why Pi is transcendental number if it will satisfy the algebraic equation as "n" tends to infinity?

Question

why Pi is transcendental number if $\pi$ also have algebraic equation like below which have root at $x =\pi/3$ as $n$ tends to infinity. $$\Biggl(\biggl(\Bigl(\bigl((x^2-2^{(n2^1+1)})^2-2^{(n2^2+1)}\bigr)^2-2^{(n2^3+1)}\Bigr)^2...-2^{(n2^{(n-1)}+1)}\biggr)^2-2^{n2^{n}+1}\Biggr)^2 = 3*2^{n2^{(n+1)}}$$

For $n = 1$ equation will be $$(x^2-2^3)^2 = 3*2^4$$ and its root $x_1$ = 1.03527618041.

For $n = 2$ equation will be $$((x^2-2^5)^2-2^9)^2 = 3*2^{16}$$ and its root is $x_2$ = 1.04420953776041 and as we increase the value of n the root tends to $\pi/3$.

For n = 12 ,$3x_{12}$ will be 3.14159264 which is some what close to $\pi$.

Another simplified version of this equation is $${\Biggl(\biggl(\Bigl(\bigl(\frac{\pi}{3*2^n}\bigr)^2-2\Bigr)^2-2\biggr)^2...-2\Biggr)^2} = 3$$ Here n is the same as number of single powered two in above equation.

Solution of the above equation will be $$\pi = 3*\Biggl(\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+....\sqrt{2+\sqrt{3}}}}}}\Biggr)*2^n$$ and $$n =\text{ number of $2$ inside largest square root}$$ So if $\pi$ can also be represented as a root of algebraic equation then why $\pi$ is transcendental .

Answer 1

Algebra for the most part is not concerned with sequences and limits. A real or complex number is said to be algebraic if it is a root of a single polynomial equation of finite degree with integer coefficients. This is something that $\pi$ is not.

If you are willing to use limits, then every number is "algebraic" : you can always find a sequence of polynomial equations with integer coefficients whose roots converge to any number. For real numbers, the polynomials can even be linear! So in a sense this is not interesting. The specific equations you outline are interesting, but the conclusion is not as it is true for any number.

Answer 2

All you've shown is that $\pi$ can be approximated by algebraic numbers. But that's not the same as being an algebraic number! When you write

$\pi$ can also be represented as a root of algebraic equation

you're misusing the term "algebraic equation." There is a sense in which $\pi$ can be thought of as the solution to an "infinite algebraic equation," but that's not the same thing.

Indeed, the same reasoning would also imply that $\sqrt{2}$ is rational since we can get better and better rational approximations to it (this is expanding on Rijul Saini's comment above). Or that the infinite sum $\sum_{i=1}^\infty 1=1+1+1+1+ ...$ is finite since for each $n$, the partial sum $\sum_{i=1}^n1$ is finite.

In general, you cannot conflate finite objects with their infinite analogues.

---

This is not to say that there's nothing interesting here. While every real number is "limit algebraic," things get more interesting when we look at functions - thinking about "infinite-degree polynomials" leads to the concept of analytic functions, which is absolutely fundamental. But that's a separate issue.