Does Pi Equal 2?? (Spoiler: no)
Вставка
- Опубліковано 3 жов 2024
- In this short, we show a thought experiment that arises by doubling the number of semicircles drawn in a semicircle but halving their radii. The process produces an infinite collection of smaller and smaller chains of more and more semicircles that always have combined circumference length equal to pi. If we think about the limiting process, it seems like maybe this technique shows that pi = 2. But Pi can definitely be shown to be greater than 3 - so where does this argument fail?
If you like this video, consider subscribing to the channel or consider buying me a coffee: www.buymeacoff.... Thanks!
Here is a related false proof that shows root 2 is equal to 2:
• Is square root of 2 eq...
This false argument was suggested to me by Jeff Stuart as a nice alternative to the more classic Pi = 4 argument.
#math #manim #visualproof #proofwithoutwords #circle #circumference #radius #pi #piday #shorts #infinite #falseproof
To learn more about animating with manim, check out:
manim.community
Length of limit of curves =/= Limit of length of curves
With GBoard you can hold the = sign to make ≠
I think my comment got removed?
@@boas_all your comments are there don't sweat it
@@boas_ it's a conspiracy! Google doesn't want us to know how to make the "not equal" sign 😱
≠
There is only half a Pie.
Conclusion; someone ate the other half.
I’ve been spotted
@@KevinLivian gettim boys
FBI OPEN UP
FBI: “sir you’re under arrest for stealing half a pie” me: “AND I’LL DO IT AGAIN”
@@KevinLivian: "but half a pi looks like tau, so really I multiplied it by 4 to make 2π= tau"
It never flattens. The derivative has to converge too. I have never seen this counterexample of that situation! Awesome
Agree. In the video he says that the semicircles "squash down and flatten". As you state, no they don't. Never. That is where his argument fails.
@@tedhorton6315let’s be a little more clear here; of course they don’t ever “flatten,” but when he said that, he meant the semi circles converge pointwise to the diameter which is true. The argument fails because you are interchanging limits without justification
Something similar happens with the schwarz lanterns: the normals must converge too!
Looks pretty flat to me. Keep going until the original 1 m diameter circule is made up of circles each smaller than the plank length.
@@JoseJimeniz this is math, not physics. we are not bound by the restraints of our reality. we openly embrace infinity and all of its counterintuitive weirdness.
Regardless of how small the semicircle is, it will never be a straight line. So π ≠ 2
True, because semicircles have length as well and breadth while a line only has length and no breadth.
Of course, but that doesn't explain why this method doesn't work. The purpose of a limit is to approximate an object as precisely as we need, not necessarily to be equal to that object. Using the same reasoning as you, I can say that 1/n is never 0 so lim 1/n ≠ 0... which is very wrong.
@@yannld9524 A diagonal length of two opposite points of a square is always going to be longer than the side, what more the length of a curve through said square with a radial length of that square's side.
So using said rational, the diameter of a circle will always be shorter than 1/2 the circumference, even if the circumference is recreated with smaller and smaller dimensions of curves within that semicircle as shown in this example.
@@randomsandwichian Good for you if you think this explanation is more convincing than the video. I was just pointing out that your first argument wasn't a good one
Yet the same people argue 0.99999 recurring is equal to 1
The famous "aproximates the area but not the permeter"
Can you elaborate?
@@pxzzvc6357every iteration, the area of the semicircles is divided by 2. But the perimeter always stays equal to π
@@pxzzvc6357Basically when things look close to each other generally area is also close. Perimeter is super deceptive though. Things that look close can have almost any perimeter including infinity.
@@pamplemooIt completely followed what you are saying.
And then I thought.. but wait, if a circle is the shape that has the ultimate (fixed ) ratio of area/permiter, then how can one stay equal while the other divides by two every step.
And then I realised that originally, the shape isn't a full circle, after one step the (total) shape is one full circle and every next iteration you keep adding full circles.
(So you are not comparing similar aggregate shapes.)
@@oscargr_ the argument the creator makes would also work if you drew every semicircle as a circle.
the argument fails still then, this isnt the flaw.
Okay so the reason why pi doesn't equal 2 here is because while the semicircles look like they are only 2 units long. It's instead similar to an accordian. Or an old plastic straw you'd get with a juice box. It looks smaller because it's compressed, but if you were to grab both ends, and pull. Then it would lengthen out to ~3.1416 units long.
Good analogy. Impossible to imagine the semicircles at the limit, but true.
@@lastchance8142 thanks. It's not my analogy originally though. I got it from Vi Hart and her video about why the proof of pi=4 doesn't work. The video is titled "Rhapsody on the Proof of Pi = 4" if you wanna check it out
Best analogy/explanation
🎉
This reminds me of the classic pi=4 "proof!"
Yep! This is the other side where pi is way too small. Maybe we should just average the two proofs to get pi = 3? 😉
Tbh thats pretty close so i'd take that@MathVisualProofs
@@MathVisualProofs engineers:
@@MathVisualProofs We're getting closer. If we averaged the results of infinitely many false pi proofs, would we get actual pi?
@@emulgatorx 😆
This is a variation of the "Staircase Paradox".
But It is not a paradox.
@@oscarmartinpico5369Mathematicians have historically called anything that’s counterintuitive as “paradoxical,” e.g. the “birthday paradox”
@Emily-fm7pt I don't think this is counter intuitive. It just misunderstands what a limit is. To take a limit we need a function which approaches something. The 'function' for this doesn't approach anything it stays exactly where it is, pi at the first approximation and pi at the nth
@@johnny40coats4 Well you at least have to accept that the area of the limiting shape approaches that of a circle, because it does get closer and closer to being a circular shape, but in doing so you’re not changing the perimeter, like you said, so you can’t use the perimeter to calculate pi. The counterintuitive part here is that a shape that looks and feels a lot like a circle can still be not even close in its perimeter.
@@Emily-fm7ptnot just mathematicians, the definition of a paradox as something that counteracts itself is only a strict definition, albeit commonly used. There exists a more lax definition as anything that doesnt make immediate sense/doesnt feel like it would work even after thought. This definition is less common however.
Technically, it never flattens, but our eyes can't see that, so they get tricked into thinking it does
exactly
Funnily enough, if you continue to create smaller semi-circles, the limit of each point on the surface of the semi-circles, does actually approach the exact line, and if the limit is taken to infinity, it genuinely does flatten, it doesn't make a bumpy approximation, the points exactly line up. The problem is the perimeter is not preserved when the process is done infinitely many times
@@-._.ChaLoe._.-The limit cannot be taken to infinity. 🤦♂️
@@nexuscross3233
Are you sure about that?
@@nexuscross3233It actually can be and the curves do converge to the diameter. But taking lengths is also calculated by a limit. Commuting two limits (the length of the limit of the curves vs the limit of the length of the curves) is a big no unless you have a theorem that allows it. And that's why a lot of early analysis is learning when you're allowed to do it :)
I've not yet studied calculus
but i think something's wrong wtih the limiting process
yeah. limits are way different that absolute values
What’s wrong is that while the height of the semi circles above line does approach zero, the number of them goes to infinity. So the total length of them does not approach 2. It remains exactly pi with each iteration.
In calculus, arc length has a different formula for area. Arc length uses circular arcs while area uses rectangles. When I asked my prof why this is so, his answer was "It just is".😃
When you have two curves and you change one to make it "converge" to the other one, this will only happen when the shape and the derivative of curve 1 are being changed. In the case in the video, the derivative was not changing, and you could see "pointy" parts in between
not always
This is like the coastline paradox.
Each iteration, the circumference of each semicircle decreases by half, while the amount of semicircles doubles.This means, since the total perimeter of these semicircles is (circumfrence of smaller semicircle)x(amount of semicircles), that each iteration, the decrement of the circumfrence of each semicircle perfectly cancels out the increment of the quantity of semicircles, causing the total perimeter to remain the same.
Here's what the math would look like
circumfrence x amount of semicircles
(π/2)x(2) = π (iteration 1)
(π/16)x(16) = π (iteration 4)
(π/1024)x(1024) = π (iteration 10)
The number π is being divided by increases by taking 2 to the power of the iteration (e.g, 2^10 = 1024) and so does the number it is being multiplied by.This allows us to come up with a formula (n representing the number of iterations):
Total perimeter = (circumfrence of smaller semicircle)x(number of smaller semicircle)
Total perimeter = (π/(2^n))x(2^n)
Total perimeter = π, since 2^n cancels out
Since π is a constant, it is unchanging, so no matter the number of iterations, the total perimeter is always π.
Finally a proof of common sense
This is okay-ish is you don’t believe in any infinity or infinitesimal or anything like that. But ultimately limits do help to solve lengths, areas and volumes. So, how to use limits without going to some contradiction like the above, is the study of Real Analysis.
@@akashpremrajan9285 this is technically still a limit. the limit for this question is when n approaches infinity. since the formula that gives us the total perimeter is a constant, even at that limit, the value of the total perimeter will still be π. none of the other proofs i saw used mathematical notation, and the most liked one "Length of limit of curves =/= Limit of length of curves" doesn't exactly "prove" the total perimeter is π using a limit. it only states a fact that, while it is true, isn't a complete proof or answer.
in order to please those who wish to use a limit:
lim n->inf π = π
farewell.
Its like diagonal of a square, and a step case approximating that diagonal.
The step case will never approach the length of the diagonal, but 2L instead.
Engineers might be loving these proofs
Nope. pi = e = 3 NOT 2
Pi = e
e = 2
So Pi = 2.
In fact, it's a very good question and I don't think most people have ever heard the answer.
If a curve C is parametrized by a (sufficiently regular) function f, then to compute the length of C we don't need f but f '. And when we approximate C (uniformly) by certain curves C_n as in the video, this corresponds to choosing a sequence of functions (f_n) that converges uniformly to f. Hence in the very convenient case where the sequence (f_n ' ) converges uniformely to f ', we will obtain that lim length(C_n) = length(C), but there's no reason for the sequence (f_n ' ) to converge uniformly to anything, there are a lot of counter examples.
Perfect.
Finally somebody who actually knows what theyre taking about. This is about uniform convergence
We need uniform convergence of the derivatives not of the function, the function itself is uniformly convergent to the line in fact.
@@giuseppegaleotti9149 Yep, that's what i said
@@yannld9524 I was answering to pendragon7600 but it didn't tag him.
I love this. I can also suggest another instance of such conundrum The explanation is, the sequence of curves does converge to the vertical line, but it doesn't converge to it *uniformly* ; this notion of uniform convergence was actually coined to help such cases. That is, all points should move and approach the line with relatively same speed, but they aren't: the middle point takes a big jump and then it stays stationary, various points at p/2^n do series of a smaller jumps before jumping right to the end on the n'th stage and stay still, and all other points perform decreasing jumps indefinitely. This is not uniform convergence, so the taking limit operation can not be swapped with the sum (of lengths of little half-circles) operation.
The problem is when you said the circles "squash down and flatten". Small circles aren't flatter than large circles. Just cause you don't have the resolution doesn't mean it's not still curvy.
The conundrum is resolved when you realize that not all infinities are equal.
Fun fact: this is basically the exact design of a fractal vise, a vise which conforms to any object's shape
I'm an undergrad, and for quite some time (4years to be exact) I've been going the same route to my uni, which incorporates some diagonals and thus considered popular among students. But there was a similar argument in my head for a while - if a diagonal is just a hypothenuse of a 1 sided triangle as the radius in this example, then by applying the same logic in limit the right-angles path would be the same length as the diagonal, which is counterintuitive. This for me was stemming from misunderstanding of the limits on my first year. The argument shatters if we consider an object moving along both lines at the same rate. As this limiting line doesn't change its length, it would still require longer time to traverse than the diagonal. This sorta reminds of integrating over the length of the line which is neat. I hope this all made sense 😅
An expression for the total circumference is (2^n)(π / 2^n), starting at n = 0. Each iteration’s semicircles have circumference π/2^n, and there are 2^n of them. The limit as n goes to infinity of this is π still
the limit is 0 actually. but true π doesn't change, its just a constant and taking the limit to 0 completely obliterates any constant down to 0.
@@XBGamerX20 What in the world are you talking about… (2^n)(π / 2^n) simplifies to π because of 2^n / 2^n, so it doesn’t matter what n is anyway, and the limit to infinity is still not affected. Do you understand algebra by chance? Lmao
@@hydropage2855 i wasn't talking about your formula. I was talking about π/2^n which gives the circumference of a single semicircle and if you take the limit to infinity the fraction becomes zero which means there's no circumference due to infinite amount of semicircles. and no circumference means no semicircle so theres nothing else besides the diameter. what you said is the most obvious thing and there's no point to take the limit of a constant. don't act like you know everything.
@@XBGamerX20 I’m not acting like I know everything lol I literally said it’s adding up all of those circumferences, and you’re talking about something totally different. The value of any given circumference goes to 0, but their sum remains as π regardless because there’s 2^n of them, making the value of n itself not actually matter as you take the amount of slices to infinity. Adding up infinite things that all go to 0, ever heard of an integral? Pretty sure those work pretty well. Also there’s absolutely a point to take the limit of a constant, the fact that the constant doesn’t depend on n anymore is the entire point, it’s why the limit is still π as n goes to infinity, because n cancels out in the first place
@@hydropage2855 you don't have to take the limit because you simplify the expression anyway and simplifying means its no longer dependable to n on any circumstance. I really don't have to continue this any further, because I'm not taking more of my time to prove that I indeed know algebra or calculus to someone on youtube.
Should it not be the infinite series of the summation of the diameters of the smaller circles equal 2?
Ye i think so
That actually will be the case. Since they original diameter was 2
π = 1 + 1 = 1 + 1/2 + 1/4 + 1/8 + ... = 2
In the Riemann Paradox and Sphere Geometry System Incorporated
The formula for circumference is 2πr, but since this is a half circle, we simply have instead πr
If we multiply the number of half circles by 2, the radius, r, of each half circle is halved. So, the algebraic result of the sum of these circumferences always adds up to π
r=1/N
ΣC_n = π * r * N = π * N/N = π
Where C_n is the nth circumference value
There is infinite distortion. This reminds me of the coastline paradox.
Even after 1milion subdivision steps, in each semicircle you can inscribe a right triangle with hypotenuse = Diameter.
The length of each semicircle C will be > than the sum of the legs, C > Sqrt(2) * Diameter.
By adding all the small C together we will get
LTotCirc > Sqrt(2) * 2
Thus PI > 2.83
No because however small the circles may be there is a diameter/radius of said circle and adding that together makes pi
Exacly and the limit is pi, not 2, since the final line length never approaches 2, but stays at pi
π = 2 in Riemann Paradox and Sphere Geometry System Incorporated
Tau ÷ π = 2
Alright 👌👍
Each semicircle with length pi/n (where n is 2^k) is longer than two sides of a triangle with sides sqrt2*R/n, sqrt2* R/n and 2R/n, so pi > 2sqrt(2) > 2. So the length of the combination of semicircles is bounded below with some number. It would converge to the straight line if there had not been such number at least after some N
It was so trippy when he zoomed in.
Let n be the number of the semi circles, while l(n) is the total arc length of semi circles.
Note that for any positive n, l(n)=π.
We consider the ε-N definition of limits. Given any ε>0, choose N=0
Suppose n≥N, note that l(n)=π still. Thus, we have lim n→∞ l(n)=π, not 2.
If the semicircles are infinitely small to match the diameter there are an infinite number of semicircles(taking the limit) so it always adds up to pi
1 + 1/2 + 1/4 + 1/8 + ... = 1 + 1 = π = 2 in the Riemann Paradox and Sphere Geometry System Incorporated
That’s some lovely ideal triangles in the half plane model of hyperbolic geometry you’ve got there
More interesting is that when you turn your phone to the side, you have a sleeping snorlax
Love that
You shall never die...
+ unless you change your mind.
Pi doesnt equal 2, its equals Tau/2. Tau is the true circle constant.
*The argument fails for the same reason dividing diagonal into steps fail.
The total arc lengths of the semicircles does not converge. It remains the same.
A constant sequence _does_ converge (to that same constant). To me that's obvious, but if you want to use the official definition of lim_(n→∞) a_n on the constant sequence a_n = C, that works too: for any ε there exists N (in fact, N = 1) so that if n > N then |a_n - C| < ε (in fact, |a_n - C| = 0).
@@theadamabrams ah, yes, true, true. Hopefully you understood the spirit of my words rather than the literal meaning. That said, your correction is valid. Thanks.
The relation between pi and diameter remains for the smaller semi circles. It's always diameter : semicircle -> n : n/2*pi
The intuition here comes from the theorem, if Sn is a sequence that converges to L, and f is continuous, then f(Sn) converges to f(L).
In this case, our sequence Sn, is a sequence of semi-circles. And L is the line. But now we need to look at what f is. f would be some sort of function, computing the length of the object.
Fundamentally, the function f, that computes arc length, with respect to path, is discontinuous. To be more precise, if I have a function f, mapping C-->R, where C is the set of all paths, and R the reals. If I take two paths c1 & c2, both infinitely close together, it is possible that the arc length f(c1) is wildly different than the arc length f(c2). Small changes in path, does not necessarily mean that there will be small changes in path length.
For intuition on why two paths can be infinitely close, but different lengths. I would tell you that "direction" is one of the most important things to consider, when computing arc lengths. The direction of the two paths are extremely different. The derivative of one, does not converge to the other. The infinite case of the semi-circles is alternating between pointing left and right, while also pointing mostly down. The line, is only pointing down.
For intuition on why discontinuity matters, you can look at simpler examples when considering infinite processes. lim{3, 3.1, 3.14, 3.141, 3.1415, ...} converges to pi. But consider the function f(x) = 0 if x is rational, 1 if x is irrational, a clearly discontinuous function. lim f({3, 3.1, 3.14, 3.141, 3.1415, ...}) = lim {0,0,0,0,0, ... } = 0 != 1 = f(pi) = f(lim{3, 3.1, 3.14, 3.141, 3.1415, ...}). The limit of the function is only equal to the function of the limit, when the function is continuous.
By the same argument, I can also state that √2=1 can't I?
Take a square, with the sides being 1 unit each. Divide two of the sides in half, and connect them in the middle, essentially creating an L shape. This new shape has the same circumference as the starting square. Repeat a few more times, and you should have a staircase, still the same circumference as the starting square. If we repeat this an infinite number of times, we'll eventually get a triangle. But the 45-45-90 triangle's sides are 1, 1 and √2. Yet, we just found our "staircase" shape's sides to be 1 unit each. Does this make sense? If it doesn't I might make a short video and link it here.
3b1b did a video on this. The partial, finite perimeters don't actually approach a value because the perimeters shown aren't approaching a straight line; they'll always be curved, no matter how many times we recur the pattern
It does approach a straight line in the limit though. This explanation is incomplete. The limit of the points is exactly on the line.
Actually the semicircles are converging to the diameter.
Same fallacy as the two sides of a square "converging" to the diagonal of that square. Nope!
To get from lower left to upper right corner. First we go across the bottom, then up the right side. Next go halfway across the bottom, then up for half the side, then finishing going across and up the right side. Two steps, still twice the side of the square, same end. Make as many tiny steps as you like, the path is always 2, never becomes square root of 2.
The fractal density is Pi - 2
Since the path is always Pi long whether the frequency is 1 or 128.
But the travel distance is 2
So actual travel Pi minus minimum path 2 = 1.14159.... is the fractal dimension of a flat< circle.
??
Cantor dust is not length zero. Same idea.
Yeah it reminds me of the pi=4 thing where you take a square and fold the corners until it almost perfectly resembles the circumference of the circle
This is why the length of a curve is defined as the integral of the derivative of the parameterization. This way, any limiting process has to ensure the derivative converges, which is not the case in this video.
This is similar to the issue with measuring coastlines accurately. When calculating smaller measurements, you will end up getting a much larger length than when estimating from a larger scale, so while it looks like two, all the tiny curves still add up to π
If you solve using limits i.e
Limit nXπ/n
N times circumference on small circle which is π/n
So answer is π
The small semi circles look flat, but they aren't.
The sum is always the same: pi. It those not matter what tiny are the ripples, there are that many as the divisions are. and there is not reason to not take them into account. And, by the way, the limit is not a line, It is a ripple, lim|n->inf| n (pi 1/n) = lim|n->inf|| pi = pi. In limits of this types, what takes out of consideretions are the fixed amount of something, whitch those not happens in this example.
If we consider that:
C1 = pi*l,
C2 = 2*(pi*l/2)
C4 = 4*(pi*l/4) and so on, we have
Cn = n*(pi*l/n) that is always pi*l, independently from the numbers of circles
This is why the coastline of California is infinite.
Lim_{n->\infty}\sum_i=1^n length(halfcircle(n)) = Lim_{n->\infty} pi = pi. whereas length(Lim_{m->\infty} \cup_i=1^m halfcircle(m)) = length(straightline). Note in one case you take the limit and then compute length and in the other you compute the length and then take the limit.
A function of the limit is not automatically the same as the limit of a function.
I got the feeling that "π" holds many secrets and it's the key to understand the mysteries of the universe
Don't even get me started on infinitesimal angles
coastline paradox lore
The term "infinitly squiggly" comes to mind. It will never be truely flate
If you do it indefinitely, the ratio under pi will approach infinity. Making the whole ratio approach 0. Meaning the circle will have a radius of 0. Which is just a straight line
Increasing limits (integrals) are supposed to DECREASE the error, until at the limit (infinity), the error is zero. In this case at Infinity you still have a very well defined error
It’s because the semi circles never actually flatten. No matter how small you go, the ratio for each semi circle will remain 2 for the diameter and pi for the circumference. So if you were to take all of the semi circles when they look like a flat line and stretch them out, they would have a length of pi.
Because if you are taking the sun of limit as x->0 of 2pi/x you are trying to integrate 1/x.whic is ln)0) or undefigned. Correctly, you view it this way. For every division of semi- circles (n) you get 2n. n* = 1/n semi circles, and 2n*/X is the expression so you meed the lim X-> 0, n*->0.you get simply 2n*/x divisions. Because as sum 2 n*/x lim n-> 0 x-> 0 the expresion is 2 [ n* is a radian factor of division of semi-circles to infinity small os the 2pi/d segments get small.]
It's easy to see that each time you double the number of semi-circles, you halve the length of each semi-circle, so the total length is still pi, and that will always be the case, no matter how many times you double it.
While the curves get infinitely smaller, there is also infinitely more curves. The ratio of how much smaller they are to how many there are stays the same.
The derivative must also converge, those curves will never flatten out.
It’s the same concept as the staircase paradox (I’m not sure if that’s its name)
true in way different context. if taken into calculus, it becomes easy to understand. the arcs of all semicircles is given by this formula
ₙ
Σₖ ₌ ₀ π/2ᵏ, where k is the number of divisions of the initial semicircle.
the circumference of the semicircle is L = πr. the perimeter of the semicircle is P = L + 2r = L + 2
taking the limit to infinity gives the following:
L = lim k→+∞(π/2ᵏ) = 0
this means that as we approach dividing the semicircle into infinite parts, the arc will become 0 and the shape will no longer be a circle but a segment, so we get that P = 2r = 2.
the perimeter is EQUAL to 2 and this is nothing more than a segment, when we take it to infinity. this however does NOT make π equal to 2 because it's a constant and the actual fact here is the arc keeps getting smaller. π will continue to exist until it hypothetically approaches infinity and the fraction becomes 0.
That's the same principle as the missing chocolate piece "paradox": what we see is different from what it is. The chocolate has many spaces from the piece you remove, the same way this has indefinitely many arcs, even if we can't see it.
This is sort of like an inverted "Coastline paradox"...
Come to think of it. This sounds like something that would have a name of it's own.
Edit: It did.
"Staircase paradox"
A fractal cannot accirately approximate a smooth surface. Neither can limits be assumed to be equality in all situations.
Diameter assumes 0 thickness, point a to point b. Semicircle divided infinitely will always have a thickness due to each semicircle having a radius.
This was misleading when mentioning the "limiting process". Since we note that the perimeter remains pi after several iterations, there is no reason to believe it will decrease at the limit.
This reminds me of the problem of measuring coastlines. As the measurements of a coastline get more precise, the longer the total coastline will be measured. This is because if you draw a straight line and deem a stretch of coast as 10 kilometers long, then measure the same coast in 100 meter increments; that same coast may appear to be 11 kilometers because of all the zigs and zags
No matter how small you draw the curves, the proportion of this curve in relation to the radius is always the same.
There will always be area underneath the curves, forcing them to have a longer total length than 2
You are "squishing" it all down into a smaller and smaller space. We're used to consider things with areas and volumes. So our tiny brains cannot really process the fact that you could fit an infinitely long line into an infinitely small area.
The limit of f(g) is equal to the limit of g(f) only if both f and g are continuous. The perimeter function is not.
Imagine two paths: one that never deviates into another dimension and another that deviates along a single dimension. If they share an origin, the second path will have a length of pi, and the straight path will have a length of 2. If the wandering path moves in positive and negative directions on the added dimension, the length becomes 2 pi.
Let me correct it . Let's increase the value of radius ( r ) to 2 and then it will be π = 4 if you notice and if we increase more , π will increase too. So , you mean the value of π is variable or equal to the diameter of a circle regardless of its fixed value π = 22/7 . Don't you think this contradiction has arisen due to your wrong assumption that the sum all of those small circles will be equal to the radius of the circle.
YES! I see the fallacy in the "proof" that Pi = 2. The "proof" neglects to take into account the fact that within sufficiently small spheres, the effect or influence of size upon shape can be made as small as one pleases in all respects.
It fails because the curves never squash down to the length of the line. No matter how many recursions you do, the curves will never be straight.
fractals are so mindblowing
Phew, you scared me for a second. I thought you were going full blown Terrence Howard on me. Anyway, the definition of a line states that it has "no width or curvature", so any equation that includes line width is erroneous.
The wiggles become smaller but don’t (mathematically) vanish. If the do, you could just smash that first half circle into the diameter and call it two. Limits work differently, with each subsequent change becoming smaller. Here the length doesnt change.
Yes, the formula for the sum of the ever decreasing semicircles is n×Π/n for the curved surface but (2n/n) for the diameter. Where n is the number of semicircles making up the diameter.
180 degree arch = pi times the radius so it’s not possible to be the same length as the circumference
Also it never flattened so can’t be the same as the flat line but the math stuff makes me sound smart
Plot twist: the original semicircle is a part of another semicircle
The 2 will multiply the π and because it is radius then the formula will be π multiply by diameter ( or π multiply by radius multiply by 2)
The thing that most people is forgetting is that while the radius of the circles are aproaching to zero, the quantity of circles are aproaching infinity, so the almost zero in lenght curve of the semicircles continue being relevant, cause there are almost infinity circles
The argument never «fails» because you didn’t make one. You just said that the total circumference of the half circles stays at pi no matter how far down you go. So why should the sum suddenly change to 2? «Squash down and flatten» is exactly what they don’t do.
Mmm we can say pi =/= 2 because if it did, the original circle would be a line in the first place, so all we did was define the length of some constant which happens to be 2, with no relation to pi at all.
The semi-circle is curved
If you try to straighten it, it will be longer than 2
Thus proofing pi > 2
( _i might spelled it wrong so you can correct me by replying to me_ )
Just because it's too small for us to see that it's a bumpy line, doesn't mean it isn't a bumpy line
But the diameter is the length of the straight line that bisects it. And a straight line is defined as having no width, depth or curvature, not even teeny weeny bumps.
No matter how many times you divide the circles, you will always be able to zoom down and see tiny half circles, never actually forming the straight line that is equal to 2
The proof that π ≠ 2 is proven by the fact that I cannot flatten a crease out of a piece of tape on a surface no matter how many ways I poke at it...
Since the perimeter of a semicircle is known to be pi * r, and the perimeter of a semicircle with r=1 must be longer than 2r, pi must be greater than 2.
The semi circles don’t flatten. Using math : If n is the iteration then the sum of the length of the arc is 2^n pi / 2^n … the factor (2^n) cancels out and the length is always pi no matter how large n is.
If you make it infinitely small, the infinity will accumulate accordingly.
i was thinking more along the lines of what was obvious like ∞×π\∞=π
It only visually approaches the line. Because the curve never flattens the circles aren't even approaching the line
The infinite bumps adds an extra 1.1428 to make it 3.1428... That is even if limiting, the bumps carries extra length and their infinitely large numbers adds more than just 2.
if you do the math just right, you can make any number equal any other number
For the same reason that a coastline actually has infinite length. The semicircles create a fractal curve
This reminds me of the coastline paradox
Well I try to explain with x,y graph coordinates - line is one dimensional (x = any number, y = always the same number) but semicircle cannot be one dimensional (x = any number, y = cannot be always the same number with different x value). In other words, the x,y coordinates have to follow equation that will never get you a straight line. I studied biology so don't bully me much if I'm wrong 😬.