Differentiating the Dot Product's Evil Twin

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SUMMARY

The discussion centers on the differentiability of the "evil twin" of the dot product, specifically addressing why it is not differentiable at parallel vectors. The key point is that the function involved, represented as ##\sqrt{|x|}##, is continuous but not differentiable at zero due to the square root. This contrasts with the standard dot and cross product formulas, which do not have square roots and thus maintain differentiability at critical angles. The explanation highlights the mathematical properties of square root functions and their implications for differentiability.

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  • Understanding of vector mathematics
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  • Basic knowledge of calculus, particularly differentiability
  • Graphing functions and interpreting their behavior
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  • Investigate the implications of non-differentiable points in mathematical functions
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Mathematicians, physics students, and anyone interested in vector calculus and the properties of mathematical functions, particularly in understanding differentiability in relation to vector operations.

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Hello,

I found this link very useful:
https://www.quora.com/Why-is-cosine-used-in-dot-products-and-sine-used-in-cross-products

I understand all of Anders Kaseorg's discussion except for ONE PONT.

At the very end, he writes: "[the evil twin of the dot product] is not differentiable at parallel vectors."

Could someone explain why? (I can see the issue with the evil twin of the cross, but not the dot)
 
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Look at his formula for the dot evil twin. It has a square root in it. Square root functions are not differentiable at zero. With a little algebra, it's easy to show that the inside of the square root is zero for parallel vectors, so the function is not differentiable there.

The actual function being used is something like ##\sqrt{|x|}##. If you graph that function on the interval ##[-1,1]## you'll see that it's continuous but not differentiable at ##x=0##. It bounces off the x-axis there.
Also look at the graph of ##|\sin x|## on ##[-1,1]## to see another relevant non-differentiable bounce.

By contrast, the formulas for the usual dot and the usual cross have no square roots, or anything else that can upset differentiability, such as the denominators in the evil twin cross. They pass zero angles (parallel for dot, perp for cross) smoothly.
 
andrewkirk said:
Look at his formula for the dot evil twin. It has a square root in it. Square root functions are not differentiable at zero. With a little algebra, it's easy to show that the inside of the square root is zero for parallel vectors, so the function is not differentiable there.

OH! I see!

THANK YOU!
 

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