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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?

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    $\begingroup$ $T_4 e^{x}$ is the fourth order Taylor polynomial of the function $e^{x}$ while $\left[T_4 e^{x}\right]_{x=t^2}$ is the eight order Taylor polynomial of the function $e^{t^2}$. These are two different functions. $\endgroup$
    – Winther
    Jan 12, 2016 at 1:46
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    $\begingroup$ @winther So $1 + t^2 + \frac {t^4} {2!}$ is the fourth order of Taylor polynomial of function $e^{t^2}?$ In other words, in some cases, we have to change the order of Taylor polynomial after substitution? $\endgroup$
    – Jay Wang
    Jan 12, 2016 at 1:48
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    $\begingroup$ Yes it is. Note that if $f(x)$ is a polynomial of degree $n$ then $f(x^2)$ is a polynomial of degree $2n$. $\endgroup$
    – Winther
    Jan 12, 2016 at 1:53
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    $\begingroup$ Your confusion seems to stem from thinking that $[T_4 e^{t^2}]$ should be equal to $[T_4 e^x]_{x = t^2}$. This is not true and writing $T_4e^{t^2}$ as you have done when making the substitution is bad notation (and also wrong). If you substitute $x=t^2$ then you obtain the Taylor polynomial of double degree, here $2\cdot 4 = 8$. $\endgroup$
    – Winther
    Jan 12, 2016 at 2:05
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    $\begingroup$ I suppose writing $T_4(t\mapsto e^{t^2})$ would be too cumbersome, but some of the confusion appears to originate from "wanting" to apply functional operators to syntactic expressions, instead of the actual functions. When "substituting $t^2$ for $x$ in the expression," you actually change the function -- composing it with the square. $\endgroup$
    – Clement C.
    Jan 12, 2016 at 2:06

1 Answer 1

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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.

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    $\begingroup$ It's worthy of note that one may also express the chain rule as $T_n((T_n f)\circ (T_n g))=T_n(f\circ g)$ in this notation, which is a nice looking expression, if you ask me. $\endgroup$
    – Milo Brandt
    Jan 12, 2016 at 2:36
  • $\begingroup$ Hey Winther, could you please give me a little bit hints over this question? $\endgroup$
    – Jay Wang
    Jan 18, 2016 at 21:28

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