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P3 Algebra Questions watch

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    Hi, I'm stuck with several things.

    1. v = ln (2x-3) dv/dx=2/2x-3

    I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

    If you could give me some clues on how to tackle 2 & 3, would help so much.

    2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

    3. The coefficients of x and x^2 in the expansion of (1+px+qx^2)^-2 in ascending pwoers of x are 4 and 14 respectively. Find the values of p and q.

    Thank you!!!
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    (Original post by TheQueen1986)
    Hi, I'm stuck with several things.

    1. v = ln (2x-3) dv/dx=2/2x-3

    I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

    Thank you!!!
    Chain rule means when you differentiate...you leave the differential of whats in the brackets on top of the fraction...and divide it by f(x)

    I think..it's been awhile since i did maths

    G
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    (Original post by TheQueen1986)
    Hi, I'm stuck with several things.

    1. v = ln (2x-3) dv/dx=2/2x-3

    I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

    If you could give me some clues on how to tackle 2 & 3, would help so much.

    2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

    3. The coefficients of x and x^2 in the expansion of (1+px+qx^2)^-2 in ascending pwoers of x are 4 and 14 respectively. Find the values of p and q.

    Thank you!!!
    ln x is the natural logarithm of x. If you are having problem with a basic function like the logarithm function then you need to go and revise quite a bit.

    Nevertheless if y = ln x then dy/dx = 1/x, that may be where your confusion came from. So if y = ln (2x-3) then we apply the chain rule i.e. y = ln t, t=2x-3
    then dy/dx = 1/t * dt/dx and dt/dx = 2 so dy/dx = 2/t = 2/(2x-3)

    2. Clue: Taylor series, take the first few derivatives and evaluate at 0.

    3. Again you need to use Taylor series, take a few derivatives treating p and q as constants and it should follow.

    If you need more help just ask or pm me.
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    (Original post by AntiMagicMan)
    ln x is the natural logarithm of x. If you are having problem with a basic function like the logarithm function then you need to go and revise quite a bit.

    Nevertheless if y = ln x then dy/dx = 1/x, that may be where your confusion came from. So if y = ln (2x-3) then we apply the chain rule i.e. y = ln t, t=2x-3
    then dy/dx = 1/t * dt/dx and dt/dx = 2 so dy/dx = 2/t = 2/(2x-3)

    2. Clue: Taylor series, take the first few derivatives and evaluate at 0.

    3. Again you need to use Taylor series, take a few derivatives treating p and q as constants and it should follow.

    If you need more help just ask or pm me.

    taylor series is not taught in P3 only binomial expansion can be used for this
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    It isn't... meh they must have changed the syllabus since I did it.
    • Thread Starter
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    OK thanks, I get about the logarithm stuff.

    But for 2 and 3, I don't know the Taylor series, never been taught that. Does anyone know a binomial way of approaching?? Thanks.
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    (Original post by TheQueen1986)
    OK thanks, I get about the logarithm stuff.

    But for 2 and 3, I don't know the Taylor series, never been taught that. Does anyone know a binomial way of approaching?? Thanks.
    For number 2- split the part in the brackets into partial fractions by long division of the term...this should give you something like "1 + 1/x+1" ....something like that...i haven't done it so that's probably wrong..but i'd imagine it was in that form.

    Then you can use binomial expansion to figure it out..i think

    G
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    (Original post by TheQueen1986)
    Hi, I'm stuck with several things.

    1. v = ln (2x-3) dv/dx=2/2x-3

    I thought ln x = 1/x so ln 2x-3 = 1/2x-3... Why?? Please somebody explain this.

    If you could give me some clues on how to tackle 2 & 3, would help so much.

    2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.

    3. The coefficients of x and x^2 in the expansion of (1+px+qx^2)^-2 in ascending pwoers of x are 4 and 14 respectively. Find the values of p and q.

    Thank you!!!
    For 2. Binomial expansion can bedone by converting into
    [(1+x)^(1/3)][(1-x)^(-1/3)]

    Then expand the two separately and multiply them.
    If you need help on the actual expansion then PM me.
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    Q3:
    (1+ px+qx^2)^-2.
    Put px + qx^2 = t. Using binomial expansion:
    (1 + t)^-2 = 1 + (-2)t + (-2)(-3)t^2/2! + ...
    = 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...
    = 1 + (-2p)x + (-2q + 3p^2)x^2 +...

    So you got the power of x and x^2. Replace then solve.
    Try this.
    • Thread Starter
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    Sorry, still have some probs.

    Q2:

    I did what you did, but I still have troubles.

    Ok I expand the numerator and the denominator. But to what x^??

    Do you mean when you multiply the two, that you multiply numerator by 1/denominator? So if the expanded denominator was something like (just making it up for now) 1+2x+3x^2 you make it 1+2x^-1+3x^-2.

    Q3:

    How did you get from:= 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...
    ----> 1 + (-2p)x + (-2q + 3p^2)x^2 +...
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    (Original post by TheQueen1986)
    2. Given that lxl <1, expand [(1+x)/(1-x)]^(1/3) in ascending powers of x and up to and including the term in x^2.
    2.

    [(1+x)/(1-x)]^(1/3) = [(1+x)^(1/3)][(1-x)^(-1/3)]

    Now using binomial expansiona dn slittping the two expressions seperately:

    (1+x)^(1/3) = 1 + x/3 + [(1/3)(-2/3)x^2]/2 + ...

    (1-x)^(-1/3) = 1 + x/3 + [(-1/3)(-4/3)x^2]/2 + ...

    Now multiplying the two expanded expressions and collecting the terms only upto x^2

    [(1+x)/(1-x)]^(1/3) = [1 + x/3 + (x^2)/9 + ...][1 + x/3 + (2/9)(x^2) + ...]

    = 1 +x/3 + (2/9)(x^2) + x/3 + (1/9)(x^2) + (1/9)(x^2)

    = 1 + (2/3)x + (4/9)(x^2)

    Therefore the answer is:

    [(1+x)/(1-x)]^(1/3) = 1 + (2/3)x + (4/9)(x^2) + ...

    ================================ ==========================
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    Does the above post makes sense to u TheQueen1986?
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    (Original post by TheQueen1986)
    Sorry, still have some probs.

    Q2:

    I did what you did, but I still have troubles.

    Ok I expand the numerator and the denominator. But to what x^??

    Do you mean when you multiply the two, that you multiply numerator by 1/denominator? So if the expanded denominator was something like (just making it up for now) 1+2x+3x^2 you make it 1+2x^-1+3x^-2.

    Q3:

    How did you get from:= 1 + (-2)(px+qx^2) + 3(px +qx^2)^2 + ...
    ----> 1 + (-2p)x + (-2q + 3p^2)x^2 +...
    Hic, it's the formula for binomial expansion in P2 but u must have condition:
    |px + qx^2| < 1. But in ur question, i didnt find that, maybe there's another way for this (not Taylor series).
    But if it uses binomial expansion, as u know
    (1 + x)^n = 1 + nx + n(n-1)x^2/2! + n(n-1)(n-2)x^3/3! + .....
    But i think it's right, cuz it's P2 question.
    • Thread Starter
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    Yeah I got Q2 and Q3 sorted completely now. Thanks guys.
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    A fellow A level student trying to finish their A2 maths before the year starts? Heh.

    Q1: Use the chain rule, don't make false assumptions:
    V = ln(2x-3)
    V = ln(t), t=2x-3
    dv/dt = 1/t, dt/dx =2
    dv/dx = 2/t,
    dv/dx = 2/(2x-3)
 
 
 
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