I often run into problems when i consider entire functions as Some sort of " polynomial of infinite degree ". Yet I do not know why most of the time.

It seems intuitive. Apart from the exp function not having a zero ofcourse.

Here is an example.

Let $f(z)$ be an analytic function such that $| f(0) | > 0$ and $| f(1)| > 1$.

Now consider the arguments below that give a false proof.

Let $n$ be the " degree " of the polynomial of infinite degree. We take the limit as n Goes to infinity.

Let us try to find ( complex) solutions $r$ such that

$$|r| > 1$$

$$ f(r) = 0 $$

We argue as follows :

Thus, let us assume $|r|\ge 1$. Since $$ f(r) = a_0 + a_1r + \cdots + a_{n-1}r^{n-1} + r^n =0 $$ then $$ r = -\frac{a_0}{r^{n-1}} - \frac{a_1}{r^{n-2}} - \cdots - a_{n-1}. $$ By taking absolute values and using the triangular inequality, we obtain $$ |r| \le \frac{a_0}{|r|^{n-1}} + \frac{a_1}{|r|^{n-2}} + \cdots + |a_{n-1}|. $$ Finally, using that $|r|\ge 1$ (so that $\frac{1}{|r|}\le 1$), we obtain the required upper bound $$ |r| \le |a_0| + \cdots + |a_{n-1}|. $$

Since when $ n$ Goes to infinity then $n-1$ Goes to infinity as well and

$$ \infty >> |a_0| + \cdots + |a_{\infty}| >= |f(1)| > 1$$ $$| r | \le |a_0| + \cdots << \infty $$

But this would imply there is boundary on the zero's of $f(z)$. In fact in the case when $f(z)$ has infinitely many zero's they all belong within an absolute value ; there making the function no longer analytic !!

This is clearly a wrong proof.

But why ???

  • 2
    $\begingroup$ As to the second line, if you considered the zeroes of the Taylor polynomial of $e^x$, the real roots approach $-\infty$ as you approach $e^x$. $\endgroup$ – Simply Beautiful Art Dec 29 '16 at 0:58
  • 7
    $\begingroup$ As is so often the case, the error is that you have simply spoken of "letting $n$ go to infinity" without explaining what that actually means rigorously or how it would imply any of the things you are claiming. $\endgroup$ – Eric Wofsey Dec 29 '16 at 1:03
  • $\begingroup$ @Simple art : yes I am aware of that , but that kinda justifies Some limit processes to exp and its zero's but ofcourse makes all formula's for the number of zero's or the value of zero's etc wrong - since there is no finite one -. At best we get a division by 0 hinting at infinity ... $\endgroup$ – mick Dec 29 '16 at 1:16
  • 5
    $\begingroup$ I don't get it. Two people have given you very good reasons why this proof-sketch cannot work, and indeed, it can't work because the conclusion you draw is false. Why do you keep trying to explain why it's a good idea? Is there an actual question here that could have an answer that would satisfy you? A proof is wrong because one step does not follow from another, or because the logic has holes in it, so you've gotten two good answers to the question you asked. So again: is there some sort of answer that would satisfy you? $\endgroup$ – John Hughes Dec 29 '16 at 1:50
  • 2
    $\begingroup$ It is unclear that what statement one is proving in the post. $\endgroup$ – Jack Dec 29 '16 at 2:03

You have started by assuming $r$ is a zero of the polynomial $$g_n(z)=a_0 + a_1z + \cdots + a_{n-1}z^{n-1} + z^n.$$ You prove some statements that are true about any such root. Then you "take the limit as $n\to\infty$", and seem to be concluding that similar statements must be true of any zero of the function $f(z)=\sum_{k=0}^\infty a_kz^k$. This doesn't follow at all though. If $r$ is a zero of $f(z)$, then it is probably not a zero of $g_n(z)$ for any $n$ (let alone for all $n$), so what you have proven about zeroes of $g_n(z)$ does not tell you anything about $r$ (at least not in any obvious way).

Moreover, it is not even true that $g_n(z)$ converges to $f(z)$ as $n\to\infty$. What is true is that $f_n(z)=a_0 + a_1z + \cdots + a_{n-1}z^{n-1}$ converges to $f(z)$, but $g_n(z)=f_n(z)+z^n$ will be very far from $f_n(z)$ if $|z|>1$ and $n$ is large. So even as a nonrigorous heuristic argument, you are doing the wrong calculation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.