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Any closed form for series like $$F(x)=\Sigma_{i=2}^{\infty}x^p,\text{p is prime}$$ or $$F(x)=\Sigma_{i=0}^{\infty}x^{i!}$$?

More generally,we can obtain a power series from decimal expansion of a number r(0< r<1 ) by replacing $$(\frac{1}{10})^i$$ with $$x^i$$ like $$\frac{1}{3}=3(\frac{1}{10})^1+3(\frac{1}{10})^2+\cdots 3(\frac{1}{10})^i+\cdots$$, we obtain : $$f(x)==\Sigma_{i=1}^{\infty}3x^i$$

when f(x) is convergent,what restriction do we have to put on r(if r is c.e number) to make f(x) have a closed form?

When is f(x) algebraic ,or transcendental?

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2  
Your first display makes no sense. Maybe you meant $$\sum_{p=2}^{\infty}x^p,p{\rm\ is\ prime}$$ –  Gerry Myerson May 24 '13 at 1:51
1  
This might be better expression: $$\sum_{p\textrm{ prime}} x^p$$ –  i707107 May 24 '13 at 2:02
    
XL, you came back here to edit your question, but did nothing to engage with the comments. Why? –  Gerry Myerson May 24 '13 at 5:34
    
@Gerry,@i707i707,thank both of you very much,your expressions are right,they are what I intend to express –  XL _at_China May 24 '13 at 5:48

2 Answers 2

up vote 2 down vote accepted

See this blog post: http://uniformlyatrandom.wordpress.com/tag/power-series/

contains a proof of the result by Fatou:

A function whose power series expansion has integer coefficients and radius of convergence 1 is either rational(in $\mathbb{Q}(x)$) or transcendental(over $\mathbb{Q}(x)$).

If $r$ is rational, then the decimal expansion will be eventually periodic. So we have rational function. (Indeed this can be done explicitly)

Otherwise, when $r$ is irrational, then the resulting function cannot be rational(plug in $1/10$, then you get irrational number). Thus, we have transcendence of $f$.

In particular, your functions $F$ in the beginning are transcendental. However, getting closed form will be extremely hard for those examples.

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@i707i707,thank you.But it is only a partial answer to this question,especially,when r is algebraic number –  XL _at_China May 24 '13 at 5:06
    
XL, if the answer is helpful, even if it's not a complete answer, I'd encourage you to vote it up. –  Gerry Myerson May 24 '13 at 5:36
    
@Gerry,thank you for your reminding –  XL _at_China May 24 '13 at 5:49
    
As I mentioned in my answer, $r$ can be either rational or irrational. When $r$ is rational, the function $f$ is rational, and when $r$ is irrational, the function $f$ is transcendental. –  i707107 May 24 '13 at 6:43
    
@i707107,those hold ,provided a series is convergent with radius r,otherwise,they may not hold,am I right?For instance,$$f(x)=\frac{1-\sqrt{1-4x^2}}{2x^2}$$ –  XL _at_China May 24 '13 at 9:54

Also, the classical Fabry gap theorem tells you that the unit circumference is the natural boundary. Meanwhile, all "elementary" functions can be analytically continued along almost every path on the plane, so give up all hopes for a closed formula of any sort...

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@fedja,thank you,only can one post be chosen as answer. –  XL _at_China May 26 '13 at 14:16

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