Quote
"Don’t lose your passion in a subject over the worry of exams."

— Emil Kerimov

Link

My flipboard magazine

Photo
Very non-rigorous proof of the integral sinx/x. But it is maybe the easiest to show first. So far I have 4 different proofs of this integral, it’s become kinda a weird hobby of mine. 

Very non-rigorous proof of the integral sinx/x. 
But it is maybe the easiest to show first. So far I have 4 different proofs of this integral, it’s become kinda a weird hobby of mine. 

Photo
Simple Derivation of a useful integral. Almost like an extended gamma function. Kinda looks almost like the inverse of the coefficients of a taylor series. 

Simple Derivation of a useful integral. Almost like an extended gamma function. 
Kinda looks almost like the inverse of the coefficients of a taylor series. 

Text

Interesting Facts

I’ve decided to try and post some interesting facts, so consider this the first installment. (Some of them will be from QI)

Fact 1: Capital of Thailand? 
The answer you’ll get even from Wikipedia is BangKok, but…… 

http://suite101.com/article/is-bangkok-the-true-name-of-thailands-capital-a122682


Which essentially summarizes to the city was never named Bangkok, ever. It’s just something people not from Thailand made up. 

Quote
"It makes the smallest of differences for me, a delta for my very large epsilon if you will."

— Emil Kerimov

Text

Two hard Math Questions part 2

8. Find all functions f(x) so that integration(dx/f(x)) * integration(f(x) dx) = c. Where c is a constant. To make it easier to write, instead of integration I’ll use these, brackets,{ }, and instead of f(x), I’ll use y. And I’ll leave out the dx’s. 

{1/y} {y} =  c

Taking derivative of both sides: 

{1/y} * y + 1/y * {y} = 0 

- {1/y} = 1/y^2 * {y}

Again taking derivative of both sides:

-1/y = (-2/y^3) * y’ * {y} + 1/y

-2/y = (-2/y^3) * y’ * {y} 

y^2 = y’ * {y}

(y^2) / y’ = {y}

Taking derivatives of both sides one final time:

(2y*y’)/y’ + (y^2) * (-1) * (y’)^-2 * y” = y

2y - (y^2) * (y’)^-2 * y”) = y

y = (y^2) * (y’)^-2 * y”)

(y’)^2 = y * y”

(y”)/ y - (y’ * y’)/(y^2) = 0

Noticing: (y’ / y)’ = y” / y - y’ (y’ /(y^2)) 

(y’ / y)’  = 0

Integrating both sides:

y’ / y = c1

y’ = c1* y

(1/y) dy =(c1) dx

Integrating both sides:

ln(y) = c1x + c2

y = e^(c1x+c2)

y = (e^c2) * e^(c1x)

e^ c2 = some constant, lets call it c2

y = c2 e^(c1x)

Emil Kerimov

Text

Two hard Math Questions part 1

So on my math exam there were two quite hard questions, that I didn’t get during the exam. Even at home I needed quite some time to do them. Here they are: 

5b) sinx * y” + (y’^2 - sin^2 x ) ^0.5 * (y`)^2  - cosx * y’ = 0 with y(0) = 0 and y’(0) = 1

(sinx * y”)/(y`^2) +   - cosx / (y’) = -(y’^2 - sin^2 x ) ^0.5

Noticing (sinx / y’)’ = (sin x)’ (1/y’) + (sin x) (1/y’)’ = cos x / y’ - sinx * y” / (y’^2)

(sinx / y’)’  = (y’^2 - sin^2 x ) ^0.5

(sinx / y’)’  =  y’ * (1 - (sinx / y’)^2 ) ^0.5

Making u = sinx / y’  => y’ = sinx / u

u(0) = sin(0) / y’(0) = 0/1 = 0 

Plugging u into our equation: 

u’ = sinx / u * (1- u^2)^0.5 

(u * (1-u^2)^-0.5) du = (sinx) dx

- (1-u^2)^0.5 = -cos x + C

using u(0) = 0

-1 = -1 + C  => C = 0

 (1-u^2)^0.5 = cos x

1 - u^2 = cos^2 x

u = sinx = sinx / y’ => y’ = 1

y = x + C

y(0) = 0  => C = 0

y = x 


Now to check that in the first step we can divide by (y’)^2 without ever dividing by 0, or in other words, finding out the domain for which the answer hold:

y’ = 1

Therefore, this answer holds for every value of x. 

Emil Kerimov

Photo
This actually happens because the brain doesn’t take in all the information that we see and hear, because that would be next to impossible, instead tries to focus on the important stuff. What happens with this though is that the brain realizes that it missed important information and can sometimes re-send all that information back up into the brain to collect the necessary data(neurons refire). Thus we understand it, however not before our reflex’s have uttered “what?”.

This actually happens because the brain doesn’t take in all the information that we see and hear, because that would be next to impossible, instead tries to focus on the important stuff. What happens with this though is that the brain realizes that it missed important information and can sometimes re-send all that information back up into the brain to collect the necessary data(neurons refire). Thus we understand it, however not before our reflex’s have uttered “what?”.

(Source: asdfghjkllove, via cinderinaa)

Photo
Okay, now I enjoy proper grammar just as much as any other bloke, but if the thought gets communicated without any ambiguity then what’s the big deal about using proper grammar?
Grammar itself is simply a social construct to avoid ambiguity in the messages between what goes on in a persons head to how other people perceive these patterns of commonly known sounds(words).
So if the purpose of grammar is fulfilled without the actual use of it in some cases, then there should be no fault in that whatsoever. So these so called “grammar-police” can suck it.  

Okay, now I enjoy proper grammar just as much as any other bloke, but if the thought gets communicated without any ambiguity then what’s the big deal about using proper grammar?

Grammar itself is simply a social construct to avoid ambiguity in the messages between what goes on in a persons head to how other people perceive these patterns of commonly known sounds(words).

So if the purpose of grammar is fulfilled without the actual use of it in some cases, then there should be no fault in that whatsoever. So these so called “grammar-police” can suck it.  

(Source: observando, via cinderinaa)

Text

Mean Girls Limit Problem

Mean Girls isn’t really a “new” movie, and I was like in grade 9 or 8 when it came out. And was amazed at the complex math that I didn’t understand in it. But after all this time, and me learning so much about math and being quite good with limits I thought that it was probably a really easy question so I googled it to check. However, I was quite interested with this problem, shown here:

Okay, so obviously the first thing I noticed was that 1 - cos^2 x = sin^2 x, that just might simplify things, but not very useful. 

Another easy thing to notice is that if we plug in “0” for x, we get 0/0 because sin 0 = 0, cos 0 = 1, and ln 1, meaning to what power we have to raise e to to get 1, being 0.

Normally people would just normally jump straight into L’Hopital’s rule which can be summarized into: 

If  exists and becomes 0/0 or infinity/infinity then  is correct to use.

If we follow our initial gut feeling to use L’Hopital’s rule the problem becomes :

lim [x->0] (-1/(1-x) - cosx)/(-2cosxsinx)  = -2/0 = inf. We can thus say that the limit is infinity or in some textbooks, doesn’t exist. Normally both would be right, however, we made a mistake here. The restriction of the first condition doesn’t come up often, but does here. 
A limit only exist if  . Which means that the limit of the number approaching from the negative and positive sides are equal. And in this case, 

lim_(x->0^-) (log(1-x)-sin(x))\/(1-cos^2(x)) = infinity\nlim_(x->0^+) (log(1-x)-sin(x))\/(1-cos^2(x)) = -infinity

Log being another way of writing ln, or natural logarithm, depending on what country you’re in, different notation is used. The reason for this difference in side limits is because of the ln x function and sin x function respectively: 

 

with the limit of approaching 0 slightly from the negative side, the ln(1-x) becomes a slightly positive 0 while the sin x becomes slightly negative and since it is being subtracted the entire top becomes a slightly positive 0, but with the limit approaching 0 slightly from the positive side, ln(1-x) becomes slightly negative and sin x becomes slightly positive and then is subtracted from the slightly negative ln(1-x). Thus the limit does not exist. 
This question requires a little be of a different approach than most limits, so in that sense, is a really good question. 

Quote
"Nothing like taking down a tank with a slingshot"

— Emil Kerimov