# Binomial expansion for approximation

I've just had to do a homework on binomial expansion for approximation:

$1.07^9$

so:

$(1+0.07)^9$

To do binomial expansion you need a calculator for the combinations button (nCr), so why would use a more complicated method, which only gives an approximation be used over just typing 1.07^9 into a calculator?

(or is this never done in real life, and it's just a homework?)

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We might need the full problem to determine the context, but I believe the point is that you get a good approximation by only taking the first few terms of the binomial expansion. This works because higher powers of $.07$ are small. E.g., adding the first $3$ of $10$ terms gives 1.8064, compared to the correct result of about 1.83845921. You don't need a calculator to find 9Cr, especially when r=0,1,2. – Jonas Meyer Jan 31 '11 at 21:36
That is about as much context as there is. in nCr, if r = 0 then nCr = 1, and if r = 1, then nCr = n, but when you get 2 it's impossible to learn off by heart, for all the possible values of n. eg 9C2 = 36. So you've used a calculator to find 9C2, and your about 0.03 out, whereas you could have just done 1.07^9? – Jonathan. Jan 31 '11 at 22:03
nC2 = n(n-1)/2 is probably worth knowing. If you don't want to memorize it (I'm with you there), keep in mind that nCr tells you how many subsets of size r there are in a set of size n. For nC2, there are n choices for the first element, (n-1) choices for the second element, and you divide by 2 because you've just counted each set twice. So for example 9C2=9*8/2 = 9*4=36. Since either n or n-1 is even, it can usually be computed quickly by hand. On the other hand, 1.07^9 would be quite tedious by hand. – Jonas Meyer Jan 31 '11 at 23:12
Another way to find $nCr$ by hand (if $r$ is not too large) is to write down Pascal's triangle row by row just by adding (each number is the sum of the two above it). en.wikipedia.org/wiki/Pascal%27s_triangle – Hans Lundmark Feb 1 '11 at 4:40

Expanding the whole thing using Binomial Theorem gives you an exact value. Not an approximation.

To get an approximation you can consider a few terms from the expansion.

For instance, for "small" $x$, $1+nx$ is a "reasonable" approximation for $(1+x)^n$.

Notice that this corresponds to picking the first two terms from the binomial theorem expansion $(1+x)^n = 1 + \binom{n}{1} \ x + \binom{n}{2}\ x^2 + \dots + x^n$.

For example

$1.0007^9 \approx 1 + 9\times 0.0007 = 1.0063$ which agrees with $1.0007^9 = 1.0063176688422737867054812736724$ upto $4$ decimal places.

Depending on how accurate you want it, you could consider more terms from the binomial expansion.

This is based on the fact that for small $x$, as the power $r$ of $x$ gets larger, the term $x^r$ becomes small quite fast.

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To expand the whole thing using Binomial Theorum, you need a calculator. You might as well use the power button and get the answer a lot quicker. If you didn't have a calculator, then it would probably be quicker to go and buy one. – Jonathan. Jan 31 '11 at 22:05
@jonathan: I think you missed the point. We are talking about an approximation. The expand the whole thing gives you an exact value. To get an approximation you don't have to expand the whole thing! Notice that $1+nx$ are the first two terms in the expansion of $(1+x)^n$ – Aryabhata Jan 31 '11 at 22:28

This looks like a badly thought-out problem to me. The first-order approximation is 1.63, but the true value is greater than 1.83. Can you give us the exact wording of the question?

As for the practicality of this, I would say that being able to estimate such expressions is a useful skill in many branches of mathematics and statistics. Not essential, but useful.

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