On 9/22/2009 11:30:59 AM, McHale wrote:
>Thanks to all three of you.
>There are still some issues.
>
>The first two ideas with the
>dummy and the local definition
>did not work on my version
>Mathcad 14.0 M010. I attached
>the two screenshots.
The way the two processors interact has changed somewhat over the years. Just use what I showed.
>Richard's way worked. I used
>the same definition in my
>first row with Fun(x), but I
>did not know that Mathcad
>needed the symbolic evaluation
>right afterwards. I thought
>this only gives insight to
>the user.
In general, you don't. I assumed you had done this deliberately, because the polylog function is not defined for the numeric processor. You can make it work by having the symbolic processor evaluate it and pass the numeric result to the numeric processor.
There are other reasons to construct expressions like this to, but in general it's not needed.
>There is another problem,
>however, since I also have to
>evaluate the 2nd-, 3rd-, and
>4th-order polylogarithms. But
>my Mathcad cannot evaluate
>these symbolically like with
>the 1st-order polylogarithm. I
>think there is no closed form,
>only the infinite series.
>
>Any idea how to handle them?
You would have to define your own function. See:
http://mathworld.wolfram.com/Polylogarithm.html>More about the background:
>
>My antiderivate of interest
>(see attached picture) belongs
>to Planck's radiation law. It
>shall be integrated vs. the
>wavelength (finite integration
>limits - my infrared-camera
>has a specified spectral
>range).
Why not just integrate it numerically?
> A substitution
>already took place, where
>x~1/wavelength.
Are you trying to convert from wavelength to wavenumber? If you are, it's not quite that simple because the spectral radiance is not just specified at a particular wavelength or wavenumber, but also per unit wavelength or wavenumber. I have a handy worksheet with various forms of the law defined so that I don't have to figure them out every time. See the attached.
>I want to use
>the antiderivative to
>calculate upper limit minus
>lower limit.
You've lost me there. Why do you need the derivative to find the integral limits?
>Or do you think I should
>integrate numerically from the
>beginning without using the
>antiderivative
Yes.
>, because the
>polylogarithms are infinite
>series anyway? The solution of
>the integration will still be
>dependent on the temperature,
That's not a problem for the numeric processor.
Richard