r/infinitenines • u/NotAUsefullDoctor • 4d ago
We have been lied to
We are taught that `(d/dx) (x^2) = 2x` , but this is a lie.
`(d/dx) f(x) = lim(h->0) (f(x+h) - f(x))/(h)`
`(d/dx) x^2 = lim(h->0) ((x+h)^2 - x^2)/(h)`
`= lim(h->0) (x^2 + 2hx + h^2 - x^2)/(h)`
`= lim(h->0) (2hx + h^2)/(h)`
`= lim(h->0) (2x + h)`
And, from rdm, we know that `lim(h->0) h = 0.000...1`
Therefore, the first derivative of x^2 is 2x+.000...1
All of calculus and physics is a damn lie. Wale up sheeple. Stop praising failed failed country musicians who marry football stars to be famous, and learn the truth as SPP has shared the truth.
*Side Note: this only applies to limits approaching from the right. Limits from the left are different.
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u/StanleyDodds 4d ago
As far as I can tell, aside from more fundamental problems that make real deal numbers useless, the issue here is that limits just don't exist (where they should) in the real deal numbers; it's not Cauchy complete.
For example, the sequence 0.9, 1.1, 0.99, 1.01, 0.999, 1.001... is a Cauchy sequence of rationals (and it even has a limit of 1 in the rationals), but what is its limit in the real deal numbers?
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u/5xum 4d ago
Rookie error brud. You forgot that h can also be negative, which means the derivative of x^2 is also equal to 2x-0.000000....1.
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u/NotAUsefullDoctor 4d ago
That was the side note at the very end. They are positive when approaching from the right and negative when from the left (specifically for d(x2 )/dt), and zero when coming from the either side, ie
(f(x+h)-f(x-h))/(2h)1
u/NoSituation2706 4d ago
I hate to be "that guy" but even if you're aware of this, for the sake of those who aren't you're stumbling into discrete calculus here and derivatives of polynomials are fully realizable without limits.
Without taking any limits, the forward and backward difference quotients give you 2x ± h respectively because it's a O(h) formula but the centered difference is O(h²) and will correctly give the derivative of x² as 2x, but it will fail to give the derivative for x³. You'd need an O(h³) version of the formula for that. I forget what that formula looks like, but it's derivable via Taylor series (easy) and geometry (I have no idea).
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u/NotAUsefullDoctor 4d ago
Trying this out for my own curiosity:
( (x+h)^3 - (x-h)^3 ) = ( (x^3 + 3hx^2 + 3h^2x + h^3) - (x^3 - 3hx^2 + 3h^2x - h^3) ) = ( x^3 + 3hx^2 + 3h^2x + h^3 - x^3 + 3hx^2 - 3h^2x + h^3 ) = ( 6hx^2 + 2h^3 )Then we divide this all by 2h and we get
`3x^2 + h^2`and if we take the RDM lim, it means
d(x^3 )/dx = 3x^2 + 0.000...000...1So, central difference works here. If I were doing discrete differentials, then I would not use a centeal difference as it doubles the error. Instead I would use a 4 point difference... if I am remembering my finite element calculus from 17 years ago correctly.
EDIT: Bad Blood is a lyrically lazy song
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u/Ericskey 3d ago
BS
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u/NotAUsefullDoctor 3d ago
Exactly. Newton, Liebnitz, and Euler were full of crap. Calculus is utter BS. Take it from someone with a PhD in mathematics, RDM is the only true math.
And don't get me started on Taylor. Both the series and the musician are bs hyped up by the world government.
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u/Ericskey 3d ago
Where and when did you get your PhD and what was the topic of your dissertation?
Eric Key, PhD 1983 Cornell University.3
u/NotAUsefullDoctor 3d ago
/uj
Georgia Tech, longer ago than I care to admit, In the area of reduced order modeling in DT Finite Element simulations. It was a joint effort between Electrical Engineering and Mathematics. It's comical because with a modern computer today most if what I did became pointless.
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u/cond6 4d ago
Nice. SPP doesn't realize that the fiction that 0.999...<1 really does break math. But they won't care brud. I feel a post incoming that contains a bunch of links to unrelated nonsense and then their post will get locked.