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Taylor Polynomial of 3rd order in 0 to f(x) = sin(arctan (x))

  1. Oct 19, 2014 #1
    The problem is as the title says. This is an example we went through during the lecture and therefore I have the solution. However there is a particular step in the solution which I do not understand.

    Using the Taylor series we will write sin(x) as:
    sin(x) = x - (x^3)/6 + (x^5)B(x)
    and
    arctan(x) = x - (x^3)/3 + (x^5)C(x)

    B and C I believe are functions restricted near 0.

    Anyway since our function is sin(arctan (x)) it is appropriate to insert arctan in every x in the sin(x) function. Therefore we get

    sin (arctan(x) ) = ( x - (x^3)/3 + (x^5)C(x) ) - ( x - (x^3)/3 + (x^5)C(x) )^3 / 6 + (( x - (x^3)/3 + (x^5)C(x) )^5) B(x)

    As I see it, it will take an enourmous amount of work to simplfy this polynom because we have long expressions with the power both 3 and 5: However in the lecture the lecturer says that the above is equal to...

    x - x^3)/3 - (x^3)/6 + (x^4)D(x), where D has the same def. as B and C have above.

    Anyway he does this in one single step. Could anyone please explain to me how he came to this conclusion? What am I not seeing? Is there a simple way to get to this conclusion? If so please demonstrate.

    Thank you!
     
  2. jcsd
  3. Oct 19, 2014 #2

    mathman

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    Science Advisor
    Gold Member

    You are interested only in the x and x3 terms, which comes out directly. He didn't need to expand the 3rd and 5th power expressions. The third power term is -x3/6 + higher order terms, while the fifth power gives only higher order.
     
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