Results 1 to 5 of 5

Math Help - Maclaurin Series

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    17

    Maclaurin Series

    Hi, can someone help me with this question?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello tottijohn
    Quote Originally Posted by tottijohn View Post
    Hi, can someone help me with this question?
    I presume that you know that the expansion of \frac{1}{1-x} is
    1+x+x^2 +x^3 + ...
    So:
    \frac{x^2}{1-2x} = x^2\Big(1+(2x) + (2x)^2 + (3x)^3+ ...\Big)
    Can you take it from here?

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2010
    Posts
    17
    Hi Grandad, thanks for the quick reply!

    Based on your guidance and reading up more on Taylor's Theorem (I initially thought this qn is on Maclaurin Integral Test), here is my final take on the question:

    x^2[1+2x+(2x)^2+(2x)^3+(2x)^4+(2x)^5+...]
    =x^2+2x^3+4x^4+8x^5+16x^6+32x^7+...

    Thus, the coefficient of x^5 is 8. Am i correct?

    -----------------------------------------------------------------

    Also, if the function given is x^2/(1-3x^3), is the expansion:
    x^2[1+3x^3+(3x^3)^2+(3x^3)^3+(3x^3)^4+(3x^3)^5+...] ???

    Lastly, I cant seem to get the maths code for the equation correct, so pardon me for type the maths equation as above. Keep getting a syntax error.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello tottijohn
    Quote Originally Posted by tottijohn View Post
    Hi Grandad, thanks for the quick reply!

    Based on your guidance and reading up more on Taylor's Theorem (I initially thought this qn is on Maclaurin Integral Test), here is my final take on the question:

    x^2[1+2x+(2x)^2+(2x)^3+(2x)^4+(2x)^5+...]
    =x^2+2x^3+4x^4+8x^5+16x^6+32x^7+...

    Thus, the coefficient of x^5 is 8. Am i correct?
    Yes, you are. (Sorry, I just noticed a 3 crept into my answer. Of course, I meant (2x)^3 not (3x)^3.
    Also, if the function given is x^2/(1-3x^3), is the expansion:
    x^2[1+3x^3+(3x^3)^2+(3x^3)^3+(3x^3)^4+(3x^3)^5+...] ???
    Again, correct!
    Lastly, I cant seem to get the maths code for the equation correct, so pardon me for type the maths equation as above. Keep getting a syntax error.
    The code for the expansion is OK:
    [tex]x^2[1+3x^3+(3x^3)^2+(3x^3)^3+(3x^3)^4+(3x^3)^5+...] [/tex]
    displays as
    x^2[1+3x^3+(3x^3)^2+(3x^3)^3+(3x^3)^4+(3x^3)^5+...]
    To write the fraction in LaTeX, write:
    [tex]\frac{x^2}{1-3x^3}[/tex]
    which displays as:
    \frac{x^2}{1-3x^3}
    Grandad
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2010
    Posts
    17
    Thanks Grandad! You are very helpful
    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