in the expansion of
the question asks for what values of x the expansion is valid
I thought it would be
as we need to avoid the quantity inside the brackets being zero
but the answer in the book is
implying that it (quantity inside the brackets) cannot be negative either? But surely it can be negative as it's a cube root, not a square root?

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 22032011 22:31
Last edited by Plato's Trousers; 23032011 at 11:17. 
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 22032011 22:34
x < 1/3 means the interval (1/3, 1/3)

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 22032011 22:35
(Original post by vc94)
x < 1/3 means the interval (1/3, 1/3) 
Farhan.Hanif93
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 22032011 22:38
(Original post by Plato's Trousers)
in the expansion of
the question asks for what values of x the expansion is valid
I thought it would be
as we need to avoid the quantity inside the brackets being zero
but the answer in the book is
implying that it cannot be negative either? But surely it can be negative as it's a cube root, not a square root?
If you want to see why this condition is necessary, plug your original function into a graph sketching package and compare that graph to the graph of the first few terms of the expansion and look carefully at where the graphs are close.
Also, 
EEngWillow
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 22032011 22:41
Ignore this
Last edited by EEngWillow; 22032011 at 22:42. Reason: *cough* Nothing to see here *cough* 
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 23032011 10:40
(Original post by Farhan.Hanif93)
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.
]
Thanks for the clarification. So you mean the two graphs only look similar around the region x=1/3 to x=1/3 ?
(Original post by Farhan.Hanif93)
That is defined as the interval of validity and is given by in this case.
Last edited by Plato's Trousers; 23032011 at 11:09. 
DFranklin
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 23032011 11:20
The expansion doesn't actually converge for x > 3. (If you're wondering what happens when x = 3, that case always requires a little more care, and I can't be bothered).

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 23032011 11:28
(Original post by Plato's Trousers)
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?
I mentioned this because I had confused your "it" (which is actually the quantity inside the brackets, as your edit suggests) with , which can be negative. My apologies, I misunderstood your post. 
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 23032011 11:37
(Original post by Farhan.Hanif93)
Yes. 
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 23032011 11:45
(Original post by DFranklin)
The expansion doesn't actually converge for x > 3. (If you're wondering what happens when x = 3, that case always requires a little more care, and I can't be bothered).
(Original post by Farhan.Hanif93)
The interval of validity of the binomial expansion of is defined as , where is a non zero constant.

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 23032011 11:46
(Original post by DFranklin)
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.
Or have I missed your point? 
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 23032011 11:49
(Original post by Farhan.Hanif93)
Yes, that's true. Wouldn't the graph of the expansion still tend to infinity for x outside the range of x < 1/3, though?
i.e. it will still exist but would be very different to the graph of (13x)^1/3 over those intervals? 
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 23032011 11:50
(Original post by Plato's Trousers)
eh? why have you suddlenly introduced x > 3 ? I thought the condition we were looking at was x< 1/3 ? 
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 23032011 11:52
(Original post by Plato's Trousers)
Does that mean you can't use a binomial expansion on something in which the term added to the 1 is greater than 1? 
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 23032011 11:55
(Original post by DFranklin)
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.
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? 
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 23032011 11:59
(Original post by Plato's Trousers)
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? 
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 23032011 12:14
(Original post by Plato's Trousers)
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?
We'll compare the expansion of (1+x)^2 and (1+x)^(2):
When your n is a positive integer, you get the following
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:
i.e. you aren't going to get to a point in this series where the rest of the terms evaluate to 0.Last edited by EEngWillow; 23032011 at 12:16. 
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 23032011 13:16
(Original post by EEngWillow)
To expand on DFranklin's post above (Hopefully still the only post above this )
We'll compare the expansion of (1+x)^2 and (1+x)^(2):
When your n is a positive integer, you get the following
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:
i.e. you aren't going to get to a point in this series where the rest of the terms evaluate to 0.
So if n is rational or a negative integer the sequnce will be infinite?
Brilliant.
Thanks all. 
EEngWillow
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 23032011 13:44
(Original post by Plato's Trousers)
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.
One point I should have added is that from the infinite expansion you can hopefully see the need for x < 1: you need a value between 0 and 1 for the terms to get smaller in value each time (take the example x = 0.5, so x^2 = 0.25, x^3 = 0.125 etc) > under these circumstances the series converges. If you had instead x = 2, then each term would get larger (x = 2, x^2 = 4, x^3 = 8) and it diverges.
EDIT: Also, thanks for the +rep, you just pushed me into the second green gem
On the otherhand, I now have NO IDEA WHATSOEVER what my electrodynamics lecturer was talking about :>Last edited by EEngWillow; 23032011 at 13:53. 
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 23032011 13:49
(Original post by EEngWillow)
To expand on DFranklin's post above (Hopefully still the only post above this )
We'll compare the expansion of (1+x)^2 and (1+x)^(2):
When your n is a positive integer, you get the following
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:
i.e. you aren't going to get to a point in this series where the rest of the terms evaluate to 0.
I didn't even consider this so it's definitely worth knowing, thanks
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