runouw.com

Posted: **May 12th, 2015, 11:57 am**

I'm too lazy to calculate it all like OJ did but backtracking was also my first thought

Posted: **September 11th, 2015, 3:08 pm**

This is so ded. I am gonna try to revive this. Here is something prety cool our calculus proffesor taught us. You only thing you have to know is Intermidiate value theorem. To make it short, the IVM says that if a function is continuous on an interval (that means, it doesn't jump from one value to another and has no holes in it) and in one of the extremes the funcion is positive and in the other is negative, then at some point it must cross the zero. Perphabs it is easier to see with a drawing:

As you can see a is negative and b is positive, so it is inevitable that a continius funcion will cross the zero at some point. Note that it can cross it multiple times, like the red function. But we are guaranteed that at some point the function becomes zero. It can't have a hole where it would cross or have a jump because it is continuous.
Another thing we will use is the fact that the sum or substraction of two continuous functions is also continuous.

So what are we going to prove? That there are two points on earth, diametrically opposed that have the exact same temperature. Pretty cool, huh?

We will have to make an assumption though: that the variation of temperature on the surface is continuous. I don't think that is too much of assumption, since metheorologists say it has a second derivative, so no probs.

Ok, so what we are going to do is actually cut the earth, vertically, in one of its meridians. So the drawing would look something like this:

Good so far? Ok, so let's define a function that meassures temperature in this new circle. We are gonna do it by meassuring the angle the point is in respects to the north pole, we are gonna call it T1. And we are calling T2 the point diametrically opposed to it on the circle. Something like this:

Now, let's look at something intresting. T1(o°) is the temperature one the north pole, and T2(o°) is the temperature on the south pole. Now, here is where it gets juicy, what is T1(180°) and T2(180°)? Well, 180°is half a turn, so it would reverse, T1(180°) is the temperature one the south pole, T2 (180°) is the temperature on the north pole. In other words:

Ok, so there are three cases possible. 1) The temperature at the south pole and on the north pole are the same (T1(0)=T2(0)), 2) the temperature on the north pole is higher than on the south pole (T1(0)>T2(0)) or 3) the temperature on the south pole is higher than on the north pole (T1(0)<T2(0))

In 1) there is nothing left to prove, since the north pole and the south pole are diametrically opposed and they have the same temperature, so yeah.

Let's focus on 2)

We will build a new function F(alpha). What this function will do is subtract T1(alpha) and T2(alpha) (aka substract the temperature of two diametrically opposed points)
Is this function positive or negative at 0°?
Well, since T1(0)>T2(0), T1(0)-T2(0) is larger than zero (we are taking a smaller number from a bigger number).
But what about F(180°)?
Well, at 180° the functions switch, so T1(180°)-T2(180°)=T2(0°)-T1(0°), which is negative.
Is F(aplha) continous? Well, it is the substraction of to continuous functions, so yeah.
But wait, at 0° it is positive, at 180° is negative, so at some point in the middle the difference between T1(alpha) and T2(alpha) must be zero, hence T1(alpha)=T2(alpha) for some alpha between 0 and 180, which means that two points diametrically oposed have the same temperature.

Case 3) is the same thing but reversed, T1(0)-T2(0) is negative, T1(180°)-T2(180°) is positive, so at some point the function must be zero, and the same happens.

Q.E.D

(Please forgive all mistakes I might have made, which I am sure are many)

Posted: **September 12th, 2015, 1:50 am**

too much math. my brain hurts.

Posted: **September 12th, 2015, 2:00 am**

A script with 100+ pages like I'm working through it could be considered as a lot of math. If you already have trouble trying to understand what makes a function continuous, I've learned this mnemonic:

A function is continous if and only if you can draw its graph without lifting your pen from the paper.

To sum it up:

T₁(α) = temperature measuring the angle α from the north pole,
T₂(α) = temperature measuring α from the south pole.

Obviously, T₁(0) = T₂(π/2) and T₂(0) = T₁(π/2). I usually prefer radian measure for angles

And in case of w.l.o.g. T₁(0) > T₂(0), the continuous function F := T₁ - T₂ is positive for α = 0 and negative for α = π/2, hence it must have a zero in the interval (0, π/2). Simple as that.

Posted: **September 16th, 2015, 12:40 am**

Chat wrote: show

Yeah, let's continue from there. The expectation value of this game is infinite because the probability for the prize level 2^n is exactly (1/2)^n, meaning if you wanna build the expectation value, you get an infinite amount of 1's in your sum. Whatever you want to offer as an input, it's always too less, your expectated win breaks all limits, people would have to queue if someone offers to play this game.

But ... they don't do it. If someone asks you "Hey, would you pay $1000 for this game?", you'd rather respond: "I guess you're crazy!"

This is a case where there's a serious gap between mathematical theory and real life experience, for two reasons. First, the offerer of this game would never be able to pay out the cash if the emblem should really come 40 times in a row or such. Imagine that you're playing this with the richest man in the world, with a capital of 100 billions. Then you would win all of his capital after 37 emblems in a row, and if you agree on a maximum win of 100 billions which is paid out after 37 emblems, then your expectated win is not infinite, but only $37, and that's indeed a huge difference.

Second, even if the offerer's capital would be infinite, we would have a rather rational reason of not betting all too much: "what the one ruined, are peanuts for the other". The fear of a possible loss weights more than the hope for an equal win, e.g. if there's a winning game with prizes $0 and $2000, both with probability 1/2, most peoples wouldn't be willing to bet $1000 for it. Same here: The expectated win may finely be infinite and yet we're rationally authorised of betting not more than $20 or $30 for this game.

Posted: **October 12th, 2015, 1:44 pm**

Hello! Lordpat back with an intresting calc problem. Also shroom, right now I am studying for being a math teacher, though in the future I might study mathematics.

Notation: "=<": larger or equal to
Anyway, let's get rolling:

"Let dn be the sequence that calculates the number of digits of n!. Calculate:

Ok, let's get rid of the elephant in the room. "the number of digis of...." and transform that into something we can work with. Now, this is actually pretty simple. We want a function that for 0<n<10, it returns 1, for 10=<n<100, sounds familiar? Well, we know that the log function in base 10 does something similar. If n is between 10^1 and 10^2, it returns something that starts with 1. something. First, but look at the fact that number of digits is always a natural number and many times log(n) will return an irrational number, so we might need to get an integer lower than it. So, let's use another function: the floor function.

What is the floor function? Very simple, it takes any number and it returns the largest integer lower than it. For example:

Remember that for values between 10 and 99 we would get 1. something, but if we apply the floor function we would get 1 alone. But wait! Numbers between 10 and 99 have two digits, not one! Yes, that is why we must add one. So, we would get that the number of digits of a number k is...

What is k? Well, it is n!, so the limit we have to calculate in the end is...

Ok, now what? This limit seems horribly hard to analyze. A factorial inside a log, all that inside a floor function, plus one? Well, there might be one theorem that can save us: the squeeze theorem (or as we call it the sandwich theorem).

Let's say we have two sequences that converge (that means, that they go to as n goes larger and larger to a limit) to the same value, and we have a function that is always between those two functions, then that function converges to that same limit. Seems confusing? Let me give you an example. Let's say we've proven rigorously than 1/n converges to zero, and we want to see if 1/n^2 goes to zero. Rather than solving the limit manually you can do the following:
We know that 1/n^2 is always less than 1/n, since the denominator is larger, but it is also larger than zero, since it is a division of positive numbers. So we would get:

Therefore, 1/n^2 also goes to zero.

Ok, so how can we do that to solve this problem? Well, let's try using the calculator and see what this tends to, just to get a clue and get where we should use squeeze. It seems that the numbers are getting closer and closer to zero, isn't it? Well, that might give us a clue then. We can put zero on the left, since both the numerator and denominator are positive for natural numbers. But what do we put on the right? Well, before putting something that goes to zero for sure, we could try to get rid of that annoying floor function. Well, we know that if we remove the floor function, the function will always be larger or equal to the one with it, since if you remember the floor function always took a number lesser or equal to log(n!). (In this case it is always less, can you think why?)

This is starting to look less awful, but we still have the problem of the factorial and the logarithm. So what do we do? Squeeze again of course. We need a function that is: a) larger than n! and b) interacts well with logs. Well, n^n seems like a good one (I am gonna leave a) as a proof to you, because I am that mean). if we do log(n^n), by properties of logarithms, the exponent goes to the left and multiplies and we get nlog(n). So we would get.

(I removed the third function to save space below, seeing how it is no longer relevant for the analysis)

We take out common factor n. Dividing the function on the numerator and the denominator by "n", we would get:

Followed me so far? Well, ther is one last step, we are almost done, log(n)<sqrt(n) for all n. (You can prove that with continious variable and derivatives). So using squeeze again we would get:

(The limit on the right aproaches zero because the largest power of the denominator (1) is larger than the largest power of the numerator (1/2). You can always divide everything by n and simplify powers if you don't believe me.)

HALLELUJAH. Since the extreemes aproach the same, all functions in between do so too. AND THERE WE GOT IT! The limit is equal to zero.
___________________________________________________________________________________________________

Phew, that was a lot. Hope you liked it! I am having a calc exam on wensday, wish me luck!

Posted: **November 27th, 2015, 8:29 pm**

I'm really ♥♥♥♥ mad about this and I need to share it somewhere okay so

Consider the equation z³ + (5 + 8i) z² = 2 z + 9 and the statements:
I) "The sum of the three roots of this equation is a (non-real) complex number."
II) "The product of the three roots of this equation is a (non-real) complex number."
Which of the following statements is true?
A) I is true and II is false.
B) I is false and II is true.
C) Both I and II are true.
D) Both I and II are false.

Spoiler: show

ahhsadflajsdjflkasdfkajdsflk this was way too concise and elegant of a derivation that I found on google to not share after seeing how much I overthought this

Posted: **November 30th, 2015, 8:29 am**

^I aprove of this.

Anyway, I just thought of something. It is like... well, a game. I want you to define the natural numbers, in your own way. It can be rigorous, I can be intuitive, it can be anything. I mean, you can just copy paste the first sentence from wikipedia, and that is okay, but that's not really what I think is the most fun. So yeah, have fun guys!

Posted: **December 13th, 2015, 4:47 pm**

I didn't remember the proof for the cuadratic formula, and so I wanted to drive away from calc and college algebra, so I went ahead and proved it. Apparently (after I looked it up) it's not the traditoinal proof, but discovered by some Cardano Vieta. EDIT: Apparently it is done by using Lagrange Resolvents, and has to do with early Galois theory, and can be generalized for 3rd degree and 4th degree polynomials. Cool.

We'll consider a=1, so the polynomial will be of the form x²+bx+c. Let s and t be the roots of x²+bx+c and s<=t (s lesser or equal than t). Then, because of polnyomial factorization, we'd get x²+bx+c=(x-t)(x-s). Then, by distributive law, we'd get x²+bx+c=x²-(t+s)*x+t*s.

This gives us two equations to work with:
1) t+s=-b
2) t*s=c

We'll square both sides of 1): (t+s)²=b². Expanding the bynomial we'd get t²+2*t*s+s²=b².
We'll multiply both sides of 2) by 4, and we'd get: 4*t*s=4c
We'll then subtract both equations and we'd get: t²+2*t*s+s²-4*t*s=b²-4c. Then we'd get: t²-2*t*s+s²= b²-4c. Now, (a-b)²=a²-2*a*b+b², we'd get:
(t-s)²=(b²-4c), applying root in both sides we'd get: |t-s|=sqrt(b²-4c). Since s=<t, we can remove the absolute value.

3) t-s=sqrt(b²-4c)

Finding t:
We add 1) to 3), and we'd get: 2t=sqrt(b²-4c)-b
so,
t=[-b+sqrt(b²-4c)]/2

Finding s:
We bustract 1) from 3), and we'd get:
-2s=sqrt(b²-4c)+b
s=[-b-sqrt(b²-4c)]/2

The only thing left would be to consider a different from one, but then you make the substitution b=b'/a' and c=c'/a'. It is left as an excersise to the reader *is shot

Posted: **December 18th, 2015, 7:33 pm**

lordpat wrote:Anyway, I just thought of something. It is like... well, a game. I want you to define the natural numbers, in your own way. It can be rigorous, I can be intuitive, it can be anything. I mean, you can just copy paste the first sentence from wikipedia, and that is okay, but that's not really what I think is the most fun. So yeah, have fun guys!

If you can count to it from one in a finite amount of time, it's a natural number.

runouw.com

The Math Thread

Re: The Math Thread

Re: The Math Thread

Re: The Math Thread

Re: The Math Thread

Re: The Math Thread

Re: The Math Thread

Re: The Math Thread

Re: The Math Thread

Re: The Math Thread

Re: The Math Thread