This was my thought process: - a parabola gets arbitrarily steep the further you move along the x axis - any rotation will exceed verticality in points far enough from x=0 - only 0 and π (guaranteed for any funcition) maintain the well-defined property
I really liked the little explanation at the end, the math made sense algebraically but framing it the other way made it click that it has to be the case
@@DrBarker Did you prove that any rotation other than n-th 90 degrees won't give us a function? If so I must say this is anti-intuitive to me. It's hard to imagine that I can't rotate ever so slightly without x at a certain point will decrease by increased y. ( y = x^n) Especially hard when considering increasing n (even integer) is similar to an anti-clockwise rotation.
Considering the slope of your non rotated function x^2. The derivative is 2x. This tells us, that for x->∞ the slope approaches infinity aswell A line with slope infinity is basically a vertical line. If you rotate a vertical line - may it only be so slightly - it's no longer vertical and is well defined for every x, thus creating infinitely many duplicate value pairs of whatever they're called.
@@RU-vid_Stole_My_Handle_Too It looked anti-intuitive to me too, at the beginning. But Dr. Barkers comment at the end is very convincing. What it amounts to is showing that the line y = m*x intersects the parabola y=x^2 at a different point (other than the origin), no matter how large m>0 is. But that is easy to see: m*x = x^2 has the non-trivial solution x=m. The slighter the roation the further out we have to seek for an intersection point.
Maybe more of a computer science perspective, but the problem seems to reduce to saying quadratics are asymptotically larger than linear functions. The alternate method really highlights this. If you take a quadratic and draw a straight line through its vertex, even a slight rotation of the line around the vertex should eventually be outgrown by the quadratic. It's an interesting bridge for me since I usually just take complexity classes at face value. Thanks for sharing!
Very nice! I think we could use this argument to show that for even-degree polynomials, we can always find a line of any gradient that passes through the curve in 2 distinct points (just not for vertical lines). When the degree is odd, this doesn't seem to work though, e.g. for y = x^3, with a rotation by a small angle clockwise.
@@DrBarker That's an interesting observation, the different symmetry of odd powered polynomials makes it so some negative slope lines don't work. I've never reasoned about function rotations like this before. A lot of weird ideas popped out like a geometric notion of the complexity hierarchy by focusing just on theta in [0, pi/2] and letting the intersecting straight line be any lower order function.
Yep. I think it should be pretty obvious for any even degree polynomial. The slope at infinity approaches infinity, so any rotation must necessarily produce an infinite number of inputs that have two outputs.
Spot on. In fact the general answer comes down to Lipschitz continuity. If a curve is Lipschitz continuous, then there exists a range of rotations with which the resulting graph is a function. Though I wonder what happens to discontinuous functions like those werid fractal functions...
@@DrBarker 296 / 5 000 Wyniki tłumaczenia Each rotation of the parabola is equivalent to the opposite rotation of the coordinate system, and therefore of the OY axis. After rotating the OY axis by an angle other than n * Pi, the image of the OY axis has the equation y = kx If the parabola has the equation y = ax ^ 2, then the image of the OY axis after rotation and the parabola have at least a common point. ax ^ 2-kx = 0 x = 0 or x = k / a
An intuitive way of understanding this would be looking at the derivative (aka slope of tangent) towards infinity After getting the inclination angle of the tangent as arctan(slope) we can just add the degree of rotation to that angle The observation here is that slope approaches positive/negative infinity when x tends to positive/negative infinity, hence the inclination angle is infinitely close to -90° and 90° at the infinities Any finite amount of rotation would make the angle cross the 90 degree threshold at one side of the curve, so it is not hard to see that only the unrotated parabolas can be a graph of a function (y = Ax² + Bx + C)
Yeah, that’s basically the idea I had too! I believe in the general case, the limit for a function to rotate ccw before losing its function status is pi/2 - arctan(max derivative value). In the parabola’s case, its max derivative approaches infinity, meaning arctan(max derivative) approaches pi/2
the derivative is 2x, which means it get's arbitrarily close to a vertical line, both with positive and negative sign, which means that any non 0 rotation will inevitably rotate the derivative beyond a vertical line, which means it ceases to be a function as there are parts where the curve turns backwards.
Exactly my reasoning. It could be used for functions that don't go vertical on both sides, for example the exponential function. Here one has to think about the direction of rotation, but it is quite obvious which the right direction is. It may also not reveal all "forbidden" rotations.
I took a very similar approach but used parametric points instead of a parametric complex number, so (t, t^2) instead of t + t^2i. and treating the point as a vector i multiplied on the right by a 2x2 rotation matrix and a lot of the things in your method looked familiar from that. Very cool!
edit: didnt saw the last part of the video before commenting sorry i thought it was by multiplying the entire function with a matrix i got another way to tackle this problem so essentially a well defined function is just that any vertical line where x=constant only intersect the graph at 1 point, so instead of rotating the entire graph of the function, we can rotate this x=constant straight line which is equivalent, and it much easier to proof that for any linear function f(x)= ax where a is a real number, this always has two solutions (intersections) with the graph of x^2, basically x^2-ax=0 only has one solution when a=0, or when approaching ±infinity, which is equivalent to this linear function being one of the axis, and the graph of x^2 rotating 90, 180, and 270 degrees, where 90 and 270 degree is a special case but can be easily shown to be not a function. I think it is the same conclusion right?
I would've approached the problem with the parabola in a more simple way: geometrically speaking there's no difference between rotating the parabola or rotating the y axis, therefore you can simply check if there are any lines y=mx such that the only intersection with the parabola is the origin, and of course the m coefficient is the tangent of the angle between the line and the x axis
I agree. y=mx meets y=x^2 at x=0 & x=m, which are two distinct points unless m=0 or "m=infinity". I'm fairly sure Dr Barker already realised this, but knew it would make a far less entertaining video!
Very nice - yes, I think working with the gradient is more practical than working with the angle of rotation. This approach is probably simpler than the complex number method in a lot of cases, although if lines through the origin don't work, then it might be tricky to find a line y = mx + c that passes through the curve in ≥ 2 points, because we'd have to vary m and c. I suppose the complex number method also gives a nice, rigorous framework to explain exactly why it is sufficient to find a line of a certain gradient passing through the curve in 2 or more points.
i feel like there could also be an arugment for this where you show something about the tangent lines not being the way they should with respect to the x or y axis. since the tangent at the limit is vertical in the x^2. but in the slightly rotatet case, i think we would have two tangents with the same non zero or non inf slope at two different points. would love if anyone could try to elaborate on this. :))
I was thinking about tangents too: For any rotation there's a tangent line that would rotate to vertical (since the tangent line gradients grow continuously without bound (across all real numbers); 2x). Since the relation would turn back, not cross that vertical tangent, it would have that one-to-many property.
If we look at the derivative of the function f(x) = x^2, we have f'(x) = 2x, which takes all real values other than ±infinity. Then we could move the tangent line slightly, and it will in general cross the curve in 2 places, unless this was at a point of inflection. So perhaps to solve the general problem, we can find the range of the derivative function f'(x), but remove points of inflection, to help give the rotations that don't lead to a well-defined function.
Rotating the parabola is analogous to rotating the line. Viewing this as a computer scientist, any quadratic is eventually going to surpass any linear growth except for a perfectly vertical line. If the line ever is larger (in magnitude, WLOG for the sign of the quadratic) than the value of the parabola (i.e. crosses it once), that line will eventually be less than the parabola (i.e. crosses it a second time). It doesn't matter how steep the line is, if it's not perfectly vertical, the parabola will cross it again.
I went with the fact that if the function reversed direction horizontally then that would mean it had to have some domain of multiple outputs. This would happen if its derivative angle passed completely vertical, in which case the derivative would pass plus or minus infinity and change signs. Since the derivative of x^2 goes to an infinity in either direction, any rotation would result in a curve back at some point, creating a domain of multiple outputs.
Interesting approach. I think checking the final unequal condition is simpler if we note that the parametrization f : *R* → *C* defined by f(t) = (t + t²i)e^(iθ) is injective, with left-inverse g(t) = Re(te^(-iθ)). Then t_1 ≠ t_2 implies f(t_1) ≠ f(t_2).
@Dr Barker another interesting topic (interesting in my estimation anyhow) is related. Naturally all well-behaved continuous closed plane curves will have at least one maximum displacement from the origin, say M. [BWOC if M DNE then d (modulus)/dt cannot vanish and the curve (if some parametrization by t (along the curve) ever has increasing modulus) cannot be closed.] Necessarily this means that rotation by a parameter θ about the origin will create at least one circle as the envelope of the curve. Expressed in polar form, r = M. Transforming the coordinates (x, y) of the original curve, we get f(x cosθ - y sinθ, x sinθ + y cosθ) = 0 for some f. Assume we can express this as u(x, y) cosθ + v(x, y) sinθ = w(x, y). Then u^2 + y^2 = w^2 defines the envelope. For instance, a spiric section under rotation is simply the sum of a constant, a biquadratic in r, and a few terms that depend on x, y and θ.It might be interesting to investigate more generally which annuli belong to the envelope of a curve under rotation.
Ugh, I just googled rotating by e^i*theta on desmos and now I have a headache. sqrt(-1) is undefined. I guess I could just multiply it by a rotation matrix. Meh.
Very interesting but I'd have to say that the cartesian plane is less than ideal. A function is no longer a function if the imposed coordinate system leads to singularities and multiple answers to a single input? Another way of saying the coordinate system and functions (in their narrow definition) are inadequate to express mappings between numbers in algebra (complex numbers). Why labor the point? You could have imposed an inverse rotation on the coordinate basis leaving the original y=x^2 unchanged - preserving the narrow definition of a function and importantly keeping the derivatives of the equation uncomplicated by an unfortunate choice of basis. TWO THUMBS UP!
Idea: Are there any function that stay being a function after any rotation? Well, obviously not, but if there were, they would have a second derivative of zero at every point, because otherwise there would be a local mountain or valley shape for which there's a rotation provoking a one-to-many situation around that zone, as in the parabola. The only family of functions having zero second derivative are straight lines, and turns out any rotation of a line turns it into a function except the rotations where it gets vertical
I'm fairly certain some cubic functions remain valid under all rotations, as they can have a turning point at gradient=0 instead of a mountain or valley.
@@GingerWithEnvy What? Cubic functions have both a mountain shape and a valley shape, one to the left and the other to the right of the turning point with zero second derivative. There are obviously rotations that make it invalid, in fact I conjecture that the only rotation turning a cubic function into a function is the 180° rotation, as in the parabola.
my solution before watching: 1. the question is essentially “how much does a parabola have to be rotated for a vertical line that intersects it multiple times to exist?” 2. this is just “how much does a line have to be rotated to be able to intersect a parabola multiple times?” 3. all lines through the origin that aren’t either vertical or horizontal will intersect the parabola exactly twice: at x=0, and x=cot(θ) QED; any parabola that has been rotated a non zero amount is not a function
Maybe this is a risky question, but if we are going with complex numbers what is stopping us from using complex angles? Rotating through the complex plane
I like to think of a parabola as an ellipse whose major axis stretches from zero to infinity. Any rotation of that ellipse makes the length of the vertical chord finite.
You can use a rational equations to get close with slant asymptotes to a rotated hiperbola,but i never thought you could do the same with a parabola,really cool
Another simple solution: "when after rotation is graph a well defined function?" question became similar to "how many intersections line y=a*x and parabola y= x^2 has?". Because we can instead of rotation just change coordinates lines. If a!=0, then 2 points {0,0} and {a, a*a}. So only case a = 0 or angle 0, pi will give us proper solution
This is a unique way to approach the problem, so instead of e.g. looking for two points on the line y = tan(theta)*x, we can look for two points on the line joining the origin to (a,a^a), at least when a is positive, and we can use symmetry to cover the other possible rotations instead of dealing with negative values of a.
Great video, but some guidance for people that don't know this stuff would help people that are learning new math through RU-vid, which would expand your potential audience. I learned complex numbers through RU-vid videos like this, but it was very difficult. It's not even initially obvious why complex numbers are related to rotation. Just throwing in a reference to another one of your videos or 3Blue1Brown's Complex Number Fundamentals would really help someone trying to learn.
For any line with a finite slope, written y = ax, we have that a * 0 = 0^2 and a * a = a^2 (that is, the line intersects the parabola at x=0 and x=a). Now if we rotate our graph such that the line is vertical, we have the points (0, 0) and (0, a * root(2, 1 + a^2)) on our rotated parabola. Thus the only rotation we can do and maintain a function relationship is 180°, which takes an infinite slope to an infinite slope. Now I wonder is there any region of the rotated parabola which is one to one. Using a similar argument to the first, you can show that one arm of the parabola will always overshadow the whole of the other, when rotated to make a finitely-sloped line vertical. But there is the point (a/2, a^2/4), which, after rotation, will be the only point in it's column.
I think it's important to phrase this as saying that we can show rotations of pi radians produce another valid function when starting with a valid function (which can be easily achieved through flipping the y axis), but other rotations do not necessarily. I say this simply because depending on your starting function they can, like linear or some cubic functions for example.
not quite a proof but an explanation of the obviousness angle of derivative gets arbitrarily close to 90 degrees, a rotation will push this to the derivative going backwards so it starts going back where it came from so it hits the same x value again
This can be done by simply analyzing the first derivative of the parabola equal to a linear graph. Its meaning is that the rise of the parabola function is getting infinite high. So any infinite small rotation of the parabola ends the quadratic equation beeing a function.
i just said that for any non 90° rotation there exists a tan of that rotaition which is a slope, then as the slope of a quadratic is a linear function any slope is achieveable, so there will be a point on the quadratic where the atan of the slope and the rotaation are greater than 90 (if the tan is real so only in the case of tan=+infinity is it not true, this is the upright positions)
It is almost like we are to the point in mathematics where we can formalize “what it means to be interesting” Subjectively, we could define interesting to be something that catches the fancy of enough PHDs maybe? I guess functions that have the same angle of rotation where they are no longer well defined would be in the the same equivalence class. Was very cool how you used the non vanishing aspect of various trig functions in the parameterization.
I suppose there are different categories of "interesting" too - like interesting to researchers, or to students, or recreational mathematicians. We should be able to find equivalence classes which correspond to subsets of [0, 2π) of angles which give the graph of a well-defined function. It would be really interesting (and probably really difficult) to see if there's a way to find a graph/function for each subset.
@@DrBarker I’m bound to be somewhere between student and recreational mathematician. The subject has an addictive character which at least a “healthy addiction” although my wife might contest that view! Congrats on finding a niche of producing rare content that even shock talents like blackpenredpen. That way you showed of finding the derivative via long division still a stunning result linking two ostensibly unrelated notions.
This conversation reminded me of the proof that all positive integers are interesting. Suppose the contrary. Then there must be some positive integers which are not interesting. There is therefore some number N which is the lowest positive integer not to be interesting, which would be a very interesting property of N. Hence the claim is proven by contradiction.
@@RGP_Maths I suppose if we could find a way of "ordering" things in maths, then we could show that all of mathematics is interesting! Sadly the whole of mathematics probably isn't something that can be well-ordered though...
Am I missing something? It seems obvious that if one has a function includes a segment with a slope of +- infinity, then it's a not a function. Imaginary numbers aren't needed.
If it includes a vertical line segment then it is not a function. If it includes only a point where the slope blows up it can still be a perfectly good function, for example arctan(x) at x=0
Nice approach, what I would have done on this perticular one is to show that the derivative is unbounded, and then that any function with unbounded derivative couldn't be rotated and still be a function. But your way is also very cool for the parabola.
You can rotate functions with unbounded derivatives. You can rotate e^x with no problems even though it has an unbounded derivative. The impossibility of rotation has more to do with x^2 not being a bijection than the derivative being unbounded.
Doesn't this fall apart if you have a function where the absolute value of Y is equal to some function of X? Or are you operating under the assumption that absolute values aren't allowed? That seems somewhat arbitrary.
|y| = x^(1/2) is not a definition of a function though. If you expand the absolute value you get y = +/- x^(1/2) which is not a function in the normal sense.
@@StormTheSquid Functions are by definition single valued. That means you have exactly one output per input. So y = +/- x^(1/2) is not a function because if you feed any strictly positive x in there you get multiple answers. y = +/- x^(1/2) is what's called an implicit function, and yes, every parabola in the video is an implicit function. In fact every graph you can draw on the plane is an implicit function so asking if the parabolas are implicit functions is pretty trivial. But that's not the question we are trying to answer :)
Wouldn't it be possible to show that for any angle theta that isn't pi, you can find two points sufficiently far along the parabola (since it gets arbitrarily close to vertical and arbitrarily close to horizontal) that that rotation will cause those two points to become vertically aligned?
@@MyOneFiftiethOfADollar I'm pretty sure you can find two points on a parabola that are connected by a line segment at any angle other than vertical by the intermediate value theorem.
Hey I was playing around with rotating all graphs and come up with a substitution. Substitute y with ycos(theta)-xsin(theta), and x with xcos(theta)+ysin(theta). This rotates all graphs around (0,0), but I've always wondered if I could choose a point to rotate a graph around (instead of it being (0,0), maybe (3,2) ect.
