Results 1 to 7 of 7
Like Tree3Thanks
  • 1 Post By domy7997
  • 2 Post By Idea

Thread: Limit with taylor

  1. #1
    Junior Member
    Joined
    Nov 2018
    From
    Italy
    Posts
    30
    Thanks
    3

    Limit with taylor

    I've tried studying the limit with taylor,(untill o(x^6)) but it result to me that the limit is equal to-5/6 and not 17/12... Any tips about this limit? I mean where do i get wrong?
    Thanks

    Sent from my Redmi Note 4X using Tapatalk
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2018
    From
    Italy
    Posts
    30
    Thanks
    3

    Re: Limit with taylor

    Quote Originally Posted by domy7997 View Post
    I've tried studying the limit with taylor,(untill o(x^6)) but it result to me that the limit is equal to-5/6 and not 17/12... Any tips about this limit? I mean where do i get wrong?
    Thanks

    Sent from my Redmi Note 4X using Tapatalk
    Here it is how i did it...
    Where did i get wrong? I'm getting crazy..

    Sent from my Redmi Note 4X using Tapatalk
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2013
    From
    Lebanon
    Posts
    958
    Thanks
    477

    Re: Limit with taylor

    for the numerator you should get

    $\displaystyle \left(x^2-x^3+\frac{11 x^4}{12}\right)-\left(x^2-x^3-\frac{x^4}{2}\right)=\frac{17}{12}x^4$

    denominator is OK
    Thanks from topsquark and domy7997
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2018
    From
    Italy
    Posts
    30
    Thanks
    3

    Re: Limit with taylor

    Quote Originally Posted by Idea View Post
    for the numerator you should get

    $\displaystyle \left(x^2-x^3+\frac{11 x^4}{12}\right)-\left(x^2-x^3-\frac{x^4}{2}\right)=\frac{17}{12}x^4$

    denominator is OK
    Mmm i'm not able to get that 11x^4/12
    Where do i get wrong?
    Thank you so much for your time...

    Sent from my Redmi Note 4X using Tapatalk
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2013
    From
    Colombia
    Posts
    1,974
    Thanks
    712

    Re: Limit with taylor

    Your coefficients of $x^4$ and $x^5$ in the series for $\ln^2{(1+x)}$ look wrong to me. I think you forgot that there are two terms for each.

    Also, $$\frac1{1-f(x)} \approx 1 + f(x) + f^2(x) + f^3(x) + \ldots$$ so

    \begin{align}
    \frac1{1-x} &= x + x^2 + x^3 + x^4 + o(x^4) \\
    \log{(1+x)} &= x - \tfrac12 x^2 + \tfrac13x^3 - \tfrac14x^4 + o(x^3) \\[12pt]
    \log^2{(1+x)} &= x^2 - x^3 + (\tfrac14 + \tfrac23) x^4 + o(x^4) \\
    &= x^2 - x^3 + \tfrac{11}{12} x^4 + o(x^4) \\
    &= x^2\big(1 - x + \tfrac{11}{12} x^2 + o(x^2)\big) \\[8pt]
    \log{(1+x^2)} &= x^2 - \tfrac12 x^4 + o(x^4) \\
    \log{(1+x^2)}-x^3 &= x^2 - x^3 - \tfrac12 x^4 + o(x^4) \\
    &= x^2\big(1 - x - \tfrac12 x^2 + o(x^2)\big) \\[8pt]
    \frac1{\log^2{(1+x)}} &= \frac1{x^2\big(1 - x + \tfrac{11}{12} x^2 + o(x^2)\big)} \\
    &= \frac1{x^2}\cdot\frac1{1 - x + \tfrac{11}{12} x^2 + o(x^2)} \\
    &= \frac1{x^2} \cdot \frac1{1 - \big(x - \tfrac{11}{12} x^2 + o(x^2)\big)} \\
    &= \frac1{x^2} \bigg(1 + \big(x - \tfrac{11}{12} x^2 + o(x^2)\big) + \big(x - \tfrac{11}{12} x^2 + o(x^2)\big)^2 + o(x^2)\bigg) \\
    &= \frac1{x^2} \bigg(1 + \big(x - \tfrac{11}{12} x^2\big) + \big(x^2\big) + o(x^2)\bigg) \\[8pt]
    \frac1{\log{(1+x^2)}-x^3} &= \frac1{x^2\big(1 - x - \tfrac12 x^2 + o(x^2)\big)} \\
    &= \frac1{x^2} \cdot \frac1{1 - x - \tfrac12 x^2 + o(x^2)} \\
    &= \frac1{x^2} \cdot \frac1{1 - \big(x + \tfrac12 x^2 + o(x^2)\big)} \\
    &= \frac1{x^2} \bigg(1 + \big(x + \tfrac12 x^2 + o(x^2)\big) + \big(x + \tfrac12 x^2 + o(x^2)\big)^2 + o(x^2)\bigg) \\
    &= \frac1{x^2} \bigg(1 + \big(x + \tfrac12 x^2\big) + \big(x^2\big) + o(x^2)\bigg) \\[12pt]
    \frac1{\log{(1+x^2)}-x^3} - \frac1{\log^2{(1+x)}} &= \frac1{x^2} \bigg(1 + \big(x + \tfrac12 x^2\big) + \big(x^2\big) + o(x^2)\bigg) - \frac1{x^2} \bigg(1 + \big(x - \tfrac{11}{12} x^2\big) + \big(x^2\big) + o(x^2)\bigg) \\
    &= \frac1{x^2}\bigg[\bigg(1 + \big(x + \tfrac12 x^2\big) + \big(x^2\big) + o(x^2)\bigg) - \bigg(1 + \big(x - \tfrac{11}{12} x^2\big) + \big(x^2\big) + o(x^2)\bigg)\bigg] \\
    &= \frac1{x^2}\big(\tfrac12 x^2 - (-\tfrac{11}{12}x^2) + o(x^2)\big) \\
    &= \frac{x^2}{x^2}\big(\tfrac12 + \tfrac{11}{12} + o(1)\big) \\
    &= \frac{17}{12} + o(1)
    \end{align}
    Last edited by Archie; Dec 29th 2018 at 02:47 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Nov 2018
    From
    Italy
    Posts
    30
    Thanks
    3

    Re: Limit with taylor

    Quote Originally Posted by Archie View Post
    Your coefficients of $x^4$ and $x^5$ in the series for $\ln^2{(1+x)}$ look wrong to me. I think you forgot that there are two terms for each.
    I controlled it 10 times... The coefficients got from the taylor series are correct for the Numerator...

    Sent from my Redmi Note 4X using Tapatalk
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2018
    From
    Italy
    Posts
    30
    Thanks
    3

    Re: Limit with taylor

    Quote Originally Posted by Archie View Post
    Your coefficients of $x^4$ and $x^5$ in the series for $\ln^2{(1+x)}$ look wrong to me. I think you forgot that there are two terms for each.
    Omg... I thought it was factorial the denominator of the log Taylor series... What a shame omg. Thx mate

    Sent from my Redmi Note 4X using Tapatalk
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limit and Taylor
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jun 8th 2009, 10:41 AM
  2. Taylor Series on a limit
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Apr 9th 2009, 07:10 AM
  3. Find limit using taylor series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 9th 2009, 06:49 AM
  4. solving limit using taylor series..
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Mar 10th 2009, 07:33 AM
  5. Limit with Taylor formula
    Posted in the Calculus Forum
    Replies: 9
    Last Post: Nov 11th 2008, 04:03 AM

/mathhelpforum @mathhelpforum