Can you find a way to define this algebraically with a single non-piecewise function?

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

Thanks

she fucking better be

Neither of these are defined at 0.

x = tan y

read the thread dipstick

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

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

Also wrong. graphs of tangent does not give a circle

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

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

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.

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

This is the only site I could find with a decent grapher for integrals.
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

I just defined it here The function is continuous and differentiable everywhere but the endpoints -4 and 4.

Hmmmm..

Can you evaluate that integral?

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

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....

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

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!

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?

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

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

x*-abs(x) ?

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

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

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.

[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]

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

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

>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]

fucking retard kill yourself

The same applies for most others in this thread desu

Take f as defined in And try g(x) = - (x^2 / x) * f(x)

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

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

Signed by my 2 cats: River & Sky

Evidence no.2

Signed W88 and L15

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?