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    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 (quantity inside the brackets) cannot be negative either? But surely it can be negative as it's a cube root, not a square root?

    :confused:
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    |x| < 1/3 means the interval (-1/3, 1/3)
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    (Original post by vc94)
    |x| < 1/3 means the interval (-1/3, 1/3)
    yes, I know what it means. I am asking why?
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    (Original post by Plato's Trousers)
    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| &lt; \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?

    :confused:
    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. That is defined as the interval of validity and is given by |3x|&lt;1 in this case.

    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, |x|&lt; \frac{1}{3} \implies -\frac{1}{3}&lt;x&lt; \frac{1}{3}
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    Ignore this
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    (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.
    ]
    Hmm... i guess I was assuming they were looking for the domain.

    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 |3x|&lt;1 in this case.
    How would you go about determining this interval of validity?

    (Original post by Farhan.Hanif93)
    Also, |x|&lt; \frac{1}{3} \implies -\frac{1}{3}&lt;x&lt; \frac{1}{3}
    um...yes. I know what |x|&lt;\frac{1}{3} means. I am not sure what you're getting at here... :confused:
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    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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    (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 ?
    Yes.


    How would you go about determining this interval of validity?
    The interval of validity of the binomial expansion of (1+ax)^n is defined as |ax|&lt;1 \implies |x|&lt;\frac{1}{|a|}, where a is a non zero constant.

    um...yes. I know what |x|&lt;\frac{1}{3} means. I am not sure what you're getting at here... :confused:
    I mentioned this because I had confused your "it" (which is actually the quantity inside the brackets, as your edit suggests) with x, which can be negative. My apologies, I misunderstood your post.
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    (Original post by Farhan.Hanif93)
    Yes.
    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.
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    (Original post by DFranklin)
    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).
    eh? why have you suddlenly introduced |x| > 3 ? I thought the condition we were looking at was |x|< 1/3 ?


    (Original post by Farhan.Hanif93)

    The interval of validity of the binomial expansion of (1+ax)^n is defined as |ax|&lt;1 \implies |x|&lt;\frac{1}{|a|}, where a is a non zero constant.
    right. I didn't know this. 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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    (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.
    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 (1-3x)^-1/3 over those intervals?

    Or have I missed your point?
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    (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?
    For one case (probably x < -1/3, although I'm not paying much attention) it will actually oscillate (behave like \sum a^n for a < -1).

    i.e. it will still exist but would be very different to the graph of (1-3x)^-1/3 over those intervals?
    You wouldn't normally say that a series that diverged to infinity (or -infinity) exists.
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    (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 ?
    Because I wasn't paying attention - sorry. Replace 3 with 1/3 everywhere I wrote it.
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    (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?
    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.
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    (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.
    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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    (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?
    The expansion in your original post has an infinite number of terms (and so the value of x matters - it doesn't work for all x).
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    (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?
    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.
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    (Original post by EEngWillow)
    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.
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    (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.
    I assume by "rational" you actually mean "fractions where the denominator isn't equal to 1 (...or 0)", since the integers are rational numbers (Not actually sure about the case where n is irrational, never tried to expand something like (1+x)^pi [EDIT2: ofc it works, just gives you a slightly more complicated looking expansion ])

    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 :>
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    (Original post by EEngWillow)
    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.
    Rep for the effort put into this post !

    I didn't even consider this so it's definitely worth knowing, thanks
 
 
 
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