Can you find a way to define this algebraically with a single

Cubic? We need more info

y(x) = -0.2x^3?

>>8627492
>>8627497
Oops sorry, I'm retarded.

Those are two quarter circles with a radius of 4.

>>8627500
Oh...

f(x) = √[r^2 - (x)^2]

I doubt there is, but if it's picewise defined, you shouldn't have any problems with many things.

>>8627506
no

>>8627512
There is a way. My teacher gave this as a challenge problem and so far no one in my class has solved it.

[eqn](\sqrt{16-x^2}-4)(\frac{|x|}{x})[/eqn]

stacy will be so impressed at your math skillz

[math]\displaystyle \frac { \left (r-\sqrt{ \left (r^2-x^2\right )}\right)\cdot x} {\sqrt {x^2} }[/math]

>>8627533
That's not algebraically defined.

Parametrize the curve as a 2-tuple valued function using trig functions then apply inverse functions on first coordinates until it is x.

>>8627533
>>8627541
Thanks

>>8627533
she fucking better be

>>8627533
>>8627541
Neither of these are defined at 0.

>>8627492
x = tan y

>>8627571
read the thread dipstick

>>8627562
This is true, so these probably aren't the right answer.

>>8627573
Oops I meant x = - tan y
Or y = -tan^1(x)

>>8627582
Also wrong. graphs of tangent does not give a circle

>>8627584
Wow I didn't know the picture in OP is a circle! Looks like I never knew what a circle looked like!

>>8627613
You're retarded dude I told you to read the thread. Third post says they are circular arcs.

>>8627613
You're awfully sarcastic for someone who thinks -atan(x) produces the graph in my pic.

4 + Integral of ( -|t|/(root(16-t^2) dt), from -4 to x.
Note the integrand is not defined at t = -4 or 4 but it does not need to be, the integral itself is still defined at those endpoints.

>>8627492
You can use a full reversal on the half parabola and twist it.

>>8627644
This is the only site I could find with a decent grapher for integrals.
http://www.integral-calculator.com/
If you set C=0 you get the graph you want. I have no idea why it does that when C should clearly be 4, it directly contradicts its graph for the integrand, but there you go.

It cannot be defined because it is discontinuous at the origin. A circle is everywhere convex. To invert it at a point is a discontinuity.

mathematically, its second derivative is undefined at zero

>>8627659
I just defined it here >>8627644

The function is continuous and differentiable everywhere but the endpoints -4 and 4.

>>8627670

Hmmmm..

Can you evaluate that integral?

>>8627685
It's antiderivative in elementary terms would be piecewise, the two arcs of the circles given by their usual functions.

>>8627691

Right, so for just the integral x/(SQRT (a^2 +- x^2)) would just be SQRT (a^2 +- x^2) but that pesky absolute value in the integral makes it undefined at zero.

Op wanted no piecewise functions. I contend that unless you can figure out a way to do it with polar coordinates, you can't define it at zero.

But I'm just a Dweeb....

>>8627703
Your reasoning makes no sense. What I gave is a valid function and it is not piecewise.

>>8627703

The absolute value of zero is fine for algebra, but how do you define it for calculus? Calculus requires a differential. No distance from zero is not a differential.
More to the point, to integrate the absolute value, you split the integral. But if you split the integral, you get a piecewise function.

No to be condescending, but one of the problems with computers in math is that everyone thinks like numerical methods, or like algorithms, and just plugs in the numbers, but integration isn't defined like that until you get to Lebesgue Integrals, and they have lots of rules about metrics and shit. The integral cannot be evaluated at a point.


I still think it has to do with either polar coordinates, or the projection of spherical coordinates, but I don't remember any of that shit, and I don't remember that shape with my spirograph, unless one of the pins came lose!

>>8627724
You do not need to refer to a function's antiderivative in order to integrate it. There are many functions that dont even have an anti derivative but are still integratable. For example the integral from 0 to 1 of the function defined by 1/n if x=m/n, and 0 if x is irrational. This has an integral of 0. An integral of an integrable function with an arbitrary bound is a perfectly reasonable function, there is no need to refer to the anti derivative or "split" the absolute value resultingly.

Taylor series.

[(x - |x|)/(2x)][4 - √(16 - x^2)] - [(x + |x|)/(2x)][4 - √(16 - x^2)]
What's a piecewise function?

>>8627659
Dont you mean differrentiable or something? You could add any continuous curve to that is 0 at 0 and it still would be continuous.

>>8627851
The problem is that your function is still not "defined" at 0. A picewise function is like a function defined by many parts. Like saying from -10 to 0 it f(x)=x and from 0 (open) to 10 its f(x)=x^2

y(x) = -x

>>8627492
x*-abs(x) ?

>>8628036
Wrong.
The tangent at -4 and 4 are both vertical.

>>8627533
If they are quarter circles then -4 and 4 would undefined in a function.

>>8627521
So it's a homework thread? Kys

Let
g(x,y) = (4*cos(t), f(t)*(4*sin(t) + 4))
for t in [-pi, 0].
We have f(t) = 1 over [-pi, -pi/2) and f(t) = -1 over (-pi/2, 0]. Clearly, f should be a shift of the sgn function:
f(t) = sgn(-pi/2 - t) = (-pi/2 - t)/|(-pi/2 - t)|
There is a discontinuity when t = -pi/2. However it is a removable discontinuity. I don't think we can do better, but I have no way to show this.

>>8627562

[math]\displaystyle \frac { \left (r-\sqrt{ \left (r^2-x^2\right )}\right)\cdot \left (x+\imath \right)} {\sqrt { \left (x+\imath\right)^2} }[/math]

>>8628109
So (x+i) / sqrt((x+i)^2) is a essentially a non-piecewise defined signum function? Cool!

>>8628109
>>8628143
It's also not a function: sqrt(x^2) = (+/-)x

>>8628143
>So (x+i) / sqrt((x+i)^2) is a essentially a non-piecewise defined signum function?

No, it's not. It isn't equal to 0 when [math]x=0[/math]. If you want to define the sign function non-piecewise you do something like this:
[math]sgn(x)=\lim\limits_{n\to\infty} \frac{2}{\pi}\arctan{nx}[/math]

>>8627659
fucking retard kill yourself

The same applies for most others in this thread desu

Take f as defined in >>8627506

And try g(x) = - (x^2 / x) * f(x)

>>8628496
With shifting the center of the circle to x=0 Y=2

File: Screenshot_20170125-162301.jpg (277KB, 1374x898px) Image search: [Google]

277KB, 1374x898px

I am posting this as evidence of my theory of unification.

Signed by my 2 cats: River & Sky

File: Screenshot_20170124-195744.png (2MB, 1440x2560px) Image search: [Google]

2MB, 1440x2560px

>>8628506

Evidence no.2

Signed W88 and L15

>>8628325
Ah, you're right. It's 1 when x = 0, so it's not a sign function. It's still nice that's it's defined at x=0 instead of being a singularity. Does it have a name?