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Math Help - determine Maclaurin development of order 3 with residual term with o notation e^sinx

  1. #1
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    determine Maclaurin development of order 3 with residual term with o notation e^sinx

    determine Maclaurin development of order 3 with residual term with o notation e^sinx


    My main problem is that i don't understand the rules of O notation


    1. e^sinx ... u= sinx

    we get e^u

    2. Develeopment of e^x = 1 + x + x^2/2 + (O(x^4))
    Develeopment of sinx = x - 1/6 x^3 + O(x^5)

    3. e^sinx = e^x-x^3/6 * e^(O(x^5))
    = e^x * e^-x^3/6 * e(O(x^5))

    I could need some help finish this off
    secondly I don't understand why we put an O notation of O(x^5)) here out of a sudden, we had an ordo of 4 as well didnt we?

    Would be grateful if someone could explain all the Ordo (O notation) rules so that i can start working on problem self without using hints like the one above. All that we have is a booklet handout with problems + a very bad book that does not cover the topic well.

    Thanks alot
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  2. #2
    MHF Contributor chisigma's Avatar
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    Setting f(x)=e^{\sin x} You can observe that is...

    \displaystyle \frac{d}{dx} \ln f(x)= \frac{f^{'}(x)}{f(x)}= \cos x (1)

    Now if is...

    \displaystyle f(x)= \sum_{n=0}^{\infty} a_{n}\ x^{n} (2)

    ... the (1) becomes...

    \displaystyle a_{1} + 2\ a_{2}\ x + 3\ a_{3}\ x^{2} + 4\ a_{4}\ x^{3} + ... =

    \displaystyle =  (a_{0} + a_{1}\ x + a_{2}\ x^{2} + a_{3}\ x^{3} + ...)\ (1-\frac{x^{2}}{2} + ...)=

    \displaystyle = a_{0} + a_{1}\ x + (a_{2}-\frac{1}{2})\ x^{2} - \frac{a_{1}}{2}\ x^{3}+ ... (3)

    ... and from (3) You derive...

    \displaystyle a_{0}=1

    \displaystyle a_{1}=a_{0}=1

    \displaystyle a_{2}= \frac{a_{1}}{2}= \frac{1}{2}

    \displaystyle a_{3}= \frac{1}{3}\ (a_{2}-\frac{1}{2}) = 0

    \displaystyle a_{4}= -\frac{a_{1}}{12}= -\frac{1}{12}

    ... so that is...

    \displaystyle e^{\sin x} = 1 + x + \frac{x^{2}}{2} + O(x^{4}) (4)

    Kind regards

    \chi \sigma
    Last edited by chisigma; February 23rd 2011 at 09:32 AM. Reason: error in computation of a3... sorry!...
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  3. #3
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    Quote Originally Posted by chisigma View Post
    Setting f(x)=e^{\sin x} You can observe that is...

    \displaystyle \frac{d}{dx} \ln f(x)= \frac{f^{'}(x)}{f(x)}= \cos x (1)

    Now if is...

    \displaystyle f(x)= \sum_{n=0}^{\infty} a_{n}\ x^{n} (2)

    ... the (1) becomes...

    \displaystyle a_{1} + 2\ a_{2}\ x + 3\ a_{3}\ x^{2} + 4\ a_{4}\ x^{3} + ... =

    \displaystyle =  (a_{0} + a_{1}\ x + a_{2}\ x^{2} + a_{3}\ x^{3} + ...)\ (1-\frac{x^{2}}{2} + ...)=

    \displaystyle = a_{0} + a_{1}\ x + (a_{2}-\frac{1}{2})\ x^{2} - \frac{a_{1}}{2}\ x^{3}+ ... (3)

    ... and from (3) You derive...

    \displaystyle a_{0}=1

    \displaystyle a_{1}=a_{0}=1

    \displaystyle a_{2}= \frac{a_{1}}{2}= \frac{1}{2}

    \displaystyle a_{3}= \frac{1}{3}\ (a_{2}-\frac{1}{2}) = \frac{1}{6}

    \displaystyle a_{4}= -\frac{a_{1}}{12}= -\frac{1}{12}

    ... so that is...

    \displaystyle e^{\sin x} = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + O(x^{4}) (4)

    Kind regards

    \chi \sigma
    Incase the OP was wondering the MEANING of the "Big Oh" notation, to use what \displaystyle \chi\sigma so kindly posted...

    \displaystyle e^{\sin{x}} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O\left(x^4\right)

    The \displaystyle O\left(x^4\right) in this example means that when you are evaluating this function for values close to \displaystyle x=0 (as you would with a MacLaurin Series - if not, you would use a Taylor Series centred around a point closer to the point you are evaluating the function at), the absolute error (remainder) is no greater than some constant multiple of \displaystyle |x^4|.
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