Results 1 to 13 of 13

Math Help - Maclaurin series

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    9

    Maclaurin series

    Hi!

    I've been trying to solve this problem for two days now without any luck.
    \displaystyle \lim_{x\rightarrow 0}\left( \frac{1}{\sin(x)} -\frac{1}{\ln(x+1)}\right)

    then i get this and now im completely stuck

    \displaystyle \lim_{x\rightarrow 0} \frac{\ln(x+1)-\sin(x)}{\sin(x)\ln(x+1)}

    when i check my notes from class i cant see that our professor has shown us how its done when x->0 or infinity or w/e...

    Thanks in advance!
    Last edited by CaptainBlack; February 16th 2011 at 06:40 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2010
    From
    Staten Island, NY
    Posts
    451
    Thanks
    2
    Applying L'Hopital's Rule twice will give you the answer.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2011
    Posts
    9
    Yeah thanks for the tip, but some in my class tried that but our professor told em he want them to use the Maclaurin.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ogward View Post
    Hi!

    I've been trying to solve this problem for two days now without any luck.
    \displaystyle \lim_{x\rightarrow 0}\left( \frac{1}{\sin(x)} -\frac{1}{\ln(x+1)}\right)

    then i get this and now im completely stuck

    \displaystyle \lim_{x\rightarrow 0} \frac{\ln(x+1)-\sin(x)}{\sin(x)\ln(x+1)}

    when i check my notes from class i cant see that our professor has shown us how its done when x->0 or infinity or w/e...

    Thanks in advance!
    What DrSteve says, but given the thread title you can expand \sin(x) and \ln(x+1) as Maclaurin series simplify etc... which will work.

    CB
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2011
    Posts
    9
    this is how far i've got...
    is it correct? and how do i continue?

    sinx=\frac{1}{1!}x-\frac{1}{3!}x^{3}+\frac{1}{5!}x^{5}-\frac{1}{7!}x^{7}+..........<br />
\\ =\sum ^{\infty }_{n=1}(-1)^{n+1}\frac{x^{2n-1}}{(2n-1)!}

    ln(x+1)=\frac{x}{1!}-\frac{x^{2}}{3!}+\frac{2x^{3}}{3!}-\frac{6x^{4}}{4!}+..........<br />
\\ \sum ^{\infty }_{n=0}unsure\frac{x^{n}}{n!}<br />
    im unsure of this last part.
    Last edited by ogward; February 16th 2011 at 08:47 AM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ogward View Post
    this is how far i've got...
    is it correct? and how do i continue?

    sinx=\frac{1}{1!}x-\frac{1}{3!}x^{3}+\frac{1}{5!}x^{5}-\frac{1}{7!}x^{7}+..........<br />
\\ =\sum ^{\infty }_{n=1}(-1)^{n+1}\frac{x^{2n-1}}{(2n-1)!}

    ln(x+1)=\frac{x}{1!}-\frac{x^{2}}{3!}+\frac{2x^{3}}{3!}-\frac{6x^{4}}{4!}+..........<br />
\\ \sum ^{\infty }_{n=0}unsure\frac{x^{n}}{n!}<br />
    im unsure of this last part.
    Does not matter you only need the first two non-zeros terms so put:

    \sin(x)=x-\dfrac{x^3}{6}+O(x^5)

    and:

    \ln(x+1)=x-\dfrac{x^2}{2}+O(x^3)

    (here the big-O notation represents the remainder in the truncated series)

    Then procceed.

    CB
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Feb 2011
    Posts
    9
    the problem is i have no idea what should be done next...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Nov 2010
    From
    Staten Island, NY
    Posts
    451
    Thanks
    2
    I think you can finish up by "substituting in" what CaptainBlack has given you, dividing numerator and denominator x^2, then taking the limit (plug in a 0 for x).
    Last edited by DrSteve; February 16th 2011 at 02:14 PM. Reason: fixed error
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Feb 2011
    Posts
    9
    this is what I get
    \lim_{x\rightarrow 0}=\frac{\frac{x^3}{6}-\frac{x^2}{2}}{\frac{x^5}{12}-\frac{x^4}{6}-\frac{x^3}{2}+x^2<br />

    click on get above.

    should I put x=0? then I get 0/0, is that good?
    Last edited by ogward; February 16th 2011 at 11:59 AM. Reason: error
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ogward View Post
    this is what I get
    \lim_{x\rightarrow 0}=\frac{\frac{x^3}{6}-\frac{x^2}{2}}{\frac{x^5}{12}-\frac{x^4}{6}-\frac{x^3}{2}+x^2<br />

    click on get above.

    should I put x=0? then I get 0/0, is that good?
    Divide top and bottom of what your input to WA gives by  x^2, and then let x go to zero

    CB
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Feb 2011
    Posts
    9
    does this look good?

    \lim_{x\rightarrow 0}= \frac{2(x-3)}{(x-2)(x^2-6)}=\frac{-6}{12}
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by ogward View Post
    does this look good?

    \lim_{x\rightarrow 0}= \frac{2(x-3)}{(x-2)(x^2-6)}=\frac{-6}{12}
    Well the number on the right is correct.

    CB
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Senior Member
    Joined
    Nov 2010
    From
    Clarksville, ARk
    Posts
    398
    Quote Originally Posted by ogward View Post
    this is what I get
    \lim_{x\rightarrow 0}=\frac{\frac{x^3}{6}-\frac{x^2}{2}}{\frac{x^5}{12}-\frac{x^4}{6}-\frac{x^3}{2}+x^2}<br />

    click on get above.

    should I put x=0? then I get 0/0, is that good?
    Divide the numerator & denominator by x . Then set x=0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: January 26th 2010, 09:06 AM
  2. Replies: 2
    Last Post: September 16th 2009, 08:56 AM
  3. Binomial Series to find a Maclaurin Series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 21st 2009, 08:15 AM
  4. Multiplying power series - Maclaurin series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 8th 2009, 12:24 AM
  5. Replies: 1
    Last Post: May 5th 2008, 10:44 PM

Search Tags


/mathhelpforum @mathhelpforum