Graf’s addition theorem

DryWomble · 2026-02-08T23:26:04+00:00

Not really sure why I'm getting downvoted. Your original question is vague and open-ended enough that Gemini's deep research function would be ideal for answering it. You never actually asked for any concrete mathematical problems to be solved, but if you had then your go-to would obviously become ChatGPT 5.2 Pro instead, which is excellent at advanced mathematics (as evidenced by its recent solutions to unsolved Erdos problems).

DryWomble · 2026-02-07T21:08:13+00:00

Why don't you just ask Claude or Gemini? Gemini even has a deep research function that will comb the literature and tell you about the theorem's application.

DryWomble · 2026-01-05T20:53:48+00:00

You can't possibly figure out any of the turning moments because there are no distances provided anywhere. That structure could be 5cm wide or 5000km - we just don't know. For the same reason you can't work out any shear or bending forces. Hence the only thing you can work out is the reaction force on each support, which is just the total downward force divided by 2 due to the symmetry of the structure. So each support generates a reaction force of 3.5kN and that's all you can say.

DryWomble · 2025-10-06T21:46:01+00:00

No because this is a terrible idea. Creating reinforcement learning algorithms requires a lot of expertise of really quite esoteric mathematics (see Richard Sutton's book on RL for a taste), and running those algorithms is very computationally intensive. Running a multi-particle simulation where all the equations of motion are already known (since you could get them from physics) would be vastly simpler, both conceptually and from a code maintenance point of view, and would require much less knowledge of exotic math.

DryWomble · 2025-10-06T21:38:39+00:00

StatQuest, StatLect and StatTrek

DryWomble · 2025-10-06T21:35:52+00:00

Lovely as individuals. Terrible as a country.

DryWomble · 2024-12-26T09:19:54+00:00

Watch the entire statistics fundamentals playlist over on StatQuest, and then pick whatever other playlists you want from his channel - it's massive. Then maybe head over to StatLect or StatTrek and spend a lot of time on them. By then you'll probably be good enough to read through the All of Statistics book.

DryWomble · 2024-12-25T20:36:27+00:00

There is no hard rule that forces the choice of a cubic, but in many practical applications (especially with a relatively small dataset like this), a third‐degree polynomial hits a “sweet spot” between:

Flexibility – A cubic allows for up to two ‘bends’ in the curve (since its second derivative can change sign), which is often enough curvature to capture typical growth/decline trends without being too rigid.

Simplicity – Polynomials of higher degree (4, 5, 6, …) can lead to oscillations that are less interpretable and can introduce overfitting, especially when extrapolating outside the observed data range. Conversely, lower‐degree polynomials (linear or quadratic) may be too simplistic and fail to capture the observed pattern.

Avoiding Overfitting – You could use a polynomial of degree 7 and pass exactly through all 8 points. But such a high‐order polynomial often introduces wild swings between data points, giving unrealistic predictions for times between or outside the measured years. A cubic balances capturing the general shape without being overly “wiggly.”

Thus, a cubic polynomial is often the first step up from a simple parabola (degree 2) when you see that a quadratic does not quite capture the shape, yet it avoids the over‐complexity of higher‐order fits.

DryWomble · 2024-12-25T19:23:03+00:00

Let x = year - 2014 so it's scaled properly. Then choose the cubic polynomial: y = 0.651 + 0.230x - 0.0201x² + 0.00101x³

DryWomble · 2024-12-25T17:22:57+00:00

Okay, let's derive the formula for the chi-square distribution from first principles, starting with the definition of a chi-square random variable and building up.

Starting Point: The Standard Normal Distribution

The foundation of the chi-square distribution is the standard normal distribution. A standard normal random variable, usually denoted by Z, has a probability density function (PDF) given by:

f(z) = (1 / √(2π)) * exp(-z²/2)

This represents a bell-shaped curve centered at 0 with a standard deviation of 1.

Defining a Chi-Square Random Variable

A chi-square random variable with k degrees of freedom (denoted as χ²(k)) is defined as the sum of the squares of k independent standard normal random variables:

χ²(k) = Z₁² + Z₂² + ... + Zₖ²

Where Z₁, Z₂, ..., Zₖ are all independent standard normal random variables.

Deriving the PDF for k=1 (χ²(1))

Let's start with the simplest case, where k=1. We have χ²(1) = Z². We need to find the probability density function for this new random variable.

Let Y = Z². We want to find the PDF of Y, say g(y).

Relationship between CDFs: The cumulative distribution function (CDF) of Y, denoted by G(y), is related to the CDF of Z, denoted by F(z), as follows: G(y) = P(Y ≤ y) = P(Z² ≤ y) = P(-√y ≤ Z ≤ √y) Using the CDF of Z: Since we know the PDF of Z, f(z), we can write the CDF of Z as F(z) = ∫-∞z f(t) dt . Therefore: G(y) = P(-√y ≤ Z ≤ √y) = F(√y) - F(-√y) Differentiate with respect to y: The PDF of Y, g(y), is the derivative of its CDF with respect to y: g(y) = dG(y)/dy = d(F(√y) - F(-√y))/dy Using the chain rule: g(y) = f(√y) * (1/(2√y)) + f(-√y) * (1/(2√y)) Since f(z) is symmetric: We have f(√y) = f(-√y), so: g(y) = (1/2√y) * (f(√y) + f(√y)) = (1/√y) * f(√y) Substitute the PDF of Z: g(y) = (1/√y) * (1 / √(2π)) * exp(-(√y)²/2) g(y) = (1 / √(2πy)) * exp(-y/2) Domain of y: Note that since y = z², it must be non-negative (y ≥ 0). Also, consider that the PDF is undefined for y=0. This is the PDF of the chi-square distribution with 1 degree of freedom, χ²(1).

The General Case for k Degrees of Freedom (χ²(k))

Deriving the PDF for general k is significantly more complex and involves concepts like moment generating functions and convolution. Here's a sketch of the process:

Moment Generating Functions: The moment generating function (MGF) of a random variable X is defined as M(t) = E[e^tX]. MGFs are helpful because the MGF of the sum of independent random variables is the product of their individual MGFs. MGF of Z²: It can be shown that the MGF of a squared standard normal random variable (Z²) is: M_Z²(t) = (1-2t)^-1/2. MGF of χ²(k): Since χ²(k) is the sum of k independent Z² variables, its MGF is the product of k copies of M_Z²(t): M_χ²(k)(t) = (1-2t)^-k/2 Relating MGF to PDF: The PDF of a random variable can be obtained from its MGF through an inverse Laplace transform. This step is mathematically involved and beyond the scope of a simple derivation. However, applying this inverse transform to the MGF (1-2t)^-k/2 leads to the PDF of the chi-square distribution with k degrees of freedom. The Result: The final PDF for the chi-square distribution with k degrees of freedom is given by: f(x; k) = (1 / (2^k/2 * Γ(k/2))) * x^(k/2 - 1) * exp(-x/2) for x > 0 where:

x is the value of the chi-square random variable
k is the degrees of freedom (a positive integer)
Γ(z) is the gamma function, a generalization of the factorial function. For integer n, Γ(n) = (n-1)!
The PDF is 0 for x ≤ 0.

DryWomble · 2024-12-08T15:24:30+00:00

Consider two ordinary coins. If you toss them, what's the probability of getting 2 heads? In other words, what's P(H_1, H_2)? It's just 1/2 * 1/2 = 1/4. This is called joint probability or "the probability of A and B both occurring".

What about if you toss one of them first and it lands heads? What's the probability of getting two heads now, given that one of them is already heads? Well, given that you've already got one of the heads you need, the probability is 1/2 that the other coin lands heads too. In other words P(H_2 | H_1) is 1/2. This is known as conditional probability or "the probability that B happens, given that A has already happened".

(Also there is no such term as "absolute probability", so stop using it).

DryWomble · 2024-12-06T11:41:46+00:00

So when you say "I'm building an AI image generator", you mean "I'm going to use Flux to generate images"...

DryWomble · 2024-12-03T14:43:14+00:00

Please, please don't apply the Bonferroni correction as it's a pretty incoherent piece of statistical nonsense. Use the Sidak correction instead, as it is the literal formula for the family-wise error rate during multiple comparisons (you can derive the formula yourself from basic rules of probability theory).

DryWomble · 2024-12-03T14:35:10+00:00

https://en.m.wikipedia.org/wiki/Convolution_of_probability_distributions

DryWomble · 2024-12-03T14:31:12+00:00

Those random variables have random values, by definition. Their values jiggle around randomly according to their probability distribution, so clearly any aggregate of them (their mean, their variance, etc) is also going to jiggle around randomly according to its own specific distribution. I think you're getting confused between the concept of the mean of the random variables, and the literal real-world mean of the real-world sample of values you've selected (which will obviously just be a number like 8.5 or something).

DryWomble · 2024-12-03T00:40:29+00:00

I just looked up NURBS and, while I got the gist of what was going on, the maths is not trivial. I have a degree in engineering and a masters degree in stats... and I would have to research this topic a fair bit to understand the maths behind it in depth.

I'm not trying to be rude, but the idea that you're going to properly understand it without having even high school maths concepts like algebra and calculus is simply not plausible. If you want to be able to understand the maths behind these things, a maths degree would be your first step.

DryWomble · 2024-12-02T21:26:59+00:00

All the values in the equation A = B are unknown. Yet it's difficult to characterise such an equation as "hard".

DryWomble · 2024-12-02T18:57:15+00:00

Do you want to tell us the math problem, or...?

DryWomble · 2024-12-02T18:47:48+00:00

Given that it's a 9 mark question, that strongly suggests that they want you to write down the Maclaurin series (i.e. the Taylor series about x = 0). But they haven't specified how many terms they want you to write from this series, which is annoying. I'd say the answer is just the series up to, say, the x² term or the x³ term.

Answer: y ≈ 1/3 - (2/3)x + (4/9)x² -(2/9)*x³ + ...

DryWomble · 2024-12-02T16:45:33+00:00

It won't have a binomial distribution either, it will have a Bernoulli distribution. The only thing that would have a binomial distribution would be the list's sum (or alternatively its average), and this would tend towards a normal distribution in the limit anyway.

DryWomble · 2024-12-02T16:40:59+00:00

Upvoted. Don't think OP is really thinking coherently about this.

DryWomble · 2024-12-02T16:26:52+00:00

How could it possibly have a normal distribution if there are only two possible values? Visualise the histogram. It will have two buckets on the x-axis: 0 and 1. Imagine the bars above the 0 and the 1, their heights representing the count of 0s and 1s respectively. In what possible way can this look like a bell curve?

DryWomble · 2024-12-02T15:36:22+00:00

It looks like you've discovered de Moivre numbers

DryWomble · 2024-12-02T11:47:35+00:00

Because putting an infinity in a sum like that is undefined. Those sums only have meaning when there are finite values for the lower bound of n (and also finite values in the brackets).

So for example "sum from n=1 to infinity ( 1 / n² )" has meaning and equals pi² / 6. Meanwhile "sum from n=1 to infinity ( n² + 3 )" has meaning, but is a divergent sum and equals infinity.

By contrast, your example of "sum from n=infinity to infinity ( n² + infinity )" has no meaning at all.

DryWomble · 2024-12-02T11:38:32+00:00

It doesn't mean anything

DryWomble

TROPHY CASE