High School How Can We Divide Lamé Coefficients When Some Are Zero?

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Dividing Lame coefficients is valid as long as the denominator is not identically zero, which limits the domain of the function. In cases where some coefficients are zero, it indicates a breakdown in the coordinate system, leading to multiple coordinate representations for a single point. To address calculations at these problematic points, switching to an alternative coordinate system is recommended. The initial multiplication of the fraction by a form of one (like h3/h3) is mathematically sound unless h3 is zero. Understanding these nuances is crucial for accurate analysis in physics.
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Sometimes in calculations authors uses
\frac{1}{h_1h_2}=\frac{h_3}{h_1h_2h_3}
where ##h_i, i=1,2,3## are Lame coefficients. For instance in spherical coordinates ##h_r=1##, ##h_{\theta}=r##, ##h_{\varphi}=r\sin \theta##. I am not sure how we can divide so easily Lame coefficients when some on them obviously can be zero for certain values of parameters. Can someone give me some explanation? Thanks a lot in advance.
 
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I have no background of the physics there but the formula seems multiplying the same number to denominator and numerator, so obviously right except the number is zero.
 
Changed problem level from A to B. The underlying concept of Lame coefficients might be advanced, but in the posted problem all that was done was to multiply a fraction by 1 in the form of ##h_3## over itself.
 
LagrangeEuler said:
I am not sure how we can divide so easily Lame coefficients when some on them obviously can be zero for certain values of parameters. Can someone give me some explanation? Thanks a lot in advance.

You can divide a function by another function provided that the denominator is not identically zero; this reduces the domain by excluding points where the denominator is zero. In the context of Lame coefficients these are points where the coordinate system breaks down, with a single point being referred to by multiple distinct coordinate tuples. If you need to analyze something at these points, the answer is to use a different coordinate system.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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