Applications of Taylor polynomials

[coneyaw]

Homework Statement

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

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 $$\lambda$$.

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

 

[dynamicsolo]

f($$\lambda$$) = $$\frac{8\pi hc\lambda ^{-5}}{e^{hc/\lambda kT}-1}$$

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 $$\lambda$$ 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...

 

[coneyaw]

Based on the information in the textbook, the Maclaurin series for e^x is sum from 0 to infinity of x to the n over n factorial. that being said, I don't know if that's the same as its Taylor series.

$$e^{hc/\lambda kT}\ =\sum^{\infty}_{n=0}\frac{[\frac{hc}{\lambda kT}]^{n}}{n!}$$ like $$e^{x}\ =\sum^{\infty}_{n=0}\frac{x^{n}}{n!}$$

Is this where I need to start?

 

[coneyaw]

The trouble I'm having is stemming from the Taylor approximation being defined as $$f(\lambda)\ =\frac{f^{(n)}(a)}{n!}(\lambda-a)^{n}$$ I don't understand what a I should use as the center "a" to start the approximation.

IIRC, to end up with the R-J result, you will only need to use the first two terms

I assume by this you mean that I will only need to find $$T_{2}$$ which would be equal to $$\frac{f^{(2)}(a)}{2!}(x-a)^{2}$$

 

[Gib Z]

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 $$f(x) = \frac{1}{x^5 ( e^{kx}-1) }$$ and then manipulating constants.

 

[coneyaw]

I'm just so lost. To see if I can clarify things, I'm going to find n derivatives of $$f(x) = \frac{1}{x^5 ( e^{kx}-1) }$$, evaluate them at f(0), then calculate my $$T_{n}$$ values?

To do that, is $$e^{x}\ =\sum^{\infty}_{n=0}\frac{x^{n}}{n!}$$ going to be useful?

 

[Gib Z]

Ahh sorry my bad, you want to find the series of $$f(x) = \frac{1}{x^5 ( e^{a/x} -1)}$$, because in the original problem, lambda is in the denominator of the exponent, and there's already a "k" so its not good to use the same pronumeral. Once you've done that, let $a=\frac{hc}{kT}$, and multiply $8\pi hc$ back in.

 

[coneyaw]

I feel like I'm getting misunderstood on my basic question.

In the book, it asks: find the taylor polynomial $$T_{n}(x)$$ for the function f at the number a, then it gives us f(x) = sin x a= pi/6, n=3

another example is f(x) = e^x a=2 and n=3, or f(x)=xe^(-2x) a=0 n=3

assuming my $$f(x) = \frac{1}{x^5 ( e^{a/x} -1)}$$ then what would i set my center a? Do I just us the Maclaurin series for it?

If I just use the Maclaurin series, is this true? $$e^{hc/\lambda kT}\ =\sum^{\infty}_{n=0}\frac{[\frac{hc}{\lambda kT}]^{n}}{n!}$$

Or more in your form: $$e^{\frac{a}{\lambda}}\ =\sum^{\infty}_{n=0}[\frac{a}{\lambda}]^{n}}{n!}$$

 

[Gib Z]

Your question seems to have changed from the Physics based one in the original post :(

Also, my bad again, I assumed we were centering about a=0 and used "a" again, sigh.

Basically I'm just saying: If you want to find the McLaurin series for that function in the first post, do what I said in post 7.

 

[coneyaw]

Its a calculus 253 course. I need to use a Taylor series approximation of Planck's law to closely match the results of Rayleigh-Jeans law.

assuming $$f(\lambda)\ =\frac{1}{x^5 ( e^{a/\lambda} -1)}$$
Then $$T_{n}(\lambda)\ =\frac{f^{(n)}(0)}{n!}\lambda^{n}\ = f(0)+\frac{f^{1}(0)}{1!}x+\frac{f^{2}(0)}{2!}x^{2}+...\frac{f^{n}(0)}{n!}x^{n}$$

 

[Gib Z]

Well ok, but we must remember the f(lambda) from the original question is not equal to the f(lambda) in your last post, but it suffices to find the MacLaurin series for the simplified one instead because setting a equal to the right constants and multiplying the function through by the right constants gives us the original function. Now, if you define T_n (lambda) as such, then the question basically says to find T_0 + T_1 + T_2...and show why that approximates the other law.

 

[coneyaw]

Ok so this is what I've drawn up based on the last post.
$$T_{n}(\lambda)\ =\frac{f^{(n)}(0)}{n!}\lambda^{n}\ = f(0)+\frac{f^{1}(0)}{1!}x+\frac{f^{2}(0)}{2!}x^{2} +...\frac{f^{n}(0)}{n!}x^{n}$$

From a few posts ago, $$f(x) = \frac{1}{x^5 ( e^{a/x} -1)}$$
So to find the first part of $$T_{n}(\lambda)$$, we set out to find f(0) by setting x=0 and get f(x)=undefined because $$0^{5}\ =\ 0\ and\ \frac{1}{0}\ =\ undefined$$

So now what?

 

[Gib Z]

The original function f(lambda) isn't defined at lambda= 0 either. Damn.

Ok start fresh:

First, find the Taylor series (centered around a general point a) of $$g(x) = \frac{1}{x^5 ( e^{b/x} -1)}$$. Then, let b = hc/kT, and multiply g(x) by 8*pi*h*c to get f(lambda).

 

[coneyaw]

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

 

[coneyaw]

$$e^{b/x}\ =\ e^{bx^{-1}}$$ which is much like the form $$e^{x^{2}}$$

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

Is going to be equal to the series representation of $$x^{-5}$$ times the series representation of $$/frac{1}{e^{bx^{-1}}}$$

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