Results 1 to 7 of 7

Math Help - Maclaurin Polynomial

  1. #1
    Junior Member
    Joined
    Jun 2008
    From
    Sanger, CA
    Posts
    58

    Question Maclaurin Polynomial

    Q: Find the Maclaurin polynomial of order 3 for the function f(x) = e^(-2x).

    I know that the Maclaurin polynomial for e^x = 1 + x + x^2/2! + x^3/3! + ...

    So my question is if I would have to plug in (-2x) in the Maclaurin polynomial of e^x so I get

    e^(-2x) = 1 - 2x + 2x^2 - 4x^3/3 + ...

    Is this right or wrong?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456
    to get the Maclaurin use the formula :

     f_{(x)}= \displaystyle \sum_{n=0} ^{\infty} \displaystyle \frac {f_{(0)}^{(n)}}{n!} x^n

    where is f_{(0)}^{(n)} nth derivative...


    so it goes

    f_{(0)}=e^{-2\cdot0}=1

    f'_{x}=(e^{-2x})'=\displaystyle -\frac {2}{e^{2x}} \mid _{x=0} = -2

    f''_{x}=(e^{-2x})''=\displaystyle \frac {4}{e^{2x}} \mid _{x=0} = 4

    f'''_{x}=(e^{-2x})''' =\displaystyle -\frac {8}{e^{2x}} \mid _{x=0} =-8

    f^{(4)}_{x}=(e^{-2x})^{(4)}=\displaystyle \frac {16}{e^{2x}} \mid _{x=0} = 16


    and so on .... and u have :


    f_{(x)}= \displaystyle \sum _{n=0} ^{\infty} \frac {(-1)^{n}}{n!} \cdot ......

    i think u can do that
    Last edited by yeKciM; August 5th 2010 at 03:24 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by katchat64 View Post
    Q: Find the Maclaurin polynomial of order 3 for the function f(x) = e^(-2x).

    I know that the Maclaurin polynomial for e^x = 1 + x + x^2/2! + x^3/3! + ...

    So my question is if I would have to plug in (-2x) in the Maclaurin polynomial of e^x so I get

    e^(-2x) = 1 - 2x + 2x^2 - 4x^3/3 + ...

    Is this right or wrong?

    It is right! (But why?)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jun 2008
    From
    Sanger, CA
    Posts
    58
    I know the formula I was just looking to see some work to see if I did the problem correctly. In another problem that I had in an exam consisted of the Maclaurin polynomial of 1/(1-x) = 1 + x + x^2 + X^3 + ...

    and the professor wanted us to find the power series representation of 1/(1-x^3) and to figure that out you needed to put an x^3 wherever there was an x in the original given equation.

    So I thought since I know the Maclaurin polynomial of e^x I thought that you were able to plug in (-2^x) wherever there was an x in that original polynomial.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456
    Quote Originally Posted by katchat64 View Post
    I know the formula I was just looking to see some work to see if I did the problem correctly. In another problem that I had in an exam consisted of the Maclaurin polynomial of 1/(1-x) = 1 + x + x^2 + X^3 + ...

    and the professor wanted us to find the power series representation of 1/(1-x^3) and to figure that out you needed to put an x^3 wherever there was an x in the original given equation.

    So I thought since I know the Maclaurin polynomial of e^x I thought that you were able to plug in (-2^x) wherever there was an x in that original polynomial.
    yes but i think yours 3rd member is wrong should be  4x^2


    just to note that when u mentioned that one  \displaystyle \frac {1}{1-x} that is from  \displaystyle \sum _{n=0} ^{\infty} x^n when  |x|<1 we had have a lot of times to find  \displaystyle \frac {1}{(1-x)^2} or  \displaystyle \frac {1}{(1-x)^3} which are just a derivate of  \displaystyle \frac {1}{1-x} and actually using that Maclouren  \displaystyle \frac {1}{1-x} = \displaystyle \sum _{n=0} ^{\infty} x^n u can have a lot of another to do based on that one, you just have to tune it so it looks like that one

    there is like 5 - 10 depending on school to school that u should know by heart and everything goes around them (at least here )



    Edit:just remembered something to note
    Last edited by yeKciM; August 5th 2010 at 03:43 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,412
    Thanks
    1328
    Quote Originally Posted by yeKciM View Post
    yes but i think yours 3rd member is wrong should be  4x^2
    No, the third term for e^x is \frac{1}{2}x^2. Replacing x by -2x gives \frac{(-2x)^2}{2}= \frac{4x^2}{2}= 2x^2 as katchat64 has.

    katchat64, yes, your method is completely correct.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    456
    Quote Originally Posted by HallsofIvy View Post
    No, the third term for e^x is \frac{1}{2}x^2. Replacing x by -2x gives \frac{(-2x)^2}{2}= \frac{4x^2}{2}= 2x^2 as katchat64 has.

    katchat64, yes, your method is completely correct.
    thank you for some reason i leave factorials as they are because in my mind i always thinking of it as sum there and then just looking for how does changes... leaving down that factorial thanks again
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Maclaurin Polynomial
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 8th 2010, 07:28 AM
  2. Maclaurin Polynomial
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 16th 2010, 07:51 AM
  3. fifth degree maclaurin polynomial
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 21st 2010, 10:43 PM
  4. Maclaurin polynomial
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 3rd 2009, 02:48 AM
  5. MacLaurin polynomial
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 28th 2009, 04:42 PM

Search Tags


/mathhelpforum @mathhelpforum