Don't Imagine, You will be Brave as Hell

Ok so I shamelessly stole this from a great foreign flick, can you guess which?

Just think about the title, seriously, just think about.

Apply it to anything and everything.

Nothing more needs to be typed.

Robin Williams Suicide

I don't get it. This guy was 63, super rich, married, and had 3 kids. Really why? I think that's the question most people are asking.

Honestly whenever I think of Robin Williams I think of a bad date I had in the 90s with some girl at the movie theatre at a Mrs. Doubtfire movie.

Anyways I don't get it. My life, and most people's lives, appear to be nowhere near as amazing as this guys, but he kills himself?

I just don't understand, Robin Williams was in Good Will Hunting, one of my all-time favorite movies. He got to make one of the greatest movies in the history of the freaking world, and he hangs himself?

I just do not get it.

Why Individual Stocks are Riskier than Other Investments

I could summarize this post with one sentence, seriously, you don't need read further after this,

Individual stocks present company specific risk.

Really, that's it, that's the biggest downside. I'm not saying one should NOT own any individual stocks, but if you do, realize that your entire investment could be completely wiped out. It does not matter how great or well known the company is, the whole "buy what you know" philosophy is just a clever catch phrase due to Lynch. 

Nobody knows what the future will bring.

Nobody is bulletproof.

Every convergent sequence is bounded, proof

Suppose we have a sequence a_n which converges to L.

Let e > 0, then there is an integer N > 0 such that for all n > N we have,

|a_n - L| < 1. Why 1? Well 1 is a number, we can use any positive number here.

Then adding and subtracting L and using the triangle inequality we have,

|a_n| = |a_n - L + L| <= |a_n - L| + L  < 1 + L for all n > N.


Set M = max{a_1, a_2, ..., a_(n-1), 1 + L}

Then, for ALL positive integers n we have,

|a_n| <= M.

This shows the sequence is bounded.

Understanding The Binomial Probability Mass Function Formula

The probability mass function for a binomial random variable X is given by,

P(x) = nCx p^x * q^(n-x)

This is the probability of exactly x successes among n trials in a binomial experiment where,
p denotes the probability of success
q denotes the probability of failure
n denotes the number of trials
x denotes the number of success
nCx is the number of ways to choose x objects from a group of n without regard to order


Ok so how does this formula make sense?

So we have n spots,

_ _ _ _ ... _ _ _

We want to choose x of them for our successes, and there are nCx ways to do that.


Ok we want x successes, by the multiplication rule that means we have,

p*p*...*p <--- x copies of p
p^x.

That leaves n-x spots, those must be the failures,
q*q*...*q <--- n-x copies of q
q^(n-x)

Put it all together, and we have

nCx p^x * q^(n-x)

Simplistic but it should help provide some insight!

Nice Convergence Proof

This is an interesting statement and proof I think.

Here is the statement.

Suppose that lim x->inf f(x) = L and that f(n) = a_n for every integer n. Then lim a_n = L.


Proof.
Suppose lim x-> inf f(x) = L. Let e > 0, then there is a postive integer N > 0 such that for all x > N, we have,

|f(x) - L| < e.

In particular, for all positive integers n > N, we have,

|f(n) - L| < e
|a_n - L| < e.

Done.


How is this useful?

Well suppose you have a limit, like say,

lim n->inf n/e^n.

It is clear that the answer is 0 because the exponential function grows faster than any polynomial, but what if we wanted to show it? Well one solution is to use L'Hopital's rule, but we can't really use L'hopitals with sequences, so we instead think of it as,

lim x->inf x/e^x = lim x->inf 1/e^x = 0 where we used L'hopitals.


It is important to note that the converse of our original statement is NOT TRUE.

Take a_n = sin(npi). Here lim n->inf sin(npi) = lim n->inf 0 = 0, but,
lim x->inf sin(xpi) DNE.

Therefore the moral here is, you can use l'hopitals with sequences but keep in mind everything should make sense if your n is actually a real number x. If it does make sense, feel free to be abusive and use l'hopitals, most people do it, and if people do it then it's ok right?:)

I hope this has helped someone out there on the internet!

A continuous function which is not differentiable, proof

Set f(x) = |x|.

First let's show it's continuous.

If x > 0, f(x) = x is continuous, if x <0, then f(x) = -x is also continuous. We just need to show it's continuous at x = 0.

Note,

lim x->0^+ |x| = lim x->0^+  x = 0  and lim x > 0^- |x| = lim x->9^- (-x) = 0, whence
lim x->0 |x| = 0 = f(0), so continuity has been shown.

This function is not differentiable at zero. To see this, note,

lim x->0^- (f(x) - f(0))/(x - 0)
lim x->0^- (|x| - 0)/(x - 0)
lim x->0^- |x|/x
lim x->0^- (-x/x)
lim x->0^- (-1)
-1

Likewise,

lim x->0^+ (f(x) - f(0))/(x - 0)
lim x->0^+ (|x| - 0)/(x - 0)
lim x->0^+ |x|/x
lim x->0^+ (x/x)
lim x->0^+ 1
1

Therefore lim x->0 (f(x) - f(0))/(x - 0) DNE, so the function is not differentiable.

This is worth knowing how to do, it might save your life some day!