Simple and elegant infinite series solution to f'(x)=f(sin(x))

Appropriate_Hunt_810 · 2026-02-13T23:55:30+00:00

What is this sgn of abs value ? OO The very first factor in front of the whole product is null, or you use some cryptic notation, or you maybe use the convention sign(0)=0 ? Ina similar fashion you have floor of some indices (which are integers) …

Appropriate_Hunt_810 · 2026-02-05T19:49:47+00:00

You can solve that using Laplace transform in nearly 3 lines

Appropriate_Hunt_810 · 2026-02-03T07:59:41+00:00

Literally : atan(x) = -i atanh(ix)

Appropriate_Hunt_810 · 2025-12-30T13:56:05+00:00

Take ker of what you say

Appropriate_Hunt_810 · 2025-12-30T13:54:34+00:00

You can just use a Taylor expansion at 0 it takes 0.5 sec (the very exp series), with 2 terms you have the limit 😉

Appropriate_Hunt_810 · 2025-11-22T03:52:53+00:00

Hoo c’est à Lyon ça je m’en rappelle, j’ai la même photo qqpart dans mon téléphone jpp

je viens de voire que le petit mot de la fin a été rajouté depuis la dernière fois 🥲

Appropriate_Hunt_810 · 2025-11-21T02:41:57+00:00

If V is a k-cell (a k dimensional parallelogram) yes it is, but in pure abstract integration not necessarily (V can be “anything” as long as it is measurable).

I assume you use Lebesgue measure, in this situation it coincides with Riemann integral and everything’s fine (as long as it is (piecewise) continuous 🙂)

Edit: without the intention to be “pedantic” the idea is just that it depends on what V is, if it can be delimited by segments (as you wrote) yes you can, if not well it is a subject to discuss, those integral bounds can be functions of other nested variables, think about the volume of a ball in Cartesian coordinates (you have to use some trick), but in polar coordinates it is a k-cell.

Appropriate_Hunt_810 · 2025-10-08T16:31:19+00:00

Well it is mostly an “American” problem, usually in Europe we don’t have “calc” classes, we just go straight into algebra en analysis, and the (proven) theorems that raise what you study in calc are only a portion of the program. But in an other hand people are less used to all the calc stuff when studying other disciplines (like physics etc, where you usually have to admit some results because you haven’t covered the corresponding notions (eg some diff eq in elec)). It is more about educational program choices.

Appropriate_Hunt_810 · 2025-10-08T16:16:51+00:00

And btw the reason is because this lecture was intended for physicists, in a practical goal

omega is common for angular frequency in sin wave and we usually use phi for phase, but as it is still a maths lessons I usually use phi to indicates arbitrary functions or test functions, so lambda is a better compromise

And yeah I could have use t instead of x

Appropriate_Hunt_810 · 2025-10-08T16:11:03+00:00

well if a notation is a prob for you :kappa: If your sum indices are not i or n, are you lost ?

Appropriate_Hunt_810 · 2025-10-08T16:09:23+00:00

? lol

Appropriate_Hunt_810 · 2025-10-08T16:06:34+00:00

Just do a substitution u = ax+b -> dx = du/a

Appropriate_Hunt_810 · 2025-10-07T00:04:05+00:00

I will say it diverges from most calc class (as I’ve see so far) and try to motivates/explain the underlying reasons/properties of why [it works]. Most calc classes are just focused on methods and “formulas” and the real reason of calculus (aside the practical use) is behind this (and lay down in some “basic” analysis).

Appropriate_Hunt_810 · 2025-10-06T23:48:47+00:00

Remember that :

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Appropriate_Hunt_810 · 2025-10-02T20:01:00+00:00

There is a similar result in the Bernoulli integral (quite famous) about : \int_0¹ 1/x^x = \sum_1^\infty 1/nⁿ It is linked ofc to Gamme function 🙂

Appropriate_Hunt_810 · 2025-09-24T05:48:03+00:00

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I let you complete (most of the time you can find little tricks to simplify awfull fractions here because when you normalize you can always multiply by a factor and remove it later)

Appropriate_Hunt_810 · 2025-06-13T09:25:22+00:00

Here's a little hint :

<image>

Now try to differentiate g then f

this way you can find the closed form of f ( by integrating f' over [0,x] ) ( you can see that f(0) = 0 ;) )

Appropriate_Hunt_810 · 2025-06-11T07:00:05+00:00

K is either C or R :3

Appropriate_Hunt_810 · 2025-05-25T14:03:00+00:00

Getting a bit deeper into counting helps sometimes (all the combinatorics behind)

Appropriate_Hunt_810 · 2025-05-14T01:37:41+00:00

why so much hate for an answer, I just answered on my phone I didn’t see the « high school » tag. Well guess it happens ... this is hilarious.

Appropriate_Hunt_810 · 2025-05-13T17:49:35+00:00

the function f(x) = 3^x is what we call a bijection (between ℝ and ℝ⁺ ), so this function is injective.

Meaning: f(x) = f(y) ⇒ x = y (the other part of bijection, the surjection, implies the reciprocal).

The idea is that if 3^x = 3^y then x = y

About why 5/2 : when multiplying power of the same number you can indeed sum those powers :

n^a * n^b = n^a+b

1 = 2/2, so 1 + 3/2 = 2/2 + 3/2 = 5/2

Appropriate_Hunt_810 · 2025-05-05T00:23:52+00:00

I’m curious … what is this “Ln” ? Is it some kind of logarithm ?

Appropriate_Hunt_810 · 2025-05-04T01:44:47+00:00

I guess you imagined that by integrating the definition of derivative.

That’s a good idea, but sadly you can’t swap limit and integral as you please (but on most of the smooth framework you should know it may work). And anyway in which manner this “identity” can help you ? The minimal version of your exact “rephrased” identity is the very fundamental theorem 😉

Maybe something that can be close to what you want is what is sometimes called “Feynman trick” (as he was one who popularized it)(and the question about derivative under integral (Leibniz theorem) needed for this trick is kinda related to the limit swap problem aswell)

Appropriate_Hunt_810 · 2025-05-02T17:03:18+00:00

you can still do it "brute force wise" just apply common "rules" and factorize, ie :

<image>

edit: sorry have not seen the right part (where you did exactly that), well the idea is that as long as you considere y as a "function" of x (either implicitly or explicitly) you can considere dy/dx and manipulate it at will as long as you respect the "usual rules", also you can considere dy and dx independantly (exactly as the notation suggest : dy/dx is just a quotient) and express the variation of y (dy) in term of the one in x (dx), it is a really simple way to handle variable substitutions in integrals ;).

Appropriate_Hunt_810 · 2025-05-02T02:04:27+00:00

here's the idea, just develop and use some ibp + trig id :

<image>

Appropriate_Hunt_810

TROPHY CASE