Applications of Taylor polynomials to Planck's Law

  • Thread starter coneyaw
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Due to too much wrong information being posted on my behalf, I am resubmitting a cleaned up version of my last post. I have 2 hours to get this problem done :(. Essentially, I don't know at all how to find the Taylor Polynomial for [tex]
g(x) = \frac{1}{x^5 ( e^{b/x} -1)}
[/tex]

Homework Statement


f([tex]\lambda[/tex]) = [tex]\frac{8\pi hc\lambda ^{-5}}{e^{hc/\lambda kT}-1}[/tex]
Is Planck's Law
where
[tex]h\ =\ Planck's\ constant\ =\ 6.62606876(52)\ \times\ 10^{-34}\ J\ s;[/tex]
[tex]c\ =\ speed\ of\ light\ =\ 2.99792458\ \times\ 10^{8}\ m\ s^{-1};[/tex]
[tex]and\ Boltzmann's\ constant\ =\ k\ =\ 1.3806503(24)\ \times\ 10^{-23}\ J\ K^{-1}[/tex]

For my calculus class, I am asked to use a Taylor polynomial to show that the values for Planck's Law gives approximately the same values as the Raleigh-Jeans Law for large wavelengths [tex]\lambda[/tex].

Homework Equations


Basically I need some help regarding leading me in the right direction. I need to know how to pursue the correct center and basically someone to give me starting conditions, then I can figure the inequality and error on my own


The Attempt at a Solution

Here is an example of one that I've done before
Isn't this similar to how I find the taylor polynomial for [tex]
g(x) = \frac{1}{x^5 ( e^{b/x} -1)}
[/tex]
http://img509.imageshack.us/img509/1417/fasdfci8.jpg [Broken]
[/URL]



You will want to replace the exponential function in the denominator with its Taylor approximation. What is the series for e^Kx? (Keep in mind that [tex]\lambda[/tex] is the variable of interest.) IIRC, to end up with the R-J result, you will only need to use the first two terms (for large wavelengths, the exponent will be small compared to 1). The "1"s cancel, after which you simplify the resulting algebraic expression...
Your misunderstanding for a lot of this may stem from your definition of the Taylor series. Check your notes again =] Once you get that right, you will see the the MacLaurin series is just the Taylor series when a=0, and the one you want to use here. You will also see what dynamicsolo meant by "first two terms".

I would personally not take dynamicsolo's route though (no offense intended) as by only taking two terms, canceling 1s and saying only the leading term will matter is not as rigorous as taking the Taylor series of [tex]f(x) = \frac{1}{x^5 ( e^{kx}-1) }[/tex] and then manipulating constants.
[tex]e^{b/x}\ =\ e^{bx^{-1}}[/tex] which is much like the form [tex]e^{x^{2}}[/tex]

So if my assumptions are right, [tex]
g(x) = \frac{1}{x^5 ( e^{b/x} -1)}
[/tex]

Is going to be equal to the series representation of [tex]x^{-5}[/tex] times the series representation of [tex]\frac{1}{e^{bx^{-1}}}\ =\ e^{-bx^{-1}}[/tex]

that being said, I could find the representation of [tex]e^{-bx^{-1}}[/tex] similarly to the above problem about [tex]e^{x^{2}}[/tex]
 
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