The smart way to do the binomial expansion (Part II)

This is a follow-up to Monday's post about the smart way to do the binomial expansion. In this one, we're going to look at how to do C4 binomial expansions - ones with crazy powers like $-3$ or $\frac{3}{2}$.

This bit is very important: you should COMPLETELY ignore the formula in the book. It will give you a headache and make all of your minus signs vanish. It's also mildly poisonous.

The good news? You can use the same table as you did for the C2 stuff on Monday. The bad news? Pascal's Triangle doesn't have a $-3$rd or $\frac{3}{2}$th row, so you need another way of working those numbers out.

Getting the C numbers

Let's think about Pascal's Triangle for a moment. The seventh row (as we saw on Monday) was 1, 7, 21, 35, 35, 21, 7, 1. It turns out, there's a simple way to generate that series without your calculator. Here's how:

Start with 1. Multiply it by 7 (the power) and divide by 1 to get 7. Multiply this by 6 and divide by 2 to get 21. Then $21 \times 5 \div 3 = 35$, and so on. Eventually you get to zero and the series stops.

Did you get the recipe there? Multiply by the power divided by one, then for each new number, drop the top and bump the bottom up.

I wouldn't give you the recipe if it wasn't useful. It works just the same for numbers that aren't positive integers. So, if $n$ was $-3$, you'd have:

$$1$$ $$1 \times -3 \div 1 = -3$$ $$-3 \times -4 \div 2 = 6$$ $$6 \times -5 ÷ 3 = -10 $$
... and so on.

If $n$ was $\frac{3}{2}$:

$$1$$ $$1 \times \frac{3}{2} \div 1 = \frac{3}{2}$$ $$3/2 \times \frac{1}{2} \div 2 = \frac{3}{8}$$ $$3/8 \times \frac{-1}{2} \div 3 = \frac{-1}{16}$$
... and so on.

Other than that, the table works just the same way as before (only it goes on forever... normally they only ask you for the first few terms).

(Notice, if your system is strong enough, that these numbers are the same as the $\frac{n(n-1)(n-2)...(n-r)}{r!}$ in the big formula.)

An example of the binomial expansion

So, let's say we need to work out $(4 - x)^{\frac{1}{2}}$ up to the term in $x^3$. If you've done binomial expansion in class at school, you'd probably groan at that because the method you've learned is ridiculous. If not, ask your teacher how s/he would do it and watch him/her groan.

For us, though, it's easy! We work out the C numbers (we only need the first four):

$$1$$ $$1 \times \frac{1}{2} ÷ 1 = \frac{1}{2}$$ $$1/2 \times \frac{-1}{2} ÷ 2 = \frac{-1}{8}$$ $$-1/8 \times \frac{-3}{2 }÷ 3 = \frac{1}{16}$$

... and throw them in the table. The A column starts at $4^{\frac{1}{2}} = 2$ and you divide by 4 each time (dropping the power by 1); the B column starts at 1 and multiplies by $-x$ each time.

C A B CAB
$1$ $2$ $1$ $2$
$\frac{1}{2}$ $\frac{1}{2}$ $-x$ $\frac{-1}{4}x$
$\frac{-1}{8}$ $\frac{1}{8}$ $x^2$ $\frac{-1}{64} x^2$
$\frac{1}{16}$ $\frac{1}{32}$ $-x^3$ $-\frac{1}{512} x^3$

... so in about four short lines of working, you get
$$(4-x)^{1/2} = 2 - \frac{1}{4} x - \frac{1}{64} x^2 - \frac{1}{512} x^3$$

In conclusion

I can barely count the number of ways this is better than the 'traditional' method, but here are a few:

  • You get at most two minus signs to deal with in any sum;
  • This method deals brilliantly with the first number not being one;
  • The C numbers are the same every time you have to do the same power (i.e., if you're doing two separate $\frac{1}{2}$ powers, you only need to work out the C numbers once
  • You can use this method for any power $n$

Can you think of any others?

* Edited 2020-01-02 to fix some LaTeX and an arithmetic error that had gone unnoticed for nearly eight years. Thanks, Rob!

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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9 comments on “The smart way to do the binomial expansion (Part II)

  • Rob

    I think you dropped a sign in the 1/512x^4 term, shouldn’t it be -ve?

    I also find it a little more intuitive if you write in an initial first index column (1-7 in this case), then it’s just a case of working right to left across the table, (Pascal’s number) x power divided by index. I think I’ve written that correctly, it is new years day and I’m a little drunk.

    Stumbled across this from your other page wrt Pascal and the Taylor series. I’d noticed the same relationship myself and googled whether it was widely known, which it doesn’t appear to be (though it’s hardly a secret).

    Good stuff though. Happy new year.

    Rob

    • Colin

      Thanks, Rob – I’ve fixed the error.

      I’m not sure I quite follow your suggestion – could you expand on it?

      • Colin

        … so to speak 😉

        • Rob

          Sure. Much easier to read now you’ve fixed the Latex stuff too.

          Not sure how well this will format, but here goes. Lob it into notepad or something similar if it looks a mess. I’ll take an easy one so as not to confuse the issue too much.

          (x+1)^4

          I    P    C    A    B      =
          1    4    1    x^4  1^0    x^4
          2    3    4    x^3  1^1    4x^3
          3    2    6    x^2  1^2    6x^2
          4    1    4    x^1  1^3    4x
          5    0    1    x^0  1^4    1
          

          (I)ndex added and I’ve taken to adding the (P)ower column over on the left too.

          It’s a little unnecessary for this one since you could borrow them from the A column, but when I’m working with rational or negative indices I tend to do the calculation for the number part and then write the result straight in(so write in say 1/2 rather than say (2)^-1.

          This way, it’s (CxP)/I for each subsequent value of C and then a product of CAB as per your example. Takes no time to do it and I find it easier to follow.

          (Edited to clarify formatting)

          • Colin

            I like it – although there’s something feels a little off about it (I feel like the I column should start at 0, even though that makes no sense at all using your method).

          • Rob

            Couldn’t reply to your comment, don’t know why so I’ll reply to myself. I believe that’s a definition of madness, but I digress.

            I take your point. Technically we start at 0, so it would make more sense, but it won’t work that way so I use the traditional initial number, we’re just numbering the rows after all, and Excel and the like start at 1. Not a fan of Microsoft products, but Gnumeric is the same.

            In any case as per your observation, it won’t work with the first index at zero.

            Anyway, I’m now checking with my course leader whether this is acceptable for my exam. I’ll do it the other way if I have to but it’s a mess and really prone to error.

            Apropos of nothing, this is for an access course, I expect I have more miles on the clock than you.

            All the best.

            Rob

          • Rob

            It does start at 0. We’re just a line out of kilter to keep the left hand C*P/I in line and make it easy.The resultant C goes to the next line.

            The I is the denominator, the factorial part. Effectively we’re progressively making up the denominator factorial as we drop down the rows (and the n(n-1) numerator too).

            I don’t think I’ve ever seen it written since in both cases it would be 1, but logically if you write the denominator for the first two terms out they would be 0! and 1!. I’ve only ever seen it written from 2! onwards.

          • Colin

            Yep, that’s exactly it. I’m not sure I can make the layout work perfectly for myself, but – here’s the thing: you get to lay it out whichever way makes sense for you. Have at it 🙂

  • Rob

    Should have been 1-8 for the index column, was referring to the Pascal row, 1, 7, 21, 35, 35, 21, 7, 1.

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