Chemistry (1)

Problem:

Solution:

Oxygen has 8 protons always
The isotope with mass number 16 has 8 neutron

The isotope with mass number 17 has 9 neutron

The isotope with mass number 18 has 10 neutron

Limit set

Problem:

Solution:

Consider $ x'^2 + y'^2 = (x - y - x\sqrt{x^2 + y^2})^2 + (x + y - y\sqrt{x^2 + y^2})^2 $

This can be expanded as follow:

$ x'^2 + y'^2 = (x - y)^2 - 2x(x - y)\sqrt{x^2 + y^2} + x^2(x^2 + y^2) + (x + y)^2 - 2y(x + y)\sqrt{x^2 + y^2} + y^2(x^2 + y^2) $

$ x'^2 + y'^2 = x^2 - 2xy + y^2 - 2x(x-y)\sqrt{x^2 + y^2} + x^2(x^2 + y^2) + x^2 +2xy +y^2 - 2y(x+y)\sqrt{x^2 + y^2} + y^2(x^2 + y^2) $

$ x'^2 + y'^2 = x^2 - 2xy + y^2 - 2x^2\sqrt{x^2 + y^2} + 2xy\sqrt{x^2 + y^2} + x^2(x^2 + y^2) + x^2 +2xy +y^2 - 2xy\sqrt{x^2 + y^2} - 2y^2\sqrt{x^2 + y^2} + y^2(x^2 + y^2) $

The whole point of the expansion is actually to simplify, let's group like terms and cancel first:

$ x'^2 + y'^2 = x^2 + y^2 - 2x^2\sqrt{x^2 + y^2} + x^2(x^2 + y^2) + x^2  +y^2 - 2y^2\sqrt{x^2 + y^2} + y^2(x^2 + y^2) $

$ x'^2 + y'^2 = x^2 + y^2 + x^2  + y^2  + x^2(x^2 + y^2) + y^2(x^2 + y^2) - 2x^2\sqrt{x^2 + y^2} - 2y^2\sqrt{x^2 + y^2} $

And then factorize:

$ x'^2 + y'^2 = (x^2 + y^2)(2 + x^2 + y^2) - 2(x^2 + y^2)\sqrt{x^2 + y^2} $

Now we realize it can all be written in terms of the norm of the vectors, so we will do it.

$ a'^2 = a^2(2 + a^2) - 2a^3 $

If it were to converge, the after update norm should be the same as before update norm, so we solve 

$ a^2 = a^2(2 + a^2) - 2a^3 $

$ 1 = 2+a^2 - 2a $

Therefore we get $ a = 1 $, in other words, if the system converge at all, it must converge on the unit circle!

An interesting sum

Problem:

$ 1^2 - 2^2 + 3^2 - 4^2 + \cdots - (2n)^2 $

Solution:

We will assume the formula:

$ 1 + 2 + \cdots + n = \frac{n(n+1)}{2} $

$ 1^2 + 2^2 + \cdots + n^2 = \frac{n(n + 1)(2n + 1)}{6} $

We split the sum into two halves:

$ 1^2 + 3^2 + 5^2 + \cdots + (2n - 1)^2 $ can be think of as

$ (2 \times 0 + 1)^2 + (2 \times 1 + 1)^2 + \cdots + (2 \times (n - 1) + 1)^2 $

Expanding them we get

$ (4 \times 0^2 + 4 \times 0 + 1) + (4 \times 1^2 + 4 \times 1 + 1) + \cdots + (4(n-1)^2 + 4(n-1) + 1) $.

Grouping terms, factoring and applying the formula, we get

$ 4\frac{(n - 1)(n)(2n - 1)}{6} + 4\frac{(n-1)(n)}{2} + n $

Simplifying we get

$ \frac{n(2n - 1)(2n + 1)}{3} $

The even number squared sum is easier, we can think of them as just a scaled version of the square sum

$ 2^2 + 4^2 + \cdots + (2n)^2 $

$ 4(1^2 + 2^2 + \cdots + n^2) $

The answer is simply $ \frac{2n(n + 1)(2n + 1)}{3} $

Therefore, the final answer is simply the difference of the two sums, and it is

$ -n(2n + 1) $

An exercise from Math StackExchange

Problem:

Give all the positive whole number solutions to the equation $ x^3−y^3 = 602 $

Solution:

Here is really just a replication of my answer on math.stackexchange.com:

Note that $ (x - y)^2 = x^2 - 2xy + y^2 $, we can write

$ x^3 - y^3 = (x - y)(x^2 + xy + y^2) = (x - y)((x - y)^2 + 3xy) $

For simplicity, let $ z = x - y $, we have

$ 602 = z(z^2 + 3xy) $

Suppose for a moment that $ z $ is known, now we can calculate

$ z^2 + 3xy = \frac{602}{z} $

$ 3xy = \frac{602}{z} - z^2 $

$ 3(x - y + y)y = \frac{602}{z} - z^2 $

$ 3(z + y)y = \frac{602}{z} - z^2 $

$ 3zy + 3y^2 = \frac{602}{z} - z^2 $

$ 3y^2 + 3zy + z^2 - \frac{602}{z} = 0 $

Despite the deceiving complexity, since $ z $ is assumed to be known, we can easily find $ y $ using the quadratic formula.

Now we have $ 602 = 2 \times 7 \times 43 $, so $ z $ can only be these options

* 1
* 2
* 7
* 43
* $ 2 \times 7 $
* $ 2 \times 43 $
* $ 7 \times 43 $
* $ 2 \times 7 \times 43 $

And the negative of these values

Out of these 16 choices, we can easily enumerate the solutions. Of course, many of these choices does not generate integer solution, just ignore them.

For example, if I choose $ z = 2 $, we get $ 11^3 - 9^3 = 602 $ and also $ (-9)^3 - (-11)^3 = 602 $.

Exercise from Quora

Problem:

Please find the problem here.

Solution:

$\begin{eqnarray*} (0! + 0! + 0!)! &=& 6 \\ (1 + 1 + 1)! &=& 6 \\ 2 + 2 + 2 &=& 6 \\ (3 + 3 - 3)! &=& 6 \\ (4 - 4 / 4)! &=& 6 \\ 5 + 5 / 5 &=& 6 \\ 6 + 6 - 6 &=& 6 \\ 7 - 7 / 7 &=& 6 \\ (\sqrt{8 + 8 / 8})! &=& 6 \\ (\sqrt{9 + 9 - 9})! &=& 6 \end{eqnarray*} $

Exercise from 9gag.com?

Problem:

http://9gag.com/gag/aVDxn1O

Solution:

Denote bottle by $ b $

$ b + b + b = 30 $, therefore $ b = 10 $.

Denote the hamburger by $ h $

$ b + h + h = 20 $, therefore $ h = 5 $

Denote the beer by $ e $

$ h + e + e = 9 $, therefore $ e = 2 $.

Now here is the fun

$ \int\limits_{2h - b}^{\infty}{\frac{b\sin x}{ex}dx} = \int\limits_{0}^{\infty}{\frac{10\sin x}{2x}dx} = 5\int\limits_{0}^{\infty}{\frac{\sin x}{x}dx} = \frac{5\pi}{2} $

The last integral is well known, for example, see:
http://math.stackexchange.com/questions/5248/solving-the-integral-int-0-infty-frac-sinxx-dx-frac-pi2

This video explains the Feynman trick!
https://www.youtube.com/watch?v=3LsXWPzlOhQ

Have fun!

Mathematical Analysis - Exercise 1.3

Problem:

Solution:

Suppose $ n = ab $ is not prime, then we can write $ 2^{ab} = (2^{a})^b $. Applying the identity we just proved in the previous problem, we have:

$ 2^{ab} - 1 =  (2^{a})^b - 1 =  (2^{a})^b - 1^b = (2^a - 1)(\cdots) $.

Therefore if $ a \ne 1 $, we have a non-trivial factor for $ 2^{ab} - 1 $.

The contradiction show $ n $ has to be prime.