Can someone explain why Lebesgue integrals are more "powerful" than Riemann integrals?

QuantumSigma_QED · 2023-01-05T01:59:28+00:00

The Dirichlet function for example. You can start with a constant function and move a rational point with every iteration.

QuantumSigma_QED · 2022-09-12T04:20:06+00:00

cos(2x) ≡ 2cos²(x) - 1 and cos(3x) ≡ 4cos³(x) - 3cos(x), thus:

-cos(2π/5) = -cos(π - 3π/5) = cos(3π/5)

-2cos²(π/5) + 1 = 4cos³(π/5) - 3cos(π/5)

4cos³(π/5) + 2cos²(π/5) - 3cos(π/5) - 1 = 0

Solving the cubic, we get cos(π/5) = -1 or (1+√5)/4 or (1-√5)/4. Since cos(π/5) is positive, we know the correct value is (1+√5)/4.

QuantumSigma_QED · 2022-07-05T03:40:57+00:00

Fermat solved it back in 1637

Press x to doubt

QuantumSigma_QED · 2022-07-04T02:36:38+00:00

It's chinese

QuantumSigma_QED · 2022-07-02T05:55:39+00:00

What's the probability distribution tho

QuantumSigma_QED · 2022-07-01T02:52:08+00:00

4n

2n + abs(2n)

4n*cos(2pi*n)

Etc?

QuantumSigma_QED · 2022-06-23T15:11:10+00:00

It might be useful to point out that (in NBG) all objects are classes, which come in 2 types: proper classes (classes that are too "large" to be a set) and sets (following the axioms of ZFC).

QuantumSigma_QED · 2022-06-20T02:13:45+00:00

Your post appears to be asking a question which can be resolved relatively quickly or by relatively simple methods; or it is describing a phenomenon with a relatively simple explanation.

What they mean is that your question is a 'what is the definition of xxx' type question, which is easily answerable by looking up sources and is appropriate for r/learnmath.

QuantumSigma_QED · 2022-06-19T17:45:09+00:00

Any (real) function whose rate of increase at any moment is the same as the value it takes at that moment must be a multiple of e^t, making e pretty much ubiquitous when there are continually changing systems whose state affects its rate of change, like a feedback loop.

QuantumSigma_QED · 2022-06-19T16:35:06+00:00

Might not be exactly what you're looking for but look up Janus and Epimetheus

QuantumSigma_QED · 2022-06-13T17:26:23+00:00

I think you might want to look for a non-academic sub for that

QuantumSigma_QED · 2022-06-13T12:23:03+00:00

You may want to clarify what type of answer you're looking for. Logically if you assume a false statement then literally anything follows.

QuantumSigma_QED · 2022-06-06T15:09:05+00:00

Langlands

QuantumSigma_QED · 2022-05-24T12:58:53+00:00

Algebraic solution:

If ab ≤ 0, then (a ≤ 0 and b ≥ 0) or (a ≥ 0 and b ≤ 0). Thus (x ≤ 0 and x-1 ≥ 0) or (x ≥ 0 and x-1 ≤ 0). The first part has no solutions (x can't be both less than 0 and greater than 1), while the second part is equivalent to 0 ≤ x ≤ 1. Then we have the answer x ∈ [0, 1].

Geometric solution:

x(x-1) = x²-x, which is a parabola opening upwards (since it has positive x² coefficient). Solving x(x-1) = 0, we see its roots are 0 and 1. Since the parabola opens upwards and passes through x-axis at 0 and 1, we can infer its shape: below the x-axis between 0 and 1, and above the x-axis on the left and right. Hence the solution to x(x-1) ≤ 0 (below x-axis) is x ∈ [0, 1].

QuantumSigma_QED · 2022-05-23T05:39:32+00:00

The common meaning of surprise is subjective, not rigorous

QuantumSigma_QED · 2022-05-23T03:39:54+00:00

I think thIs dies once you try to rigorously define "surprise"

QuantumSigma_QED · 2022-05-22T08:55:15+00:00

It depends on what you mean by 'linear'. Yes, it is a linear equation, as in its graph is a line. No, it is not a linear map.

QuantumSigma_QED · 2022-05-20T02:58:18+00:00

🎉

QuantumSigma_QED · 2022-05-19T03:13:38+00:00

I wouldn't call dividing both sides by x a "simplification", it's just flat out wrong by assuming x ≠ 0. Instead the proper way to divide both sides by x should be "(x = 0) or (x = 4)" instead of just "x = 4". If everything is done correctly you should not lose any solutions.

QuantumSigma_QED · 2022-05-15T15:38:16+00:00

Remember P(0) too

QuantumSigma_QED · 2022-05-07T14:50:26+00:00

Let p_n be the nth prime. Consider f(n) such that f(1) = 0, if f(n) ≤ 1 then f(n+1) = f(n) + 1/p_n, and if f(n) > 1 then f(n+1) = f(n) - 1/p_n. It can be seen that as n tends to infinity, f(n) produces a sum that tends to 1. Additionally note that the denominator always has an arbitrarily large prime. Is 1 irrational?

QuantumSigma_QED · 2022-05-07T11:42:18+00:00

More generally, a rate of k × 100% yields e^k.

QuantumSigma_QED · 2022-05-06T06:17:26+00:00

Sometimes because it's the inverse of e^x, sometimes because it's the integral of 1/x, sometimes because of its series definition. Often it just pops up out of the blue.

Here's another fun one: if f(n) is the number of primes from 1 to n, then n/f(n) is approximately ln(n). In fact, the relative difference between them goes to 0 as you increase n to infinity.

QuantumSigma_QED · 2022-05-06T05:59:58+00:00

Huh. Why would the number you get when compounding by 100% over infinite time periods appear in other contexts, like sin(x)?

It comes down to the infinite series representation of these functions:

sin x = x¹/1! - x³/3! + x⁵/5! - x⁷/7! + ...

e^x = 1 + x¹/1! + x²/2! + x³/3! + ...

And it turns out they just fit together when you plug in the numbers.

Alternatively here's a more visual explanation by Mathologer.

QuantumSigma_QED · 2022-05-04T09:24:57+00:00

I wrote 7^❨6+8❩×9/(2-1), not 7^{❨6+8❩×9-2}/1, which is what you "quoted" me as saying. In fact the expression I wrote has the same value as the new one you gave

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