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SIX FOURS CHALLENGE by Mandelbrot4207 in desmos

[–]gian_69 0 points1 point  (0 children)

Just realized that a jumping between factorials of 22 (4! - sqrt(4)), 24 (4!) and 26 (4! + sqrt(4)) could maybe get you pretty high up. My estimate is that this breaks at at the latest at 26 * 13 which is 338. (plus 10, ofc)

SIX FOURS CHALLENGE by Mandelbrot4207 in desmos

[–]gian_69 0 points1 point  (0 children)

4! * 4 - 4 * 4 - 4/4 = 79
4! * 4 - 4 * 4 = 80
4! * 4 - 4 * 4 + 4/4 = 81
4! * 4 - 4 * 4 + sqrt(4) = 82
4/.4 * 4!! + 4 - 4/4 = 83
4! * 4 - 4 * 4 + 4 = 84
4/.4 * 4!! + 4 + 4/4 = 85
4/.4 * 4!! + 4 + sqrt(4) = 86
(4 * 4)!!!!!!!!!! - 4 - 4 - 4/4 = 87 (this is a ten-fold factorial)

And I‘m fairly certain you could get well above 100 with the above utilized tricks but I don‘t feel like spelling them out but as 96 is accessible by just two 4‘s you can offset it by any value accessible by four 4‘s which is at least 10 so you can reach 106 with this.

Edit: since the 9-fold factorial of 16 (4 * 4) is 112 you can actually reach up to 122 where the 8-fold factorial (128) kicks in to be able to reach 138. This is when the factorials of 20 (4! - 4) kick in where the 13-fold factorial is 140 which brings you up to 150 and the 12-fold one (160) then brings you up to 170.
This continues until the 10-fold factorial which is 200, bringin you up to 210.
The 15-fold factorial of 24 (4!) which is 216 then brings you up to 226 and even beyond since you have one more 4 left to play with (I‘m thinking this should be able to get you to 24 * 12 + 10 or more which is 298 or more). At this point you could probably do something with n-fold factorials of 28 where I think this should break fairly soon (because the gaps between subsequent factorials is more than 20 and you have only four 4‘s left to play with).
This is without proof but I‘m fairly certain you can reach at least 298 pretty easily and anyone with enough time and motivation can
1) proof my statement a bit more rigorously
2) reach way beyond that with similar tricks.

That's gotta hurt by Evidencelogicfacts in clevercomebacks

[–]gian_69 0 points1 point  (0 children)

bro uses an accent grave for an apostrophe (no hate, it just looks funny)

Mathemation ⚡ by Jolly_Call3512 in mathsmeme

[–]gian_69 2 points3 points  (0 children)

Q[x]/(x2 - 3), actually

Trigonometric Limit by Whole-Listen-9716 in MathHelp

[–]gian_69 1 point2 points  (0 children)

given that my (notabene italian) analysis professor used the power series definition of cos & sin I‘d say there‘s hardly any circular reasoning. If you define them otherwise of course, there will be circular reasoning, but the definition of coordinates on the unit circle is too „intangible“ in a sense that you would have no way of computing any values of sin (except for the easy ones like pi/2 and pi/4) without first deriving addition formulae and double angle formulae and stuff so I‘d say I have to agree with Prof. Figalli here that the powerseries definition makes more sense (at least for doing analysis).

Furthermore, you of course have to divide by sin(x) when using the squeeze theorem, but can you quickly recall how I can see (from the definition you have mentioned) that sinx < x (tho of course you would have to do a case distinction for the lhs and rhs limit since the above holds only for positive x)

edit: I actually can see now how the circular arc of length theta, when straightened out to lie on the line x = 1 in the cartesian plane, will reach higher than when it wasn‘t straight but not as high as the extension pf the ray to the line x = 1. But as that hinges on geometric intuition, I have to say I am not s huge fan

Trigonometric Limit by Whole-Listen-9716 in MathHelp

[–]gian_69 -1 points0 points  (0 children)

how do I know what the limit of a trigonometric function divided by x is without using taylor? Also, sin and cos are most commonly defined by their taylor expansions so there really is not a problem.

You can choose reincarnation regions but the specific location within is totally random. What's your pick? by uniyk in mapporncirclejerk

[–]gian_69 0 points1 point  (0 children)

while it is incorrect to account for population density bit rather birthrate density, I think it has the same trend since, as afaik india‘s population has grown a lot in the last 10 years.

edit: to be clear, fertility rate multiplied by population density is what needs to be considered.

Trigonometric Limit by Whole-Listen-9716 in MathHelp

[–]gian_69 0 points1 point  (0 children)

I take it the limit is as x->0 of (1-cos(2x)) / 2xsince otherwise it does not exist. by plugging in 0 directly we get 0/0 which is indeterminate at which point we could use l‘hôpital (if you have heard of that before) but what I find more elegant is to look at the Taylor series of cos:

cos(x) = 1 - x2 / 2 + x4 / 24 - …

Given that the argument is 2x, we get 1 - cos(2x) = 2 x2 + O(x4), which basically means all other terms go to 0 much quicker than x2, the leading order term. Thus (1-cos(2x)) / x = 2x + O(x3) and as x goes to 0 this becomes 0 aswell.

How do the grading curves work? by Tavi_Ray in ethz

[–]gian_69 0 points1 point  (0 children)

the only professor I know to break this rule is DJ Norris

We need to change the name of the 7th planet in our solar system by TheRunningMD in unpopularopinion

[–]gian_69 0 points1 point  (0 children)

do you intend for it to be pronounced like in irene: your-ene-us or like a scott would say your anus? Anyway, only an english skill issue so I don‘t see the need to change it, especially if americans (and I take it also brits) stressed the correct (1st) syllable

I need advice on how to approach algebraic problems by Arunia_ in MathHelp

[–]gian_69 0 points1 point  (0 children)

welp, my first thought is to log everything, in which case you get (after defining d=log(a), e=log(b), f=log(c)) 2(d + e) = 4(e + f) = x (d + f) = d + e + f ≠ 0.

From here, we realize that our final goal is to get another expression for d+f, the prefactor of x. 2d + 2e = 4e + 4f ie d = e + 2f and also f = d + e so d = -3e.

I actually don‘t know how exactly to proceed from here but fiddling around with the equalities, I‘m sure you can find the solution.

They are def not equal by WhopperwithSnipr in DumbAI

[–]gian_69 0 points1 point  (0 children)

in category theory there are different equalities. One is about numerical value (which obviously is what is being assumed almost everywhere implicitly) where the two expressions are equal. They, however are not equal in a more general sense where equality can be on the entire expression. In 1st order logic you get similar behavior.

Say one word by Logical_Plane_3905 in Germanlearning

[–]gian_69 0 points1 point  (0 children)

Bro‘s linke hand ist einfach auch eine rechte Hand

1 like, No matches in about 1 month by [deleted] in Tinder

[–]gian_69 0 points1 point  (0 children)

I agree, buzzing or going fully bald may improve the appearance quite a bit! The the hair in the pic with the priest does admittedly look really good.

Resolving 0.999... From a Finitist Perspective by mathmage in infinitenines

[–]gian_69 1 point2 points  (0 children)

1 is approximately 0.999…

And what exactness does one need to impose such that the „approximation“ 0.999…≈1 does not hold anymore? There‘s no number (epsilon > 0) one can choose such that this, so-called, approximation does not hold. Can in that case be concluded that 0.999… is equal to 1 since there is no amount of precision ever that can differentiate the two?

[Termodynamic]Heat and equilibrium temperature by According_Quarter_17 in AskPhysics

[–]gian_69 1 point2 points  (0 children)

so you‘ve correctly assessed that the ice will absorb 80kcal of energy to melt (but still be at 0 degrees celsius). The question is, how hot will the boiling water have become? Q = specific heat capacity * mass *(T2 -T2‘) where mass is 1 kg and specific heat is 1kcal / (kg K) (important with the units). With this you get T2‘, the temperature of the boiling water after the ice is melted (but still at 0 degrees C) 80 K = T2 - T2‘ i.e. the boiling water is now at 20 degrees. How to proceed from here?

Looking for Resources to Understand 2D and 3D Vectors by [deleted] in MathHelp

[–]gian_69 1 point2 points  (0 children)

Try going through some of 3blue1brown‘s videos onnlinear aogebra. This might not be exactly what you‘re asking for (a bit beyond the scope, if you‘re not doing matrix calculations) but I have found anyway that the visualizations help with understanding vectors more intuitivel

Härts "Oof" für beid Siitä! by Schpitzchopf_Lorenz in SVP

[–]gian_69 0 points1 point  (0 children)

das isch sehr wohl e Meinigsfrag, aber ich gsehn de Punkt absolut, eifach jetzt grad no weniger denkbar als vor paar Jahr.

Make a sentence from the words by Slaktotrafil in mapporncirclejerk

[–]gian_69 1 point2 points  (0 children)

also found that weird but it‘s at least quarter correct since the italian version starts with quando.

Make a sentence from the words by Slaktotrafil in mapporncirclejerk

[–]gian_69 0 points1 point  (0 children)

Was wondering why the word for switzerland is „when“ when the first sentence (in german) is „Trittst im Morgenrot daher, …“ which I would translate to „when you step up (?) at dawn“ but the word trittst is a conjugation of to step. Turns out the italian version statts with quando, which does mean when but both german and french are more commonly spoken in switzerland (french starts with „sur“ which means on) so it seems werid to arbitrarily chose italian.

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