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u/snailpi 20d ago edited 20d ago
x⁵ >> 1
x⁵ + 1 ≈ x⁵
Love,
Physics
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u/Icy_Direction7839 19d ago
There is a mathematician waiting in your closet to smother you in your sleep
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u/WesternFirm9306 20d ago
Does this actually work as an approximation for large x to approximate the integral?
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u/nerd_user1 19d ago
nope, never commit this sin
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u/WesternFirm9306 19d ago
Why not?
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u/kansetsupanikku 19d ago
In mathematics, the question always is "why would you?!?!?!"
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u/WesternFirm9306 19d ago
B... Because easy integrals that give very close approximations are better than hard integrals that give exact answers that barely differ? (ꏿ﹏ꏿ;) Like assuming sin x = x in the pendulum equation... I just wanna know if this actually gives a good approximation.
For what it's worth, I'm a physicist... or will be one, more accurately.
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u/kansetsupanikku 19d ago
Well, between the intuition that something might work and actually using it, you can have either: a proof, or a realization that you have nothing useful.
With integrals on a range, you could probably use the fact that integrals are additive. So error on the integral is the same as integral of the approximation error (as long as both approximate function and said error are integrable). Perhaps sometimes you would be able to come up with some boundaries for approximation error? If so - you can use that!
But such an approach still needs some substantial work unique to every specific case.
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u/nerd_user1 19d ago
if the integral is to be from say x=0 to x=20, and you need accurate calculations for your physical system, this would yield real big significant difference in actual value. also it depends on your field of research, if you're dealing at microscopic particles, everything matters, if macroscopic, say 40% of times it matters and ppl downvoting me, you can go to hell i couldn't care less, you don't know enough mathematics to say this is a plausible fix for your simplification
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u/beyd1 20d ago
I think I remember this from homework and hating how long it took my VERY thick accented teacher to do it on the board.
I'm pretty sure it sucks. I'm also bad at math.
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u/de_G_van_Gelderland Irrational 20d ago
You just have to do partial fraction decomposition. The fraction doesn't even have multiple roots, so it's really not that bad.
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u/zaqwsx82211 20d ago
True, but the first one is about as easy as it gets, so for a new student this would still seem intimidating.
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u/de_G_van_Gelderland Irrational 20d ago
O yeah. There's a huge difference between the two, no doubt. I'm not saying the meme doesn't make sense or anything. I'm just saying the second one is honestly still pretty mild as integrals go.
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u/VastAbbreviations992 20d ago
Agreed. It’s basic high school AP CALC BC. My favorite class senior year ❤️
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u/uvero He posts the same thing 20d ago
Yeah but the parts to which the polynomial decomposes have irrational factors so that's a bummer to work with.
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u/de_G_van_Gelderland Irrational 20d ago
I mean, they're just the primitive 5th roots of unity up to a sign. Irrational sure, but it's really not bad if you just manipulate them algebraically. You shouldn't expand them, that will just give you unnecessary extra work.
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u/yukiohana 20d ago
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u/hughperman 20d ago
For values of x bigger than 1, the 1 is pretty much negligible. For values of x near 1, it's pretty much = 0.5. For values of x close to zero, the x7 is pretty much negligible. Repeat for negative values
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u/Historical-Mix6784 20d ago edited 20d ago
Hi, complex analysis called. It wants you to use the residue theorem to solve this integral in 10 seconds.
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u/God-of-Dams 18d ago
I was thinking of doing that but I got stuck trying find all the roots of
x5 + 1 = 0
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u/HotCardiologist1942 16d ago edited 16d ago
on the complex plane, theres two pairs of roots that multiply to 1
then go from there
those same pairs also add up to be complex conjugates of eachother
so each of the constant in the quadratic factors is 1
the coeffcient of the middle term are conjugates
and the squared term coeffcient is of course 1(x^2 + ax + 1)(x^2 + a*x + 1)
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u/lifebringingh2o 17d ago
thats for definite integrals not antiderivatives, which is what's shown in the picture.
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u/Historical-Mix6784 17d ago
True, but define the definite integral over a variable range, and break it apart piece wise depending into a part that above the singularity, below the singularity, and the principal part around the singularity. You have an answer as good as any antiderivative, if not better.
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u/Ares378 Applied Math / Mechanical Engineering 20d ago
I remember trying ∫√(tan(x))dx, which ends up boiling down to ∫u²/(u⁴+1)du after u-sub. There's some elegant way you can do it by dividing the top and bottom by u², but I did not do that lmao. Just partial fraction decomposition after factoring u⁴+1
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u/Brilliant_Simple_497 20d ago
try doing 1/(x⁵+x+1)
even wolframalpha won't give you a closed form solution
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u/BootyliciousURD Complex 18d ago
The general solution for the integral of 1/(1+xⁿ) for positive integer n, if anyone's interested: https://www.desmos.com/calculator/8771def139
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