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

[Alcubierre]

Homework Statement

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.

Homework Equations

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

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.

 

[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 =)

 

[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

 

[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 $\lambda\to 0$

 

[Chestermiller]

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.

 

[Alcubierre]

But why is h(x) that way?

 

[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

 

[Chestermiller]

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)

 

[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?

 

[Chestermiller]

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

 

[Alcubierre]

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

 

[Chestermiller]

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

Definitely use the know series.

Chet

 

[Alcubierre]

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

 

[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...

 

[Chestermiller]

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

 

[Alcubierre]

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

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?

 

[Chestermiller]

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?

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

 

[Alcubierre]

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

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.

 

[Chestermiller]

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?

 

[Alcubierre]

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?

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.

 

[Chestermiller]

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.

The solution I gave, x = 4.97 is identical to the result given in the wiki article. Also, you made an algebra mistake in solving for λ as a function of T. Try again.

 

[Alcubierre]

You're right. It should be,

\frac{hc}{\lambda kT} = 4.97

\frac{\lambda kT}{hc} = \frac{1}{4.97}

\frac{k}{hc} = \frac{1}{0.014}

\frac{\lambda T}{0.014} = \frac{1}{4.97}

\lambda T = 0.06958

\lambda_{max} = \frac{0.06958}{T}

Got it.

 

[kar32009]

Hi
Can anyone provide the complete solution to this problem only using Taylor's series?? Please.

 

[Chestermiller]

Hi
Can anyone provide the complete solution to this problem only using Taylor's series?? Please.

Please so us your efforts on this so far.

 

[kar32009]

Please so us your efforts on this so far.

 

[Chestermiller]

Are you saying that there was something wrong with what was done previously in this thread?

 

[kar32009]

Are you saying that there was something wrong with what was done previously in this thread?

Not at all. Question is to compare the results with what we get with Taylor's series. So is it enough to conclude that Taylor's series gives only an approximate result to Wien's law?

 

[Chestermiller]

Not at all. Question is to compare the results with what we get with Taylor's series. So is it enough to conclude that Taylor's series gives only an approximate result to Wien's law?

Sure. That's what was found.

 

[kar32009]

Cool. With 7, 8 and 9 terms expansion in Taylor's series, I found the maximum wavelength to be .001605/T, .002247/T and .002546/T respectively. So as the terms increase, it approaches to Wien's law.

Sure. That's what was found.
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