<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] When you square the right-hand side, you will get another (infinite) polynomial $a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots$ . Now go through the coefficients $a_i$ and ask yourself which terms in the product $\big ( 1+ {1 \over 2} x - {1 \over 8} x^2 + \ldots\big) \big ( 1+ {1 \over 2} x - {1 \over 8} x^2 + \ldots\big)$ will contribute to $a_i$.
For example, there is only one product that contributes to $a_0$, namely $1 \cdot 1$, since the degree of $x$ in every other term is at least 1. Next, to get $a_1$, you can multiply $1$ from the first factor with ${1 \over 2} x$ from the second factor, or ${1 \over 2} x$ from the first and $1$ from the second. Thus $a_1 = 1\cdot {1 \over 2} + {1 \over 2} \cdot 1 = 1$. Continue similarly for the rest.
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