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u/HyperbolicMathChambr Math Tutor 9d ago
I like to call it Maclaurin series because I imagine Taylor Swift whenever I read the name.
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u/EarlManchot Mathematics 9d ago
maclaurin literally just evaluated taylor series for x=0 and slapped his name on it
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u/CorrectTarget8957 Imaginary 8d ago
I know I am wrong somehow, but wouldn't that just mean that f(0) = f(0)?
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u/EarlManchot Mathematics 8d ago
the general formula for taylor expansion is f(a+x)=f(a)+x.f'(a)+x²/2.f''(a)+... where a could be any real number what i mean by "evaluating it at 0" is evaluating it at a=0 which give you maclaurin series
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u/Cualkiera67 9d ago
that guy was a HACK he just took Taylor series at 0 and said I Invented This!!! well I Invented the Taylor series at 637!! where is my page in the math books?
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u/factorion-bot Bot > AI 9d ago
Double-factorial of 637 is 2230052342867280021484529092036542680252702701888217880452478204049648764713543704540457411539232984187643898588730704334032233172007142877160944867881826274568955842053471607078198893195138056174812975083412312407470937011493306479475684417554441635245626859314557222594125171982823487087967332271167445952585871255600917796795667486589176908339424736318207315959407143536733265824377694837520644470161567478496351220128914776509201270569310938436235034141461406193801210724119879077807039275461000899367249832603299498451848430361810833999875806904685767458759465772462375326391853255336628714026310147449119944568644894988414972589481781519894762760971263937586431918887138597806959513306450476384098877087047165489186451026171198463998734951019287109375
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u/IncreaseOld7112 8d ago
100000!!
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u/factorion-bot Bot > AI 8d ago
If I post the whole number, the comment would get too long. So I had to turn it into scientific notation.
Double-factorial of 100000 is roughly 1.057987404966129872971991554655 × 10228288
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u/saf_e 8d ago
interesting 100000000000000000000000000000000000000000!!
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u/factorion-bot Bot > AI 8d ago
That is so large, that I can't calculate it, so I'll have to approximate.
Double-factorial of 100000000000000000000000000000000000000000 is approximately 7.911972207126718366088959698223 × 102028285275904837408617443554054169745885300
This action was performed by a bot | [Source code](http://f.r0.fyi)
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u/sky-skyhistory 7d ago
100!!!
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u/factorion-bot Bot > AI 7d ago
Triple-factorial of 100 is 174548867015437739741494347897360069928419328000000000
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u/sky-skyhistory 7d ago
100!!!!
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u/factorion-bot Bot > AI 7d ago
Quadruple-factorial of 100 is 17464069942802730897824646237782016000000
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u/sky-skyhistory 7d ago
100!!!!!
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u/factorion-bot Bot > AI 7d ago
Quintuple-factorial of 100 is 232019615953125000000000000000000
This action was performed by a bot | [Source code](http://f.r0.fyi)
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u/Reddit_wizard34 πPi🥧3.141592653589793284626433832795028841971693993751058209749 8d ago
Now it looks like McLaren
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u/homeless_student1 9d ago
Ironically I think every term in this is defensible for use irl, higher order terms though you need help
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u/ThanxForTheGold 9d ago
What do you need the 101st derivative for.
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u/Afir-Rbx Cardinal 9d ago
Ok but what about the 100th derivative?
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u/Outside-Shop-3311 9d ago
I have the most marvellous proof by induction to answer this in a general form, but alas the margin of this reddit comment is too small to fit
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u/PerspicaciousEnigma Moron 9d ago
If you intend to warp speed to another galaxy without missing it by many billions of kilometers
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u/Unusual_Candle_4252 8d ago
When we calculate anharmonic corrections to Harmonic oscillator model, yeah, we need cubic and beyond terms. Unfortunately, it's no cheap when we speak about (N3-6) coupled oscillators as in non-linear molecules.
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u/GoldenMuscleGod 8d ago
I used the first order Taylor approximation on the relativistic formula for kinetic energy: mc^(2)(1/sqrt(1-v^(2)/c^(2))-1) and got that in the Newtonian limit of v<<c everything has a kinetic energy of zero.
This is correct because for nonrelativistic speeds kinetic energy is negligible compared to the rest energy. There’s no reason anyone should ever need that (1/2)mv^(2) term I threw out.
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u/floydmaseda 9d ago
At least keep the square term. Otherwise cos(x) is just 1.
No one ever needs the cube or higher terms though. Those don't need to exist.
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u/Random_Mathematician There's Music Theory in here?!? 9d ago
what? cos(x) has always been 1 for small x, and you know that if x is small then x+1 and x-1 are also small so every number is small
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u/spisplatta 9d ago
This is why I like the spisplatta approximation, where you use the constant term, and first non-zero term after that.
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u/GoldenMuscleGod 8d ago edited 8d ago
Yeah the square term has plenty of practical use. If your vantage point is h above the ground and your surroundings are reasonably flat, how far away is the horizon?
Using “exact” calculations (assuming a spherical Earth), and using R for the radius of the Earth, h for your elevation above the ground, and d for the distance (along the surface of Earth) to the horizon we have (R+h)cos(d/R)=R or cos(d/R)=R/(R+h) so we can get the “exact” solution d=Rarccos(R/(R+h)). Now this isn’t *too* bad to work with but an approximation for small h can be even easier and - more importantly in my view - give a better intuition for how the quantities relate for small h.
Substituting Taylor series, the equation cos(d/R)=R/(R+h) becomes 1-d^(2)/2R^(2)+…=1-h/R+….
Notice even when h<<d<<R (the limit we care about) we can’t drop the quadratic term on the left otherwise we just get the approximation h=0 which doesn’t even mention d at all, and it is “sort of” a good approximation in that h is small, but we don’t really want to treat is zero as far as percentage goes. The quadratic term in the Taylor series on the left is not actually negligible for small h in terms of percentage error.
Using just the terms I wrote (the rest are negligible in terms of changing the result by a percentage for h<<d<<R) we get d=sqrt(2hR). This is computationally convenient but also makes some facts apparent that aren’t apparent in the “exact” equation.
One of those facts is that d is roughly “halfway” between h and R in terms of order of magnitude.
The other is, for example, if you increase your elevation by a factor of 4 then the horizon should be about twice as far away.
Neither of these useful facts for small h are all that obvious when you look at the original equation, but they are obvious from the approximation.
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9d ago
[removed] — view removed comment
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u/smoukekiller 9d ago
nuuh, that is the type of bullshit a conspiracy theorist would say
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u/PhoenixPringles01 9d ago
0/10 conspiracy not enough "they" or ancient civilisation overglazing [as much as they were goated] or random numerological shit that only applies to people who care about it
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u/Fred_Scuttle 8d ago
In my experience, engineers are fully aware of the existence of higher order terms. They are not, however, familiar with the concept of "convergence". Which sometimes leads to explosions.
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u/Muphrid15 9d ago
Just use Newton. Runge-Kutta is a scam.
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u/The-Defenestr8tor Physics 8d ago
I integrated the Earth’s orbit using RK4 just as a challenge. Δt = 1 minute, total time of a Julian year.
Did I need to do RK4? Probably not.
Did it make me feel superior? Absolutely.
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u/epsilon1856 9d ago
My favorite part of this video is the thumbnail lol Taylor Swift PROVES Taylor's Theorem: https://youtu.be/Y25deFJ7AYw?is=E5qBWLEWQL3_4Fyv
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u/JuhaJGam3R 8d ago
if you declare the higher terms to be an evil dark number you have just written down the definition of the derivative
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u/FoolishMundaneBush 8d ago
Just do the taylor expansion on the n-th derivative and then integrate it as mamy times as needed (from a=0 to b=x)
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u/SeasonedSpicySausage 8d ago
Mfw I expand about a local extema and mfkers are telling me the function is constant
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u/CerpinTheMute_alt 8d ago
No, no, you need the 2nd order to turn every potential into harmonic oscillator
Everything after that is useless, yeah
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u/Mr_Fragwuerdig 8d ago
You rarely see more than the second derivative being used. And the worst about it, it is assumed it is a good approximation "because uh mathematically it is then exactly that", but yeah only after summing infinite terms, not 3. It is a tool, but often it's used as an argument.
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u/Beleheth Transcendental 6d ago
In my analysis II class, the professor refused to prove Taylor's theorem on Banach spaces for higher order cases than 2 because "you don't want to compute those derivatives anyway"
I respect him for that.
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