I am trying to understand Taylor Series Expansion. I tried to used it to approximate a function say $f(x)$. So I know that based on my understanding the $f(x)$ can be written as:

$f(x) =\sum_{n=0}^{\infty} \frac{f^n(c)}{n!} (x-c)^n$

They call it expansion of the function "Around c". My question is how do we choose where to expand "around"? Do we usually expand at c=0? When do we expand at c=0? how about expand at c=1? and how about expand at c=-10, or c = 88 or c=7321838?

Because I don't quite understand the way of choosing the c. It says c is the the point where we want to start approximating from.

I am asking this question, because I was reading something about the moment -generating function which is denoted as $M_z(\frac{s}{\sqrt{n}}) = E(e^{z\frac{s}{\sqrt{n}}})$ and the author says oh we will perform a Taylor Series Expansion of the moment-generating function (i.e. $M_z(\frac{s}{\sqrt{n}}) = E(e^{z\frac{s}{\sqrt{n}}})$ ) around s = 0 .

But I don't understand why s = 0? just like I don't understand why expanding "around 0" or "expanding around c = 382931" or "expanding around c= -5".

Could someone kindly explains.

  • $\begingroup$ Your formula is wrong. It must be $f^n(c)$ $\endgroup$
    – GReyes
    Dec 7, 2019 at 8:19
  • $\begingroup$ Also, you must start from $n=0$. $\endgroup$
    – GReyes
    Dec 7, 2019 at 8:22
  • $\begingroup$ it has been corrected now, thanks for pointing out the error. But would be great if someone could explains. $\endgroup$
    – john_w
    Dec 7, 2019 at 8:24
  • 3
    $\begingroup$ Typically, you center the series at a point where you know the value of the function and its derivatives, and you want to estimate the value of your function at a nearby point. Say, for linear approximation you take $f(x)\sim f(c)+f'(c)(x-c)$. You can use this to find approximately $\sqrt{82}$ knowing the exact value of $\sqrt{81}=9$. In the case of the moment generating function, you expand around zero because you want to link the function to the momenta of your random variable. $\endgroup$
    – GReyes
    Dec 7, 2019 at 8:32

1 Answer 1


Say, you want to expand $x^2$ in terms of $x$. Not that it requires serious treatment, but just as a simple example, it would require the above formula to fit a polynomial on the curve of $x^2$ vs. $x$. To create the polynomial, you need to choose c first. Should you choose to keep $c=0$, you will end up getting $$x^2=x^2$$ as the expansion, and putting $x=a$ would give you value $x^2=a^2$. Should you choose $c=a$, the expansion will look like $$x^2=a^2+2a(x-a)+(x-a)^2$$which eventually is the same thing for it is $(x-a+a)^2=x^2$. But you see an easy-to-discern answer of what the value of the function at $x=a$ will be (This is a simple function, but sometimes you will see so). Sometimes, you want to evaluate the value of a function near a point at which you know its value. Suppose, in this case; you had to find the value at $x=0.0...01$(you know the value at $x=0$), then you would want to use expansion corresponding to $c=0$. If you had to do it for $x=1.00...1$, then $c=1$ is more tempting. In both these cases, you could neglect the squared terms for convenience. I repeat that it is a simple example, and everything above makes sense for more complicated ones.


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