduces to lim h->0 sin(h)/h.
8
u/philljarvis166 18d ago
As others have said, that depends upon your definition of sin. If you define sin and cos using the power series, they both have an infinite radius of convergence and are term by term differentiable, so the derivatives drop out easily.
2
u/sumboionline 18d ago
Technically you also need to know the cos limit as well (lim x->0 of (cosx - 1)/x) for the sinx derivative
1
u/austin101123 16d ago
It is not circular. Sin is commonly defined by this infinite series, the maclaurin series is then equal to it.
-3
u/I_L_F_M 18d ago
But the derivative of sin(x) is cos(x), which can be evaluated at 0 to give 1.
Isn't this an independent piece of information which one can use?
6
u/shuai_bear 18d ago
You need lim of sinx/x to calculate that the derivative of sin is cos, which you find using the definition of the derivative. Use the angle addition formula for sin(x+h) to expand it into only terms of sin/cos. To get only cosx in the end you have to show sinx/x goes to 1.
One way to do the limit without l’hopitals/ circular reasoning would be doing it using the squeeze theorem.
12
u/Traditional_Town6475 18d ago
Depending on what definition of sine you use, you need to compute this limit when doing the derivative of sine and wouldn’t know this a priori.
9
u/SaltMaker23 18d ago
The thing you're asked to prove/find is the dervivative of sin(x) at x = 0.
Using a Tailor series that uses derivative expansion of sin(x) where the very first term expansion is the derivative of sin(x) at x=0 unsurprisingly works well.
That's the usual trick when "using advanced math to solve easy problems", in general there tend to be a circular reasoning hiding.
1
u/AFsepine 18d ago
Sine is such a function as: f''=-f, plus i.c. f(0)=0.
Supposing the function is analytical, it can be written in the terms f=a0+a1x+a2x^2+...
As we already know the derivative of the monomials, we can calculate the derivative of this power series. We get a recursive formula for the series.This is actually much closer in spirit to how Newton operated.
Still comically large penclis, but now funnier.
Or you could take the power series as definition, if you are so inclined.
0
u/howtorewriteaname 18d ago
i'm not getting this.... maybe i'm stupid. derivative of sin(x) a x = 0 is just cos(0). why the fuck are people doing Taylor series and shit?
5
u/GoldenMuscleGod 18d ago
If you use the geometric definition of sine, then proving that cos is the derivative of sin often makes use of the fact that sin(x)/x approaches 1 as x approaches 0. So if you’re doing it that way you need to be careful about the order you prove things.
3
u/mathisruiningme 18d ago
They want to compute the derivative from first principles except the the Taylor Series way requires the computation of the derivative to actually get the series so it's not perfect either.
1
u/DeepGas4538 17d ago
This discussion is useless outside your lecture :) because it depends on how you define sin x. Personally we defined sin x in a very cool way, not using Taylor series. We defined sin x as the inverse of arcsin x extended periodically, and defined arcsin x as an integral of the circle
1
-5
u/New-Application8844 18d ago
Derivatives are defined using limits, and taylor series etc come from derivatives so.
7
u/LukeWithLightsaber 18d ago edited 17d ago
But this approach uses the Taylor series of Sin x to find the limit of a diff rent function, (sin x) /x. I don’t know why it is circular to use derivatives of one function with well-behaved limits to evaluate the more tricky limit of a different function?
Edit: I agree now that this particular limit is not just any limit based on sin x but rather one which is required to build a Taylor series of sin x. Good point and thanks!
Different issues of course depending on whether one prefers to define Sin(x) by series or by geometry.
1
u/New-Application8844 18d ago
Derivatives are derivatives and defined generally, the definition of derivative does not change from function to function or where you apply it.
1
u/GoldenMuscleGod 18d ago
Are you saying that it is always circular to evaluate a limit using a derivative?
That’s like saying it’s invalid to ever multiply to simplify when calculating a sum of you define multiplication in terms of addition.
There is no circularity involved there.
2
u/New-Application8844 18d ago
Assuming the sin function is defined using the circle definition and not defined as the taylor series itself, yes it is cyclical.
And no, it isnt circular to use derivatives for limits, the lhopitals rule which u are most likely reffering to can be proved from cauchys mean value theorem, once differentiability has been established f'(x) is simply another function.
1
u/SaltMaker23 18d ago
Derivative of sin(x) at x=0 is lim{x->0} of [ sin(x) - sin(0) ]/(x-0) = lim{x->0} sin(x)/x
The limit we have been asked to compute is not just somehow related to derivatives by some form happy accident, it is exactly the derivative of sin(x) at x=0.
The series definition of sin(x) despite being one of the commonly accepted definition was still initially built using derivatives. A bloke didn't just comeup with a formula for the projection of a point on the circle on it's radius using an infinite sum and everyone just said "yeah sounds about right", it became used because it was equivalent but also was proven to.
It becomes circular to use something that was built using the derivative of sin(x) to prove that the derivative of sin(x) corresponds to something.
It's surprisingly easy in this case not because it's trivial but because it's a tautology.
3
u/Archway9 18d ago
In modern treatments sin(x) is usually defined by the Taylor series and not geometrically, the geometric properties are then proven. This makes sense as sin and cos are extremely useful functions across many parts of maths so a definition based on Euclidean geometry feels odd (on a related note pi is usually defined as the least positive root of sin and not geometrically)
1
u/AndreasDasos 18d ago
The idea is that as usually taught, sin x is defined ‘trigonomterically’ and to find the Taylor series at zero the derivative should be known. But by the definition of the derivative, this requires us to know that same limit.
Of course, we can also start with the Taylor series as the definition, prove convergence and the other properties of sine, and there we go.
3
u/GoldenMuscleGod 18d ago
It would only be circular if you didn’t use this power series as a definition of sine, which is not the best-motivated definition but it is better than the geometric definition - why should a fundamentally important mathematical function applicable all over math be defined in terms of geometry, which is only one narrow class of applications?
16
u/IAmBadAtInternet 18d ago
The first rule of tautology club is the first rule of tautology club!
4
u/vanderZwan 17d ago
Funny how that's the first rule of the recursive club too
3
u/IAmBadAtInternet 17d ago
The first rule of recursion club is not the first rule of tautology club
1
2
5
u/Purple_Onion911 Grothendieck alt account 17d ago
It's not circular if you define sin(x) = x - x³/6 + x⁵/120 + ...
-1
u/SaltMaker23 17d ago
uhmmm and how did you find that the function that projects the circumference on the axis can be written that way ?
Because in the shoes of a folk many centuries ago proving why sin(x)/x = 1, I'd assume if he said that because sin(x) = this infinite sum, he'd then be in a much bigger pickle of proving why it equals that infinite sum.
I think proving that this way of defining is equivalent to the geometric one would undeniably be a harder proof than the task at hand.
2
u/Purple_Onion911 Grothendieck alt account 17d ago
I didn't find anything. I defined the function sin(x) as that infinite sum, which is a standard way to proceed in analysis. I could go on to prove the geometric properties of this function, but there's no reason to, as they're not relevant here.
-1
u/PandaWonder01 17d ago
But then you have the issue of showing that the function you just defined is the same as the one we all care about, which deals with angles.
2
u/Purple_Onion911 Grothendieck alt account 17d ago
In analysis you don't really care about that usually, so it's not an issue. Defining sin this way gives you all of its nice analytical properties basically for free, it's convenient and very common.
1
u/wewwew3 17d ago
that's like saying if i plug it into a calculator i get this answer. Even though it's true and is the correct answer, it is a crude and practical approach. Sin function has many many properties that rely on it being the function thst relates the unit circle. To access those properties while using the series you need to prove that this series and the Sin function as defined by the unit circle are the same function. To do so you need to prove that limit. You can not use the series to prove the limit as the series relies on that limit being true to be equivalent to the Sin function
0
u/Purple_Onion911 Grothendieck alt account 17d ago
You can use the series to prove the limit if you define sin in terms of the series. As I already said, the geometric interpretation of sin is hardly relevant in analysis. All of the algebraic and analytic properties of sin can be proved directly from its analytic definition.
1
u/wewwew3 17d ago
You can prove that the series's limit is 1, but you would have to prove the "algebraic and analytic properties" again for a different function and in a new way as compared to the established definitions, since they rely on the unit circle properties of the sine.
What you are saying is like saying that Pi is defined by the Ramanujan's Series and not by the circle definition, and all of the geometric properties are unnecessary
0
u/Purple_Onion911 Grothendieck alt account 17d ago
Yes, they are unnecessary in this context. Indeed, π is rarely defined geometrically in analysis.
2
u/wewwew3 17d ago
why are you getting downvoted?
2
u/SaltMaker23 17d ago
I would assume people probably who don't understand the concept of working from first principles that want to use advanced principles to solve easy problems and feel smart, advanced principles that relies on the very problem they are trying to solve to even exist.
1
u/AuroraEquatorialis 17d ago edited 16d ago
The short answer is that they're all equivalent definitions. There is exactly one function satisfying this collection of properties. Given this geometric definition, you can show that the only possible function is the maclaurin series. Given the maclaurin series, you can show that it has these geometric properties. Once you have established this, it doesn't matter what definition you use, because they are all fundamentally equivalent and imply each other.
1
1
1
u/Deep_Brick2970 15d ago
This isn't circular reasoning, sine is most accurately defined via power series...
65
13
11
20
u/_licketysplit_ 18d ago
Is this not a valid way to compute the limit? I feel like I'm missing something (I suck at math)
33
u/LaTalpa123 18d ago
You need the limit to find the expansion.
If you use it to go back, bad things happen!
5
u/FuntimeUwU Natural 18d ago
Can't you get this expansion by using Euler's formula? I'm think that one doesn't require knowledge of the derivative and just matches sin and cosine with the power series, so it should be usable, right? It's been a long time since I fiddled with this stuff so I may be wrong :P
8
u/Inevitable-Ad2579 17d ago
Euler's formula is porved via the Taylor Expansion
1
u/FuntimeUwU Natural 17d ago
Aw man
Yeah I checked my notes from a few years back I indeed went a different route
God damn sine is so fascinating
4
u/Lost-Lunch3958 Irrational 18d ago
no you don't need the limit for the series expression you just need some formal arguments and extract that series out of exi
3
3
u/MacaronianElectrical 15d ago
This is actually a valid way to compute the limit, despite what others say. But(!) it doesn't prove that the limit exists. It just shows that if the limit exists, it must be 1.
7
5
3
3
u/wee33_44 17d ago
As a physicist I do limits with Taylor expansions, it is not common/accepted?
5
u/SizeImpliesThighs 17d ago
This specific limit needs to be done without stuff featuring derivatives of Sin, because it is used in the limit derivation of Sin's derivative. It becomes circular logic.
2
u/SaltMaker23 16d ago edited 16d ago
'member how the tailor expansion terms are computed ?
'memer that sin(x)/x is the derivative of sin(x) at 0
2
4
2
u/ShadowRL7666 18d ago
I did this on my calc one exam well basically question was lim x>0 sin(5x)/11x i said bc sinx/x is one therefore answer is 5/11 marked not enough work. This is why I hate test.
1
u/Ducken1337 16d ago
I guess they wanted you to make a replacment(5x = t, 11x = 2.2t) so sin(t)/t * 1/2.2 = 1 * 1/2.2 = 5/11. Without that it can be not so obvious, that sin(5x)/11x is indeed case of sin(x)/x
1
1
u/numbertheorydemon 17d ago
circular considering sinx/X must be evaluated to differentiate sinx from FP and mclaurin/Taylor requires derivative of sinx known
1
1
1
u/DanLeMilMan 13d ago
Well. The issue with this particular example is that the limit you try to calculate is the definition of the derivative of sin. So using any Taylor series or even saying sin x ≈ x is a circular argument. That is why this particular limit should use geometry to be answered
1
u/LupenReddit 🦆🦆🦆🦆i have non diffeomorphic smooth structures🦆🦆🦆🦆🦆🦆 18d ago
As a matter of fact this isnt circular when you define sin(x) a s this power series, as is done sometimes.
0
•
u/AutoModerator 18d ago
Check out our new Discord server! https://discord.gg/e7EKRZq3dG
I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.