- #1

Taiga

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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!