Strange/Interesting Mathematical Phenomena

[Nathanael D. Striker]

What are your favorite mathematical concepts that are strange due to them violating preconceived notions or are just interesting? To get this topic started, let me show you a proof of one of my favorites: the probability of hitting a rational number in the set [0,1] is zero.

 

Before we begin, let us define the Lebesgue Measure. In essence, it is a way to measure subsets of Rⁿ (for a more detailed definition, click on the above link). For our purposes, we are interested in the following property: the Lebesgue Measure of a closed interval [a,b] is m([a,b]) = b-a, where a

 

The Lebesgue Measure of [0,1] is trivial to find as it is a direct application of the property in question: m([0,1]) = 1.

 

The Lebesgue Measure of Q is not as straight-forward to find. In order to motivate this result, let us consider the following: Q = U(q_i). What this says is that Q can be thought up as the union of an infinite number of rational numbers q_i (there are a countably infinite number of rational numbers). This separation of Q allows us to use our property above: m(Q) = m(U(q_i)) = sum(m(q_i)) = sum(0) = 0, where sum() is the summation function.

 

So, we have that m([0,1]) = 1 and m(Q) = 0. This is enough to show that the probability of hitting a rational number in the set [0,1] is zero, but it begs the question of what makes m([0,1]) = 1 if m(Q) = 0. Well, we can define the irrational numbers as R/Q, which denotes the set of elements of the real numbers that are not in the set of rational numbers. As we are concerned with the set [0,1] and not R, we can redefine the irrational numbers as [0,1]/Q. Now, the m([0,1]) = m(Q ∩ [0,1]/Q) = m([0,1]/Q) = 1. Therefore, the Lebesgue Measure of the irrationals in the set [0,1] is 1.

 

The above result not only suggests that the probability of hitting a rational number in the set [0,1] is 0, but it also suggests that the probability of hitting an irrational number in the set [0,1] is 1. Now this makes sense since P(S) = 1, and we have defined S to be S = [0,1] in our example.

 

And with that, let us discuss the topic at hand: What are your favorite mathematical concepts that are strange due to them violating preconceived notions or are just interesting?

[~~~~]

I flunked A level maths, sorry.

[Ryusei the Morning Star]

e^( π*i)+ 1= 0

[Nathanael D. Striker]

  On 11/3/2017 at 4:44 PM, White said:

e^( π*i)+ 1= 0

And do you know why that's true? I'm sure others would like to know.

[Ryusei the Morning Star]

  On 11/3/2017 at 4:47 PM, Hades said:

And do you know why that's true? I'm sure others would like to know.

Want me to prove it?

 

It relates to complex Taylor series

[EHN.]

  On 11/3/2017 at 5:23 PM, White said:

Want me to prove it?

 

It relates to complex Taylor series

 

Taylor/Maclaurin is a bit of a whimsy proof. It's like yeah LHS = RHS but it doesn't really give an intuitive explanation of why.

[Nathanael D. Striker]

  On 11/3/2017 at 5:23 PM, White said:

Want me to prove it?

 

It relates to complex Taylor series

Why yes, if you are willing. Though, Euler's Equation would give a simpler proof.

[EHN.]

  On 11/3/2017 at 6:15 PM, Hades said:

Why yes, if you are willing. Though, Euler's Equation would give a simpler proof.

 

I'm pretty sure he knows and wanted to give a general proof for Euler's formula too. I might be wrong though.

[Ryusei the Morning Star]

 

 

 

The Maclaurin series expansion for cos(x) is given by

 

 

(you can plug i*x for x in the above series)

 

cos(i*x)= 

 

Maclaurin series for sin(x) is

 

 

 

(you can plug i*x for x in the above series)

 

-i*sin(ix)=

 

 

cos(ix) - i sin(ix)=e^x e^(π*i)= cos (i* π*i))- i(i* π*i)= cos ( -π) - i sin ( -π) = 1-0=1

 

 

Pretty simple overall

[Nathanael D. Striker]

  On 11/3/2017 at 6:23 PM, EHN. said:

I'm pretty sure he knows and wanted to give a general proof for Euler's equation too. I might be wrong though.

 

And there exists a straight-forward proof for that as well.

 

Let's define y = cos(x) + i*sin(x).

Taking the derivative: dy/dx = -sin(x) + i*cos(x) = i[cos(x) + i*sin(x)] (i^2 = -1) = i*y

Rearranging both sides: dy/y = i*dx

Integrating both sides: ln(y) = i*x + c

Turning both sides to a power of e: y = exp(i*x+c) (exp(ln(y)) = y due to inverse)

So, we have exp(i*x+c) = cos(x) + i*sin(x). At x = 0, exp© = 1 ==> c = 0. Therefore, exp(i*x) = cos(x) + i*sin(x)

 

Proving exp(i*pi) + 1 = 0 can be shown by letting x = pi in the above equation.

[Ryusei the Morning Star]

  On 11/3/2017 at 6:36 PM, Hades said:

And there exists a straight-forward proof for that as well.

 

Let's define y = cos(x) + i*sin(x).

Taking the derivative: dy/dx = -sin(x) + i*cos(x) = i[cos(x) + i*sin(x)] (i^2 = -1) = i*y

Rearranging both sides: dy/y = i*dx

Integrating both sides: ln(y) = i*x + c

Turning both sides to a power of e: y = exp(i*x+c) (exp(ln(y)) = y due to inverse)

So, we have exp(i*x+c) = cos(x) + i*sin(x). At x = 0, exp© = 1 ==> c = 0. Therefore, exp(i*x) = cos(x) + i*sin(x)

 

Proving exp(i*pi) + 1 = 0 can be shown by letting x = pi in the above equation.

Or that

[xlArisenRoselx]

Is it bad that these are all just characters to me? 

[EHN.]

  On 11/3/2017 at 6:36 PM, Hades said:

And there exists a straight-forward proof for that as well.

 

Let's define y = cos(x) + i*sin(x).

Taking the derivative: dy/dx = -sin(x) + i*cos(x) = i[cos(x) + i*sin(x)] (i^2 = -1) = i*y

Rearranging both sides: dy/y = i*dx

Integrating both sides: ln(y) = i*x + c

Turning both sides to a power of e: y = exp(i*x+c) (exp(ln(y)) = y due to inverse)

So, we have exp(i*x+c) = cos(x) + i*sin(x). At x = 0, exp© = 1 ==> c = 0. Therefore, exp(i*x) = cos(x) + i*sin(x)

 

Proving exp(i*pi) + 1 = 0 can be shown by letting x = pi in the above equation.

 

Yes I know. I'm just really confused how you can think that complex integration with several steps is any more straight-forward than equating 2 sides of a series expansion.

 

Either way my first point still stands with both variants of the proof. 

[Ryusei the Morning Star]

  On 11/3/2017 at 6:53 PM, EHN. said:

Yes I know. I'm just really confused how you can think that complex integration with several steps is any more straight-forward than equating 2 sides of a series expansion.

 

Either way my first point still stands with both variants of the proof. 

Complex integration doesn't need infinite srs

 

My personal fav tho is way to evaluate any indefinite integral I came up on the toilet

 

Basically got to this result from Integration by parts

∫f(x)g'(x)dx= f(x)g(x) - ∫f'(x)g(x)dx

Therefore is stands that

∫f(x)g(x)dx=f(x)∫g(x)dx-∫f'(x)(∫g(x)dx)dx

(this can be verified by differentiating both sides)

=>

f(x)g(x)=f'(x)∫g(x)dx+f(x)g(x)-f'(x)∫g(x)dx

Given: ∫f(x)g(x)dx=f(x)∫g(x)dx-∫f'(x)(∫g(x)dx)dx

set g(x)=1

∫f(x)1dx=∫f(x)=f(x)∫1dx-∫f'(x)(∫1dx)dx=xf(x)-∫f'(x)*x dx

Repeat the "product rule for integration) for ∫f'(x)*x dx

which gets you:

xf(x)-f'(x)∫x dx+∫f''(x)(∫x dx)dx = xf(x)-x²f'(x)/2+∫f''(x)x²/2 dx

Now expand

∫f''(x)*x²/2 dx

(you might notice a trend here)

xf(x)/(1)-x²f'(x)/((2)(1))+f''(x)x³/((3)(2)(1))-∫f'''(x)x³/((3)(2)(1)) dx

You can expand that too, but the pattern is pretty clear at this point

Terms alternate in sign, starting on + for the first term. Let's call this n=0

You multiply the nth derivative of the function you want to integrate f(x) by xⁿ⁺¹ then divide by (n+1)!

∫f(x)dx=∑n=0∞ ((-1)ⁿfⁿ(x)xⁿ⁺¹)/(n+1)!

Benefits of this over say Talyor? It has an infinite radius of convergence given you can infinitely differentiate a function at a point.

Example: Finding an expansion of ln(x+1)

f(x)=1/(x+1)

∫1/(x+1) dx:

(-1)⁰(x)(1/(x+1)) / (1!) = x/(x+1)

+

(-1)¹(x²)(-1/(x+1)²) / (2!) = x²/(2*(x+1)²⁾

+

(-1)²(x³)(2/(x+1)³) / (3!)= x³/(3*(x+1)³⁾

Might notice another pattern

ln(x+1)= ∑n=0∞ (x/(x+1))ⁿ/n

Let's test it:

ln(1)=0

ln(0+1)= ∑n=0∞ (0/(0+1))ⁿ/n = 0

[Dad]

  On 11/3/2017 at 6:49 PM, Kyumi said:

Is it bad that these are all just characters to me? 

 

Not at all.  Everyone hasn't reached this level of mathematics yet.  Learn it at your own pace.

[Ryusei the Morning Star]

  On 11/3/2017 at 6:49 PM, Kyumi said:

Is it bad that these are all just characters to me? 

Never feel that way

[KatnissGod]

I feel dumb for just scrolling through and thinking everyone just started slamming their faces on keyboards

[EHN.]

It's all just a matter of context. When you're a kid you think basic algebra looks impossible but when you start learning it you realise it isn't that bad. It's exactly the same situation here. 

[Ryusei the Morning Star]

  On 11/3/2017 at 9:10 PM, EHN. said:

It's all just a matter of context. When you're a kid you think basic algebra looks impossible but when you start learning it you realise it isn't that bad. It's exactly the same situation here. 

Thoughts on my integration idea

[Nathanael D. Striker]

  On 11/3/2017 at 6:53 PM, EHN. said:

Yes I know. I'm just really confused how you can think that complex integration with several steps is any more straight-forward than equating 2 sides of a series expansion.

 

Either way my first point still stands with both variants of the proof.

It was just integrating i, so it just a simple integration. If it was something like (3+4i)^2, then that would need a little more work. And tbh, I do a lot of integration, so it just comes more naturally.

[JessicaMuddy]

I thought this one was really cool. VSauce is always worth a watch.

 

[Nathanael D. Striker]

  On 11/3/2017 at 11:54 PM, 【 M U D D Y 】 said:

I thought this one was really cool. VSauce is always worth a watch.

 

https://www.youtube.com/watch?v=J51ncHP_BrY

Well, I haven't seen that before. Very interesting indeed. Thank you for that inclusion.

[LordCowCowCowCowCowCowCowCow]

I have nothing to add but I want to say I'm weirdly happy to see this kind of thread here of all things. Good on you, nerds.

Math was always something I couldn't get into due likely to the teachers I had for it. It's something that really needs a good teacher to interest anyone.

[Dad]

Believe it or not, teachers/professors do play a big part in how you learn.  Personally, I haven't reached this level of math but I find it interesting.  I'm hoping that when I do get here, my professor is amazing.

[mitchermitcher]

1 + 2 + 3 + 4 + 5 + 6 + ......... = -1/12