Approximation involving an exponential function

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Homework Help Overview

The discussion revolves around the approximation of the exponential function exp[gbH/(2kT)] in the context of a derivation related to certain laws. The original poster seeks clarification on the reasoning behind the approximation gbH/2 << 1 leading to the expression being simplified to 1 + gbH/(2kT).

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster questions the origin of the term gbH/(2kT) and its significance in the approximation. Some participants reference the Taylor expansion of the exponential function as a basis for the approximation, while others suggest visualizing the approximation through a tangent line to the curve.

Discussion Status

Participants are exploring different methods to understand the approximation, including the Taylor series and graphical interpretations. There is engagement with the concepts, but no explicit consensus has been reached regarding the best approach to clarify the original poster's confusion.

Contextual Notes

The discussion includes references to the conditions under which the approximation holds, specifically the assumption that gbH/2 is much less than 1, which is central to the validity of the Taylor expansion used.

mccoy1
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Homework Statement



I was following a derivation of some laws and I didn't get how they approximate some portion of the expression. That portion/part is exp[gbH/(2kT)]. The book says gbH/2 <<1 and therefore exp[gbH/(2kT)] = 1+gbH/(2kT).

Homework Equations


The Attempt at a Solution


I agree with the value 1, but where did gbH/(2kT) come from? Please help. My understanding is that if gbH/2 is way less than 1, then e.g exp[1.0*10^-15/KT)] = 1.
 
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Because it's the Taylor expansion of exp(x) near 0

exp(x) = 1 + x + (x^2)/2 + (x^3)/6 + ...

You can cut the series at any term you would like, however you can't equal it to 1 because there will be no parameter left to give values to...
 
You can also get this approximation by replacing the curving graph by a tangent line.
 
atomthick said:
Because it's the Taylor expansion of exp(x) near 0

exp(x) = 1 + x + (x^2)/2 + (x^3)/6 + ...

You can cut the series at any term you would like, however you can't equal it to 1 because there will be no parameter left to give values to...

Haa, thank you very much. That didn't pop in my head. Thanks a lot.
 
HallsofIvy said:
You can also get this approximation by replacing the curving graph by a tangent line.

Thanks for that.
 

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