The red function is arcsine of sine of x.
Some memers would say that it is not differentiable at some points but that is BS because green is its derivative and it looks accurate. That is literally what its derivative would look like.
Other memers would then say that clearly green has no derivative but then red is the derivative of green.
Stop this non differentiable meme already, it is bullshit.
In my first sentence I meant to say the pink function.
I see, so it's funny to pretend to be a retard and say that everything and everyone is a meme.
Don't forget to call people "cuck" too
But I am not a retard. Look at pic related.
f(x) = arcsine(sine(x)) is supposedly non differentiable. Well, I just fucking computed its derivative and then graphed it. It looks 10/10 at every point. Exactly what it should look like.
Notice. The graph of the pink function switches between two inclinations immediately and in the derivative this is reflected by the graph having a discontinuity exactly where that happens, switching to the value of the other continuity, creating that weird rectangle graph.
Now this shit is not even continuous so memers would tell you that you can't even calculate the limits (and they are right) but you can still find its geometric derivative and I did not use any fancy tricks. Just typical derivation techniques.
Look at the green graph, it is always like the graph of f(x) = c where c is some constant. In other words, it has inclination 0 and look at its derivative a straight line.
AND BEFORE YOU CLAIM I JUST PULLED THAT STRAIGHT LINE OUT OF MY ASS, SEE PIC RELATED.
Those are the 3 functions being graphed. Each one followed by its derivative.
How come I can differentiate non differentiable functions and EVEN discontinuous functions?
I think calculus has many holes. Obviously discontinuity does not mean you can't differentiate and obviously the definition of differentiable is fucking wrong.
Do you see the vertical lines on graph? That means that the derivative is undefined at that point, retard
>That means that the derivative is undefined at that point
But that is a bit dumb. That function is behaving like a derivative.
The reason the derivative moves like that is because the original function changes inclination rapidly so it makes sense that this would cause the value of the derivative to change immediately, causing a discontinuity.
If our derivative works exactly like a derivative why would you say that the original function is not differentiable?
At the point where the original function changes inclination the derivative simply could be either the positive or the negative inclination and while that means undefined, it should not mean non-differentiable.
you are right op
t. math phd
The definition of a derivative is just a form of a limit.
Limits only exist if you can approach the limit from any direction and yield the same answer.
Therefore, [math]ArcSin[Sin[x]][/math] is not differential at certain points.
The best you can say is that certain points can only yield one sided derivatives.
>The best you can say is that certain points can only yield one sided derivatives.
No, this just means that we need to allow discontinuities in our derivatives that reflect the sudden change.
It is differentiable everywhere except those corners.
At those corners what is the derivative? If you can give me a reasonable answer then I will admit you are right. Otherwise the function is not differentiable on the corners because it has no derivative on the corners.
Yes it does. Read up on limits. A well behaved limit is one that only yields one answer. That is not the case for the derivative you mention. The best you can get are one sided limits for certain points.
Therefore, the function is not differentiable everywhere.
If it has a vertical line it isn't a function
3/10 I replied.
>If you can give me a reasonable answer then I will admit you are right
The answer is 0. Look at pic related. We should assume (add a fundamental axiom that then allows us to prove this) that the derivative at the points where the functions suddenly chage is 0, picturing a transition where the derivative goes from one inclination to a straight horizontal line to the other inclination.
Then more complex functions like the Weierstrass function would have derivative equal to 0 everywhere and that would actually be satisfactory because you could just keep zooming and see how those steep points have derivative 0.
How has math advanced so far without my advancement in discontinuous analysis? This shit is top tier stuff. Now you can differentiate everything.
Are you legitimately retarded OP? It's differentiable everywhere except the corners. Plug the x-coordinate of a corner into the derivative and see. Jfc I hope you're in high school.
That is why I'm saying we should redefine derivatives.
For limits I am fine, I see no problem because it makes sense but when derivatives are more of a geometrical thing the limit definition is just not good enough.
>a sharp corner has a gradient
wot a wanker
>Trying to perform some kind of average of two points on a discontinuous function to find the value at the discontinuous point
kys OP
It is not an average, look at pic related.
There the equation of the tangent would look like y=c
completely horizontal so all spikes would now have derivative 0 in my new system of discontinuous analysis.
I think this is revolutionary because even though you cannot currently differentiate my functions at certain points, you can ALWAYS integrate them.
My discontinuous analysis will finally unite the theory of integrals and derivatives, making everything integrable and differentiable.
You are saying that the gradient of the pointy bit is zero? Why?
And the way to represent that zero best on the green function is with a vertical line, rather than a point at zero?
Where are you studying?
With that answer, the derivative is no longer analogous to the rate of change, or slope.
I mean sure, you can define your own version of a derivative that assigns the average of the two slopes on every corner, but what is the point? This doesn't serve any purpose.
There are a whole lot of possible tangents, so long as gradient is between 1 and -1
That's why your vertical green line takes values from 1 to -1 at that point.
You already could integrate both functions you posted. This is nothing revolutionary.
This is all addressed in measure theory.
Plug the point into the derivative. You're being fooled by the way the graphing software and how it handles undefined points. And you should feel ashamed for being this retarded and not checking by hand
>You are saying that the gradient of the pointy bit is zero? Why?
It is intuitive to me. Isn't it to you? What else would it be?
In a continuous function if the inclination changes from positive to negative you know at some point the derivative has value 0.
Why not change our axioms so that we can prove that this also happens to discontinuous functions?
>And the way to represent that zero best on the green function is with a vertical line, rather than a point at zero?
That is just the grapher trying to weirdly connect the points. This would actually look like many segments alternating from positive to negative.
>Where are you studying?
Currently taking Calc 2 but that does not matter, I will change the game with my discontinuous analysis.
But it is, it is intuitive. Also, I am not average. The derivative of all corners would be 0, this way the derivative of the Weierstrass function would always be 0.
This would make weirdly everywhere-pointy functions like the Weierstrass one to be analogous to straight horizontal lines and I think that will probably unveil new theorems in analysis.
I know but you cannot differentiate them. Why not both?
Now you can do both, with discontinuous analysis. Patent pending.
>completely horizontal so all spikes would now have derivative 0 in my new system of discontinuous analysis.
what about at pic related?
>intuitive
no, my intuition says pointy things don't have gradients
>discontinuous functions
it isn't discontinuous
> I will change the game with my discontinuous analysis.
no, you won't. this isn't a new problem
Once again, what purpose does this serve? Having the derivative at corners be some arbitrary number like 0 instead of undefined does nothing for mathematicians. Name one application this might have in any field of math.
Still 0 because from the beginning derivatives just approximate the tangent with a secant line. Most functions don't even have tangent lines because a straight line will touch some other part of the graph somewhere.
Here the same happens. I am not worried by the derivative y=0 touching the function elsewhere because it is just an approximation.
>it isn't discontinuous
But its derivative is. That is what I mean. Instead of saying that derivatives don't exist at certain points, simply allow derivatives to be discontinuous.
0 is not arbitrary. I just explained how this is a theorem for differentiable functions so now I want to extend it to non-traditionally-differentiable functions.
I don't know the purpose but we would be able to do more calculus and that will probably do something somewhere.
>but when derivatives are more of a geometrical thing the limit definition is just not good enough
I don't know what to tell you. The derivative has always been defined as:
[eqn]\lim_{\Delta x\rightarrow 0} \frac{f(x + \Delta x) - f(x))}{\Delta x}[/eqn]
>The derivative has always been defined as:
And I'm saying that definition is shit.
That is simply the way to compute it in my theory of discontinuous analysis.
a line of zero gradient would cross the curve at the point it was a tangent.
a secant line would have a gradient bounded above zero by the gradient of the less steep line. why does it suddenly jump to zero?
i would not take calc 3 if you can't see this
But any secant line is at the very least the second lines gradient. Which is positive.
>0 is not arbitrary
Where is your motivation for choosing 0???
What about this function:
f (x) = x for x < 2
f (x) = 2x for x >=2
What's the derivative at x=2?
>allow derivatives to be discontinuous.
we do allow that you massive twat
Tried something similar with my picture at The fool dug his heels in and said still zero
What OP is saying is "looks like all continous functions are differentiable everywhere except on a 0-measure set"
But he is wrong about that.
Wish he was just saying that, he's saying much worse.
Oh my, well at the very least this should confirm that his choice of 0 was arbitrary.
If you draw a line of derivative 0 it slices right through the graph, it isn't tangent at all.
differentiate it as a piece wise function? Am I missing something?
>I want the slope at any discontinuity to be zero because I dunno. the calculator gave me an error
No, I am not talking about your standard calculus.
I am saying calculus is wrong and should be changed with my calculus.
In my calculus that function has derivative 0 everywhere given that it is an everywhere spike.
Then you aren't talking about derivatives. You are just messing with definitions.
I don't see what the point is anyway. You can still plot the derivative of arcsin[sin[x]] where the limit does exist. Where it doesn't exist, you just get vertical lines.
>In my calculus that function has derivative 0 everywhere given that it is an everywhere spike.
Ok you got me triggered.
Looks like you're right.
>Ok you got me triggered.
But this is intuitive.
What else has derivative 0 everywhere? A straight line.
Now, in my calculus you need to think differently of traditionally non differentiable functions.
Lets take a look again at arcsine of sine x. What it look like? It looks like if a straight line was forced to 'break'.
This does not happen to curves because curves don't even look like straight lines but you have to say that the Weierstrass function looks really straight, just broken in many points.
In pointy functions like arcsine of sine you have spreaded out breaks and we should define these breaks to have derivative 0.
Now what happens with everywhere pointy functions? Well, they are straight lines who got really fucking broken. Absolute destroyed. Does the Weierstrass function not look like that? A straight that was broken a lot?
If you look at it you can see how bits of it look straight but also broken. It is a straight broken function.
Maybe that will be a class of functions in my analysis, straight broken functions.
In my analysis I am simply pointing out the similarities between broken functions and straight functions, that they both look and are straight.
You could define a straight line in terms of the Weierstrass function. A straight line would simply be a Weierstrass function who had every broke point fixed and unbroken.
Please stop fucking talking.
You'll always be a mediocre student until you understand what rigour is.
Fucking hell, i hope this is bait OP.
>What else has derivative 0 everywhere? A straight line.
like the line y = x
lol u
>You'll always be a mediocre student until you understand what rigour is.
>mfw when 3.9 GPA
>mfw straight As all around
>mfw top of my class in Calc 2, the current calc course I am taking
It is funny that you all think that I suck at traditional calculus when in reality the average person would not even try to question its foundation.
When my discontinuous calculus solves the riemann hypothesis and earns me the fields medal I want every poster in this thread to attend the ceremony and suck my dick as the medal is handed to me.
You just fucking wait. In case I end up changing the name of my theory just know that my initials are L and C so that you know when you have to come attend to suck the dick.
Dunning-Kruger
Ignorance plus arrogance.
I see my mistake. Before I had been mentioning only straight horizontal lines but this time I fucked up.
My inspiration for saying that pointy bits have derivative 0 is because of the theorem that says that if a differentiable function changes its inclination from positive to negative then at some point it has to be 0.
For non differentiable function this sudden change happens at the pointy bits so I say the derivative is 0.
In everywhere pointy functions this derivative is 0 everywhere, just like a horizontal function.
And now geometrically I am thinking of that as breaking a horizontal straight line so that it keeps its straightness but also moves up and down.
I've done analysis to masters level at a UK top 5 university. Your redefining differentiation is of *at best* zero utility.
It's great that you play around with things, but you have to expect to be usually wrong.
>the definition of the derivative is shit
you're just amazingly retarded. generalizations of the derivative exist (subderivative, etc) but you're being idiotic here, the only reason you don't understand it is because you're stubborn and dumb
Look, I understand it very well.
I am just saying that in my theory I will have a more fundamental definition that will allow for my claims of the derivative being 0 at pointy bits to be proven.
Then the limit definition of a derivative will simply be the tool to compute derivatives, but I will probably have to mess with that a bit so that it all makes sense.
But we have given you two examples where a pointy bit's "inclination" changes from positive to less positive, but still not negative. So why do you counter-intuitively say these have derivative zero?
undefined doesn't mean 0, especially not because it would make some theorem work when it should fail
no, it won't. you need to cut it. what you're trying to do isn't worthwhile. we know how these things work, we know it very well.
>My inspiration for saying that pointy bits have derivative 0 is because of the theorem that says that if a differentiable function changes its inclination from positive to negative then at some point it has to be 0.
>For non differentiable function this sudden change happens at the pointy bits so I say the derivative is 0.
For fuck's sake retard, 0 isn't a special value. What would be the value of the derivative of pic related at x=0 ?
You know this could give strictly monotonic functions points with zero derivative.
That would be very odd.
"Guys, this function does nothing but increase, but still has points with zero derivative."
The theorem I am quoting works for traditionally differentiable functions.
I will change the foundation so that it works everywhere.
Okay, maybe in your case I am not seeing something. Maybe the derivative should have inclination at those points.
Okay I think I have a new definition.
Let A be the left limit of the derivative and B be the right limit of the derivative at pointy bits.
The derivative of the pointy absolute(A) - absolute(B)
This works because for example in the absolute value function itself at x=0 the derivative would be
1 - 1 = 0
and now this will work for your cases.
Now my theory is stronger. I am one step closer to that fields medal. Easy money.
>When my discontinuous calculus solves the riemann hypothesis
Stop shitposting and leave.
already been asked at least twice, the thick wanker still says zero
No, my theory has evolved.
>I will change the foundation so that it works everywhere
and then you will create a contradictory piece of shit, that's what everyone's trying to tell you
>The derivative of the pointy absolute(A) - absolute(B)
fuck
>Let A be the left limit of the derivative and B be the right limit of the derivative at pointy bits.
>The derivative of the pointy absolute(A) - absolute(B)
That works for functions that are derivable by part.
Now what is the definition of your derivative for the Weierstrass function, where you can't define the left-derivative and right-derivative?
>and then you will create a contradictory piece of shit, that's what everyone's trying to tell you
It will not be contradictory because when I get a PhD I wil probably be very good at foundations so I will create a good system.
>fuck
The derivative of the pointy with will be absolute value of A minus absolute value of B.
So now in functions with complete change like the absolute value function it self at x=0 the derivaitve will be 0.
When the function only changes slightly but immediately like from inclination 7 to inclination 4 (like in your case) the inclination will be 7 - 4 = 3
Geometrically that looks very good. Think about it.
Use the mean of A and B if you want this nonsense to have a modicum of utility. I think this was suggested to you before.
>Now what is the definition of your derivative for the Weierstrass function, where you can't define the left-derivative and right-derivative?
Here I will need my theory to grow again.
I will give a rigorous basis to say that the derivative THERE is always 0.
Everywhere pointy bits must have derivative always 0 because as I said I am imagining many strong breaks.
I stand by that claim. My calculus will change the world.
>Use the mean of A and B if you want this nonsense to have a modicum of utility
I don't really like that. I don't feel it is intuitive.
Like Newton I am doing what feels intuitive and after my intuition takes me to new heights I will start making a formal system for it.
>function with a derivative that's 0 everywhere isn't constant
sounds like your analysis is pretty crappy
After a bombardment of us repeating a flaw you finally see it and make an adjustment, a really terrible one that would still give counter-intuitive results
But in my analysis, everywhere pointy functions are equivalent to straight horizontal lines.
I think there are many revoloutionary theorems hiding in this idea. To me it just makes so much sense.
I don't think so and if it does I just have to try again. I am already closer to a Fields medal than anyone in this thread and I am just a freshman in mathematics.
This is too easy man.
if A is the gradient to the left, and B the gradient to the right
then your new version pic related will have a positive gradient at the pointy bit, and negative elsewhere
your intuition is shit
>I don't think so
better try again then
Maybe my idea of adding absolute value was dumb.
It will be something looking like A - B or some other relation with the left and right limits. Maybe the mean as someone else is saying.
I am busy right now so unfortunately I cannot go too deep into my own theory to fix this issue.
I still think this has value.
Is this the result of that "self worth first" American education I keep hearing about?
Don't derivatives of square wave functions typically yield dirac delta functions?
Like the derivative should be 0 everywhere but the rising and falling edge where it would be infinite, an impulse.
It doesn't seem logical that a derivative of the square wave sin(2πf)/|sin(2πf)| would be arcsin(2πf)
>I still think this has value.
Even though people far more qualified are certain it doesn't?
Are you the special one who will prove the mathematical establishment wrong?
The track record at the moment is you 0, everyone else 10
Exactly, this whole thing, if fully corrected and done well, is still reinventing the wheel. Trouble is OP was oblivious to the gaping pitfalls.
Dirac functions, Fourier analysis, etc
seriously what the fuck is going on in this thread? arguing about differentiation?
I am proposing a new foundation for a different kind of analysis. Discontinuous analysis where derivatives will allowed to be discontinuous functions so that functions that have pointy bits will have derivatives.
I am also thinking of continuous discontinuous analysis where pointy bits will be thought as expanding the real line geometrically so that there exist infinitesimals filling the point of the pointy bit so that if you zoom in infinitely into that point you can see that actually it is continuous.
This will probably solve the Riemann Hypothesis.
Your function is not wrong it's energy's curvations.
Stop shitposting.
so, hows your first month in calc I?
shut up
> functions that have pointy bits will have derivatives
this already exists, friend; it's called the weak derivative
>I have no idea what measure theory is: the post
>top of my class in Calc 2, the current calc course I am taking
Did everyone miss this line? Why are these still replies to this guy in this thread?
Sage goes in all fields btw