Taylor/McLaurin series of $f(x)=(\sqrt{1-2x+x^3}-\sqrt[3]{1-3x+x^2})\cos\pi x$

Question

The function is given as:

$$f(x)=(\sqrt{1-2x+x^3}-\sqrt[3]{1-3x+x^2})\cos\pi x$$

And the problem states:

a) Find the McLaurin expansion up to order of $x^3$

b) Find the Taylor expansion when $x\to2$ up to order of $x^2$.

My question is:

Is there a smarter (more efficient) way to calculate this other then just applying the formula. I'm asking because finding the first three derivatives of the function is pretty complicated so it makes me think that there must be a way to simplify this problem.

Any ideas?

Thanks

Answer 1

Hint:

You can make algebraic operations with Taylor's expansions. All you have to do is to truncate the powers of $x$ which are beyond the given order. For instance, at order $3$:

• $\sqrt{1+u}=1+\frac12u-\frac18u^2+\frac 1{16}u^3+o(u^3),\;$ so \begin{align} \sqrt{1+\underbrace{-2x+x^3}_u}&=1+\tfrac12(-2x+x^3)-\tfrac18(-2x+x^3)^2(\!\!\bmod x^3)+\frac 1{16}(-2x+x^3)^3(\!\!\bmod x^3)+o(x^3) \\ &=1-x+\frac12x^3-\tfrac18(4x^2)+\frac 1{16}(8x^3)+o(x^3)\\ &=1-x -\tfrac12x^2+x^3+o(x^3) \end{align} You can do the same with
• $\sqrt[3]{1+u}=1+\frac13u-\frac19u^2+\frac 5{85}u^3+o(u^3),\;$ and $\;u=-3x+x^2$.

Answer 2

Yes, you can find the series separately and add and multiply the results to get the combined series for the whole function.

Of course you do not need the whole series for any of the functions just enough terms to give you up to third degree.