zkfcfbzr

zkfcfbzr, 16 days ago

Mark Rober for me, seems like he wants to be Mr. Beast Lite or something lately.

zkfcfbzr, 18 days ago

solutionI’m sure there’s some short elegant solution here that uses beautiful vector math. Instead I went for the butt ugly coordinate geometry solution. Let u = <a, b> and v = <c, d> See diagram. WLOG, assume u has a smaller angle than v. We can find cos(θ) by constructing a right triangle as shown, and by finding a new vector w, in the same direction as u, which has the correct length to complete our right triangle. Once done, we will have cos(θ) = |w| / |v|. Let’s consider each vector to be anchored at the origin. So we can say u lies on the line y = (b/a)x, and v lies on the line y = (d/c)x. To find w, let us first find z - the third side of the triangle, which we know must be perpendicular to u, and pass through (c, d). The line perpendicular to y = (b/a)x and passing through (c, d) is y = (-a/b)(x-c) + d. So this is the line containing the third side of our constructed triangle, z. To find w, then, let’s find its point of intersection with y = (b/a)x, the line containing w. (b/a)x = (-a/b)(x-c) + d → Setup (b/a)x = (-a/b)x + (ca/b) + d → Distribute on right x(b/a + a/b) = (ca/b) + d → Add (a/b)x to both sides, factor out x on left x(a² + b²) = ca² + dab → Multiply both sides by ab x = (ca² + dab) / (a² + b²) → Divide both sides by (a² + b²) x = (ca² + dab) / |u|² → a² + b² is |u|² y = (b/a)x = (b/a)(ca² + dab) / |u|² → Plug solution for x into (b/a)x, don’t bother simplifying as this form is useful in a moment So w = <(ca² + dab) / |u|², (b/a)(ca² + dab) / |u|²>. Now we need |w|. |w| = sqrt((ca² + dab)² / |u|⁴ + (b²/a²)(ca² + dab)² / |u|⁴) → Plugging into normal |w| formula |w| = sqrt(((ca² + dab)² / |u|⁴) * (1 + b² / a²)) → Factor out (ca² + dab)² / |u|⁴ |w| = (ca² + dab) / |u|² * sqrt(1 + b² / a²) → Pull (ca² + dab)² / |u|⁴ out of the root |w| = (ca² + dab) / |u|² * sqrt((a² + b²) / a²) → Combine terms in root to one fraction |w| = (ca² + dab) / |u|² * |u| / a → Evaluate sqrt |w| = (ca + db) / |u| → Cancel out |u| and a from numerator and denominator So this is the length of our adjacent side in the constructed right triangle. Finally, to find cos(θ), divide it by the length of the hypotenuse - which is |v|. cos(θ) = |w| / |v| = (ac + bd) / (|u||v|) And note that ac + bd is u•v. So we’re done. cos(θ) = u•v / (|u||v|), or |u||v|cos(θ) = u•v. Note that since this formula is symmetric with respect to u and v, our assumption that u’s angle was smaller than v’s angle did not matter - so this should hold regardless of which is larger.

zkfcfbzr, 17 days ago

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zkfcfbzr, 16 days ago

Making one additional note here, because of your parenthetical message, letting you know that my short backlog of problems I found interesting has actually run dry with the last one I posted - so my series is unfortunately on hiatus. I’ll still post if I come across other interesting problems but that honestly doesn’t happen too often - even the ones I have posted were collected (and saved by a friend) over several years.

zkfcfbzr, 21 days ago

responseThis is close, but you’ve got an off-by-one error where you calculate the probability of making it to the nth round as 1/2^n. Consider that that would imply you have a probability of 1/2 for the game to make it to round 1, or 1/4 to make it to round 2, etc - it should be 1/2^(n-1). Correcting this would give you the correct answers for P(B) and P(C). As for the dependence question, I’m not sure I followed your arguments there - but saying both are dependent is not correct.

zkfcfbzr, 21 days ago (edited 21 days ago)

Games are always played to completion, though if you wanna make it (barely) more challenging you can add in a 5% chance for both players to get bored and give up on each round (before flipping), leading to a loss. Though it seems unlikely - after flipping a quarter 20 times and getting Tails every time, I’d be inclined to keep flipping if anything.

responseThese are correct. It is possible to reason out which of B and C is independent of A without going into the numbers.

zkfcfbzr, 21 days ago

responseThis is the correct answer, although P(A|B) should actually be 2/3 rather than 7/12 - I think you meant 1/2 + 1/8 + 1/32 + …? The reasoning is good, either way. Since past flips won’t affect future flips, if the player has made it to turn 5, an odd turn, then their future prospects are no different than they were on turn 1, another odd turn - so A and C are independent. Similarly, A and B are dependent because your chances of winning and losing effectively flip: If you’ve made it to an even turn, then you now win if it takes an odd number of flips from there to get Heads. So it should be an almost paradoxical-seeming situation: You win 1/3 games overall, you win 2/3 games that make it to turn 2, you win 1/3 games that make it to turn 3, you win 2/3 games that make it to turn 4, 1/3 that make it to turn 5, etc.

zkfcfbzr, 21 days ago

In probability, two events are said to be independent if one event happening has no effect on the probability of another event happening. So coin flips, as an example, are independent - because when you flip a coin and get Tails, that doesn’t affect the probability of the next coin flip also coming up Tails.

So in this context, asking if A and B are independent is asking: Does knowing the game lasts at least four turns change the probability of winning? And similarly for A and C. Does knowing the game lasts at least 5 turns change the probability that the game will end in a victory?

Rest of responseTo be clear about your other answers, saying P(B) = 1/16 and P(C) = 1/32 are not correct - I was saying if you adjusted your formula, from 1/2^n to 1/2^(n-1), then your answers would be correct. So the probability of the game lasting at least four turns is 1/2^(4-1) = 1/2^3 = 1/8, and the probability of it lasting at least 5 turns is 1/2^(5-1) = 1/2^4 = 1/16. But I think you were more asking about why this didn’t affect your win rate - that’s because there’s a subtle difference between “making it to the nth round” and “having the game end on the nth round” - and that difference is that once you make it to a round, you then have a 1/2 probability to end the game - which makes the probability of ending on the nth round the 1/2^n you used.

zkfcfbzr, 21 days ago

Honestly, I still just google for relevant reddit threads. Lemmy’s the only place I actively participate in, but this is one of the use cases it hasn’t been able to replace reddit for for me either yet.

zkfcfbzr, 21 days ago

Sure, but it’s still a lot more reliable than something like the amazon review section, or a lengthy AI-generated article comparing the two products you just happened to google together that somehow manages to say nothing at all.

zkfcfbzr, 23 days ago (edited 22 days ago)

solutionThis series can be rewritten as: 2 * lim (n → ∞) Σ (k = 0 to n) k / 2^k The final term of this series will of course be n / 2^n. So let’s try to write each term with 2^n as its denominator: 2 * lim (n → ∞) Σ (k = 0 to n) k * 2^(n-k) / 2^n → Multiply by 2^(n-k) / 2^(n-k) = lim (n → ∞) 1/2^(n-1) Σ (k = 0 to n) k * 2^(n-k) → Factor the denominator out of the sum, combine it with the 2 from in front of the limit What will this sum look like, though? Its first few terms are: 02^n + 12^(n-1) + 2*2^(n-2) + … + (n-1)2^1 + n2^0 (my construction below works backwards, from the last term here to the first) This can be written as a double sum: Σ (k = 0 to n-1) Σ (j = 0 to k) 2^j (which is 2^0 + 2^0 + 2^1 + 2^0 + 2^1 + 2^2 + 2^0 + 2^1 + 2^2 + 2^3 + …, so we have n copies of 2^0, n-1 copies of 2^1, etc) So we have: lim (n → ∞) 1/2^(n-1) Σ (k = 0 to n-1) Σ (j = 0 to k) 2^k = lim (n → ∞) 1/2^(n-1) Σ (k = 0 to n-1) (2^(k+1) - 1) → Evaluate innermost sum, just sum of powers of 2 from 0 to k = lim (n → ∞) 1/2^(n-1) * (2^(n+1) - 2 - n) → Evaluate next sum, which is powers of 2 from 1 to n, and don’t forget the -1 part = lim (n → ∞) (2^(n+1) - n - 2) / 2^(n-1) → Combine leftover terms into one fraction in the limit = lim (n → ∞) 4 - n/2^(n-1) - 1/2^(n-2) → Distribute denominator to each term = 4 - 0 - 0 → First term is just 4, other terms go to 0 = 4 I’ve been checking and thinking about this at least a bit each day, and today’s the first day a solution came to me. Edit: Checking your solution now, it’s obviously much nicer than this - but at this point I’m just happy I solved it at all. Obviously I knew it was 4 from just calculating partial sums from day 1, but proving it proved challenging.

zkfcfbzr, 26 days ago

Background / incredibly minor spoilers:Unlike the other problems I’ve posted, this is the only one (so far) that I came up with on my own - back in high school while playing around with my calculator I entered this function at some point, and got curious about the minimums. I was in pre-calculus at the time, so it took me a while - but I did eventually work out the solution. Needless to say I was pretty happy the answer was something as cool as it was. I actually didn’t ever work out (or try to work out) the extra credit portion until I wrote up the problem for this post, so it was a nice revisit of a very old problem. The language asking for an “increasing” function is mostly to avoid smartasses who might submit the original equation as a technically correct solution. Basically I want the simplest possible function that passes through the described points.

zkfcfbzr, 26 days ago

sin(x) is defined everywhere - it’s sin(x)^sin(x) that has undefined spots: A negative number can only be raised to a negative power if that negative power can be expressed as a rational number with an odd denominator. That doesn’t describe most real numbers, so sin(x)^sin(x) is undefined for most values that make sin(x) negative. It does still describe infinitely many values, though - and the extra credit graph kinda just connects them up.

zkfcfbzr, 26 days ago (edited 26 days ago)

No worries! The idea isn’t too bad - it all comes down to that you can’t take the square root of a negative number, but that you can take the cube root of a negative - or in general, you can’t take even roots of negatives, but you can take odd roots of negatives. Remember that if your exponent is a/b, it’s like taking a power of a, and a root of b. So something like (-25)^(1/2) is asking for the square root of -25, which isn’t defined (since we’re not considering complex numbers in this problem). On the other hand, something like (-125)^(1/3) is perfectly fine: It’s just -5, since -5 * -5 * -5 = -125. That’s why negative numbers to exponents which can be written as a rational number with an odd denominator are defined, regardless of the sign of the exponent. If a is negative and a^b is defined, then |a^b| will always equal |a|^b - like how |(-125)^(1/3)| = |-125|^(1/3) = 125^(1/3) = 5

When you write that rational exponent, its numerator will be even or odd, too. If the exponent is even, then the answer will end up positive: Negative numbers have negative odd roots, and negative numbers to even powers are positive. On the other hand, if the exponent is odd, the answer will end up negative: Negative numbers to odd powers are still negative. So if your expression is x^x and is defined, and x is negative, you know it’s going to equal either |x|^x or -|x|^x. Here’s an example of each case:

(-1/3)^(-1/3) = -1.4422495 = -(1/3)^(-1/3) (a case for x^x = -|x|^x)

(-2/3)^(-2/3) = 1.31037 = (2/3)^(-2/3) (a case for x^x = |x|^x)

And that’s why the two extension functions I mention wrap the first sin(x) in parenthesis, and swap its sign around. The actual y = x^(sin(x)^sin(x)) has infinitely many points that lie on both of those functions (within the intervals where sin(x) is negative - so within the intervals where desmos seemingly graphs nothing) already - desmos just isn’t graphing them, and these new functions are filling in the gaps so everything gets drawn. The gaps happen for irrational numbers and for any rational number with an even denominator - but if you actually put a point at every defined value, you would get a graph that looks like this (only taking principle real roots). If you allowed graphing both real roots when available, you’d get this. If you wanted to go to negative x-values too you’d get this.

All the points you see on that final graph correspond to (x, y) values that do satisfy just the original equation, y = x^sin(x)^sin(x) - them not displaying in the original graph is a combination of desmos not graphing individual discontinuous points, and desmos only graphing primary roots in equations set up as functions. So it’s a case where those points exist on the graph, but lack any continuity: It’s just infinitely many non-connected points, that are nonetheless dense enough that you could zoom in forever and still always see infinitely many defined points in any given region. The key is that between any two real numbers, there are always infinitely many rational numbers, infinitely many of which can be written with odd denominators - so there are points everywhere, just not everywhere enough.

zkfcfbzr, 25 days ago (edited 25 days ago)

SolutionFor both of these, it’s sufficient to consider only sin(x)^sin(x). It stands to reason that the maximums will happen when sin(x)^sin(x) is at its maximum value, and its minimums will happen when sin(x)^sin(x) is at its minimum value. The x^ part really just makes the function prettier to look at. The function lacks any continuity when sin(x) is negative, so we can consider just the regions where sin(x) is positive. For the maximums, this is very easy: The largest sin(x) can be is 1, and so the largest value for sin(x)^sin(x) is also 1. So the monotonically increasing function which passes through each local maximum is y = x^1, or just y = x. For the minimums it’s a bit trickier: The smallest sin(x) can be is 0, and while 0^0 is indeterminate, it happens to approach 1 in this case - our minimum is somewhere else (side note: sin(x)^sin(x) approaching 1 for these values is why our max function also exactly bounds the flailing arms of the function). So we turn to derivatives. We can do this using either sin(x)^sin(x) or x^x - I chose to use sin(x)^sin(x) here and x^x (sort of) in the extra credit. Using sin(x)^sin(x) here has the added benefit of also showing the maximums are when sin(x) = 1. y = sin(x)^sin(x) y = e^(ln(sin(x)) * sin(x)) y’ = sin(x)^sin(x) * (cos(x)/sin(x) * sin(x) + ln(sin(x)) * cos(x)) → Chain rule, product rule, and chain rule again y’ = sin(x)^sin(x) * (cos(x) + ln(sin(x)) * cos(x)) → Simplify y’ = sin(x)^sin(x) * cos(x) * (1 + ln(sin(x))) → Factor We can set each of these factors to 0. sin(x)^sin(x) will never be 0. cos(x) will be 0 when x = π/2 + 2πn (2πn instead of πn because we’re skipping over the solutions where sin(x) is negative). These are the maximums, because when x = π/2 + 2πn we have sin(x) = 1. 1 + ln(sin(x)) = 0 ln(sin(x)) = -1 sin(x) = e^-1 = 1/e And that’s actually sufficient for our purposes - we could of course say x = arcsin(1/e), but our goal is to calcualte the value of sin(x)^sin(x). Since sin(x) = 1/e, we have sin(x)^sin(x) = (1/e)^(1/e). So our monotonically increasing function which hits all the minimums is y = x^((1/e)^(1/e)), or: y = (e-th root of e)-th root of x. Graphed solution visible here. For the extra credit, we can do the same basic idea: Find the maximum of |sin(x)|^sin(x). Since we know we’re in a region where sin(x) is negative, we can model this as x^-x, and the solution will be valid as long as x is something between 0 and 1, within sin(x)‘s positive range. y = x^-x = e^(ln(x) * -x) y’ = x^-x * (-1 - ln(x)) x^-x won’t ever be 0, and -1 - ln(x) will be 0 when ln(x) = -1, or x = 1/e again, same as before. So our maximum value happens at (1/e)^(-1/e), which simplifies to e^(1/e). In other words, our new maximum function is x^(e^(1/e)), or x to the power of the e-th root of e. Graph of all three solution functionsAnd, finally - I mentioned that we should also be able to expand the function to its less-defined regions using y = x^-|sin(x)|^sin(x). I don’t care enough to post the full work here, but the reciprocal of the three answer solutions above are what works here: y = 1/x, y = 1/x^((1/e)^(1/e)), and y = 1/x^(e^(1/e)). Graph. Note that the upper function here only applies in regions where the original function is already defined, and so the other parts of this extension are invalid - you can toggle the main function at the top off to see only the parts that truly apply. Nothing particularly interesting happens if you extend things to values of x less than 0 - the graphs (more or less) just reflect across the y-axis. Bonus graph with everything + phase shifts and pretty colors

zkfcfbzr, 25 days ago (edited 25 days ago)

I’ve posted my solution.

The question of why the bounds swap isn’t too relevant to the problem, since there are no maxes or mins below x = 1 - but the reason they swap is because numbers between 0 and 1 get larger with smaller powers, and smaller with larger powers. Think of how 0.5^2 = 0.25 - so the more a (positive) power is biased toward “rooting” a number in this interval (biased toward 0), the larger the result will be - so the minimum of g now produces a max value, and the max of g yields a minimum value.

Edit: If it helps, you can also think of numbers between 0 and 1 as being numbers larger than 1, but with a negative power. So something like 0.5^2 is equivalent to (2^-1)^2 = 2^-2. So larger powers become lesser powers that way.

Really, it’s for the same reason the graphs of 2^x and 0.5^x are reflections of each other

zkfcfbzr, 26 days ago

Kind-of solutionF = ma and m = 1, so a = F = 1/r^2. I used a = -1/r^2 so the force experienced by the particle at r = 5 would be negative. a = dv/dt. dv/dt = -1/r^2 v dv/dt = -1/r^2 dr/dt → v = dr/dt. Multiply both sides by v, but write it as dr/dt on the right v dv = -1/r^2 dr → Multiply both sides by dt 0.5v^2 = 1/r + C → Integrate v^2 = 2/r + C → Multiply both sides by 2, find C = -2/5 by plugging in r = 5, v = 0 v^2 = 2/r - 2/5 5arcsin(sqrt(r/5)) - sqrt(5r-r^2) = sqrt(2/5) t + C → Solve the differential equation y’^2 = 2/y - 2/5 with Wolfram Alpha, find C = 5π/2 by plugging in r = 5, t = 0 5arcsin(sqrt(r/5)) - sqrt(5r-r^2) = sqrt(2/5) t + 5π/2 → Can find that the particle reaches the origin at t = 5sqrt(5)π / 2sqrt(2) ≈ 12.418 seconds I made heavy use of internet resources on this one, not just for solving the differential equation but also for technique - I got a bunch of wrong answers before getting this one that I finally think is right, at least up until the moment of origin contact Graph (the hidden ellipse is because I thought the path may be elliptical, but it doesn’t appear to be)

zkfcfbzr, 15 days ago (edited 14 days ago)

Not sure how much you’d care, but I came back to this and found (by hand) a function which closely approximates the solution - it’s not exact but it’s also not super far. Graph.

I think I could also solve the differential equation by hand at this point (getting the same solution as before) - I haven’t, but I’m pretty sure I could if I wanted to for whatever reason. I’m doubtful it’s possible to get an exact solution in terms of just x but if you ever manage I’d love to see how.

zkfcfbzr, 27 days ago (edited 27 days ago)

solutionThe terms can be rewritten as: (2/1) * (3/2) * (4/3) * … * ((n+1)/n) * … Each numerator will cancel with the next denominator. In total everything cancels, so the answer is the empty product, 1. …Wait… Uhm, ignore that. Rather, consider the products we get when multiplying. We get: 2/1. 6/2. 24/6. Etc. That is, we have: Π (n = 1 to k) (n+1)/n = (k+1)! / k! = (k+1)k!/k! = k+1 k+1 clearly goes to infinity as k → ∞, so our product diverges to infinity.

zkfcfbzr, 27 days ago (edited 27 days ago)

replyYes - I mostly left the first part in for the humor (it was legitimately my first stab at the problem), but it gives the same result, just in a way that’s a little harder (to me, at least) to see. The cancellation is unbalanced: Each numerator cancels with the next denominator, which necessarily brings with it the next numerator - you’ve always got the next numerator in line, as-is, after any number of canceled pairs. So while everything cancels out in the limit, the product up to n equals n+1, and so the limit of the product is ∞.

zkfcfbzr, 27 days ago (edited 27 days ago)

commentWas wondering about that hint - read it after my solution then tried coming up with another that made the product like (1 + 1/n)^n, but the best I was able to manage was proving that the product is larger than e - an impressive feat since it takes a whopping two terms to get that large… Thought it might be something with writing the product like lim (n → ∞) Π (k = 1 to n) (1 + (n/k)/n), but was never able to figure out a way to do anything with that either.

zkfcfbzr, 27 days ago

solutionHere’s a pretty minimum effort hand-wavy proof. A function is differentiable on an interval if it’s differentiable at every point in that interval. Similarly, a function is continuous on an interval if it’s continuous at every point in that interval. So I’ll show that differentiability at a point implies continuity at that point, from which differentiability on an interval implying continuity on an interval will automatically follow. A function is differentiable at x = a if lim (x → a) (f(x) - f(a)) / (x - a) exists, so let us assume it does exist. In that limit, the denominator equals 0 at the limit point. Yet, we know by assumption that the limit exists. This implies the numerator also equals 0 at the limit point, because the only way for our denominator to equal 0 at the limit point, and still have the limit exist, is for the expression to be indeterminate, rather than undefined. The only indeterminate form with 0 in the denominator is 0/0. So we can assume lim (x → a) (f(x) - f(a) = 0. As f(a) is a constant value, this implies lim (x → a) f(x) = f(a) - so the function is continuous at x = a.

zkfcfbzr, 27 days ago

Just noting this is in the wrong problem thread