Consider the power series $\Sigma_{n=0}^{\infty}c_n(x-a)^n$.It is required to prove that it converges at $x=a$. Here is a link that attempts at this.
My issue is the appearance of $0^0$ and then the conclusion that it equals $1$. I always thought it was undefined.
What now? There might be other proofs of convergence of this power series, but isn't this result obtained,well,undefined? Does this proof need some fixing?
Hints are fine. Thanks in advance!
Asked
Active
Viewed 27 times
2
Ambica Govind
- 713
-
This may be helpful! – Mikasa Jan 21 '22 at 14:54
-
Ah this is rather a matter of convention. Usually, of course, this is undefined at $x = a$. But You can just define that $0^0= 1$ in that case such that everything works out for every value. Similar thing is done in measure theory where one defines $\infty \cdot 0 = 0$. In general this of course makes no sense but in some NARROW SCENARIOS this type of convention is acceptable. – Hyperbolic PDE friend Jan 21 '22 at 14:55
-
It will be a duplicate, but $0^0$ is generally taken to be $1$, especially in power series such as $e^x=\sum\limits_{n=0}^\infty \frac{x^n}{n!}$ and binomial expansions $(a+b)^n =\sum\limits_{k=0}^n {n \choose k}a^kb^{n-k}$ and combinatorial uses – Henry Jan 21 '22 at 14:55
1 Answers
0
Keep in mind that
$$\sum_{n=0}^{+\infty} c_n (x-a)^n$$
is a notation that gathers the sum $c_0 + c_1 (x-a) + c_2 (x-a)^2 + ...$ to avoid the iterative writing of terms and the "...". So:
$$\sum_{n=0}^{+\infty} c_n (x-a)^n = c_0 + c_1 (x-a) + c_2 (x-a)^2 + ...$$
So, there's no $0^0$ indetermination in power series, it's just a matter of notation to make a compact sum. A power series always converges at its center $x=a$.
Luciano
- 426