B Identity Theorem for power series

[Hill]

Consider this proof:

Is it a valid proof?

When we divide by $z$, we assume that $z \neq 0$. So, we cannot put $z=0$ on the next step. IOW, after dividing by $z$ we only know that $$c_1+c_2z+c_3z^2+...=d_1+d_2z+d_3z^2+...$$ in a neighborhood of $0$ excluding $0$.

 

[fresh_42]

We know that $\sum_{j=0}^\infty c_jz^j=\sum_{j=0}^\infty d_jz^j$ for all $z\in U(0)$ in a neighrhood of $0.$ By the argument you posted, we see that $c_0=d_0$ and therefore that
$$
\sum_{j=1}^\infty c_jz^{j-1}=\sum_{j=1}^\infty d_jz^{j-1}\quad \forall\,z\in U(0)-\{0\}
$$
Therefore,
$$
c_1=\lim_{z \to 0}\sum_{j=1}^\infty c_jz^{j-1}=\lim_{z \to 0} \sum_{j=1}^\infty d_jz^{j-1}=d_1\quad \forall\,z\in U(0)-\{0\}
$$

 

[Hill]

We know that $\sum_{j=0}^\infty c_jz^j=\sum_{j=0}^\infty d_jz^j$ for all $z\in U(0)$ in a neighrhood of $0.$ By the argument you posted, we see that $c_0=d_0$ and therefore that
$$
\sum_{j=1}^\infty c_jz^{j-1}=\sum_{j=1}^\infty d_jz^{j-1}\quad \forall\,z\in U(0)-\{0\}
$$
Therefore,
$$
c_1=\lim_{z \to 0}\sum_{j=1}^\infty c_jz^{j-1}=\lim_{z \to 0} \sum_{j=1}^\infty d_jz^{j-1}=d_1\quad \forall\,z\in U(0)-\{0\}
$$

Yes, of course. I don't doubt the theorem. I question the quoted proof.

 

[fresh_42]

Yes, of course. I don't doubt the theorem. I question the quoted proof.

I think it is only sloppy. My argument with the limits works because the functions are analytical (per construction) and therefore continuous, and the equation holds in $\{0\},$ too, because it is a limit point.

 

[Hill]

I think it is only sloppy. My argument with the limits works because the functions are analytical (per construction) and therefore continuous, and the equation holds in $\{0\},$ too, because it is a limit point.

Your argument works, undoubtfully. The quoted proof, OTOH, would fail a high school exam, I think.

 

[Hill]

The theorem holds even if the two series are equal only on an infinite sequence of points converging to $0$.

 

[DaveE]

Math problems with the word "neighborhood" make my brain instantly think of limits and ε-δ arguments. It is common, albeit sometimes sloppy, shorthand for people to say $z=0$ when they really mean $\lim_{z \rightarrow 0}$.

 

[Hill]

I agree, but in this case the author seems to mean actually putting $z=0$, because a bit later, when talking about strengthening this result, he says,

The verification is essentially the same, only instead of putting z = 0, we now take the limit as z approaches 0, either along the segment of curve or through the sequence of points.

 

[FactChecker]

Consider this proof:
View attachment 344552
Is it a valid proof?

When we divide by $z$, we assume that $z \neq 0$. So, we cannot put $z=0$ on the next step. IOW, after dividing by $z$ we only know that $$c_1+c_2z+c_3z^2+...=d_1+d_2z+d_3z^2+...$$ in a neighborhood of $0$ excluding $0$.

The fact is true, so any "proof", no matter how weak, might seem valid. But that "proof" never uses their equality in the neighborhood, so you should be skeptical. Try applying that "proof" to $f(z) \equiv 0$ and $g(z) = z$ at $z=0$.

 

[PeroK]

Just to add something simple but useful. The proposition is identical to saying that if, for all $z$ in a neighbourhood of 0, we have:
$$c_0 + c_1z + c_2z^2 +c_3z^3 \dots = 0$$Then all the coefficients are zero.

And, as any neighbourhood of zero contains a real, open interval of zero, then the result follows immediately from the real case.

 

[Erland]

We know that $\sum_{j=0}^\infty c_jz^j=\sum_{j=0}^\infty d_jz^j$ for all $z\in U(0)$ in a neighrhood of $0.$ By the argument you posted, we see that $c_0=d_0$ and therefore that
$$
\sum_{j=1}^\infty c_jz^{j-1}=\sum_{j=1}^\infty d_jz^{j-1}\quad \forall\,z\in U(0)-\{0\}
$$
Therefore,
$$
c_1=\lim_{z \to 0}\sum_{j=1}^\infty c_jz^{j-1}=\lim_{z \to 0} \sum_{j=1}^\infty d_jz^{j-1}=d_1\quad \forall\,z\in U(0)-\{0\}
$$

For this to work, the functions defined by the power series must be continuous at $0$, as you pointed out in a later post. This continuity needs to be proved, and this is done by using the property that a power series converges uniformly on compact subsets inside its circle of convergence (or is there another way to prove that?). So to be complete, the OP's book should have dealt with these issues before the theorem quoted.