I'm trying to find a simpler form for $y=\sqrt{(-1 +\sqrt{1+8x^2} + 4x^2)/8} \cdot (3 + \sqrt{1+8x^2})/4|x|$
This is a "V"-shaped curve that:
- goes through $x=0, y=1$
- asymptotically approaches $y=\frac{|x|+\sqrt{2}}{2} + \frac{1}{8x+\sqrt{8}}$ as $|x|\to \infty$
but beyond that, I'm stuck.
Any suggestions?
Let $x$ be a positive quantity, so we have; $$f(x)=\left(\frac{3+\sqrt{8x^{2}+1}}{8x}\right)\sqrt{\frac{4x^{2}-1+\sqrt{8x^{2}+1}}{2}}$$
Another thing to consider is that for $T_{n-1}=n(n-1)/2$ or the $n$-th triangular number, we have $$\sqrt{8T_{n-1}+1}=2n-1$$
Now keeping this in mind, If we let $x^2=T_{n-1}=n(n-1)/2$,
$$\begin{align} f(x) & = \left(\frac{3+\sqrt{8T_{n-1}+1}}{8\sqrt{T_{n-1}}}\right)\sqrt{\frac{4T_{n-1}-1+\sqrt{8T_{n-1}+1}}{2}} \\ & = \left(\frac{3+\left(2n-1\right)}{8\sqrt{T_{n-1}}}\right)\sqrt{\frac{4T_{n-1}-1+\left(2n-1\right)}{2}} \\ & = \left(\frac{1+n}{4\sqrt{T_{n-1}}}\right)\sqrt{2T_{n-1}+\left(n-1\right)} \\ & = \left(\frac{1+n}{2\sqrt{2}\sqrt{n\left(n-1\right)}}\right)\sqrt{n\left(n-1\right)+\left(n-1\right)} \\ & = \left(\frac{1+n}{2\sqrt{2}\sqrt{n\left(n-1\right)}}\right)\sqrt{n\left(n-1\right)+\left(n-1\right)} \\ & = \left(\frac{1+n}{2\sqrt{2}\sqrt{n\left(n-1\right)}}\right)\sqrt{\left(n-1\right)\left(n+1\right)} \\ & = \frac{1}{2}\sqrt{\frac{\left(n+1\right)^{3}}{2n}}\tag{1} \end{align}$$
Now what we have to do is substitute the value of $n$ in $x$. We have defined $n$ as;
$$\frac{n(n-1)}{2}=x^2$$ $$\Rightarrow n^{2}-n-2x^{2}=0$$
Evaluate this as a quadratic equation and find $n$ using the quadratic formula:
$$n=\frac{1}{2}\left(1+\sqrt{8x^{2}+1}\right)$$
Substitute this in $(1)$ and we finally get; $$\small{\left(\frac{3+\sqrt{1+8x^{2}}}{8x}\right)\sqrt{\frac{4x^{2}-1+\sqrt{1+8x^{2}}}{2}}=\frac{1}{4\sqrt{2}}\cdot\sqrt{\frac{\left(3+\sqrt{8x^{2}+1}\right)^{3}}{1+\sqrt{8x^{2}+1}}}}$$
EDIT: I just realized, in your question there is a near miss, we can denest: $$\sqrt{4x^{2}\color{red}{+}1+\sqrt{8x^{2}+1}}=\frac{1+\sqrt{8x^{2}+1}}{\sqrt{2}}$$