C4 Binomial problem

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  1. ch0wm4n's Avatar
    1 01-04-2012  16:03
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    C4 Binomial problem
    Im not sure how to go about answering this question/ where im going wrong.

    5 A) Write down the first 3 terms in the binomial expansion of (1+3x)^n in ascending power of x.

    5 B) By choosing suitable values of n and x, use the series to find the cube root of 1003 to 9 signifigant figures.

    For part A i obtained the first 3 terms to be 1 + 3nx + \dfrac{n(n-1)(3x)^2}{2!}

    For part B I assume n is 1/3 and x = 334, as 1003 - 1 = 1002 and 1002 / 3 = 334 so i substitute x as 334 later.

    So i get (1+3x)^ \dfrac {1}{3}

    I expand it to 3 terms and get 1 + x - x^2

    I put substitute 334 as x and get -111,221. :confused:, as i get more terms of x the answer seems to deviate further?
    Last edited by ch0wm4n; 01-04-2012 at 17:00. Reason: Made a typo :(
  2. Mr M's Avatar
    2 01-04-2012  16:13
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    Re: C4 Binomial problem
    You have an incorrect sign in your expansion.
  3. just george's Avatar
    3 01-04-2012  16:14
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    Re: C4 Binomial problem
    I think it hasnt worked because that binomial expansion is only valid for |x|<1

    Im going to have to think about how to actually do it though :L

    edit:
    (Original post by Mr M)
    You have an incorrect sign in your expansion.
    just a typo i think - the answer -111,221 comes from the expansion with the correct sign
    Last edited by just george; 01-04-2012 at 16:15.
  4. Mr M's Avatar
    4 01-04-2012  16:19
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    Re: C4 Binomial problem
    (Original post by just george)
    I think it hasnt worked because that binomial expansion is only valid for |x|<1

    Im going to have to think about how to actually do it though :L

    edit:
    just a typo i think - the answer -111,221 comes from the expansion with the correct sign
    Ok well your point about |x|<1 is important.

    How would finding the expansion of 1.003^(1/3) help?
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  5. raheem94's Avatar
    5 01-04-2012  16:22
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    Re: C4 Binomial problem
    (Original post by Mr M)
    You have an incorrect sign in your expansion.
    (Original post by ch0wm4n)
    Im not sure how to go about answering this question/ where im going wrong.

    5 A) Write down the first 3 terms in the binomial expansion of (1+3x)^n in ascending power of x.

    5 B) By choosing suitable values of n and x, use the series to find the cube root of 1003 to 9 signifigant figures.

    For part A i obtained the first 3 terms to be 1 + 3nx + \dfrac{n(n-1)(3x)^2}{2!}

    For part B I assume n is 1/3 and x = 334, as 1003 - 1 = 1002 and 1002 / 3 = 334 so i substitute x as 334 later.

    So i get (1+3x)^ \dfrac {1}{3}

    I expand it to 3 terms and get 1 + x + x^2

    I put substitute 334 as x and get -111,221. :confused:, as i get more terms of x the answer seems to deviate further?
    (Original post by just george)
    I think it hasnt worked because that binomial expansion is only valid for |x|<1

    Im going to have to think about how to actually do it though :L

    edit:
    just a typo i think - the answer -111,221 comes from the expansion with the correct sign
    Here's how to do it:

    (1+3x)^n = 1 + 3nx + \dfrac{n(n-1)(3x)^2}{2!}

    Multiply both sides by 10

    \displaystyle10(1+3x)^n = 10\left(1 + 3nx + \dfrac{n(n-1)(3x)^2}{2!} \right)

    Substitute x=0.001 and n=1/3

    \displaystyle10(1+3(0.001)^{ \frac13 } = 10\left(1 + 3nx + \dfrac{n(n-1)(3x)^2}{2!} \right)

    \displaystyle10(1.003)^{\frac13} = 10\left(1 + 3nx + \dfrac{n(n-1)(3x)^2}{2!} \right)
    \displaystyle(10^3 \times 1.003)^{\frac13} = 10\left(1 + 3nx + \dfrac{n(n-1)(3x)^2}{2!} \right)

    \displaystyle(1003)^{\frac13} = 10\left(1 + 3nx + \dfrac{n(n-1)(3x)^2}{2!} \right)

    OP, you got the wrong answer because this expansion is only valid for,
    |3x|&lt;1
    \displaystyle |x|&lt; \dfrac13

    Hope it helps .
    Last edited by raheem94; 01-04-2012 at 16:25.
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  6. just george's Avatar
    6 01-04-2012  16:24
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    Re: C4 Binomial problem
    (Original post by Mr M)
    Ok well your point about |x|<1 is important.

    How would finding the expansion of 1.003^(1/3) help?
    Ah thats it 1003^\frac{1}{3} = 1000^\frac{1}{3} \times 1.003^\frac{1}{3}

    then you can go from there ch0wm4n?
    Last edited by just george; 01-04-2012 at 16:25.
  7. Mr M's Avatar
    7 01-04-2012  16:24
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    Re: C4 Binomial problem
    You have made over 1,500 posts raheem so you must know you have just broken the Forum Rules.
  8. raheem94's Avatar
    8 01-04-2012  16:30
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    Re: C4 Binomial problem
    (Original post by Mr M)
    You have made over 1,500 posts raheem so you must know you have just broken the Forum Rules.
    Actually, i do know the rules, i didn't meant to break them.

    That's why i didn't substituted the values in the RHS so that the OP does it himself.

    I just showed him how to manipulate the LHS to get the answer, my post is only so big, looking like a solution, because i tried to show each step so that he doesn't get confused.

    I apologise if you think i was wrong.
  9. ch0wm4n's Avatar
    9 02-04-2012  18:24
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    • Posts: 568
    Re: C4 Binomial problem
    (Original post by raheem94)
    Here's how to do it:

    (1+3x)^n = 1 + 3nx + \dfrac{n(n-1)(3x)^2}{2!}

    Multiply both sides by 10

    \displaystyle10(1+3x)^n = 10\left(1 + 3nx + \dfrac{n(n-1)(3x)^2}{2!} \right)

    Substitute x=0.001 and n=1/3

    \displaystyle10(1+3(0.001)^{ \frac13 } = 10\left(1 + 3nx + \dfrac{n(n-1)(3x)^2}{2!} \right)

    \displaystyle10(1.003)^{\frac13} = 10\left(1 + 3nx + \dfrac{n(n-1)(3x)^2}{2!} \right)
    \displaystyle(10^3 \times 1.003)^{\frac13} = 10\left(1 + 3nx + \dfrac{n(n-1)(3x)^2}{2!} \right)

    \displaystyle(1003)^{\frac13} = 10\left(1 + 3nx + \dfrac{n(n-1)(3x)^2}{2!} \right)

    OP, you got the wrong answer because this expansion is only valid for,
    |3x|&lt;1
    \displaystyle |x|&lt; \dfrac13

    Hope it helps .

    (Original post by just george)
    Ah thats it 1003^\frac{1}{3} = 1000^\frac{1}{3} \times 1.003^\frac{1}{3}

    then you can go from there ch0wm4n?

    (Original post by Mr M)
    Ok well your point about |x|<1 is important.

    How would finding the expansion of 1.003^(1/3) help?
    Thanks for the help guys .
    Post rating:
    1
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