Nice math proof arctan(x)+arctan(1/x)=sign(x)pi/2

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In summary, the conversation discusses a proof involving Brewster's angle in optics and its relation to the equation arctan(x)+arctan(1/x)=sign(x)pi/2. The discussion includes a diagram and explanation of the concept of negative refractive indexes. There is also mention of using the sign(x) function and its definition as a function that is negative on the left side of the axis and positive on the right side.
  • #1
johnsmi
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Hi everybody, I am new here and like physics very much
I came across a post here about the proof for: arctan(x)+arctan(1/x)=sign(x)pi/2
and wanted to share a differnt point of view (not completely scientific but could be aranged)
OK. so the proof uses the Brewster angle in Optics.

Brewster summarization:
Brewster's angle is the angle where the wave is completely transferred from
one matter (n_1) to another (n_2): tan(Theta_B)=n_2/n_1
=>Theta_B=arctan(n_2/n_1)
Now Imagine two waves one from n_1 hitting the surface at brewster angle and another on the other side of the surface at it's own brewster angle (the fraction of n's is inverse)
it is pretty "clear" (unexplained here but true) that the first one will continue at the same angle as the second one hit the surface and therefore according to snell's law the angle between the first hitting wave and it's transferred wave will be pi/2. (same for the second wave)
===>arctan(x)+arctan(1/x)=sign(x)pi/2
Yeah, I know its not complete and it does not work for negative x's (no negative refractive index as far as I know) but it's a little 'out of the box'
 
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  • #2
Hi John,

I'm new here, too. I like the idea of using an application to prove a pure math theorem, but I don't think I have a very good visual of your explanation. Can you post a diagram so I can see what's going on?
 
  • #3
Here is a sketch:
http://img9.imageshack.us/img9/3328/45971616gg3.jpg

Two different matters (refractive indexes n1 & n2)
The xy plane is the border between matter 1 and 2.

K1,2 are the wave vectors and ThetaB12,21 are the Brewster angles for waves from 1 to 2 and from 2 to 1 respectively.

Now try and read the last post and see if it makes it any clearer.
If not I don't mind trying again.
 
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  • #4
I'm afraid there are negative refractive indexes, now all you have to do is generalise it. :-p

http://en.wikipedia.org/wiki/Negative_refractive_index#Negative_refractive_index

Superlens

The first superlens with a negative refractive index provided resolution three times better than the diffraction limit and was demonstrated at microwave frequencies at the University of Toronto by A. Grbic and G.V. Eleftheriades.[9] Subsequently, the first optical superlens (an optical lens which exceeds the diffraction limit) was created and demonstrated in 2005 by Xiang Zhang et al. of UC Berkeley, as reported that year in the April 22 issue of the journal Science,[10] but their lens did not rely on negative refraction. Instead, they used a thin silver film to enhance the evanescent modes through surface plasmon coupling. This idea was first suggested by John Pendry in Physical Review Letters.
 
  • #5
Yeah your right I forgot.

In one of my courses (EM fields) we actually tried to build (theoretically) a matter made of little dielectric balls and calculated their properties in order to get a negative refractive index.
negativerefraction.jpg


And here is what it would look like: (the one on the right)
445346a-i2.0.jpg


By the way that is one of the ways they intend to build invisible cloaks where the EM waves detour the object
 
  • #6
I wonder if anyone would be interested in a proof in all cases?

Never the less it is interesting. I always groan when my maths program throws out the signum function though. :eek:
 
  • #7
you prefer looking at it as a step function?:wink:
 
  • #8
johnsmi said:
you prefer looking at it as a step function?:wink:

Given a preference, not looking at it at all would be favourite. :smile:
 
  • #9
Hi there I noticed that you used the sign(x) function, I came across it the other day and had trouble trying to figure it out I asked some Math C teachers I know but they didn't know either, I am wondering if you could please explain it to me
 
  • #10
First of all I don't know if you noticed but the post is from two and a half years ago (quite some time).

Now, if you want to know about the sign function an easy way to think of it is a funtion which is negative one on the left hand side of the axis and positive otherwise.
In other words:
if x<0 sign(x)= -1
else sign(x)= +1

if anymore explaining is needed i'd be more than happy to help though I am sure you could find a whole topic about it in Wiki

Good luck
 

Related to Nice math proof arctan(x)+arctan(1/x)=sign(x)pi/2

1. What is the significance of the "arctan(x)+arctan(1/x)=sign(x)pi/2" proof in mathematics?

The proof serves to demonstrate the relationship between the inverse tangent function and the sign of a number, showing that the sum of the inverse tangent of a number and the inverse tangent of its reciprocal equals the sign of the number multiplied by π/2.

2. How can this proof be applied in real-world situations?

This proof has various applications in physics, engineering, and other fields where trigonometric functions are used. For example, it can be used in calculating the phase difference between two signals in electrical engineering or in determining the angle of inclination in a right triangle in surveying and navigation.

3. What are the key steps in this proof?

The key steps involve manipulating the inverse tangent function using trigonometric identities, substituting the reciprocal of x for 1/x, and then using the properties of the inverse tangent function to arrive at the final equation of arctan(x)+arctan(1/x)=sign(x)pi/2.

4. Is this proof applicable to all values of x?

Yes, this proof is applicable to all real numbers except for zero, as the reciprocal of zero is undefined.

5. Can this proof be extended to other trigonometric functions?

Yes, this proof can be extended to other inverse trigonometric functions, such as arccosine and arcsine, by using similar techniques and identities. However, the resulting equations may vary depending on the specific trigonometric function involved.

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