How to change the value of $a$ in a Taylor series yet still get the correct approximation?

Question

Set up

For example, if $f(x) = x^3 + 4x + 6$, then the second degree Taylor polynomial is

$$ P_2(a) = (a^3 + 4a + 6) + (3a + 4)( x - a) + \left(\frac{6a}{2!}(x - a)^2\right) $$

If I want to approximate $f(1.2)$ then I can use $P_2 (1.2)$.

You can see that this is close by graphing the functions, and noting that about $f(1.2)$ they're very close.

Question

From other questions I get the idea that I'm able to change the value of $a$ , by setting it to zero or whatever, and still approximate the function as needed.

But looking at this example, I don't know how I could change the value of $a$ , and still approximate the original function at $x = 1.2$.

Sometimes people say that you're able to change the value which the function is being evaluated about, but I don't understand how this is so.

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definitions

Taylors formula is $f(x) = P_n(x) + R_n(x)$ where $P_n(x)$ is

\begin{equation} \begin{aligned} P_n(x) = f(a) + \frac{f'(a)}{1!}(x - a) + \frac{f''(a)}{2!}(x - a)^2 & + \ldots + \frac{f^{(n)}(a)}{n!}(x - a)^n \\ \end{aligned} \end{equation}

And $R_n (x) $ is (where $\xi$ is between $a$ and $x$) \begin{equation} \begin{aligned} R_n(x) = \frac{f^{(n + 1)}(\xi)}{(n + 1)!}(x - a)^{(n + 1)} \end{aligned} \end{equation}

Answer 1

I think you've gotten a little confused between your $x$ and your $a$. The Taylor series is a function of $x$, parameterised by $a$. So, for example, the 2nd-order Taylor polynomial of your cubic is:

$$T_{2;a}(x) = (a^3 + 4a + 6) + (x - a)(3a^2 + 4) + 6a\frac{(x - a)^2}{2}$$

If we evaluate this series at $a = 0$, we get the (linear) polynomial:

$$\begin{eqnarray}T_{2;0}(x) & = & (0^3 + 4\cdot0 + 6) + (x - 0)(3\cdot0^2 + 4) + 6\cdot 0 \frac{(x - a)^2}{2} \\ & = & 6 + 4x\end{eqnarray}$$

Using this function, the approximation for $f(1.2)$ is $T_{2;0}(1.2) = 6 + 4\cdot 1.2 = 10.8$.

If we evaluate the series at $a = 1$, we get the polynomial:

$$\begin{eqnarray}T_{2;1}(x) & = & (1^3 + 4 \cdot 1 + 6) + (x - 1)(3 \cdot 1^2 + 4) + 6 \cdot 1 \frac{(x - 1)^2}{2} \\ & = & 11 + 7(x - 1) + 3(x - 1)^2 \\ & = & 7 + x + 3x^2\end{eqnarray}$$

Using this function, the approximation for $f(1.2)$ is $T_{2;1}(1.2) = 11 + 7(0.2) + 3(0.2^2) = 12.52$. Note that this is a lot closer to the true value of $f(1.2) = 12.528$.

If we take $a = 1.2$, then we'll get a Taylor polynomial that is exactly right for $f(1.2)$, but whose approximation gets worse further away from it.

Answer 2

So, to resume my comment and your further question to it, let me try and fix the matter, just from the conceptual point of view. For rigorous details you can refer to any of the many sources available on the topic.

You have a generic function $f(x)$, be it a polynomial or not.
Suppose there is a range (domain) of $x$ where the function is defined and differentiable whichever number of times.

Fix there a value for $x$, $x=a$ (or $x=x_0$ as it is usually written).
$a$ can be fixed wherever (in the said domain) you like/need, in the limit you can go to $\pm\infty$ (you'll probably learn about in the future).

Then you can calculate $f(a)$, and you can say (as the roughest approximation) that "for $x$ very near to $a$" $f(x)\approx f(a)$: you are approximating $f(x)$ with a polynomial of degree $0$ (a constant), at $x=a$. The error will be null at $x=a$, but at $x=b$ it will be $f(b)-f(a)$. That will be as great as much $f(x)$ "deviates" from being constant (at $x=a$).

The situation does not conceptually change if you approximate $f(x)$ with a polynomial of higher degree. If you approximate it in $x=a$ you will have some error at $x=b$, unless of course the approximation be a polynomial of same degree as $f(x)$ (if it is a polynomial).