<s>[INST] Would the order of Taylor Polynomial change after substitution? I found the order of Taylor Polynomial is kind of confusing. 
For example, we know:
$$T_4e^x = 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \frac {x^4} {4!}$$
After substitute $x$ as $t^2$, we would have:
$$T_4e^{t^2} = \underbrace{1 + t^2 + \frac {t^4} {2!} + \frac {t^6} {3!} + \frac {t^8} {4!}}_\text{It seems to be 8th order?}$$
or:
$$T_4e^{t^2} = 1 + t^2 + \frac {t^4} {2!}$$
Could anyone tell me the relationship between the order of Taylor Polynomials and the degree of variables inside the polynomial?
In this case, which $T_4e^{t^2}$ is correct?
 [/INST] You mistake is a quite common one to make and stems from thinking $T_4 e^x$ can be though of as a function of $x$. When writing
$$T_4 e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}$$
we see that we have $x$ on both sides so it looks like we could just substitute $x=t^2$ to get
$$T_4 e^{t^2} = 1 + t^2 + \frac{t^4}{2!} + \frac{t^6}{3!} + \frac{t^8}{4!}$$
This is not correct and the reason is that $T_4$ is not a function, but rather a functional that takes in a function $e^x$ and outputs a polynomial of degree $4$.
What is true however, and which you can show with some work, is that $[T_n f(x)]_{x = t^2} = T_{2n} f(x^2)$ so substituing $t=x^2$ in a Taylor polynomial of $f(x)$ (centered at zero) gives the Taylor polynomial of $f(x^2)$ to double degree.
</s>