— Emil Kerimov

My flipboard magazine

### 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.

— Emil Kerimov

### 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

### 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

### 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,

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.

— Emil Kerimov