Approximating the Cosine Integral?

1. Jan 12, 2005

Manchot

Does anyone know of a semi-quick way of approximating Ci(x)? I tried to find an asymptotic expansion for it, but had little luck. Truth be told, I'm not even sure exactly what the definition of asymptotic expansion is. I discovered it while learning about ways of approximating harmonic numbers and Stirling's approximation, and only know that it works for large values. Basically, I'd like it to be valid for high numbers, and this is why I'm trying to use this type of series. Can anyone offer any insight? Thanks a bunch!

Edit: By "semi-quick" I meant in terms of simple, nice, elementary functions. I don't care if there are a finite number of terms or not, because to me, more terms = more accuracy.

Last edited: Jan 12, 2005
2. Jan 12, 2005

dextercioby

As u might have guessed,u cannot express Ci(x) through elementary functions.Asymptotic behavior,series expasion and a lot more can be found here and the next pages.

Daniel.

3. Jan 12, 2005

Manchot

Yeah, I knew that it couldn't be expressed in a finite number of elementary functions. But what prevents there from being an asymptotic series which approximates it?

4. Jan 12, 2005

dextercioby

What do you mean??For large values of the argument,you can approximate the function with its asymptotic series and that's it...

Daniel.

5. Jan 12, 2005

Manchot

Maybe I'm being unclear about what I mean. I'm trying to find what its asymptotic series is. Now, I see on the pages you linked to (page 232) something labelled "Asymptotic series," but it doesn't say what they refer to. Should I assume that 1/z (1-2!/x²+4!/x^4...) is Ci(x)'s asymptotic series?

6. Jan 13, 2005

dextercioby

No.The formula u've indicated (no.5.2.34,page 233) is not for Ci(x) (else it would have been written at the left of the equality sign),but for another function,namely f(z).U needto use formulas 5.35,page 233 and 5.2.9,page 232 to get your desired result,which DOES NOT EXIST,since "sine" and "cosine" do not have well defined asymptotic behavior and hence don't have asymptotic series...

Daniel.

7. Jan 13, 2005

Manchot

Well, then, if there is no asymptotic expansion, does anyone know a relatively easy way to calculate it? (without using tables?) Any help is greatly appreciated.

8. Jan 13, 2005

dextercioby

Check the same site for its power series.And pay attention to its convergence radius,which means you cannot compute the function using its power series for arbitrary arguments.

Daniel.