# Taylor Series expansion of a function around a point but what point

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.

• Your formula is wrong. It must be $f^n(c)$ Dec 7, 2019 at 8:19
• Also, you must start from $n=0$. Dec 7, 2019 at 8:22
• it has been corrected now, thanks for pointing out the error. But would be great if someone could explains. Dec 7, 2019 at 8:24
• 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. Dec 7, 2019 at 8:32

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.