The discussion focuses on manipulating the equation ##(-h^2/2uc)*(dp/dx)*ln((1+c*(x/h)^2)/(1+c))## into the desired form ##-(h^2/2u)*(dp/dx)*(1-(x/h)^2)##. The key technique involves using the approximation ##\ln(1+a) \approx a##, which is valid when ##a## is small. Participants also mention the use of logarithmic properties, specifically ##\ln(x/y) = \ln(x) - \ln(y)##, to simplify the expression. The approximation is derived from the first term of the Taylor series expansion for the natural logarithm.
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shreddinglicks said:Homework Statement
I want to manipulate an equation to suit a desired form.
Homework Equations
##(-h^2/2uc)*(dp/dx)*ln((1+c*(x/h)^2)/(1+c))##
becomes
##-(h^2/2u)*(dp/dx)*(1-(x/h)^2)##
The Attempt at a Solution
I have no idea, I'm not even sure how the natural log disappears. [/B]
Dick said:They are using the approximation ##\ln(1+a) \approx a## valid when ##a## is small. Try that along with other rules about logs.
shreddinglicks said:Is there a link to somewhere online showing that approximation you gave? Just curious.
Dick said:I don't know any good links. But approximations like this generally come from taking the first terms of the Taylor series. ##\ln(1+x)=x-
\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots## Keeping just the first term gives the approximation. Similarly, ##\sin(x) \approx x## etc.