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For an odd integer $k\ge1$ , let $\mathbb{F}$ be the set of all entire functions f such that $f(x)=\vert{x^k}\vert$ for all $x\in(-1,1)$. Then the cardinality of $\mathbb{F}$ is

1.0

2.1

3.strictly greater than 1 but finite

4.infinite

f(x) = \begin{cases} x^k, & \text{if $x\in(0,1)$ } \\[2ex] -x^k, & \text{if $x\in(-1,0)$ } \end{cases}

consider convergent sequence {$\frac{1}{n}$} in (0,1)

Let g(z)=$z^k$ ,since $f(\frac{1}{n})=g(\frac{1}{n})$ on (0,1)

hence $f{(z)}=z^k$ on (-1,1) by using identity theorem.

similarly on (-1,1) i can show that $f{(z)}=-z^k$

but then how to conclude for final answer?

dipali mali
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1 Answers1

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If $f(z)=z^k$ on $(0,1)$, then $f(z)=z^k$ on $ \mathbb C$.

A similar argument gives $f(z)=-z^k$ on $ \mathbb C$.

Conclusion: no such function exists.

Fred
  • 77,394
  • I am not getting . why such function cannot exist.? Both $z^k$ and $-z^k$ are entire functions .Because for $f(z)=z^k, so f(\frac{-1}{2})=-\frac{1}{2^k}$ , but mod should be positive hence such function does not exist or i have to use identity theorem? – dipali mali Jul 29 '17 at 04:57