Multiplication and division are equal in value and occur from left to right when separated by corresponding symbols, but the parenthesis directly associates with the 2 (without a symbol) thus separating the entire segment 2(2+1) as the divisor.
Edit:...which is why multiplying by reciprocal works for division when symbolically separated.
Edit 2:
So in this case, multiplying by the reciprocal would yield 6*.166666666666..... which equals 1.
EDIT 3:
Well crap. I'm mostly wrong. It is apparently no longer the convention to do the parenthesis association with the adjacent number. Though it USED TO BE the convention. Every math teacher I had through all my school used the old convention, not the new one.
I don't understand why the new convention is used; you'd never separate the expression 6÷2x to be 6÷2*x
Traditional order-of-operations guides say nothing about implicit multiplication. If converted to a regular multiplication i.e. 6/2*(2+1) most would agree the answer is 9 (multiplication and division have the same precedence and are evaluated left to right). However, some treat implicit multiplication as an operator with higher precedence than regular multiplication, resulting in 6/(2*(2+1)) = 1. Since there is no convention regarding implicit multiplication either answer could be taken as correct.
By the way, if you enter the expression in Windows Calculator it simply refuses to accept the 2 before the brackets unless you put in a multiplication sign, thus avoiding the ambiguity entirely.
Something I saw on math.se and I was tooling around with it, and I figured I'd let you guys try it as well.
Let w be a 2mth primitive root of unity.
Is it the case that the following sequence has a period of 2m for almost all initial values?
xn+2 = (w4xn-(w3+w2)xn+1-wxnxn+1) / (-w-(w3+1)xn+w2xn+1)
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
Here's something I found interesting.
Tetration is repeated exponentiation. So 4x is equal to xxxx.
Between e-e and e1/e if you take the limit as n->infinity nx converges to a value, specifically W(-ln x)/- ln x, where W is the Lambert W function.
What is the area under this curve?
Build a man a fire, warm him for a day,
Set a man on fire, warm him for the rest of his life.
In a recent video the youtuber blackpenredpen tackles a double integral. A screenshot is the best way to show which:
This is essentially calculating the volume of an origin-centered half-cylinder that's capped by the surface x3+xy2.
The result is a rational number. Feel free to try to solve it. However, it baffles me a bit how it can be, given that a circle is involved.
In his most recent video, blackpenredpen solves this by using algebra. He ends up with the equation
4sin2(18°) + 2sin(18°) - 1 = 0
This quadratic equation gives two solutions:
sin(18°) = (-1 ± sqrt(5)) / 4
He discards the - version because it would result in a negative number, which clearly can't be the value of sin(18°). However, this got me thinking: The equation, and its solution, is still mathematically correct. The - version is just discarded because "it can't be the correct value", but it still has to hold some mathematical meaning. What is this meaning?
I'm definitely not an expert here, so I'd be glad to hear one.
Let me give a simpler example here to prove my point:
Let 2a=6, then
=> 4a2=36
<=> a2=9
<=> a=3 or a=-3
and with the assumption we can discard the negative value.
The "new answers" are introduced when you use a function which does not have a one-to-one correspondence (not being bijective), for example squaring.
It breaks the equivalence (<=>) and just becomes an implication (=>), meaning you can't get back to the assumption from that point on. Thus, it means that you might have to apply the assumption again to get the correct value.
The full solution with the additional (wrong) answers has meaning when you start from after the non-bijective function has been applied. For example, 4a2=36 => a=3 or a=-3.
In the video, the part which introduced new answers was applying sin() to 2θ=90°-3θ.
Squaring introduced one new answer, since for each y there is a maximum of two x solutions. Applying sin() introduces an infinite amount of new solutions, which later get reduced down to two solutions by applying other functions.
The other solution, (-1-sqrt(5))/4 ≈ -0.809, is the solution of for example sin(-54°) or sin(306°).
All these answers satisfy the equation sin(2θ)=sin(90°-3θ).
In other words: sin(2θ)=sin(90°-3θ) => sin(θ)=(-1±sqrt(5))/4, and adding the assumption θ=18° gives you sin(θ)=(-1+sqrt(5))/4.
Warning: Might glitch to creditsI will finish this ACE soon as possible
(or will I?)
Masterjun is correct. Looking at the solution formally you have:
A) The value of sin 18 exists and is unique.
B) The value of sin 18 satisfies the given quadratic equation.
C) The only numbers that satisfy the equation in (B) are the two numbers.
D) sin 18 is a positive number.
E) Only one of the numbers in (C) is positive.
Therefore, you can find the value of sin 18 using these statements.
When you write down an algebraic equation to solve a problem, you have to understand that it is simply a constraint that the correct value, when it exists, must satisfy. So, the algebraic equation is a different problem and might not be strong enough to solve the problem without further statements. The existence of spurious solutions just indicates that the equation you used is too weak to solve the problem alone, they don't have special meaning.
There is a massive confusion about this in analysis. For example, you give a sequence and ask for the value it converges to. When the equation is given in the form of a recurrence, it's possible to write an equation and find a value, and is what most people do.
However, this is incomplete, because the equation just says that, if the value it converges to exists, then the equation holds for that value. It says nothing about the existence. The sequence can diverge and you would still get a meaningless value as answer.
Example: consider the sequence with an+1 = 2/an with a1=1.
If you follow the procedure to find the limit you get L=2/L, L= +-sqrt(2).
The problem is, the sequence is just (1,2,1,2,...) and does not converge at all. So, in this case all solutions to the equation are useless and don't help you solve the problem.
Joined: 12/28/2013
Posts: 396
Location: Rio de Janeiro, Brasil
This channel posted some weeks ago this 'joke proof' of the 4 color theorem. It seems so simple, but I've tried it for a while and I couldn't find the mistake yet:
https://www.youtube.com/watch?v=Ryoipkksktg
Don't worry if you can't find the mistake. I only found it because I know a little bit the story from the four-color theorem. This is essentially Kempe's flawed proof of 1879, it took mathematicians eleven years to find the mistake. It's very complicated for an April fools joke.
I'll refer to the written version of the "proof" instead of the video, because it is easier.
Look at Figure 2b. The counterexample follows when you flip the red-green ring to pass below (circling the b-y blocks below instead instead of the y block above). Similarly, you can flip the blue-green ring and make it pass above, with the r-y blocks inside instead of only the y block. That's possible when the green block attached to the blue block marked 'b' also touches the one marked 'r'.
If the rings circle this way, it is possible that the yellow-blue chain attached to the 'y' block above touches the yellow-red chain attached to the 'y' block below. When you have a blue block from the first chain touching a red one from the second, you cannot swap colors as claimed, and the argument fails.
If you cannot understand it with me writing, there's a drawing on Fig. 9 in this article. It's extremely tricky, no wonder it took so long to find the error :)
Joined: 12/28/2013
Posts: 396
Location: Rio de Janeiro, Brasil
It's indeed very hard to find the mistake, but I was surprised it took mathematicians more than a decade!
I did assume it was something easier to spot, as it was stated as an April fools video. This is the simplest structure of a counter example I could make, based on the drawing of the author of the video.
I'm not sure how to reduce the size, so have this huge picture.
The place where the proof fails exactly is at 11:54: "... we can recolor the left blue square to green, the right one to yellow, and make the center blue."
To be less pedantic, there are four-color colorings of the second 5-neighbor case (at 11:16) which satisfy all the conditions supposed by this case, except that after switching blue/green, the right blue square cannot be switched to yellow without affecting the yellow square on the bottom-left. In particular:
* The red/yellow and red/green connections can cross, because they both share red.
* In this case, it is possible for a green region in the red/green connection to be contained within the red/yellow connection, connected through green/blue to the left blue square.
* Then recoloring the left blue square changes this green region to blue. This breaks the red/green connection above and possibly enables a blue/yellow connection between the right blue square and the yellow square.
* In that case, there is no way to recolor the right blue square to yellow without changing the yellow square to blue.
An example that shows this is in the image below.
Regions AEFCH surround the ? region with the same colors as in the video. AB is a blue/green region surrounded by red/yellow connection HDGE. Similarly, CD is a blue/yellow region surrounded by red/green connection HBGF.
The "proof" says we can switch the blue/green region AB, then the blue/yellow region CD, in order to color the ? region blue. However, if we switch the blue/green region AB, then B becomes blue. This creates the blue/yellow connection CDBE. Then switching the blue/yellow region containing C switches E to blue, making it impossible to color ? blue.
That was super interesting! I didn't manage to solve it, but my suspicions were on the right case (5 neighbours, blue colours opposite, ie p4wn3r's Figure 2b case). I worked out that the two lines could cross, which the video diagram doesn't make clear, but I didn't spot that recolouring the blue-green region could break the red-green line, thus disallowing the blue-yellow recolouring.
Does the proof say anything useful at all that can be salvaged? For example, is the 4-neighbour argument sound? If it is, can we use it to say that in any minimal non-4-colourable graph, every vertex has at least 5 neighbours?
(Of course, I know that this statement is vacuously true, because no such graph exists; but that requires a proof of the 4-colour theorem. I'm interested in what we can say from the arguments in this failed proof.)
The 4-neighbor argument is sound. You can also modify the arguments to prove that every planar graph can be colored with five colors, proving something nontrivial. Perhaps even come up with an algorithm to color a graph using at most five colors, which might even be more efficient than those that color in four, since the proof of the four-color theorem is so complex.
The commutative property of multiplication (of real numbers) is often deemed a kind of self-evident fact. However, from a quick google search I haven't found a formal definition that states that the rule is only valid when changing the order of a finite amount of terms. The property does not necessarily hold if changing the order of infinitely many terms in an infinite product.
For example, the Wallis product is:
2/1 * 2/3 * 4/3 * 4/5 * 6/5 * 6/7 * 8/7 * 8/9 * ... = π/2
However, if we rearrange those terms a bit, we get:
2/3 * 2/1 * 4/3 * 4/5 * 6/5 * 8/7 * 6/7 * 10/9 * 12/11 * ... = sqrt(2)*π/2
Clearly the commutative property for multiplication does not hold for infinitely many terms being exchanged.
This showcases how precarious it is to do operations on infinite series. Something that seems completely self-evident with finite series might not be true for infinite ones, and may lead to incorrect results. I was wondering if there are formal rules on which operations are valid or invalid for infinite series.
You don't need a rule for convergence of infinite products, you can borrow the one for series. Do this for your both products: take the natural logarithm of each member. The property of the logarithm transforms the products into summations.
Then, it is easy to see why the final result depends on the order. When the terms are less than 1, the log is negative, when they are more then one, the log is positive. So, this is an alternating series. If the series of the absolute value of the logs diverges, it is possible to rearrange the series to give any value at all. See here.
This is precisely what happens to the Wallis product. When you take the log, the terms greater/smaller than 1 become -log(1-1/2n)/-log(1+1/2n), which is just +-1/2n for large n. So, for terms of the same sign, you are summing something that grows like the harmonic series.
Since the harmonic series diverges, the Wallis product should converge only conditionally, and it should be possible to change the final value to anything using Riemann's construction.
I came up with a nice exercise today.
Let A, B and C be the angles of a triangle, which satisfy the relation
cot(A) + cot(B) + cot(C) = sqrt(3)
Prove that this triangle is equilateral. Bonus points if you can do it without calculus!
Let u=cot(A) and v=cot(B). Then:
cot(C) = cot(180°-(A+B)) = -cot(A+B) = -(cot(A)cot(B)-1)/(cot(A)+cot(B)) = -(uv-1)/(u+v).
Plugging into the above equation gives:
u + v - (uv-1)/(u+v) = sqrt(3),
or, after multiplying by 4(u+v) and rearranging:
4u2 + 4v2 + 4uv - 4sqrt(3)u - 4sqrt(3)v + 4 = 0.
To solve this equation, we will complete the square three times. We can write:
(2u+v)2 + 3v2 - 2sqrt(3)(2u+v) - 2sqrt(3)v + 4 = 0.
Substituting w=2u+v:
w2 + 3v2 - 2sqrt(3)w - 2sqrt(3)v + 4 = 0.
Then complete the square in w and v:
(w-sqrt(3))2 + 3(v-(sqrt(3)/3))2 = 0.
Since squares of real numbers are always non-negative, this gives us w=sqrt(3) and v=sqrt(3)/3. Then 2u+v=sqrt(3) and so u=sqrt(3)/3. This gives cot(A)=cot(B)=sqrt(3)/3, and so A=B=60°, and C=60°. Therefore the triangle is equilateral. QED
By the way, it's a lot easier with calculus; you can just plug in C=180°-(A+B) and differentiate, or use Lagrange multipliers. You can even simply say that if any two angles are not equal, say, A<B, then since A<180°-B, the derivative of cot(x) (which is -csc2(x)) at A is less than the derivative at B and so you can decrease the value of cot(A) + cot(B) + cot(C) by increasing A and decreasing B. Therefore cot(A) + cot(B) + cot(C) is bounded below by the value at A=B=C=60°; that is, cot(60°) + cot(60°) + cot(60°) = sqrt(3).
Edit: Fixed some signs.
Nice solution, Fractal!
My original solution without calculus was very different, which is cool.
First, suppose the triangle is acute. In this case, the angles A,B and C are between 0 and pi/2. At this range, the cotangent function is convex, and we can prove this easily using Jensen's inequality.
From the convexity of cot, we have
cot[(A+B+C)/3] = sqrt(3)/3 <= [cot(A)+cot(B)+cot(C)]/3
And from Jensen's inequality, the equality holds if and only if A=B=C, giving an equilateral triangle.
The hard part involves proving that no obtuse triangle can satisfy the identity. Let A >= pi/2, rewrite the sum as
S = cot(B) + cot(C) - cot(B+C)
Since B and C are smaller than pi/2, we can use Jensen:
cot[(B+C)/2] <cot>= 2*cot[(B+C)/2] - cot(B+C)
We now write B+C = pi/2 - x, and use the formula for the sum of cotangents to find:
S >= 2*(1+cot(x/2))/(1-cot(x/2)) + 1/cot(x)
Now, 1/cot(x) >=0 and (1+cot(x/2))/(1-cot(x/2)) >= 2, so for an obtuse triangle, S >= 2 > sqrt(3).
Therefore, no obtuse triangle can satisfy the identity, and the only solution is the equilateral one we found in the acute case. QED.
The Numberphile YouTube channel is quite famous and popular. Perhaps slightly less known and popular is the Mathologer channel.
I find the latter to be many times of much higher quality than the former, in that the presentation is often clearer and more understandable on many of the more complicated subjects. For example, Mathologer's explanation of the Kakeya needle problem is orders of magnitude clearer and easier to understand than the more obscure, technical and even boring explanation given in the Numberphile channel.
Sometimes it happens in the other direction, though. While Mathologer's video on "the most irrational number" was very good, recently Numberphile posted a video on that very same subject, and I actually found it even more illustrative and enlightening.
On that note, both argue that phi, ie. the golden ratio, is "the most irrational number". The latter video tangentially, and perhaps serendipitously, also gives an argument for what could be considered "the second most irrational number", which would be sqrt(2). (It's all about their continued fractions, and how fast they approximate the actual value.)
Indeed, it appears that if you need a "very irrational number" (that has the properties that their continued fractions have), sqrt(2) seems like an easy and good choice.
The whole parts of the denominators in phi's continued fraction are [1, 1, 1, 1, ...], which is the minimal such sequence, so it is indeed the most irrational number in that sense. While root 2 is also fairly irrational in that sense, it's not "the second most irrational number", nor is that concept well-defined. Consider that for any number but phi, you can obtain a more irrational number by, say, prepending a 1 to the list of denominators in its continued fraction representation.
This is, of course, going quite a lot on the more philosophical side of things, but if you do that, you are essentially adding a rational number to an irrational one. Whether the end result is "more irrational" is a matter of definition. (Can you make an irrational number "more irrational" by adding a rational number to it?)
There's absolutely no philosophy involved. If we accept, as you suggest, that the number is "more irrational" the harder it is to provide a rational approximation to it, then there is an infinite amount of numbers whose convergence properties are just the same as the golden ratio.
Convergence properties only depend on asymptotic behavior (how the sequence behaves for large n). As long as its continued fraction is some finite ordering of numbers and after that 1,1,1,1,1,1,..., the convergence properties are the same. That's quite easy to prove if you know the definition of the limit of a sequence.
So, Nickolas's remark is correct. Assuming the criteria in the videos you presented, there is no "second most irrational number", because an infinite amount of irrationals are tied on first place.