Home » Blog » Find a polynomial of minimal degree such that sin (x – x2) = P(x) + o(x6)

Find a polynomial of minimal degree such that sin (x – x2) = P(x) + o(x6)

Find the polynomial P(x) of minimal degree such that

    \[ \sin (x - x^2) = P(x) + o(x^6) \qquad \text{as} \qquad x \to 0. \]

Using the Taylor expansion of \sin x we know as x \to 0 we have

    \begin{align*}  \sin (x-x^2) &= (x-x^2) - \frac{(x-x^2)^3}{3!} + \frac{(x-x^2)^5}{5!} + o(x^6) \\[9pt]  &= x - x^2 - \frac{x^3}{6} (1-x)^3 + \frac{x^5}{120} (1-x)^5 + o(x^6) \\[9pt]  &= x - x^2 - \frac{x^3}{6} (1 - 3x + 3x^2 - x^3) \\  & \qquad +\frac{x^5}{120} (1 - 5x + 10x^2 - 10x^3 + 5x^4 - x^5) + o(x^6) \\[9pt]  &= x - x^2 - \frac{x^3}{6} + \frac{x^4}{2} + \left( -\frac{1}{2} + \frac{1}{120} \right)x^5 + o(x^6) \\[9pt]  &= x - x^2 - \frac{x^3}{6} + \frac{x^4}{2} - \frac{59}{120} x^5 + \frac{x^6}{8} + o(x^6) \end{align*}

(This is where we see how nice o-notation can be. All of the terms in the polynomials larger than x^6 will get absorbed into the o(x^6). This simplifies computations tremendously when we don’t care about the higher order terms.) Therefore, we have

    \[ P(x) = x - x^2 - \frac{x^3}{6} + \frac{x^4}{2} - \frac{59}{120} x^5} + \frac{x^6}{8} + o(x^6). \]


  1. Anonymous says:

    Why does the expression inside the little o not change when you replace x in the sin expansion? Is it because the new sin expression approaches 0 as x approaches 0 both before and after the substituion?

    • vlfom says:

      The expression inside little o can be arbitrary and in this case the problem statement implied that we want to have a solution specifically with o(x). It could also be o(x-x^2) instead [and then we could stop the calculations at the first line as it’s clearly would be o((x-x^2)^6)].

    • Mohammad Azad says:

      This is also my problem with this approach. I think this topic is poorly explained in Apostol’s book and little o can mean a lot of things at the same time. Rant aside, I got the same answer by differentiating sin(x-x^2) six times (using maxima) and evaluating the Taylor polynomial of degree 6 which gives us o(x^6) without problems.

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