Approximating pi using the $\int_0^1\sqrt{1-x^2}dx$ The first step would be to find the series expansion for $\sqrt{1-x^2}$ using the binomial theorem
$$\sqrt{1-x^2} = \sum_{n=0}^\infty {1/2\choose n}(-x^2)^n $$
Expanding and simplifying the first terms
$$\sqrt{1-x^2} = 1 -\left(\frac{1\cdot x^2}{2 \cdot1!} + \frac{1 \cdot 3 \cdot x^4}{2^2 \cdot 2! } + \frac{1 \cdot 3 \cdot 5 \cdot x^6}{2^3 \cdot 3!} +... \right)$$
Here I run into a problem. How come all the terms are negative while $1$ isn't? This would mean I can't add it to the infinite sum. So continuing:
$$\sqrt{1-x^2} = 1- \sum_{n=0}^\infty \frac{(2n+1)!!}{2^n n!}x^{2n} = 1-\sum_{n=0}^\infty \frac{(2n+1)!!}{(2n)!!}x^{2n} $$
When I plot this on desmos my approximation is a) shifted by $1$ unit down b) inaccurate. On the image below I have used 50 terms so I doubt the error is due to a lack of terms.

Where have I made a mistake. Thank you

For $n\ge2$,$$[x^{2n}]\sqrt{1-x^2}=(-1)^n\binom{1/2}{n}=(-1)^n\frac{1}{n!}\frac12\prod_{k=1}^{n-1}(\tfrac12-k)=-\frac{(2n-3)!!}{n!2^n},$$so$$\sqrt{1-x^2}=1-\tfrac12x^2-\sum_{n\ge2}\frac{(2n-3)!!}{n!2^n}x^{2n}.$$See how that looks in Desmos.
