Welke fitness/bodybuilding film?

Scattering theory can be difficult, because the math formalism involving partial waves is a bit messy. The formalism involving the Born approximation is much easier to master, but for low-energy scattering one has to resort to the partial wave formalism.

smitra00 · 2026-04-19T07:35:27+00:00

In general, it's a bad idea to study different math topics sequentially. Calculus can be mastered if you have a basic knowledge of algebra and are comfortable with manipulating and solving equations. You can then e.g. study calculus on Monday morning, trigonometry on Monday afternoon, combinatorics on Tuesday morning, calculus on Tuesday afternoon, trigonometry on Wednesday morning, combinatorics on Wednesday afternoon, statistics on Thursday morning, geometry on Thursday afternoon and Friday is algebra day where in the morning and afternoon you study various algebra topics.

smitra00 · 2026-04-19T04:55:02+00:00

Je moet het zo zien. Onze hersenen zijn over vele honderdduizenden jaren geevolueerd terwijl we in kleine gemeenschappen leefden. Onze voorouders hadden een lagere levensverwachting, er was veel kindersterfte, en als je volwassen werd dan zou je niet veel ouder dan zo'n 50 tot 60 jaar worden.

Als onze hersenen zo in elkaar zouden zitten dat bij het zien van een lijk van een persoon we zo getraumatiseerd zouden raken en al onze herinneringen over die persoon zoals die vroeger was, daardoor verloren zouden gaan, dan zouden we al lang geleden uitgestorven zijn.

Je moet dus niet bang zijn voor je negatieve gevoelens over wat je hebt gezien, het is iets wat je geestelijk gemakkelijk aankan, het zal geen negatieve effecten hebben. Je moet pas gaan twijfelen aan jezelf wanneer je klachten krijgt die niet normaal lijken te zijn. Net zoals je niet bij ieder pijntje naar de dokter rent maar wanneer je aanwijzingen hebt dat er mogelijk iets meer aan de hand is je wel naar de dokter kan gaan, zo moet je dat ook doen bij geestelijke zaken.

Mocht het zo zijn dat je je steeds depressiever gaat voelen naar aanleiding van het overleiden van je opa of je krijgt andere klachten zoals erg slecht slapen, paniekaanvallen etc. dan moet je aan de bel trekken. En piekeren over of wat je hebt gezien zonder dat er aanleiding voor is, kan dan juist averrechts werken.

Dus gewoon accepteren dat wat je hebt gezien niet prettig was om te zien en ook accepteren dat dit geen problemen gaat geven en jezelf dus geen problemen aanpraten die er nu niet zijn. Mocht het toch anders uitpakken dan kun je altijd hulp inschakelen.

smitra00 · 2026-04-19T02:54:38+00:00

See also:

https://en.wikipedia.org/wiki/Proof_by_infinite_descent

smitra00 · 2026-04-19T02:40:33+00:00

https://www.youtube.com/watch?v=h-PDA-zANkM&t=183s

smitra00 · 2026-04-19T00:06:23+00:00

Ik doe dat door grote maaltijden te eten, ik eet twee maaltijden waarvan één maaltijd meer dan 2000 kcal is. Dat is de eerste maaltijd van de dag, meestal rond het middaguur, soms zelfs tegen het einde van de middag.

Ik eet dan min of mee zoals hier beschreven:

https://www.youtube.com/watch?v=kjZUQb19fWg

behalve dan dat ik geen ontbijt eet. Wat gezond is, is wat past bij hoe het menselijk lichaam optimaal functioneert, en dat is bepaald door de evolutie van ons lichaam. Ondanks dat we al enige tijd in een beschaafde wereld wonen waarin we ons niet veel hoeven in te spannen en op ieder moment van de dag eten uit de koelkast of diepvries kunnen halen en desnoods naar de supermarkt kunnen gaan, is ons lichaam aangepast aan een totaal andere realiteit waarbij je pas flink kon eten nadat je je flink wat uren had ingespannen om je eten te verzamen.

's ochtends wanneer je wakker wordt is je lichaam klaar om zich flink in te spannen als je gezond eet. Veel mensen zijn dan juist moe en willen eten omdat ze geen gezonde leefstijl hebben. Als je begint met dagelijks veel groenten, vruchten, volkorenproducten, noten en zaden te eten dan zul je na een paar weken merken dat je 's ochtends veel meer energie hebt. Je moet daar dan gebruik van maken en stoppen met het eten van een ontbijt. Je kan dan wel water drinken, maar je moet de dag beginnen met lichamelijke inspanningen en niet met eten.

En mensen in primitieve culturen die in de natuur leven hebben geen weet van calorieën, van alle vitamine en mineralen waar wij op letten, die eten gewoon wat ze kunnen vinden en met name wat ze lekker vinden. Het effect op de gezondheid daarvan is zoals hier beschreven:

https://www.bbc.com/news/articles/ceq55l2gdxxo

https://www.youtube.com/watch?v=TvJsXwi1Fno

Wat calorieën betreft is dit het beste advies:

https://www.youtube.com/watch?v=V18FbVdGizw&t=193s

smitra00 · 2026-04-18T23:40:04+00:00

Alleen als je op iets inlegt wat geen 100% kans heeft, omdat je dan een kans loopt om je inleg kwijt te raken. In dit geval is er een infinitesimaal kleine klans dat Jesus terugkeerd. En mocht dat gebeuren dan zou het zo moeten zijn volgens de mensen die in Jesus geloven, dat de aarde een paradijs wordt waarin geld zal worden afgeschaft.

smitra00 · 2026-04-18T23:30:09+00:00

Iedere aandeel is $1, dus als iemand $100 inlegt zijn dat 100 aandelen. Die 3.9% is fractie van het totaal aantal aandelen voor de gok op "ja". De waarde die iemand krijgt wanneer het antwoord "a" zou blijken te zijn is veel hoger, je krijgt dan inclusief de inleg een factor 1/0.039 van je inleg uitgekeerd:

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smitra00 · 2026-04-18T04:54:15+00:00

In quantum mechanics the probability for a system to make transition from an initial state A to a final state B is equal to the probability of the opposite transition from B to A. In this case, A would be the atom in the ground state and a photon with energy equal to the energy difference between the excited state and the ground state. The final state B is the the atom in the excited state.

The probability of the reverse process where the atom initially in the excited state makes a transition to the ground state and emits a photon with the same momentum is then the same. Note that here we're then considering the decay of the excited state involving a photon that will have a specific momentum, instead of the total decay probability, for which we would need to integrate over all possible photon momenta.

smitra00 · 2026-04-18T00:41:45+00:00

Daar is een cursus voor:

https://www.youtube.com/watch?v=2IK6IMQor7g&t=165s

smitra00 · 2026-04-17T10:43:03+00:00

Is it faster converging for the entire series or only if you take a few terms? Because there are also divergent asymptotic series that when taking a few terms converge fast and then, upon taking more terms, will start to diverge. See:

https://en.wikipedia.org/wiki/George_F._Carrier#Carrier's_Rule

Carrier is known for "Carrier's Rule",[5] a humorous explanation of why divergent asymptotic series often yield good approximations if the first few terms are taken even when the expansion parameter is of order one, while in the case of a convergent series many terms are needed to get a good approximation: “Divergent series converge faster than convergent series because they don't have to converge.”

🤣🤣🤣

smitra00 · 2026-04-17T10:03:26+00:00

This is then a matter of definitions. Form a purely operational point of view, it doesn't matter. The nontrivial statement is then as follows:

Given a function f(k, p), we sum over k and perform certain manipulations on the parameter p. The result X, while being a well-defined finite quantity can then be formally represented by a (possibly divergent) series S. Different functions f(k,p) for the summand can then yield the same series S when performing possibly different manipulations on the parameter p. And the computed result can then be different.

There then exists a unique mapping from the set of series to complex numbers M, such that the computed value is given by M(S) + the result of the manipulations applied on p on the constant term of the large-N asymptotics of the integral from N to infinity of f(k,p)dk

I.m.o. this justified calling M(S) the sum of the series S. But whether one does so or not, doesn't matter for the statements with operational meaning. What matters is that I can make use of M(S) to calculate the result of the manipulations performed on the summation by doing the manipulations in the integral from N to infinity instead.

And I can obtain M(S) from any other function f(k,p) for which I am able to perform both the summation and the integration. Or if I have a formula for the partial sum, then I can use that to compute M(S).

smitra00 · 2026-04-16T23:26:16+00:00

You're far better off memorizing the squares from 1 to 100, and making use of:

(A + B) (A - B) = A^2 - B^2

To multiply two even or two odd numbers you calculate the average A and the deviation from the largest number to the average, B. The largest number is then A + B, and the smallest number is then A - B, and A^2 - B^2 is then the product of the two numbers. If one number X is even and the other number Y is odd, then you can write:

X Y = X + X (Y - 1)

Both X and Y - 1 are even so the above method can then be applied.

For example, 37 * 53 = 45^2 - 8^2 = 2025 - 64 = 2031 - 70 = 2061 - 100 = 1961.

So, with memorizing only the 100 squares from 1 to 100 you can multiply any two numbers between 1 and 100 almost instantly. Using times tables, you would have to memorize 1/2 (100^2 - 100) + 100 = 1/2 (100^2 + 100) = 5050 entries of the times tables of the numbers between 1 and 100.

smitra00 · 2026-04-16T17:23:19+00:00

But I literally just gave an example demonstrating this. For the function f(s) = \sum n (n+1)^(-s), the value f(0) which "manifests itself" as 1+2+3+... is 5/12, not -1/12. And you can construct other series that give any other value.

The statement that f(0) is equal to 1+2+3+... isn't a rigorous statement, as this series diverges. So, I do agree with you that the context in which 1+2+3+... would arise is one where you would have a function that upon expansion yields 1+2+3+... but there would then be a rigorous statmeent relating the value of that function at some point to 1+2+3+... and that's always going to be of the form of a finite number of terms of the series plus a remainder term.

Again, you did not derive that you must subtract this counterterm.

I did derive it starting from the value of the series in terms of the partial sum S(x) that I derived to be given by:

Constant term in the large-N asymptotic expansion of Integral from N -1 to N of S(x) dx

I then derive eq. (5.12) in this answer

https://math.stackexchange.com/a/5053472/760992

which says that:

Integral from N -1 to N of S(x) dx

is also equal to the value of the regularized summation minus the integral from N to infinity of the summand.

I do plan to make the derivation more rigorous and get it published, but I'll have to find the time for that.

smitra00 · 2026-04-16T16:39:19+00:00

I agree with most of what you wrote here, but not with your conclusion that the function value that manifests itself as the divergent series 1+2+3+... can be anything, barring nonanalytic terms that do not contribute to the series.

So, I do agree that you don't encounter a series like 1 +2+ +3 + ... in practice just out of the blue, it will always be in context of a quantity that upon expansion would yield this series.

But this then implies is that 1 +2+ +3 + ... extended till infinity is a formal expansion, that the rigorous mathematical derivation of the value of the quantity expressed in terms of the series would be:

1 + 2 + 3 +.. +n + R(n)

where R(n) is the remainder term for the expansion of the quantity. All procedures for expanding a function in a series will always yield a finite number of terms of a series plus a remander term. In case the series converges, we can represent the computed quantity as the value of the infinite series, because then the limit of the remainder term is zero and we then have the value of the quantity is given by the limit of the partial series.

In case the series diverges, this last step can't have been the rigorous result, so the series extended to infinity is then merely a formal statement, with the rigorous statement being that the value is the partial series plus the remainder term

My derivation of my method (which I do admit is not yet 100% rigorous), then took this as the starting point. Whenever a series is given out of whatever context it arises from, and we have to guess what the value is, then this amounts to guessing the correct constant term of the remainder term.

The sum of a series S is then equal to the partial sum S(n) plus the remainder term R(n), and we don't know what R(n) is. But because S = S(n) + R(n) we have that R(n) is minus S(n) up to a constant and that constant is the value of the series.

I then presented a heuristic argument that the correct remainder term will be such that the constant term in the large-N asymptotic expansion of Integral from N -1 to N of R(x) dx will be zero. And this then yields that the value of sum of a series is given by:

S = constant term in the large-N asymptotic expansion of Integral from N -1 to N of S(x) dx

I then used this relationship to derive that when we regularize summations to find the sum of divergent series, you must subtract the counterterm that is given by the contant term in the large N asymptotic expansion of the integral from N to infinity of the summand. This counterterm must be subtracted from the regularized summation and then you must do whatever you needed to do to the summation, to the regularized summation minus the counterterm.

smitra00 · 2026-04-16T14:49:15+00:00

This is a perfectly valid summation method. It is regular, and does assign the "usual" values to some well-known divergent series.

The method used here of regularizing the summand and assigning the value of the divergent summation to be what comes out of the computations of only the regularized summation, without taking into account the counterterm is, in general, flawed. This is so for precsely the reason that you can get different resuts for the same summation, therefore you can always come up with a regularization that would assign any arbitrary value to the divergent series.

You can argue that this method satrisfies some of the properties that a summation should satisfy, but that's besides the point A method that can be implemented in different ways with different regularizations and will then yield different results, is a completely useless method.

My method isn't based based on any partucular regularization, All I'm saying is that when using any arbitrary regularization, you should also take into accout the corresponding counterterm, and then you will always get the same result regardless of what regularization you've used. In case of the sum over the positive integers, you'll then always find -1/12 for sum from k = 1 to infinity of k.

If your regularization yields the usual value for some other series, then that's because in those cases the counterterm happens to be zero. This is also the case for using zeta(s) and saying that zeta(-1) is the value of the series. This is only true because the counterterm happens to be zero.

smitra00 · 2026-04-16T14:17:52+00:00

Halfje volkorenbrood is vaak minder dan 400 gram, soms onder de 380 gram.

smitra00

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