An approximate solution to the Van der Pauw equation

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SUMMARY

The forum discussion centers on approximating the function f in the Van der Pauw equation for resistivity, specifically when the ratio R_1/R_2 is approximately 1000. The relevant equation is given as ρ = (π * d / ln[2]) * (R_1 + R_2)/2 * f(R_1/R_2), with f defined by a specific relation involving arccosh. The approximation for f when R_1 and R_2 are nearly equal is provided, but participants express uncertainty regarding its accuracy for large ratios. The discussion suggests that while the equation is transcendental, the Taylor series method may still be applicable for finding approximate solutions.

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smithg86
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Homework Statement


question: approximate the function f if the ratio R_1 / R_2 is about 1000.

given:
the van der pauw equation for resistivity:
\rho = ( pi * d / ln[2] ) * (R_1 + R_2})/2 * f(R_1 / R_2),

where f is a function of the ratio R_1/R_2 only and satisfies the relation:

(R_1- R_2)/(R_1+R_2) = f arccosh{exp(ln[2]/f) / 2}.

from van der pauw's 1958 paper:if R_1 and R_2 are almost equal, f can be approximated by the formula

f ~ 1 - [ (R_1 - R_2) / (R_1 + R_2) ]^2 * ln[2] / 2 - [(R_1 - R_2)/(R_1 + R_2)]^4 * {(ln[2])^2 / 4 - (ln[2])^3 / 12}.

Homework Equations


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The Attempt at a Solution


I have no idea how to start this.
 
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smithg86 said:
from van der pauw's 1958 paper: if R_1 and R_2 are almost equal[/color], f can be approximated by the formula

f ~ 1 - [ (R_1 - R_2) / (R_1 + R_2) ]^2 * ln[2] / 2 - [(R_1 - R_2)/(R_1 + R_2)]^4 * {(ln[2])^2 / 4 - (ln[2])^3 / 12}.

That approximation doesn't seem like it would be very accurate when R_1=1000R_2!:wink:


I have no idea how to start this.

What does \frac{R_1-R_2}{R_1+R_2} equal when \frac{R_1}{R_2}=1000?
 
What does \frac{R_1-R_2}{R_1+R_2} equal when \frac{R_1}{R_2}=1000?[/QUOTE]

It's 999/1001. But I don't know how to apply that to find an approximate solution. How do you go about finding an approximate solution to that, or any other function?
 
I can't seem to solve the equation explicitly for f (I think it's transcendental). Can I still use the Taylor series method?
 
Yes, it is transcendental, but I think you can still use a Taylor series. However, at the moment I can't think of exactly what expression your going to want to expand to make things easiest...I'll give it some more thought in the morning, but in the meantime hopefully someone else will jump in.
 

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