Results 1 to 11 of 11

Math Help - Taylor Series about -1

  1. #1
    Junior Member
    Joined
    Apr 2011
    Posts
    60

    Taylor Series about -1

    I have a result for part a) which is
    1+(3/2*x)+((27/16)*x^2)+((27/16)*x^3)+......
    using this and the fact that 1-3x=4-3(x+1) i need to find a taylor series for
    g(x) =[16/(1-3x)^2]

    I know its by substitution but I just don't know where to start.

    I did post part a of this question but I was confident in what I was doing - this I haven't got clue as once I have resolved this I need to check first four terms in the Taylor series by finding first, second and third derivatives.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,401
    Thanks
    1273

    Re: Taylor Series about -1

    Please post the whole question.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2011
    Posts
    60

    Re: Taylor Series about -1

    please see attched question and my answer to part a -
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Taylor Series about -1

    If your answer a is correct (is that sure? Or? ...) than you can do (like you said) a substitution, if you say: x+1=t
    Then you get: 16/(4-3t)^2, this is exactly the same as f(x).

    EDIT:
    I checked, question a seems right, I guess you've used the binomial serie?
    Last edited by Siron; July 9th 2011 at 07:17 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2011
    Posts
    60

    Re: Taylor Series about -1

    Quote Originally Posted by Siron View Post
    If your answer a is correct (is that sure? Or? ...) than you can do (like you said) a substitution, if you say: x+1=t
    Then you get: 16/(4-3t)^2, this is exactly the same as f(x).

    EDIT:
    I checked, question a seems right, I guess you've used the binomial serie?
    Yes I have used binomial expansion that's what we r to use
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Taylor Series about -1

    Have you solved the exercices? Or? ...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Apr 2011
    Posts
    60

    Re: Taylor Series about -1

    Quote Originally Posted by Siron View Post
    Have you solved the exercices? Or? ...
    Not yet will have a go again.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Taylor Series about -1

    This is what I think, for your first exercice you have calculated the Taylor serie about 0 for f (so practically an maclaurin serie).
    \frac{16}{(4-3x)^2}=\frac{1}{\left(1-\frac{3}{4}x \right)^2}=1+\frac{3}{2}x+\frac{27}{16}x^2+\frac{2  7}{16}x^3 (this is just an approximation)
    Now you have to use this answer and the fact that 1-3x=4-3(x+1).
    If you say: x+1=t then you get this \frac{16}{(4-3t)^2} for the function g.
    Do you notice any relation with f? ...
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Apr 2011
    Posts
    60

    Re: Taylor Series about -1

    yes I do notice that but I have to find the series about -1 and this is what I don't understand-
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Apr 2011
    Posts
    60

    Re: Taylor Series about -1

    Quote Originally Posted by vidhi96 View Post
    yes I do notice that but I have to find the series about -1 and this is what I don't understand-
    I know the answer should be= (5/32)=(3x/32)+(27(x+1)^2/256)+(27(x+1)^3/256)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Apr 2011
    Posts
    60

    Re: Taylor Series about -1

    I have tried and but I can't seem to get my head around this as this is really a new thing for me- can you please help any further
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Getting stuck on series - can't develop Taylor series.
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: October 5th 2010, 08:32 AM
  2. Replies: 0
    Last Post: January 26th 2010, 08:06 AM
  3. Formal power series & Taylor series
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 19th 2009, 09:01 AM
  4. Replies: 9
    Last Post: April 3rd 2008, 05:50 PM
  5. Series Expansion/Taylor Series Help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 12th 2005, 10:23 AM

Search Tags


/mathhelpforum @mathhelpforum