Why is the derivative not defined like this? It's clearly more accurate this way as h goes to zero faster...

found the brainlet.

The only wrong thing with OP's definition is that it is only calculating the right derivative.

indeed, "a little knowledge is a dangerous thing"

I appreciate this idea, however I would like to propose changing it to
[eqn]f'(x)=\lim_{\epsilon\to 0}\frac{f(x+\epsilon^2)-f(x)}{\epsilon^2}[/eqn]
since [math]\epsilon[/math] represents very small numbers it clearly goes to zero faster

Since [math]h^2[/math] goes to zero faster than [math]h[/math], is it true that
[math]f''(x) =\lim_{h \to 0} \frac{ f(x+h^2+h)-f(x+h) - f(x+h^2) + f(x)} {h^3} [/math]?

If you want to get h to zero the fastest, you'll need to minimize time dilation caused by massive gravitation forces, so you'll first want to move out of your mother's basement.

>this fucking thread

yes

Pretty sure this is the exact definition in my book.

woah shit this is crazy

because its easy to find a counter example where:
[eqn]f^\prime(x) = \lim\limits_{h\text{ } \rightarrow \text{ } 0} \frac{f(x+h^2)-f(x)}{h^2}= \lim\limits_{h\text{ }\rightarrow \text{ } +0}\frac{f(x+h)-f(x)}{h} \neq \lim\limits_{h \text{ }\rightarrow \text{ }-0}\frac{f(x+h)-f(x)}{h}[/eqn]

holds.

fucking hell

see you on r/Veeky Forums

kek

Why not
[eqn] f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x-h)}{2h} [/eqn]
it converges faster than the standard definition.

Why even use any limits in the first place when there exists a completely algebraic method for finding linear tangent approximations to a polynomial?

Consider the function f(x) = a + bx + c(x^2) +d(x^3)

Translate function by r; f(x+r)

Use the binomial theorem to expand all the terms of (x+r)^n

Substitute x for x - r

collect terms of (x-r)^n

You should get f(x) = a + br + cr^2 + d(r^3) + (b + 2cr + 3d(r^2))(x-r) + (2c + 6dr)(x-r)^2 + d(x-r)^3

At this point you select an r to be the x coordinate of the point of tangency, and truncate the parametized version of f(x) to T(sub1) = a + br + cr^2 + d(r^3) + (b + 2cr + 3d(r^2))(x-r)

Clearly you can see that this equation has the slope that the "classical derivative" is getting at, but also this equation is actually for the line itself that is tangent to the function.

If you werent convinced that youve been getting cucked by the analysts this whole time, T(sub2) = a + br + cr^2 + d(r^3) + (b + 2cr + 3d(r^2))(x-r) + (2c + 6dr)(x-r)^2 is the equation for the tangent parabola to the function at r

T(sub3) = a + br + cr^2 + d(r^3) + (b + 2cr + 3d(r^2))(x-r) + (2c + 6dr)(x-r)^2 + d(x-r)^3 is the equation for the tangent cubic at r.

This method generalizes to polynomials of any natural number degree, is completely algebriac, and is relatively straightforward to justify if you understand the binomial theorem.

Oh yeah, its also valid over a general field, not just the "real" numbers

...

The history of calculus is messed up because we keep trying to push trancendental explanations for patterns we have noticed before we understood them. Turns out there was an algebraic, more powerful, and straightforward method all along

Polynomials are only a small part of all functions.

Yea you can stop there.

If you want to see a lecture on the algebriac method of a derivative, heres a handy link:

youtube.com/watch?v=oW4jM0smS_E&index=4&list=PLIljB45xT85DWUiFYYGqJVtfnkUFWkKtP

If you want to see an algebriac proof of Cavaleiri's formula for integration:

youtube.com/watch?v=Js2mwsHc4p4&index=11&list=PLIljB45xT85Bfc-S4WHvTIM7E-ir3nAOf

If you want to learn how to do trig without a prior theory of transcendental functions and "real" numbers:

youtube.com/playlist?list=PL3C58498718451C47

Aint it great how accessible this guys material is?

If youre not at least evaluating his arguments for yourself, then basically youre a brainlet with your head in the sand.

Is that all you got cuck? Polynomials are the foundations of circular functions, and calculations of e^x.

Combine this method with rational trig and youve made a lot of the fundamental difficulties with mathematics obsolete for 99% of people who get exposed to them.

Your notation is clearly fucked but if I have to guess what you mean, it can be solved by taking just odd powers, like h^3 which is still better.

>Combine this method with rational trig and youve made a lot of the fundamental difficulties with mathematics obsolete for 99% of people who get exposed to them.
Wo cares about brainlets? Turns out you were the cuck all along :^)

>not seeing how powerful a method for finding a tangent polynomial of an arbitrary natural number degree to another polynomial is.

one of the biggest things here is that rational trig and algebraic calculus are VALID OVER GENERAL FIELDS, something the classical renditions of the subjects cannot claim

Because this method can be introduced straight out of the binomial theorem, it means you can have an entirely internally justified theory when you present it to people for the first time. Analysts cant claim that either.

Not to mention the fact that you never have to appeal directly to an idea of limits or infinitesimals, which is a recurring problem in the history of calculus.

[math]\text{Given a function } f : \mathbb{R} \longrightarrow \mathbb{R} \text{, let } \big(f^\prime(x)_n\big) \text{ be a sequence.} [/math]
[math]\text{The k-th element in } \big(f^\prime(x)_n\big)\text{ is given by } y \text{, such that } y \text{ satisfies the following}[/math]
[eqn]
\forall \epsilon > 0 \quad \exists \delta >0, \text{ such that } \bigg|\frac{f(x+h^k)-f(x)}{h^k}-y \bigg|

ebin

don't care if Real numbers exist, they're useful

If such [math]y[/math] exist and the series converges.

naturally

OP, if you write h^5 or h^3 or h or whatever doesn't matter, because the limit expression is the same in either case.

You're of course implying that expression captures a numerical approximation scheme. And this is all well-studied

en.wikipedia.org/wiki/Numerical_differentiation

What's more, you can do algebriac differentialtion for well behaved functions like polynomial. That's called "Fermat observation" and Wildburger sure has good videos on it (mixed with finitist propaganda, but who cares)

I watched 8 of these... it doesn't really appeal to me. All he's doing is squaring things. You have a triangle with side lengths a,b and c, and opposite angles u,v,w. He defines A=a2, B=b2, C=c2, U = sin2(u), V = sin2(v), W=sin2(w). Then squaring the law of sines gives A/U=B/V=C/W. Squaring the law of cosines gives
(A+B-C)2 = 4 A B (1-W)

What's the big deal?

you seem to be in introductory calculus. there is a further exploration of this topic in Real Analysis:

The limit is defined in terms of arbitrarily small quantities strictly greater than zero. There is no growth rate associated with a limit. hence we do not speak of it's "convergence rate" or anything similar.

with this in mind, we don't need any sorry of power function on h in the definition of the derivative.

This just shows that OPs definition is more general, and will lead to a deeper analytical theory.

Not using (f(x+h) - f(x-h))/2h

wut a dumass

The big deal is that it becomes computable by hand, that all 5 rational trig formulas are and that you never have to consider an angle in the first place.

If you cut out circular functions, make it algebraic, quadratic, and valid over the a general field, then its more accessible to learn, more capable of encoding useful information, and more applicable to other areas of mathematics.

Specifically consider this, the measure of the separation between two lines in rational trig, spread, can be thought of as (sine of theta)^squared

however you do not have to consider the angle or the sine calculation at all, because for two vectors and the spread s between them is given:

s = ((ad - bc)^2/(a^2 + b^2)(c^2 + d^2)

Notice that it is hand computable, and that geometrically you are basically normalizing the determinant between two vectors by the lengths of those vectors (except this is the square version).

>s = ((ad - bc)^2/(a^2 + b^2)(c^2 + d^2)

dude, isn't this pretty much just
cos(t) |a| |b| =
squared?

The advantage is you never have to consider an angle

>year [math]\frac{24205}{12} + 1 + 2 + 3 + 4 + \dots[/math]
> not using [math]f'(x) = \underset{h \rightarrow \infty}{lim} \left(f(x + \frac{1}{h \uparrow \uparrow \overset{h}{\cdots} \uparrow h}) - f(x)\right) h \uparrow \uparrow \overset{h}{\cdots} \uparrow h[/math]

it's not supposed to be appealing, it's supposed to be rigorous, as opposed to using a number system that's inconsistent at best.

literally 0 people will read your post if you don't tex your fucking shit

Kek. Just don't reach Graham's number.