What is the significance of the $-a$ in the definition of a power series: $c_n(x-a)^n$

Examples of power series that I've seen in my textbook:

$\sum_{n=0}^{\infty} x^{n}$

$\sum_{n=0}^{\infty} n!x^{n}$

$\sum_{n=0}^{\infty} (-1)^nx^{2n}/2^{2n}(n!)^2$

$\sum_{n=0}^{\infty} (-3)^nx^{n}/\sqrt{n+1}$

$\sum_{n=0}^{\infty} n(x+2)^{n}/3^{n+1}$

The only one that conforms to the equation of $c_n(x-a)^n$ is the last one, in which I suppose $a=-2$ and $c_n=n/3^{n+1}$

Are those the correct values of $a$ and $c_n$ for the last series I gave?

Also, why do all the other series seem to not take on the form of $c_n(x-a)^n$ in that there is no $-a$ term within brackets along with the $x$ (it's usually just the $x$)?

I just don't think I'm understanding what a power series is on a conceptual level and I think my confusion stems from this equation. If anyone could help me better understand this I'd appreciate it a lot!

  • 6
    $\begingroup$ The others also conform to that form, only $a=0$. $\endgroup$
    – Arthur
    Apr 1 '19 at 19:10
  • $\begingroup$ Ahh I see, don't know why I didn't think of that. It's strange that the $-a$ would be included in the definition despite the majority of power series not making use of that aspect, but I suppose it makes the formula more all-encompassing. Thank you! $\endgroup$ Apr 1 '19 at 20:06

Think back to middle school or high school algebra class. Imagine I have some function $f(x)$, say $f(x)=x^2$. How do I shift the graph of function to the right by $5$? I need to do $g(x)=f(x-5)=(x-5)^2$. It's counterintuitive that the function moves right when we subtract $5$, but you can check that this is correct: the new graph's value at $x=5$ should be the old graph's value at $x=0$, and indeed $g(5)=f(5-5)=f(0)$. Similarly, if I want to shift the function left by $5$, I should do $h(x)=f(x+5)=(x+5)^2$. Of course there is nothing special about the $f(x)$ that I chose or the value of $5$.

Given a power series $\sum_{n=0}^\infty c_n (x-a)^n$, the value of $a$ is called the center of the power series. When $a=0$, we say it is centered at zero. Based on the previous paragraph, we see that subtracting $a$ shifts the graph to the right by $a$. (If $a<0$, then the graph is shifted to the left by $-a$.)

So if your last example $\sum_{n=0}^\infty \frac{n(x+2)^n}{3^{n+1}}$ should look like the power seris $\sum_{n=0}^\infty \frac{nx^n}{3^{n+1}}$ but shifted to the left by $2$ units. You can see for yourself here: https://www.desmos.com/calculator/gznudxcn3h. You can try playing around with the value of $a$.

  • $\begingroup$ My god, thank you so much! I had no idea that it was really as simple the $-a$ affecting the graph the same way we learned in pre-calculus, but I suppose it makes sense since all power series are representing some function. Thank you! $\endgroup$ Apr 1 '19 at 20:05

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