Now, assume, just in theory, of course I'm not trying to imply anything here, that when they show you the daily value of this fund, instead of calculating it from the assets there, they just use the following Python code, with a Gaussian of mean 7500 and standard deviation 3000, which is summed to the value each day:
The question is: what is the probability that this code will increase the value of the fund each day it runs? Do you think this pattern is reasonable for a fund consisting of an asset whose price fluctuates like this?
Please, don't try to connect this problem to current events. This is a purely academic question :D
Let y and z be as shown in the image above based on the subdivisions of the bottom edge. Then:
(2x)2+y2=145 (A)
(5x)2+(y+z)2=689 (B)
(4x)2+(z+3)2=377 (C)
Now let's declare that x,y,z are natural numbers. (Just do this. Don't think about it.)
Equation (A) says that 145 = 29*5 is a sum of two squares. Checking shows that 145 = 82+92 = 122+12 (you can also use 29 = 32+22 and 5 = 22+12, and the Brahmagupta-Fibonacci identity to get the above sums of two squares). The first gives x=4, y=9, and then in (C) z=8, and (B) checks out. The second gives x=6, but then (C) has no solution for z, so reject that. Therefore x=4.
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If you are unsatisfied with the above argument, specifically the declaration that x,y,z are natural numbers, note that no other argument I've found is interesting. You are free to solve it in real numbers by squaring equation(s) a few times to get rid of radicals. WolframAlpha will happily tell you that the only solutions are x=±4.
For this second one, find all solutions in the Reals, prove you have all of them.
Here's a pair of interesting problems I ran across recently.
Instead of solving explicitly for x, it's easier to square x + 1/x to get x^2 + 1/x^2 + 2 = 293. Notice that the goal formula reduces to x - 1/x, and if you square that you get x^2 + 1/x^2 - 2, which is 293 - 4 = 289 from our above formula. Then just take the square root of 289 = 17.
For this second one, find all solutions in the Reals, prove you have all of them.
Working is the best way to achieve goals in speedruning. Hardworking is a pain.
Speaking of math, I got a small challenge for you! Using epsilon-delta definition, prove that lim(x->0) of sin(x)/x = 1
The angle BAC is x in radians. The circle has unit length.
The area of the triangle ABC can be computed using AB as base, with height sin(x). So, the area is just sin(x)/2.
The area of the circular sector ABC is the square of the radius (1) times the angle in radians divided by 2, so it's just x/2.
The lines CD and BD are tangent to the circle. We have BD = CD, and since ACD and ABD are right angles, we have that angle BDC = pi - x
Computing the area of triangle BDC using BD as base, we have the height as BD*sin(x), so the area of triangle BDC is BD²*sin(x)/2.
Now, notice:
Since the circular sector is larger ABC is larger than the triangle ABC, we have x/2 > sin(x)/2, which implies x > sin(x)
The circular "cut" (don't know the name of this in English, sorry) ending at B and C is smaller than the triangle BDC, we have (x-sin(x)/2 < BD²*sin(x)/2
The point C is always at a greater height than D. Since the height of C is sin(x), we have BD<sin(x), so BD²sin(x)/2 < sin³(x)/2 < x³/2 .
Chaining all our inequalities, we get 0<x-sin(x)<x³
Finally, let's get to limits:
For every 0 < x < delta, we have
|sin(x)/x - 1| = |(sin(x)-x)/x| < |-x³|/|x| = x² < delta²
Therefore, suppose we have a given epsilon>0. If we choose delta = sqrt(epsilon), then for every x satisfying 0<x<delta, we have
|sin(x)/x - 1| < delta² = epsilon. Since we were able to choose a delta to make the inequality true, then lim sin(x)/x for x->0 is 1, by definition.
(Notice that this proof is significantly harder than the usual one that calculus textbooks give, using the squeeze theorem. To prove using the definition, we need to bound the difference x-sin(x) using an easy function so we can invert the epsilon-delta dependency, while the proof using the squeeze theorem will work even with "harder" functions as long as they tend to 1 when going to 0).
Working is the best way to achieve goals in speedruning. Hardworking is a pain.
Red curve is sin(x)/x, black curve is 1/|x|. 1/|x| explodes to infinity at 0, and your goal is to show that sin(x)/x tends to 1 at 0. Using this bound you can't even conclude that sin(x)/x is finite.
Proving limits using the definition is like this: think of the epsilon as a "challenge" to how close the function is to the target value, and the delta is the "answer" to the challenge. The way I did it was to show that sin(x)/x is between 1 and 1 - x^2. Look at the graph:
Blue line is y=1, red is sin(x)/x, and green is 1-x^2. Purple one is y=0.95, which corresponds to an epsilon of 0.05.
The goal is to find a delta such that the red curve always sits between 0.95 and 1.05. The 1.05 part is trivial, since sin(x)/x<=1 for its entire domain. The hard part is the 0.95 line shown.
In theory, you could just solve for delta finding the points where the red line crosses the purple, but to find them you'd need to solve sin(x)=0.95*x, which is a transcendental equation.
Now comes the idea. The definition of limit does not require you to be precise. You don't need to find the largest delta possible. You can find a smaller one. That's why we use the green curve instead, because the equation 1-x^2 = 0.95 can be easily solved.
Of course, later in calculus you learn the squeeze theorem, and the fact that 1-x^2 <= sin(x)/x <= 1 is in fact enough to prove that the limit of sin(x)/x is 1 at 0, because it's bounded above and below by two functions that tend to 1 at 0. Even better, since you don't need to solve an equation, as in the definition of limit, you can use weirder functions for a lower bound in the green line, requiring you to do less geometry to prove the bound. I hope that helped.Of course, later in calculus you learn the squeeze theorem, and the fact that 1-x^2 <= sin(x)/x <= 1 is in fact enough to prove that the limit of sin(x)/x is 1 at 0, because it's bounded above and below by two functions that tend to 1 at 0.
(where x is a positive angle DAB in radians)
Area ΔABE < Area of sector DAB < Area ΔADF
(1/2) sin x cos x < (1/2) x < (1/2) tan x
cos x < x/sin x < 1/cos x
Since limit as x->0+ of cos x is 1, by the squeeze theorem,
lim x->0+ x/sin x = 1
That is, lim x->0+ sin(x)/x = 1
Replacing x with -x, we get lim x->0- sin(x)/x = 1.
So lim x->0 sin(x)/x = 1
(On the other hand, in Late Transcendentals calculus, 1 - x^2/6 <= sin(x)/x <= 1 follows directly from the definition of sin(x) = x - x^3/3! + x^5/5! - ... )
The circular "cut" (don't know the name of this in English, sorry) ending at B and C is smaller than the triangle BDC, we have (x-sin(x)/2 < BD²*sin(x)/2
