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

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!

• The others also conform to that form, only $a=0$. Apr 1 '19 at 19:10
• 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! 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 $$a$$.
• 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! Apr 1 '19 at 20:05