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

[Taiga]

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 - x³/6 + x⁵B(x)
and
arctan(x) = x - x³/3 + x⁵C(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 - x³/3 + x⁵C(x) ) - ( x - x³/3 + x⁵C(x) )³ / 6 + ( x - x³/3 + x⁵C(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 - (x³)/3 - (x³)/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!

 

[HallsofIvy]

I don't see why you are keeping the "B" and "C" terms since they have degree higher than 3 and so are irrelevant to the third order Taylor's polynomial.

As you say \(\displaystyle sin(x)= x- x^3/6+ ...\) where the "..." indicates terms of degree higher than 3 and \(\displaystyle arctan(x)= x- x^3/3+ ...\).

So \(\displaystyle sin(arctan(x))= (x- x^3/3)- (x- x^3/3)^3/6\)

Now, \(\displaystyle (x- x^3/3)^2= x^2- 2(x)(x^3/3)+ (x^3/3)^2= x^2\) "plus terms of degree higher than 3".

So \(\displaystyle (x- x^3/3)^3= x^2(x- x^3/3)= x^3\) "plus terms of degree higher than 3".

\(\displaystyle (x- x^3/3)^3/6= x^3/6\).

\(\displaystyle sin(arctan(x)= x- x^3/3- x^3/6= x- (3/6)x^3= x- (1/2)x^3\) "plus terms of degree higher than 3".