# Maximizing Planck's law using Taylor polynomial for e^x

1. Feb 25, 2013

### Alcubierre

1. The problem statement, all variables and given/known data

The energy density of electromagnetic radiation at wavelength λ from a black body at temperature $T$ (degrees Kelvin) is given by Planck's law of black body radiation:

$f(λ) = \frac{8πhc}{λ^{5}(e^{hc/λkT} - 1)}$

where $h$ is Planck's constant, $c$ is the speed of light, and $k$ is Boltzmann's constant. To find the wavelength of peak emissions, maximize $f(\lambda)$ by minimizing $g(\lambda) = λ^{5}(e^{hc/λkT} - 1)$. Use a Taylor polynomial for $e^{x}$ with n = 7 to expand the expression in parentheses and find the critical number of the resulting function. (Hint: Use $\frac{hc}{k}$ = 0.014.) Compare this to Wien's law:
$\lambda _{max} = \frac{0.014}{T}$. Wien's law is accurate for small λ. Discuss the flaw in our use of Maclaurin series.

2. Relevant equations

$e^{x} = \sum_{n = 0}^{∞} \frac{x^{n}}{n!}$

3. The attempt at a solution
I have no idea where to begin. I started with setting x to $\frac{0.014}{\lambda T}$ and expanding the series to the 7th term but I don't know the direction to go.

2. Feb 26, 2013

### tia89

I hope you have the possibility to use some calculator to do all this, because otherwise it ill be quite boring.

I did not try it but I think you should first of all expand the parenthesis in $g(\lambda)$ and eventually simplify the $\lambda$s where you can. Then you derive $g(\lambda)$ and find the value of $\lambda$ for which this derivative is $0$. This $\lambda$ (check that it is really a minimum) is the one which represents the maximum of $f(\lambda)$.

Then compare with the value they give you from Wien law.

I hope this helps you =)

3. Feb 26, 2013

### Alcubierre

And when would the use of a Taylor series for e^x come in play? After that or is your method an alternate way of finding the solution

4. Feb 26, 2013

### tia89

I'm sorry I probably explained something not so clearly... when I say "expand the parenthesis in $g(\lambda)$" I actually mean use the Taylor expansion of the exponential in the parenthesis (up to the order they ask you to).
In practice what you have to do is to write $g(\lambda)$ using the expansion as
$$g(\lambda)=\lambda^5\left( 1+\frac{hc}{\lambda k T}+\frac{1}{2}\frac{(hc)^2}{(\lambda k T)^2}+\dots +\frac{1}{7!}\frac{(hc)^7}{(\lambda k T)^7} -1 \right)$$
then cancel $1$ with $-1$ and simplify the $\lambda$ where you can; then derive this expression and so on as I said before... boring but it should work.

An alternate method would be to derive first $g(\lambda)$ and then expand in Taylor series the exponentials you still have. I think anyway that the result is the same, and in this last way it could get even longer to do (you will have to expand two exponentials instead of one).

Beware also the fact (and it should be the object of last question on the flaw of using McLaurin expansion, i.e. Taylor expansion around $0$) that indeed all this stuff holds if $\lambda$ is small enough... therefore for low temperature you will have problems, as this expansion works only for $\lambda\to 0$

5. Feb 26, 2013

### Staff: Mentor

Let $h(x)=\frac{e^x-1}{x^5}$, expand the numerator in a 7 term taylor series, take the derivative with respect to x, and set the derivative equal to zero. Solve for x.

6. Feb 26, 2013

### Alcubierre

But why is h(x) that way?

7. Feb 26, 2013

### tia89

Simply disregard the constants (you can think them to be 1 for now) and define $x=\frac{1}{\lambda}$... in this way you have the same thing as before

8. Feb 26, 2013

### Staff: Mentor

Because $g(\lambda) = λ^{5}(e^{hc/λkT} - 1)=(\frac{hc}{kT})^5(\frac{kTλ}{hc})^{5}(e^{hc/λkT} - 1)=(\frac{hc}{kT})^5h(x)$

9. Feb 26, 2013

### Alcubierre

Okay I get that so far and I think I got the answer but just to better understand, why did you do the reciprocal?

10. Feb 26, 2013

### Staff: Mentor

No reason. It just seemed easier to work with and differentiate.

11. Feb 26, 2013

### Alcubierre

Do I use a known series or differentiate h(x) 7 times

12. Feb 26, 2013

### Staff: Mentor

Definitely use the know series.

Chet

13. Feb 27, 2013

### Alcubierre

I can't solve for x. This is what I have

14. Feb 27, 2013

### tia89

I would say that
$$\frac{e^x-1}{x^5}=\sum_{n=1}^{\infty}\frac{x^{n-5}}{n!}$$
(I start from 1 because I already subtracted -1)
Then you can derive directly in this form (derivative of a sum is sum of derivatives) and then write all the terms up to n=7.
The problem would then be to compute $h'(x)=0$ but for that I think you would need some calculator, as it appears to be not so trivial... unless you can simplify something but you will have to try...

15. Feb 27, 2013

### Staff: Mentor

$$h=\frac{1}{x^4}+\frac{1}{2x^3}+\frac{1}{6x^2}+1/(24x)+\frac{1}{120}+\frac{x}{720}$$

$$h'=-\frac{4}{x^5}-\frac{3}{2x^4}-\frac{1}{3x^3}-\frac{1}{24x^2}+\frac{1}{720}=0$$

All the first four terms are negative. I looked at each of these terms individually to see what it would take for each to cancel the 1/720. By this rationale, I determined that x had to be at least 6.21. I tried a value of 10 for x, but it was too large. So x must lie between 6.21 and 10 (probably much closer to 6.21). Once x is bounded like this, it is simple to use the half-interval technique to solve for the value of x.

I hope I did the "arithmetic" correctly.

Chet

16. Feb 27, 2013

### Alcubierre

Does it matter that your series is starting at -4? My first try was actually exactly what you did but I thought that it had to start at n = 0 up to n = 7 hence why I used those terms. But now that I am writing this, I realize that what I did and what you did is essentially the same thing, right?

17. Feb 27, 2013

### Staff: Mentor

I really don't know what you did in your first try. The problem statement said to use a 7 term truncated series for ex. That is all that I did.

18. Feb 27, 2013

### Alcubierre

My first try was what you did using those seven terms, but then I realized that those particular seven terms that I used (the same you used) is the series starting at -4 rather than starting at 0. That's why I erased it and wrote what is in the picture.

But anyway, as to the value of x, it is 8.72124. So, how does this look

$x = \frac{hc}{\lambda kT}$

$8.72124 = \frac{hc}{\lambda kT}$

Substitute $\frac{hc}{k}$ as 0.014 and solve for lambda, yielding

$\lambda_{max} = \frac{T}{0.122097}$

(For the substitution I used x = 1/λ and hc/k = 0.014)

It doesn't look right to me, although the reciprocal of that seems reasonable.

19. Feb 27, 2013

### Staff: Mentor

Maybe we need to include more terms in the series. If you differentiate the non-series version of the function h, and set the derivative equal to zero, you get:
$$(x-5)+5e^{-x}=0$$
There is a solution to this equation at a value of x slightly less than x = 5. A good first order approximation to this solution is
$$x=5-5e^{-5}=4.97$$
Is this value any closer to what you expected?

20. Feb 27, 2013

### Alcubierre

That seems reasonable because with that value I get 0.06958 but my predicament is when I substitute the constants back to x and simplify because I get λ = T/0.06958 but I must compare it to Wien's law which is λ = #/T, the number symbol representing the number in. Perhaps I am not understanding the question when it asks me to compare my answer with Wien's law.

EDIT: What you did makes sense now: http://en.wikipedia.org/wiki/Wien's_displacement_law#Derivation_from_Planck.27s_Law

EDIT 2: In my first post I wrote Wien's law to be 0.014/T when in actuality it is 0.002898/T, I hope you caught that and my apologies for the careless mistake.

Doing it using partial derivatives is so much more convenient wow.

Thank you very much for all the help, tia89 and Chet.

Last edited: Feb 27, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted