Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like so find the function $n(z)$ that solves the following equation:

$$ n(z) = \frac{.2 + .24 z \int_1^{\infty} \frac{n(z)}{z^2}dz}{.5 + .24 z \int_1^{\infty} \frac{n(z)}{z^2}dz} $$

That is, $n(z)$ is defined in terms of it's own integral. Even if there's no nice analytic solution to this, Numerical approximations of some sort are good enough. In fact this is just a parametrized instance of a more general equation I am interested in. What methods should I be using to find such a solution? Does Mathematica or Matlab has nice built in routines for this kind of thing?

For your further information I am interested only in $n(z)$ on $[1,\infty]$, and it is required to be between 0 and 1, so those integrals lie between 0 and 1, so $n(z)$ should be monotonically increasing. It seems like it should be well behaved, but on the face of it Mathematica doesn't like it.

Background: it is the steady state of a dynamic system in a network; think epidemic diffusion. $n(z)$ is the infection rate among agents with z links, and in the steady state it is

$$ n(z) = (g + (1-g)t v z )/(g + r + (1-g) t v z ), $$ where $g$ is the new infection rate, $r$ is the cure rate,and $v$ is the new infections from your peers in a networks, and $f(z)$ is the distribution of that networks; the proportion of peers withg $z$ links. One such distribution often of interest is power-law, which is

$$ f(z) = 2 z^{-3} $$ and t is the infection rate among a random peer, given by $$ t = \int_1^{\infty} z n(z) f(z)dz, $$ which, along with a paramtetrization of the above rates, gave the problem above.

Other distributions of interest are geometric, where $f(z) = Log(4) 2^{-z}$. So we are solving for the steady state level of infection among guys with z peers.

share|cite|improve this question
up vote 6 down vote accepted

Well you know how $n(z)$ looks like. Note that $\displaystyle \int_{1}^{\infty} \frac{n(z)}{z^2} dz$ is just a number. Call it $a$.

So now we have

$$n(z) = \frac{0.2+0.24 a z}{0.5+0.24 a z} = 1 - \frac{0.3}{0.5+0.24 a z}$$

where $a = \displaystyle \int_{1}^{\infty} \frac{n(z)}{z^2} dz$.

Use the above equation to get an equation solely in $a$ which you can solve for numerically.

I have done the solving part below.

$a = \displaystyle \int_{1}^{\infty} \frac{n(z)}{z^2} dz = \int_{1}^{\infty} \frac{1}{z^2} dz - \int_{1}^{\infty} \frac{0.3}{0.5 z^2 + 0.24 a z^3} dz = 1 - \int_{1}^{\infty} \frac{15}{25 z^2 + 12 a z^3} dz$

$$a = \frac{2}{5} + \frac{36}{125}a \log(\frac{25}{12a}+1)$$

You can now try to solve this numerically or plug it in mathematica as I did, which gives me $a \approx 0.673338$.

As you expected, $a \in (0,1)$.

So the function now is $$n(z) \approx 1 - \frac{15}{25+8.080056z} = \frac{10+8.080056z}{25+8.080056z}$$

and the function $n(z) \in (0,1)$, $\forall z \in \mathbb{R}^+$. In fact when $z \in [1,\infty)$, $n(z) \in (\frac{6}{11},1)$.

share|cite|improve this answer
Great answer. This is also very helpful for my more general problem, which I won't expound on here, save to say that the .2, .24 and .5 are products of parameters, and the $z^2$ is a particular distribution - so long as I don't ask Mathematica to treat everything symbolically, this method extends nicely to other instances of the parameters and other distributions. So Numerically, I think I can meet all my needs. Thanks! – Dennis Feb 9 '11 at 23:08
@Dennis: Nice problem and I felt happy when I wrote down the solution :). Once I found out the solution, I said to myself... "Wow! Cool solution!" :). Could you throw light on the more general problem? I am just asking out of my own interest. If the problem is too complicated or requires a lot of explanation or if you do not wish to reveal, no problem. – user17762 Feb 9 '11 at 23:18
Well it is the steady state of a dynamic system in a network; think epidemic diffusion. n(z) is the infection rate among agents with z links, and it follows the following process: – Dennis Feb 9 '11 at 23:24
@Dennis: Thanks for throwing light on the general problem in the question. – user17762 Feb 10 '11 at 1:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.