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fp2 maclaurin

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.......
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.
Reply 2
Original post by Mr M
Dip? If you mean replace x in your standard expansion with 1.5x then yes.


that simple.. nice.
Reply 3
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.......


ln(2+3x)=dln(2+3x)dxdx=312+3xdx\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
3211(32x)dx=\displaystyle \frac{3}{2} \int \frac{1}{1-(-\frac{3}{2}x)} dx=
320(1)n(32x)ndx\displaystyle \frac{3}{2} \int \sum_0^{\infty} (-1)^n \left(\frac{3}{2}x\right )^n dx
Whit that 32x<1\left |\frac{3}{2}x\right |<1
Integrating pointwise you will get the series
(edited 11 years ago)
Reply 4
Original post by ztibor
ln(2+3x)=dln(2+3x)dxdx=312+3xdx\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
3211(32x)dx=\displaystyle \frac{3}{2} \int \frac{1}{1-(-\frac{3}{2}x)} dx=
320(1)n(32x)ndx\displaystyle \frac{3}{2} \int \sum_0^{\infty} (-1)^n \left(\frac{3}{2}x\right )^n dx
Whit that 32x<1\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
Reply 5
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
Reply 6
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
sinx=n=0(1)nx2n+1(2n+1)!=n=0(1)n+1x2n+1(2n+1)!\displaystyle -sinx=-\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=\sum_{n=0}^{\infty}(-1)^{n+1}\frac{x^{2n+1}}{(2n+1)!}
Assuming that
y=n=0anxn\displaystyle y=\sum_{n=0}^{\infty} a_n x^n
y"=n=2n(n1)anxn2=n=0(n+1)(n+2)an+2xn\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
n=0(n+1)(n+2)an+2xnn=0anxn=n=0(1)n+1x2n+1(2n+1)!\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
Unparseable latex formula:

\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
(n+1)(n+2)an+2=an+(1)n+1xn+1(2n+1)!\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
(edited 11 years ago)
Reply 7
Original post by ztibor
Arrange to y"-y=-sinx
sinx=n=0(1)nx2n+1(2n+1)!=n=0(1)n+1x2n+1(2n+1)!\displaystyle -sinx=-\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=\sum_{n=0}^{\infty}(-1)^{n+1}\frac{x^{2n+1}}{(2n+1)!}
Assuming that
y=n=0anxn\displaystyle y=\sum_{n=0}^{\infty} a_n x^n
y"=n=2n(n1)anxn2=n=0(n+1)(n+2)an+2xn\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
n=0(n+1)(n+2)an+2xnn=0anxn=n=0(1)n+1x2n+1(2n+1)!\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
Unparseable latex formula:

\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
(n+1)(n+2)an+2=an+(1)n+1xn+1(2n+1)!\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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