3
$\begingroup$

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?)

  • 2
    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=0,1,2.2011-01-31
  • 0
    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?2011-01-31
  • 0
    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.2011-01-31
  • 0
    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). http://en.wikipedia.org/wiki/Pascal%27s_triangle2011-02-01

2 Answers 2

4

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.

  • 0
    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.2011-01-31
  • 1
    @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$2011-01-31
-1

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.