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    When you have expanded the binomial expansions expressions. You have to write this part - "Expansion is valid as long as |4x| < 1" \Rightarrow |x| < \frac{1}{4}

    How do you know what to write for the |4x| < 1 and then the bit on the right of the arrow? Help me understand it.
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    The expansion of (1 + y)^n (for any* real n) is valid for |y| < 1. So if your question asks you to expand (2 + 5x)^-7, first take out the 2:

    (2+5x)^{-7} = 2^{-7}(1 + \frac{5}{2} x)^{-7}

    and then note it's in the form (1 + y)^n, which is valid for |y| < 1, so:

    (1 + \frac{5}{2} x)^{-7} \text{ valid for } |\frac{5}{2} x| &lt; 1

    which you should normally give in the form |x| < (or maybe even >) something, so just multiply by 2/5 to get |x| &lt; \frac{2}{5}.

    * An exception is when n is a positive integer, when it's actually valid for all y.
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    So basically if its (1 + y)^n, we always write |y| < 1 and then simplify it. Any good way to remember this?

    And if it seems like a binomial expansion question from C2, then y is a subset of R.
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    (Original post by edward_wells90)
    So basically if its (1 + y)^n, we always write |y| < 1 and then simplify it. Any good way to remember this?

    And if it seems like a binomial expansion question from C2, then y is a subset of R.
    Thats basically it. The poster above may have a cunning way as he is far smarter than me. However I would just remeber it like that.

    Do a few "real" examples. With numbers, like you would get asked in a C4 exam. And then move on. Good luck with your exam buddy.
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    Thanks, both of you.

    I've got another question:

    When we get asked to find an approximation of a value and it says to substitute a suitable value of x. Do we always let x=0.01 ?

    For example:
    6. Find the first four terms in the expansion of (1 - 3x)^\frac{3}{2}. By substituting in a suitable value of x, find an approximation to 97^\frac{3}{2}.

    I've got the first four terms to be:
    1 - \frac{9}{2} x + \frac{27}{8} x^2 - \frac{27}{8} x^3.

    What do i do next? I've let 97^\frac{3}{2} = (1 - 3x)^\frac{3}{2} to get x = -32 and subbed this value of x into the expansion but this makes no sense.
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    (Original post by edward_wells90)
    Thanks, both of you.

    I've got another question:

    When we get asked to find an approximation of a value and it says to substitute a suitable value of x. Do we always let x=0.01 ?

    For example:
    6. Find the first four terms in the expansion of (1 - 3x)^\frac{3}{2}. By substituting in a suitable value of x, find an approximation to 97^\frac{3}{2}.

    I've got the first four terms to be:
    1 - \frac{9}{2} x + \frac{27}{8} x^2 - \frac{27}{8} x^3.

    What do i do next? I've let 97^\frac{3}{2} = (1 - 3x)^\frac{3}{2} to get x = -32 and subbed this value of x into the expansion but this makes no sense.
    Check your expansion. Its wrong. Then try your method again mate.
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    k i tried it again and i got this:

    1 - \frac{9}{2}x + \frac{135}{8}x^2 - \frac{135}{8}x^3

    How do i do the second part of the question?
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    (Original post by edward_wells90)
    k i tried it again and i got this:

    1 - \frac{9}{2}x + \frac{135}{8}x^2 - \frac{135}{8}x^3

    How do i do the second part of the question?
    How did you get that? It dosnt look right to me. Infact I have done it myself and think its wrong. Although I'm a bit out of practise so might be wrong. Show me your working.
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    haha this time i did it here on latex and got a different answer:

    (1 - 3x)^\frac{3}{2}

    = 1 + \frac{3}{2} (-3x) + \frac{\frac{3}{2}\frac{1}{2} (-3x)^2}{2} + \frac{\frac{3}{2}\frac{1}{2}\fra  c{-1}{2} (-3x)^3}{6}

    = 1 - \frac{9}{2}x + \frac{27}{8}x^2 + \frac{27}{16}x^3
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    (Original post by edward_wells90)
    haha this time i did it here on latex and got a different answer:

    (1 - 3x)^\frac{3}{2}

    = 1 + \frac{3}{2} (-3x) + \frac{\frac{3}{2}\frac{1}{2} (-3x)^2}{2} + \frac{\frac{3}{2}\frac{1}{2}\fra  c{-1}{2} (-3x)^3}{6}

    = 1 - \frac{9}{2}x + \frac{27}{8}x^2 + \frac{27}{16}x^3
    thats right
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    (1 - 3x)^\frac{3}{2} = 1 - \frac{9}{2}x + \frac{27}{8}x^2 + \frac{27}{16}x^3

    Okay. But how do i do the second part of the question:

    "By substituting in a suitable value of x, find an approximation to 97^\frac{3}{2}."
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    Someone please help:

    I got up to this:
    (0.99)^\frac{3}{2} = 1 - \frac{9}{200} + \frac{27}{80000} + \frac{27}{16000000}

    (0.99)^\frac{3}{2} = 0.955339187

    I see that i'm nearly there but im stuck cuz i have to get 0.99 to become 97 somehow.
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    If x=1

     97 = 100-3x



\displaystyle 97 = 100(1-\frac{3}{100}x)

    Can you do it now?
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    or if you let x=0.01 the expansion is 1000(1-3x)^\frac{3}{2}
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    I let x=0.01

    and i got this:
    (1 - 3x)^\frac{3}{2} = 1 - \frac{9}{200} + \frac{27}{80000} + \frac{27}{16000000} = 0.955339187

    (1 - 3x)^\frac{3}{2} = (1 - 3(0.01))^\frac{3}{2} = (1 - 0.03)^\frac{3}{2} = (0.97)^\frac{3}{2}

    (0.97)^\frac{3}{2} = 0.955339187

    I can't seem to get my head around this. I have to make 0.97 become 97 (but it is inside the bracket), so i have to times something by 100, but i dont understand how, or what to do!

    If i multiply both sides by 100 then the left hand side would become:
    100 (0.97)^\frac{3}{2} but that isn't 97^\frac{3}{2}, right?
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    Via your method, you will need to multiply through by  100^{\frac{3}{2}}=1000 because

    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    100^{\frac{3}{2}} \times 0.97^{\frac{3}{2}} = (100\times0.97)^{\frac{3}{2}



    = 97^{\frac{3}{2}}


    The way I pointed out in the above post might be more 'logical.'
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    (Original post by edward_wells90)
    I let x=0.01

    and i got this:
    (1 - 3x)^\frac{3}{2} = 1 - \frac{9}{200} + \frac{27}{80000} + \frac{27}{16000000} = 0.955339187

    (1 - 3x)^\frac{3}{2} = (1 - 3(0.01))^\frac{3}{2} = (1 - 0.03)^\frac{3}{2} = (0.97)^\frac{3}{2}

    (0.97)^\frac{3}{2} = 0.955339187

    I can't seem to get my head around this. I have to make 0.97 become 97 (but it is inside the bracket), so i have to times something by 100, but i dont understand how, or what to do!

    If i multiply both sides by 100 then the left hand side would become:
    100 (0.97)^\frac{3}{2} but that isn't 97^\frac{3}{2}, right?
    I did say 1000(1-3x)^\frac{3}{2} not 100(1-3x)^\frac{3}{2}
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    u have to put 97^{\frac{3}{2}} over 1000
    working out is in the attachment
    Attached Images
     
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    (Original post by tami*)
    u have to put 97^{\frac{3}{2}} over 1000
    working out is in the attachment
    thanks alot, this helped. Expressing 0.97 as \frac{97}{100} made all the difference.
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    I need some more help on 'stating the range of values of x for which the expansions are valid':

    How do i do it for:
    1. (1 + \frac{x}{2})^\frac{1}{2} (1 - x)^\frac{-1}{2}

    2. 2 (1 + x)^{-1} + 3(1 - x)^{-1} - 2(1 + \frac{x}{2})^{-1}

    -----------------------------------------------------

    For question 1, if i do each of the 2 bracket separately i get:

    for the first bracket:
    |\frac{x}{2}| &lt; 1
    \ldots |x| &lt; 2

    for the second bracket:
    |-x| &lt; 1
    \ldots |x| &gt; -1

    If i combine them together, i get -1 &lt; x &lt; 2 but this is wrong.

    -----------------------------------------------------

    For question 2, there are three brackets:

    for the first bracket:
    |x| &lt; 1

    for the second bracket:
    |-x| &lt; 1
    \ldots |x| &gt; 1

    for the third bracket:
    |\frac{x}{2}| &lt; 1
    \ldots |x| &lt; 2

    if i combine them, then there are no set of valid values of x lol.
 
 
 
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