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    I get that the rationalised the denominator but how does ln(1- (1-x etc etc = -ln(1+root1 etc etc

    basically I don't understand the final line
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    (Original post by creativebuzz)
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    I get that the rationalised the denominator but how does ln(1- (1-x etc etc = -ln(1+root1 etc etc

    basically I don't understand the final line
    1-(1-x^2) = 1-1 + x^2 = x^2

    So \displaystyle \frac{1-(1-x^2)}{x(1+ \sqrt{1-x^2})} = \frac{x^2}{x(1 + \sqrt{1-x^2})} = \frac{x}{1 + \sqrt{1-x^2}}

    Now you have \displaystyle \ln \left( \frac{x}{1 + \sqrt{1-x^2}}\right)

    And you know that \ln a = -\ln \frac{1}{a}

    So \displaystyle -\ln \frac{1 + \sqrt{1-x^2}}{x} = \ln \frac{x}{1 + \sqrt{1-x^2}}
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    (Original post by Zacken)
    1-(1-x^2) = 1-1 + x^2 = x^2

    So \displaystyle \frac{1-(1-x^2)}{x(1+ \sqrt{1-x^2})} = \frac{x^2}{x(1 + \sqrt{1-x^2})} = \frac{x}{1 + \sqrt{1-x^2}}

    Now you have \displaystyle \ln \left( \frac{x}{1 + \sqrt{1-x^2}}\right)

    And you know that \ln a = -\ln \frac{1}{a}

    So \displaystyle -\ln \frac{1 + \sqrt{1-x^2}}{x} = \ln \frac{x}{1 + \sqrt{1-x^2}}
    Ah I see! Last question, would you mind spotting where I went wrong my working out

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    (Original post by creativebuzz)
    Ah I see! Last question, would you mind spotting where I went wrong my working out
    Have you posted the right picture? I genuinely have no clue what's going on.

    You want to solve the system;

    x+2y+z = 2 and 6x + y - 4z = 16, you decided to let y = \lambda. This gives you two equations in two variables:

    x +z = 2 - 2\lambda and 6z - 4z = 16 - \lambda, you can now solve for x and z in terms of \lambda.
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    (Original post by Zacken)
    Have you posted the right picture? I genuinely have no clue what's going on.

    You want to solve the system;

    x+2y+z = 2 and 6x + y - 4z = 16, you decided to let y = \lambda. This gives you two equations in two variables:

    x +z = 2 - 2\lambda and 6z - 4z = 16 - \lambda, you can now solve for x and z in terms of \lambda.
    Oops sorry ignore the first picture, the last two are the relevant ones!

    (I was meant to send

    x + 2y + z = 2

    6x + y - 4x =16

    6x + 12y + 6z = 12
    6x + y - 4z = 16

    11y + 10z = -4

    and then the rest of my working out is in the last picture I sent you)
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    (Original post by creativebuzz)
    Oops sorry ignore the first picture, the last two are the relevant ones!

    (I was meant to send

    x + 2y + z = 2

    6x + y - 4x =16

    6x + 12y + 6z = 12
    6x + y - 4z = 16

    11y + 10z = -4

    and then the rest of my working out is in the last picture I sent you)
    Why are you solving for lambda? Solve for x and z in terms of lambda.

    I can't give a more detailed reply because I'm not home, sorry!
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    (Original post by Zacken)
    Why are you solving for lambda? Solve for x and z in terms of lambda.

    I can't give a more detailed reply because I'm not home, sorry!
    Didn't I do that in the rest of my working out?

    Also, quick question, how do you do find the mod arg form of modz = root3

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    I know that the mod is root3 but I can't seem to get the argument
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    (Original post by creativebuzz)
    Didn't I do that in the rest of my working out?

    Also, quick question, how do you do find the mod arg form of modz = root3

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    I know that the mod is root3 but I can't seem to get the argument
    It says \arg w = \frac{7\pi}{12} right on top. \arg means argument. So the argument of w is \frac{7\pi}{12}.
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    (Original post by Zacken)
    It says \arg w = \frac{7\pi}{12} right on top. \arg means argument. So the argument of w is \frac{7\pi}{12}.
    o
    m
    g:facepalm:
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    (Original post by creativebuzz)
    o
    m
    g:facepalm:
    Perhaps taking a break and relaxing for a while is in order?
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    (Original post by Zacken)
    Perhaps taking a break and relaxing for a while is in order?
    Lol yup!

    Do you know how I can draw the graph V = 0.5tsin8t?
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    (Original post by creativebuzz)
    Lol yup!

    Do you know how I can draw the graph V = 0.5tsin8t?
    Well, starts at (0,0) - has roots at 8t = \pi, 2\pi, 3\pi, \cdots \Rightarrow t = \frac{\pi}{8}, \frac{2\pi}{8}, \frac{3\pi}{8}, \cdots bascally every integer multiple of \frac{\pi}{8}. The amplitude gets bigger and bigger but it always bounded between the lines y = \pm 0.5t, maxima and minima exactly where you would expect them. That should be enough to draw the graph.
 
 
 
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