<s>[INST] Taylor expansion square Consider the following expansion $$\sqrt{1+x} = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \dfrac{1}{16}x^3 .. $$

Show this equation holds by squaring both sides and comparing terms up to $x^3$.

I wonder, how can I square the right hand side? 
 [/INST] You can put $t=\frac 12 x-\frac 18 x^2 + \frac 1{16}x^3$ then, you get:
$$\sqrt{\sqrt{1+x}}=\sqrt{1+t}=1+ \frac 12 t--\frac 18 [t^2]_3 + \frac 1{16}[t^3]_3 ... $$
where $[P(x)]_3$ for a polynom $$P(x)=a_0+a_1 x +a_2x^2+a_3x^3+a_4x^4 ... +a_n x^n $$ is:
$$[P(x)]_3=a_0+a_1x+a_2x^2+a_3x^3.$$
For example , since: $t=x(\frac 12 -\frac 18 x + \frac 1{16}x^2)$  we have :
$$t^2=x^2(\frac 12 -\frac 18 x + \frac 1{16}x^2)^2$$ then : $$[t^2]_3=x^2(\frac 14 -\frac 18 x)$$
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