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.

enter image description here


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.


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}

  • $\begingroup$ Try centering at $a = 1$ to increase the accuracy near $x = 1$. $\endgroup$ Commented Jul 23, 2017 at 22:17
  • $\begingroup$ @AustinMohr thanks, but i'm not sure that I follow. Are you saying that if I want to approximate the value of the original function at $x = 1$ (so $f(1)$), then I should use $a = 1$ , ? $\endgroup$
    – baxx
    Commented Jul 23, 2017 at 22:23
  • 1
    $\begingroup$ Taylor polynomial is the polynomial that best approximates f(x) at x=a, but as much as you get far from x=a, the approximation gets worst. In your example you are approximating a polynomial of degree 3 (which goes to $-\infty$ to the left, and to $+\infty$ to the right) with a polynomial of degree 2 (which goes to $+\infty$ on either side) ... $\endgroup$
    – G Cab
    Commented Jul 23, 2017 at 22:37
  • $\begingroup$ @GCab yes , in the example that I've given I'm using a quadratic to approximate a cubic (maybe a bad example!), but what I'm not sure about is that I've seen and heard about the value of $a$ being altered so as to make the calculations easier and such. I'm sorry that I'm being vague - so are you saying that in fact the value of a cannot be changed, that if I want to find $f(1.2)$, then I must set $a = 1.2$? thanks $\endgroup$
    – baxx
    Commented Jul 23, 2017 at 23:01
  • $\begingroup$ I posted an answer to try and clear your doubts: wish it succeeds to. $\endgroup$
    – G Cab
    Commented Jul 24, 2017 at 12:25

2 Answers 2


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.

  • $\begingroup$ Thanks. So if I evaluate the series at $a = 1$, that will cancel out some terms, but it will also leave another function (in that, there will be the terms $6 + 4x$ remaining)? So we can evaluate that the Taylor Polynomial (i'm not saying series as I haven't done those yet!) at $ a = 1$, and by doing that we're getting a function, not a value. Then, we can evaluate the returned function at a particular value of $x$, and that will approximate the original function at that value. Does that make sense? $\endgroup$
    – baxx
    Commented Jul 23, 2017 at 23:32
  • $\begingroup$ I think you're getting it, although your first $a = 1$ should be $a = 0$. The Taylor series is just what happens when you just keep adding terms - it's kind of the $\infty$-degree Taylor polynomial. $\endgroup$
    – ConMan
    Commented Jul 23, 2017 at 23:36
  • $\begingroup$ OK - and the way that would be expressed would be as evaluating "*around $a = 0$", ? At zero, this is the exact value of the function, and (as can be seen on a graph or whatever) the accuracy reduces as we move further away. I thought that the way that the approximation was reduced was by decreasing $n$ - but it seems that we're using $a$ to increase / decrease the accuracy of the approximation here? $\endgroup$
    – baxx
    Commented Jul 24, 2017 at 0:27
  • $\begingroup$ Yes, it's evaluating around $a = 0$. It only improves the accuracy close to $x = a$ - as good as $a = 1$ is for evaluating $f(1.2)$, it's not as good for evaluating $f(-0.2)$. $\endgroup$
    – ConMan
    Commented Jul 24, 2017 at 0:40

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).

  • $\begingroup$ thanks for following up - much appreciated. I was getting confused / mixed up with how the value of $a$ could be moved, yet we would approximate a different value of $x$ from that $a$. However, I see that if we take a value of $a$ that's very close to the value of $x$ that we want to approximate, we can use a lower degree approximation and still be quite accurate. As we move $a$ further away from $x$ we need greater degrees, so that the information about the functions curve it preserved. Thinking about concavity past the second degree isn't clear, but I can 'believe' it more now. Cheers $\endgroup$
    – baxx
    Commented Jul 24, 2017 at 12:38
  • $\begingroup$ not always a greeter degree gives better approximation! it gives for sure at infinitesimal distance, but if you go at finite (although small) distance, you might have better approximation on the right and worst on the left ! In your sketch, add also the tangent line and see what happens vs. the second deg. you already have. $\endgroup$
    – G Cab
    Commented Jul 24, 2017 at 13:38

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