0
$\begingroup$

given transcendental function $$F(x)=\sum_0^{\infty}a_i x^i,a_i\in \mathcal{N} \bigcup 0,\exists M \space a_i \leq M^i$$.

is there algebraic function $$A(x)=\sum_0^{\infty}b_i x^i,b_i\in \mathcal{N} \bigcup 0,$$,such that $a_i =b_i$ if $a_i = 0$;$a_i \leq b_i$ otherwise?

Improve this question
$\endgroup$
2
  • $\begingroup$ The only exponential function I know, $F(x)=e^x$, the Maclaurin series coefficients are not whole numbers. How do you define "exponential function"? [I'm assuming expotential is a typo for exponential] $\endgroup$
    – Gerry Myerson
    Commented Mar 24, 2013 at 3:52
  • $\begingroup$ Sorry,a typos,I have edited the post,the exponential should be transcendental.Very sorry,Gerry,thank you very much $\endgroup$
    – XL _At_Here_There
    Commented Mar 24, 2013 at 4:19

2 Answers 2

Reset to default
7
$\begingroup$

By Hadamard's theorem, a lacunary series $\sum_k c_k z^{\lambda_k}$ with finite radius of convergence where $\inf_k \lambda_{k+1}/\lambda_k > 1$ can't be analytically continued outside its circle of convergence, and in particular can't be the Maclaurin series of an algebraic function. So if $F(z)$ is such a series, there can be no algebraic $A(z)$.

Improve this answer
$\endgroup$
0
2
$\begingroup$

Yes: $g(x):= \sum_{k \geq 0} X^{k!}$ is transcendental; and so is $F(x)=g(x) + \frac{1}{1-x}$. Put $M=2$ and $A(x)= \frac{2}{1-x}$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Does every transcendental function have at least one algebraic function corresponding the transcendental one? I have just edited the post to clarify what I intend to ask.Thank $\endgroup$
    – XL _At_Here_There
    Commented Mar 24, 2013 at 6:15
  • 1
    $\begingroup$ The example $g(x)$ shows there isn't always an algebraic function with the same set of zero coefficients. You can also say this by a cardinality argument. $\endgroup$
    – Douglas Zare
    Commented Mar 24, 2013 at 6:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .