anonymous
  • anonymous
Interesting!!
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Just found out these formulas, the proofs for them are very easy. \[\cosh(ix)=\cos(x)\] and\[\cos(ix)=\cosh(x)\] where cosh(x) is the hyperbolic cosine function given by \[\cosh(x)=\frac{e^x+e^{-x}}{2}\] Now this is easily proved using euler's identity \[e^{ix}=\cos(x)+i \sin(x)\] and consider \[e^{i(-x)}=\cos(-x)+i \sin(-x)\] \[\implies e^{-ix}=\cos(x)-i \sin(x)\] \[\cosh(ix)=\frac{e^{ix}+e^{-ix}}{2}=\frac{\cos(x)+i \sin(x)+\cos(x)-i \sin(x)}{2}=\frac{2\cos(x)}{2}=\cos(x)\] Now consider the argument ix and -ix \[e^{i(ix)}=\cos(ix)+i \sin(ix)\] \[\implies e^{-x}=\cos(ix)+i \sin(ix)\] \[e^{i(-ix)}=\cos(-ix)+i \sin(-ix)\] \[\implies e^x=\cos(ix)-i \sin(ix)\] Now for the hyperbolic function \[\cosh(x)=\frac{e^x+e^{-x}}{2}=\frac{\cos(ix)-i \sin(ix)+\cos(ix)+i \sin(ix)}{2}=\frac{2\cos(ix)}{2}=\cos(ix)\]
anonymous
  • anonymous
I found them very interesting, is there a name for such pairs?
UsukiDoll
  • UsukiDoll
I used to know a name...but I forgot... I had to use this in PDEs though.

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Michele_Laino
  • Michele_Laino
the trigonometric functions and the hyperbolic trigonometric functions are defined for real values only. For example the writing cos(ix) has no meaning
Empty
  • Empty
You've found something quite interesting, it turns out both of these functions are intimately related if you look into the complex plane! In fact, @Nishant_Garg the two formulas you have given are actually basically the same thing, check this out: \[\cosh(ix) = \cos(x)\] We can then plug in \(x \to ix\) to get: \[\cosh(i *i x) = \cos(ix)\] \[\cosh(-x) = \cos(ix)\] However cosh is an even function, so: \[\cosh(x) = \cos(ix)\] You also might have noticed there is a similar situation with sine and sinh. I'll leave that to you to figure out, it's a little trickier. A more general form of this would be to combine both as: \[e^x = \cosh x + \sinh x\] \[e^{iy} = \cos y + i \sin y\] into \[e^{x+iy} = ( \cosh x + \sinh x) ( \cos y + i \sin y)\] To get one big bad function of a complex variable!
Michele_Laino
  • Michele_Laino
I think that if we are in complex analysis, then we we are not in real analysis and vice versa
anonymous
  • anonymous
amazing, hyperbolic functions are mind blowing
Michele_Laino
  • Michele_Laino
all real functions can be defined in complex plane, nevertheless when we make that transition we go from real analysis into the complex one
Empty
  • Empty
I think it's a good question to ask, why is Euler's constant "e" related to the quadratic forms at all in the first place @Nishant_Garg ? Weird huh? :D
Michele_Laino
  • Michele_Laino
for example: cos(ix) is a complex function, so it is represented by a complex number whereas cosh(x) is a real function, so has the subsequent writing meaning: cosh(x)=cos(ix) ?
Empty
  • Empty
@Michele_Laino I think the fact that we've used "i" from the start says we're out of the realm of real analysis to begin with.
Michele_Laino
  • Michele_Laino
ok! nevertheless what do you think about the subsequent formula: cosh(x)=cos(ix) ? @Empty
Michele_Laino
  • Michele_Laino
I understand that cos(ix) is a real number for each value of x?
Empty
  • Empty
cos(x) is a real number for each value of x, however cos(ix) is not, we can consider an imaginary circular angle to be a hyperbolic angle! This is very much like when you look at a conic section and see that the hyperbola is a 90 degree cut perpendicular to the cut that creates the circle! https://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/60cbe5b1-6b6c-48bd-9208-df73201f498a.gif
anonymous
  • anonymous
I tried for the sine let's see \[\sinh(ix)=\frac{e^{ix}-e^{-ix}}{2}=\frac{\cos(x)+i \sin(x)-(\cos(x)-i \sin(x))}{2}\]\[\sinh(ix)=\frac{\cos(x)+i \sin(x)-\cos(x)+i \sin(x)}{2}=\frac{2i \sin(x)}{2}=i \sin(x)\] and \[\sinh(x)=\frac{e^x-e^{-x}}{2}=\frac{\cos(ix)-i \sin(ix)-(\cos(ix)+i \sin(ix))}{2}\]\[\sinh(x)=\frac{\cos(ix)-i \sin(ix)-\cos(ix)-i \sin(ix)}{2}=\frac{-2i \sin(ix)}{2}=-i \sin(x)\] \[\sinh(x)=-i \sin(ix)\]\[i \sin(ix)=-\sinh(x)\]\[i \sin(ix)=i^2 \sinh(x)\]\[\sin(ix)=i \sinh(x)\]
Michele_Laino
  • Michele_Laino
right! nevertheless cosh(x) is a real number for each value of x @Empty
Empty
  • Empty
Wait what have I said haha I agree with you, cosh(x) and cos(ix) will both have a completely real range if x is completely real domain, because they're equal.
Empty
  • Empty
@Nishant_Garg Perfect! :D So is there an elliptic or parabolic pair of trig functions hiding out there somewhere? :P
anonymous
  • anonymous
No idea, that would be interesting though!! a trig functions for every conic section
Empty
  • Empty
Why is it that these quadratic forms are represented by e^x but what about cubic forms? It seems sort of weird and special for some reason and I don't know why.

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