Why does raising cos to 0.1 power expand the graph like this? I am trying to intuitively understand why the polar equation $r=(\cos(\theta))^{0.1}$ has the following shape compared to the circular $r=\cos(\theta)$. I see that the figure should have a greater area, as the values of $\cos$ between $1$ and $0$ will be increased when being raised to powers less than one. However, as to why the graph possesses the following curvature, I am at a loss to understand from the perspective of the qualities of a root function. I have tried so far thinking of the rate at which $x^{0.1}$ increases. I would appreciate an explanation as to why the shape is like this.
Thanks

 A: I think that we can perceive what is going on considering the more general case of
$$f_n(x)=\sqrt[n]{\cos (x)} \qquad \text{for}\qquad -\frac \pi 2 \leq x \leq \frac \pi 2$$
Built around $x=0$, its series expansion is
$$f_n(x)=1-\frac{1}{2 n}x^2+\frac{3-2 n }{24 n^2}x^4-\frac{16 n^2-30 n+15 }{720n^3 }x^6-$$ $$\frac{272 n^3-588 n^2+420 n-105 }{40320 n^4 }x^8+O(x^{10})$$ which reproduces quite well the function except around the bounds.
The second derivative being $\frac {-1}n$ at $x=0$ explains the tightening of the curve.
Similarly
$$A_n=\int_{-\frac \pi 2}^{\frac \pi 2} \sqrt[n]{\cos (x)}\, dx=\sqrt{\pi } \,\frac{\Gamma \left(\frac{n+1}{2 n}\right)}{\Gamma
   \left(\frac{2 n+1}{2 n}\right)}$$ increases from $2$ (for $n=1$) up to $\pi^-$ if $n\to \infty$ (is the circle coming ?).
Asymptotically
$$A_n=\pi \left(1-\frac{\log (2)}{n}+\frac{\pi ^2+3 \log ^2(4)}{24
   n^2}+O\left(\frac{1}{n^3}\right)\right)$$
A: When we have $0 \le x \le 1$ , we will have $0 \le x^{n} \le 1$ when $n$ is Positive Integer.
More-over , we will have $0 \le x^{1/n} \le 1$ when $n$ is Positive Integer.
When $n \rightarrow \infty$ , $x^{n} \rightarrow 0$
Eg $(0.2)^{2}=0.04$ , $(0.2)^{3}=0.008$ , $(0.2)^{4}=0.0016$ , $(0.2)^{5}=0.00032$ , $(0.2)^{6}=0.000064$ .... $(0.2)^{13}=0.0000000008192$ ....
It is going closer to $0$ with increasing Power.
When $n \rightarrow \infty$ , $x^{1/n} = \sqrt[n]{x} \rightarrow 1$ (Inverse Case)
Eg $(0.2)^{1/2}=\sqrt{(0.2)}=0.447$ , $(0.2)^{1/3}=\sqrt{\sqrt{(0.2)}}=0.5848$ , $(0.2)^{1/4}0.66874$ , $(0.2)^{1/5}=0.72$ , $(0.2)^{1/6}=0.76$ , .... $(0.2)^{1/64}=0.975$ ....
We can keep taking $\sqrt{....}$ to keep going closer to $1$.
In your Case , $0 \le |\cos(\theta)| \le 1$
Hence Power $(0.1)$ is Power $(1/10)$ which is taking $\sqrt[10]{....}$
The value will go closer to $1$.
Change the Power $(0.1)$ to Power $10$ & the value will go closer to $0$.
The EndPoint values will be $0$ & $1$ , everything else will move closer to either $0$ ( Power $[n]$ ) or $1$ ( Power $[1/n]$ )

Plot here shows 3 Curves ( generated by wolfram alpha & overlaid on each other & then altered a little ) : $A=\cos(\theta)$ (centre) , $B=\cos^{0.1}(\theta)$ (outer) , $C=\cos^{10}(\theta)$ (inner)
Each Curve has the Same EndPoint values ($0$ & $1$) , while the intermediate values are between these two.
Take 2 Example Points on $A$ , where the green lines intersect. With Power $10$ , the Corresponding Points (at that $\theta$ value) will go towards $0$ , hence get smaller. With Power $0.1$ , the Corresponding Points (at that $\theta$ value) will go towards $1$ , hence get larger.
Crux of the Intuition : This shift towards $1$ or $0$ will occur all over the Curve $A$ at all $\theta$ values ! Yet , the EndPoint values must be Same ! Hence the Curve $B$ going near $1$ will have to twist back to $0$ ! This is what you are seeing !
Curve $C$ has the Same in Inverse : It starts at $1$ , yet it has to quickly go towards $0$.
Change the Powers to $100$ & $0.01$ [ then to $1000$ & $0.001$ , then to $10000$ & $0.0001$ ] to see this more sharply.
