Why doesn't the $x$ term in a Taylor polynomial (or series) change? What is the reasoning behind this?

Question

Why doesn't the $x$ term in a Taylor polynomial (or series) change? What is the reasoning behind this?

I am referring to $x$ in the following Taylor polynomial (or series) about $a$.

$$f(a) + f'(a)(x - a) + \dfrac{ f''(a)(x - a)^2 }{ 2! } + \dfrac{ f'''(a)(x - a)^3}{3!} + ... + \dfrac{ f^{n}(a)(x - a)^n }{ n! }$$

I am aware of how the Taylor polynomial is derived. However, when we have other polynomials, such as the quadratic $f(x) = x^2 + 4x + 2$, $ f$ changes depending on $x$. This is different to the Taylor polynomial, where $x$ does not seem to change at all! Rather, we have

$$f(x_0) + f'(x_0)(x - x_0) + \dfrac{ f''(x_0)(x - x_0)^2 }{ 2! } + \dfrac{ f'''(x_0)(x - x_0)^3}{3!} + ... + \dfrac{ f^{n}(x_0)(x - x_0)^n }{ n! }$$

Why is this the case? What is the reasoning behind it?

Thank you.

Answer 1

The user qbert's post allowed me to figure out what my confusion was. My confusion came from the fact that the examples I had dealt with were all functions of $x$ alone ( $f(x)$ ). Let us find the value of $\sin(x)$ using the Taylor polynomial of $\sin(x)$ about $0$.

$$\sin(x) = \sin(0) + \cos(0)(x - 0) - \dfrac{ \sin(0) }{ 2! } (x - 0)^2 - \dfrac{ \cos(0) }{ 3! } (x - 0)^3 + \dfrac{ f^{(4)}(c) }{ 4! } (x - 0)^4 $$

Where the term $\dfrac{ f^{(4)}(c) }{ 4! } (x - 0)^4$ is the Lagrange remainder at some point $c \in (0, x)$.

$$ \Rightarrow \sin(x) = x - \dfrac{ x^3 }{ 3! } + \dfrac{ f^{4}(c) }{ 4! } x^4$$

However, if we were to find the value of $\sin(2x)$ using the Taylor polynomial of $\sin(2x)$ about $0$, we would get the following.

$$\sin(2x) = \sin(0) + \cos(0)(2x - 0) - \dfrac{ \sin(0) }{ 2! } (2x - 0)^2 - \dfrac{ \cos(0) }{ 3! } (2x - 0)^3 + \dfrac{ f^{(4)}(c) }{ 4! } (2x - 0)^4 $$

Where the term $\dfrac{ f^{(4)}(c) }{ 4! } (2x - 0)^4$ is the Lagrange remainder at some point $c \in (0, x)$.

$$ \Rightarrow \sin(2x) = 2x - \dfrac{ (2x)^3 }{ 3! } + \dfrac{ f^{4}(c) }{ 4! } (2x)^4$$

Answer 2

Although I am not sure I understand your question, perhaps this is confusion over what Taylor's theorem is actually saying.

Taylor's theorem gives you an approximation for a function in terms of a polynomial function. Of course this approximation may give you different outputs depending on the x input, just like any function.

So perhaps the source of confusion comes from saying $f(x)$ is approximately equal to the polynomial $P(x)$ and $f(2)$ is roughly $P(2)$.