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Let $f:\mathbb{C}\to\mathbb{C}$ be entire function and $g:\mathbb{C}\to\mathbb{C}$ be $g(z)=f(z)-f(z+1)$. Which of the following statements are true?

a. If $f(1/n)=0$ for all positive integers $n$, then $f$ is a constant function.

b. If $f(n)=0$ for all positive integers $n$, then $f$ is a constant function.

c. If $f(1/n)= f(1/n + 1)$ for all positive integers $n$, then $g$ is a constant function.

d. If $f(n)= f(n + 1)$ for all positive integers $n$, then $g$ is a constant function.

I am stuck on this problem. Can anyone help me please?

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  • $\begingroup$ By "positive $n$", do you mean "positive integers $n$"? $\endgroup$ – Cameron Buie Dec 17 '12 at 15:07
  • $\begingroup$ Dear pankaj, you have now asked three questions, all of which show no own work. Try investing a little time in the material before you come to us, and if you had any thoughts, please tell us what you've tried. $\endgroup$ – akkkk Dec 17 '12 at 16:44
  • $\begingroup$ @akkkk Who might "us" be? $\endgroup$ – WimC Dec 17 '12 at 17:30
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Hint 1: Use the Identity Theorem

Hint 2: $\left\{\frac{1}{n}:n\in \mathbb{N}^*\right\}$ has an accumulation point ($0$)

Hint 3: $\mathbb{C}$ is connected

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  • $\begingroup$ How can I apply Identity theorem?? $\endgroup$ – pankaj Dec 17 '12 at 16:28
  • $\begingroup$ I know identity theorem.. How can I apply it?? $\endgroup$ – pankaj Dec 17 '12 at 16:41
  • $\begingroup$ still not getting sir.. $\endgroup$ – pankaj Dec 17 '12 at 16:48
  • $\begingroup$ There's hardly anything to add without giving the answer away at this stage. $\endgroup$ – akkkk Dec 17 '12 at 16:48
  • $\begingroup$ ok.......... thanks sir.... $\endgroup$ – pankaj Dec 17 '12 at 16:53
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Can you think of any entire function that has zeros at regular intervals (consider periodic functions)? Simple transformations of one of those should provide a counterexample to both (b) and (d)--in particular, let $f$ be one such (appropriately transformed) function, and show that neither $f$ nor $g$ is constant.

Note: By definition of $g$, we have $f(\frac1n)=f(\frac1n+1)$ for all positive integers $n$ if and only if $g(\frac1n)=0$ for all positive integers $n.$

Now, an entire function is defined and analytic on an open connected set--namely, all of $\Bbb C$--so by the Identity Theorem, if two entire functions agree on a set $S$ that has some accumulation point in $\Bbb C$, then they are in fact the same function. Observing that any constant function (and in particular, the constant $0$ function) is entire, and that the set of reciprocals of positive integers has an accumulation point at $0$, what can you conclude about an entire function that has a zero at $\frac1n$ for every positive integer $n$? This should let you conclude that (a) is true, and (along with the Note) that (c) is true.

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Only the cases (a) and (c) are correct since the sequence has an accumulation point. The other cases do not leads to the mentioned results. Use the Identity Theorem. See this link:

http://en.wikipedia.org/wiki/Identity_theorem

In the theory of analytic functions, one have the analytic continuation principle: if two functions are equal on a set with an accumulation point, these functions are equal for all values. if $(r_{n})$ is a sequence of distinct numbers such that $ƒ(r_{n})=0$ for all $n$ and this sequence converges to a point $r$ in the domain of $D$, then $ƒ$ is identically zero on the connected component of $D$ containing $r$.

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  • $\begingroup$ How can I apply Identity theorem?? $\endgroup$ – pankaj Dec 17 '12 at 16:29
  • $\begingroup$ See the edited answer. The identity theorem is used to prove the result described above. You just need to apply it to your case. $\endgroup$ – China Dec 17 '12 at 16:41
  • $\begingroup$ I know identity theorem.. How can I apply it?? $\endgroup$ – pankaj Dec 17 '12 at 16:41
  • $\begingroup$ still not getting sir.. $\endgroup$ – pankaj Dec 17 '12 at 16:47
  • $\begingroup$ If you are still not getting it, you don't know the identity theorem. $\endgroup$ – akkkk Dec 17 '12 at 16:47

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