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    so how do you expand ln(2 + 3x), i can clearly see you can split it like
    ln( 2(1+1.5x) ) = ln2 + ln(1+ 1.5x). but then what?

    do i just dip the 1.5x in the formula? it doesnt make sense to me.......
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    (Original post by cooldudeman)
    so how do you expand ln(2 + 3x), i can clearly see you can split it like
    ln( 2(1+1.5x) ) = ln2 + ln(1+ 1.5x). but then what?

    do i just dip the 1.5x in the formula? it doesnt make sense to me.......
    Dip? If you mean replace x in your standard expansion with 1.5x then yes.
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    (Original post by Mr M)
    Dip? If you mean replace x in your standard expansion with 1.5x then yes.
    that simple.. nice.
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    (Original post by cooldudeman)
    so how do you expand ln(2 + 3x), i can clearly see you can split it like
    ln( 2(1+1.5x) ) = ln2 + ln(1+ 1.5x). but then what?

    do i just dip the 1.5x in the formula? it doesnt make sense to me.......
    \displaystyle ln(2+3x)=\int \frac{dln(2+3x)}{dx} dx =\int 3\frac{1}{2+3x} dx
    Use here the formula for infinite geometric series
    \displaystyle \frac{3}{2} \int \frac{1}{1-(-\frac{3}{2}x)} dx=
    \displaystyle \frac{3}{2} \int \sum_0^{\infty} (-1)^n \left(\frac{3}{2}x\right )^n dx
    Whit that \left |\frac{3}{2}x\right |<1
    Integrating pointwise you will get the series
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    (Original post by ztibor)
    \displaystyle ln(2+3x)=\int \frac{dln(2+3x)}{dx} dx =\int 3\frac{1}{2+3x} dx
    Use here the formula for infinite geometric series
    \displaystyle \frac{3}{2} \int \frac{1}{1-(-\frac{3}{2}x)} dx=
    \displaystyle \frac{3}{2} \int \sum_0^{\infty} (-1)^n \left(\frac{3}{2}x\right )^n dx
    Whit that \left |\frac{3}{2}x\right |<1
    Integrating pointwise you will get the series
    yeah i kinda see it now. thanks.

    could you or anyone else help with differential equations using Taylors series, its the last part in the book with this topic. i don't understand it at all and i have done 1st and 2nd diff orders.

    can you just go through this question
    d2y/dx2 = y - sinx, use the taylor method to find a series solution, until x^3 given that when x=0, y=1 and dy/dx = 2
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    (Original post by cooldudeman)
    yeah i kinda see it now. thanks.

    could you or anyone else help with differential equations using Taylors series, its the last part in the book with this topic. i don't understand it at all and i have done 1st and 2nd diff orders.

    can you just go through this question
    d2y/dx2 = y - sinx, use the taylor method to find a series solution, until x^3 given that when x=0, y=1 and dy/dx = 2
    Start the homogeneous part: y"-y=0
    From this you wil get the general solution
    For particular y_p use the variation of parameters c1(x) and c2(x)
    You will get for c1(x) and c2(x) some sinx*e^x type integral
    Write the solution of them with Taylor series for e^x and sinx, cosx
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    (Original post by cooldudeman)
    yeah i kinda see it now. thanks.

    could you or anyone else help with differential equations using Taylors series, its the last part in the book with this topic. i don't understand it at all and i have done 1st and 2nd diff orders.

    can you just go through this question
    d2y/dx2 = y - sinx, use the taylor method to find a series solution, until x^3 given that when x=0, y=1 and dy/dx = 2
    Arrange to y"-y=-sinx
    \displaystyle -sinx=-\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=\su  m_{n=0}^{\infty}(-1)^{n+1}\frac{x^{2n+1}}{(2n+1)!}
    Assuming that
    \displaystyle y=\sum_{n=0}^{\infty} a_n x^n
    \displaystyle y"=\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}=\sum_{n=0}^{\infty}(n+1)(n+2) a_{n+2} x^n
    shifting n by 2
    Substituting into the original equation
    \displaystyle \sum_{n=0}^{\infty}(n+1)(n+2) a_{n+2} x^n-\sum_{n=0}^{\infty} a_n x^n=\sum_{n=0}^{\infty}(-1)^{n+1}\frac{x^{2n+1}}{(2n+1)!}
    arranging
    \displaystyle \sum_{n=0}^{\infty}\left [(n+1)(n+2) a_{n+2} - a_n] x^n=\sum_{n=0}^{\infty}\left [(-1)^{n+1}\frac{x^{n+1}}{(2n+1)!}] x^n
    Equating the coeffitients of like terms
    \displaystyle (n+1)(n+2) a_{n+2}=a_{n}+(-1)^{n+1}\frac{x^{n+1}}{(2n+1)!}
    This is a recurremve relation and you van get a0 and a1 from the
    initial conditions, then a2 and so the first 3 terms of y.
    Good work
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    (Original post by ztibor)
    Arrange to y"-y=-sinx
    \displaystyle -sinx=-\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=\su  m_{n=0}^{\infty}(-1)^{n+1}\frac{x^{2n+1}}{(2n+1)!}
    Assuming that
    \displaystyle y=\sum_{n=0}^{\infty} a_n x^n
    \displaystyle y"=\sum_{n=2}^{\infty} n(n-1)a_n x^{n-2}=\sum_{n=0}^{\infty}(n+1)(n+2) a_{n+2} x^n
    shifting n by 2
    Substituting into the original equation
    \displaystyle \sum_{n=0}^{\infty}(n+1)(n+2) a_{n+2} x^n-\sum_{n=0}^{\infty} a_n x^n=\sum_{n=0}^{\infty}(-1)^{n+1}\frac{x^{2n+1}}{(2n+1)!}
    arranging
    \displaystyle \sum_{n=0}^{\infty}\left [(n+1)(n+2) a_{n+2} - a_n] x^n=\sum_{n=0}^{\infty}\left [(-1)^{n+1}\frac{x^{n+1}}{(2n+1)!}] x^n
    Equating the coeffitients of like terms
    \displaystyle (n+1)(n+2) a_{n+2}=a_{n}+(-1)^{n+1}\frac{x^{n+1}}{(2n+1)!}
    This is a recurremve relation and you van get a0 and a1 from the
    initial conditions, then a2 and so the first 3 terms of y.
    Good work
    ok thank you for all your help. appreciate it.
 
 
 
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