# Finding the functions that describes a Viral effect

I can't seem to find a function that makes sense to me in this.

Edit:

Lets look at it like a party.

Say there is a party in 4 weeks. 10 people are going to the party. They each invite 10 others to the party. 50% of the people they invite will attend the party. This results in 50 new attendees in the 1st week. Each of those 50 invite 10 other people. Since 50% of the invites that go out are accepted now 250 new people have accepted the invitation in the 2nd week.

week 0 = 10

week 1 = 50

week 2 = 250

week 3 = 1,250

week 4 = 6,250

Total people who will be attending = 7,810

I'd like to be able to find out two things.

1. How many new people are invited in any given week.
2. How many total people are invited by the time of any given week.
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The 10 initial people each tell 10 other people so that gives 10 + 100 = 110 people. Then half of the new people, 50, tell 10 other people so this gives 500 more people. So after one cycle 610 people know. Now you're saying each of the 500 people tell 10 people, and then 50% of those new people tell 10 people? –  Eric O. Korman May 23 '12 at 14:50

It's difficult to interpret your question but I think you want the following. Let $f(n)$ be the number of new invites at week $n$. $f(0)=10$. The number of new invites depends on the number of people invited the previous week. This is a "viral" situation and is almost always characterized by exponential growth. Indeed, the number of new invites follows the recursion relation

$$f(n+1)=\frac{1}{2}\cdot 10 f(n)=5f(n)$$

Solving this we obtain $f(n)=A\cdot 5^{n}$, where $A$ is an undetermined constant. But we know that $f(0)=10$. Plugging $0$ into $f(n)$ we see that $f(0)=A\cdot 5^{0}=A=10$. Therefore

$$f(n)=10 \cdot 5^{n}$$

The total number of people invited at by the end of a given week $n$ is simply a sum over the new invites. Let $T(n)$ be the total number of invites, then

$$T(n)=\sum_{i=0}^{n}f(i)=10\sum_{i=0}^{n}5^{i}=\frac{5}{2}(5^{n+1}-1)$$

In particular, $T(4)=7810$ as expected.

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thanks very much sir. This is very helpful! –  ninja08 May 23 '12 at 20:05