How do I manipulate this to the form desired?

  • Thread starter Thread starter shreddinglicks
  • Start date Start date
  • Tags Tags
    Form
shreddinglicks
Messages
225
Reaction score
7

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]
 
Physics news on Phys.org
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]

They are using the approximation ##\ln(1+a) \approx a## valid when ##a## is small. Try that along with other rules about logs.
 
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.

I see:

Use ##ln(x/y) = ln(x) - ln(y)##

The rest is simple.

Is there a link to somewhere online showing that approximation you gave? Just curious.
 
shreddinglicks said:
Is there a link to somewhere online showing that approximation you gave? Just curious.

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.
 
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.

Thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top