Binomial theorem real world use

Question

I understand binomial theorem helps expand and calculate two terms raised to nth power $(a+b)^n$ easily. Can someone explain briefly how they are used and applied in a real world application? I see lot of mentions about their use in weather forecasting, IP subnetting, economic forecast etc. But couldn't find anything more than names of applications that uses it.

Would love to see its usage explained at a high level in any one of the real world application.

Answer 1

One case is when $a$ is substantially larger than $b$ and you are doing a rough (maybe mental) calculation. You can write $(a+b)^n=a^n(1+\frac ba)^n$. If $\frac ba \ll \frac 1n$ you can write $(1+\frac ba)^n \approx 1+\frac {nb}a$ because the next term is smaller by another power of $\frac ba$.

If you enlarge the radius of a sphere by $1\%$ how much does its volume increase? The new volume is $\frac 43 \pi (1.01r)^3=\frac 43 \pi r^3 (1+0.01)^3\approx \frac 43 \pi r^3(1.03)$ so the increase is $3\%$

Answer 2

I don't think this is an answer to your question so feel free to downvote..

I've thought about this question a lot...but the truth is that the binomial theorem really isn't the important part. It's the binomial coefficient that's the more important part. The binomial theorem is a result of the binomial coefficient!

So how is that? We have $(a + b)^n$ which means $(a + b)(a + b)\dots(a+b)$.

Let's look at $(a + b)^2$:

$$ (a + b)(a + b)= a(a + b) + b(a + b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2 $$

Now let's do $(a + b)^3 = (a + b)(a + b)^2$:

$$ a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3= a^3 + 3a^2b + 3ab^2 + b^3 $$

Ok, so why is it that the binomial coefficient keeps showing up in the binomial theorem?

The reason has to do with combinatorics. Think of it this way: if I have $(a + b)^n$, then I can get $a\cdot a\cdot a\ ...\cdot a$, $n$ times, by choosing $a$ each time from the $n$ number of $(a + b)$'s or I can get some combination of $a$'s and $b$'s, e.g. $a^kb^{n - k}$ (where $0 \leq k \leq n$) by choosing $k$ of the $(a + b)$'s to be $a$ and the rest $b$.

So now the question is if I choose $a^kb^{n - k}$, how many ways can I get that by choosing among the $n$ possible $(a + b)$'s? The answer is the binomial coefficient. Think of it like this, there are $n$ distinct $a$'s and $n$ distinct $b$'s. How many different $k$-groups of $a$'s can I get by choosing among the $n$ distinct $a$'s? Easy: $\binom{n}{k}$. Meaning that if I choose $a^k$ there will be $\binom{n}{k}$ different combinations of the different $(a +b)$ groups that can give me $a^k$ with the "rest" being $b^{n - k}$, giving the binomial theorem:

$$ (a + b)^n \rightarrow\ ?\ +\ ?\ +\ ...\ +\ \binom{n}{k}a^kb^{n - k}\ +\ ?\ ...\ +\ ? $$

Since we can obviously do that for every possible combination of $a$ and $b$ we finally get:

$$ (a + b)^n = \sum\limits_{k = 0}^{n} \binom{n}{k}a^kb^{n - k} $$

Here's one last concrete example, let's say I have $(a + b)^5$ and I want to find the coefficient for $a^3b^2$. It's going to be $\binom{5}{3} = 10$. We can see this by enumerating all of the possible ways of choosing $a^3b^2$:

$$ aaabb + aabab + aabba + abaab + ababa + abbaa + baaab + baaba + babaa + bbaaa\\ $$

$$ =10a^3b^2 $$