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I was dealing with the derivation of Stefan's law from Planck's law. In this derivation one gets to the following equation
[itex]\frac{xe^{x}}{e^{x}-1}=5[/itex]
At this point most places I looked provide an aproximate numerical solution to this transcendental equation and carry on. But I wwas thinking about this. Suppose there is x that satisfies the equation. Then,
[itex]xe^{x}=5e^{x}-5[/itex]
[itex]e^{x}(5-x)=5[/itex]
[itex]e^{x}=\frac{5}{5-x}=\frac{1}{1-\frac{x}{5}}[/itex]
The numerical solution they provide is less than 5, so we can use the geometrical series sum.
[itex]e^{x}=\sum(\frac{x}{5})^{n}[/itex]
Now, if I write the exponential series, and using that two power series must have the same coefficients i arrive at my problem. I thought a bit about it and since I can neither figure it out nor know exactly how to look for it I decided to ask it here. Maybe it is a stupid doubt and I am forgeting something. Either way, I would apreciate some help. Thanks.
[itex]\frac{xe^{x}}{e^{x}-1}=5[/itex]
At this point most places I looked provide an aproximate numerical solution to this transcendental equation and carry on. But I wwas thinking about this. Suppose there is x that satisfies the equation. Then,
[itex]xe^{x}=5e^{x}-5[/itex]
[itex]e^{x}(5-x)=5[/itex]
[itex]e^{x}=\frac{5}{5-x}=\frac{1}{1-\frac{x}{5}}[/itex]
The numerical solution they provide is less than 5, so we can use the geometrical series sum.
[itex]e^{x}=\sum(\frac{x}{5})^{n}[/itex]
Now, if I write the exponential series, and using that two power series must have the same coefficients i arrive at my problem. I thought a bit about it and since I can neither figure it out nor know exactly how to look for it I decided to ask it here. Maybe it is a stupid doubt and I am forgeting something. Either way, I would apreciate some help. Thanks.