I have a task of visualizing few implicit functions. Firstly lets say I have the following function of $N$:

$$\epsilon = \sqrt{\frac{8}{N}\ln \left( \frac{4(2N)^{50}}{0.05} \right)}$$

Now this is very easy to visualize, just insert some values of $N$ from some range to get corresponding $\epsilon$. But what about in this case:

$$\epsilon = \sqrt{\frac{1}{2N}\left( 4\epsilon(1+\epsilon ) + \ln\left(\frac{4(N^2)^{50}}{0.05}\right) \right)}$$

How do I visualize this function? Do I simply need to firstly, pick some value for $N$ and then see which value of $\epsilon $ satisfies the equality? And then I repeat this process for some range of $N$ values?

Thank you for any help =)


1 Answer 1


Your equation is of the form \begin{equation} \epsilon = \sqrt{a\epsilon^2+b\epsilon+c} \end{equation} where $a,b$ and $c$ are functions of $N$. Also, we have that $\epsilon>0$.

Squaring, you get \begin{equation} \epsilon^2 = a\epsilon^2+b\epsilon+c \end{equation} so we have \begin{equation} \epsilon^2(a-1)+b\epsilon+c=0 \end{equation} Solving with respect to $\epsilon$ leads to \begin{equation} \epsilon=\frac{-b\pm\sqrt{b^2-4(a-1)c}}{2(a-1)} \end{equation} These are two functions of $N$. You can see the relation between $\epsilon$ and $N$ as the union of the two graphs. In your particular case, \begin{equation} a=2/N\\ b=2/N\\ c=\frac{1}{2N}\ln\left(\frac{4N}{0.05}^{\!100}\right) \end{equation} so your two functions are \begin{equation} \epsilon(N)=\frac{-\frac{2}{N}\pm\sqrt{\frac{4}{N^2}-4(\frac{2}{N}-1)\frac{1}{2N}\ln\left(\frac{4N}{0.05}^{\!100}\right)}}{2(\frac{2}{N}-1)} \end{equation} which simplifies to \begin{equation} \epsilon(N)=\frac{-1\pm\sqrt{1-\frac{N-2}{2}\ln\left(\frac{4N}{0.05}^{\!100}\right)}}{2-N} \end{equation} You can plot these two functions together in the same graph.

  • $\begingroup$ Aah..of course, nice :) thank you :) $\endgroup$
    – jjepsuomi
    Oct 21, 2014 at 14:00

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