binomial expansion question

|x| < 1/3 means the interval (-1/3, 1/3)

in the expansion of
$(1-3x)^{-\frac{1}{3}}$

the question asks for what values of x the expansion is valid

I thought it would be $(-\infty,\frac{1}{3})\cup(\frac{1}{3},\infty)$

as we need to avoid the quantity inside the brackets being zero

but the answer in the book is $|x| < \frac{1}{3}$

implying that it cannot be negative either? But surely it can be negative as it's a cube root, not a square root?

I think you've misunderstood what the question is asking for. It's asking you for the values of x for which the expansion is a good approximation of the function you're expanding binomially.
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That is defined as the interval of validity and is given by $|3x|<1$ in this case.

Also, $|x|< \frac{1}{3} \implies -\frac{1}{3}<x< \frac{1}{3}$

So you mean the two graphs only look similar around the region x=-1/3 to x=1/3 ?

How would you go about determining this interval of validity?

um...yes. I know what $|x|<\frac{1}{3}$ means. I am not sure what you're getting at here...

The expansion doesn't actually converge for |x| &gt; 3. (If you're wondering what happens when |x| = 3, that case always requires a little more care, and I can't be bothered).

The interval of validity of the binomial expansion of $(1+ax)^n$ is defined as , where $a$ is a non zero constant.

Bit misleading this - it's not that the graph is different, it's that the expansion doesn't converge, so you can't even draw the 2nd graph for |x| > 1/3.

i.e. it will still exist but would be very different to the graph of (1-3x)^-1/3 over those intervals?

Does that mean you can't use a binomial expansion on something in which the term added to the 1 is greater than 1?

You can't use an expansion with an infinite number of terms under those circumstances. But if n is a positive integer then the expansion of (1+x)^n is valid for all x.

um..not sure I get your first sentence?

I understood the binomial expansion to work for all x as you say, but that only has a finite number of terms (Pascal's triangle)? How could you make this have an infinite number of terms?

um..not sure I get your first sentence?
I understood the binomial expansion to work for all x as you say, but that only has a finite number of terms (Pascal's triangle)? How could you make this have an infinite number of terms?

To expand on DFranklin's post above (Hopefully still the only post above this )

$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \frac{n(n-1)(n-2)(n-3)}{4!}x^4 + ...$

We'll compare the expansion of (1+x)^2 and (1+x)^(-2):

When your n is a positive integer, you get the following

$(1+x)^2 = 1 + 2x + \frac{2(1)}{2}x^2 + \frac{2(1)(0)}{6}x^3 + \frac{2(1)(0)(-1)}{24}x^4 + ...$
$= 1 + 2x + x^2 + 0x^3 + 0x^4 + ...$

As you can see, every term after the x^2 term is going to include that multiplication by 0, and hence the series terminates.

When you have the negative index though:

$(1+x)^{-2} = 1 + -2x + \frac{-2(-3)}{2}x^2 + \frac{-2(-3)(-4)}{3!}x^3 + ...$
$= 1 - 2x -3x^2 -4x^3 + ...$

i.e. you aren't going to get to a point in this series where the rest of the terms evaluate to 0.

Right! I see that now. Interesting. So because of the requirement for n to decrease by an integer in each term, you will only get zeros in the higher terms if n is a positive integer? (Obvious, really. I had just never thought about it)

So if n is rational or a negative integer the sequnce will be infinite?

Brilliant.

Thanks all.

To expand on DFranklin's post above (Hopefully still the only post above this )

$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \frac{n(n-1)(n-2)(n-3)}{4!}x^4 + ...$

We'll compare the expansion of (1+x)^2 and (1+x)^(-2):

When your n is a positive integer, you get the following

$(1+x)^2 = 1 + 2x + \frac{2(1)}{2}x^2 + \frac{2(1)(0)}{6}x^3 + \frac{2(1)(0)(-1)}{24}x^4 + ...$
$= 1 + 2x + x^2 + 0x^3 + 0x^4 + ...$

As you can see, every term after the x^2 term is going to include that multiplication by 0, and hence the series terminates.

When you have the negative index though:

$(1+x)^{-2} = 1 + -2x + \frac{-2(-3)}{2}x^2 + \frac{-2(-3)(-4)}{3!}x^3 + ...$
$= 1 - 2x -3x^2 -4x^3 + ...$

i.e. you aren't going to get to a point in this series where the rest of the terms evaluate to 0.