# Taylor expansion with non integer powers

I am not sure how to deal with non integer powers in the following problem: I am asked to find the truncated Taylor expansion of order $5$ (around $0$) of:

$1+x^5(2+x^{1/2})^3$

This already doesn't make sense to me, because in some sense this is already a polynomial, right? Why would it need a taylor expansion?

My progress so far is simply expanding the expression to:

$1+x^5(8+12x^{1/2}+6x+x^3)$

• A polynomial is something of the form $\sum_{n=1}^N a_n x^n$, i.e. all the powers are integers. – Benjamin Lindqvist Jan 1 '16 at 20:04
• No, this is not a polynomial. As you write yourself in the title, there are non-integer powers. And what you are referring to by 'Some sense' is not defined. – Thomas Jan 1 '16 at 20:05
• But, as you have already kind of noted, you only need to expand $x^{1/2}$, the rest is just multiplying polynomials. – Benjamin Lindqvist Jan 1 '16 at 20:06
• So that equals $1+8x^5+12x^{11/2}+6x^6+x^8$. The only part that's not a polynomial is the term $12x^{11/2}$. So if you just expand that part to the fifth polynomial and throw on the other terms of degree up to five you should have it. – Gregory Grant Jan 1 '16 at 20:06
• It represents the best polynomial approximation of whatever degree. – Benjamin Lindqvist Jan 1 '16 at 20:07

Then you may just write $$1+x^5(8+12x^{1/2}+6x+x^3)=1+8x^5+12x^{11/2}+6x^6+x^8=\color{red}{1+8x^5+o(x^5)}$$ as being the sought expansion.
Use asymptotic analysis: $(2+x^{1/2})^3=(2+o(1))^3=8+o(1)$, hence $$1+x^5(2+x^{1/2})^3=1+8x^5+x^5o(1)=1+8x^5+o(x^5).$$