Maths C4 - Binomial Expansion... Help???

So I don't really understand the last part of this question where I have to state the range of values for x. Why is the answer stated as a modulus inequality? and why is it <1 ??

The binomial series expansion $(1+x)^n = 1 + nx + ...$ converges when $|x|<1$. This is something you need to know for C4 but don't have to prove.

Do you understand what $|x|<1$ means and what it means for the series to converge?


Then if you are expanding $(1-x)^n$ then this is the same as $(1+(-x))^n$

So using the C4 result above, this series must converge when $|-x|<1$.

Which is the same as $|x|<1$.


Similarly the series expansion of $(1+4x)^n$ converges when $|4x|<1$, which is the same as $\displaystyle |x|<\frac{1}{4}$

I understand that |x| < 1 is the same as -1 < x < 1

But I'm not quite sure what you mean when you say "series to converge"

Time for a long explanation

The binomial series is

$\displaystyle (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ...$


This is an inifinite series so for example if we consider $(1+0.25)^{-1}$. Without expanding we can calculate that this is equal to $0.8$.

If we use the binomial series expansion formula:

$(1+0.25)^{-1} = 1 - 0.25 + 0.0625 - ...$

This is never going to be equal to 0.8 because the series goes on forever. But the sum will get closer and closer to the true value (0.8) as you sum more terms (try this using a calculator). This is what is known as a convergent sequence - I think the textbook would refer to a convergent expansion as "valid".

It is convergent because the $x$ in this case is $0.25$, which satisfies $|x|<1$.

And if you have convergence then you can use the binomial expansion to approximate something of the form $(1+x)^n$. The more terms you work out, the better approximation you get.


If instead for the same expansion you try an $x$ outside the range $|x|<1$ like $2$ then

$(1+2)^{-1} = \frac{1}{3}$

But using the series expansion:

$(1+2)^{-1} = 1 - 2 + 4 - ...$

And if you keep going, you'll see that the sum moves further away from the true value of $\frac{1}{3}$. This is a divergent sequence and is pointless if you want to approximate something of the form $(1+x)^n$ using the series.

Ohhh I see!!

So does that mean if |x| is more than or equal to 1 then it'll always become a divergent sequence (depending on how many terms of the binomial expansion you use) as it will exceed it's true value.

But if |x| is less than 1 then it will always be a convergent sequence as it'll tend towards it's true value but won't exceed it no matter how many terms of the binomial expansion you use.

Correct?

You've got the general idea correct but your use of the word "exceed" isn't completely valid.

A sequence can converge towards a number from below e.g. this sequence is converging to 3 from below:

2, 2.5, 2.8, 2.9, 2.95, 2.99, ...

Or a sequence could converge to 3 from above:

4, 3.5, 3.2, 3.1, 3.05, 3.01,...

Or a sequence could alternate above and below 3 but converge to 3:

4, 2, 3.5, 2.5, 3.2, 2.8, 3.1, 2.9, 3.05, 2.95, 3.01, 2.99,..

And actually the series I showed in my last post where $x=0.25$ was an example of alternating convergence. If you work out more terms using your calculator, you'll see it goes below and above 0.8 but still converges to 0.8.