Pythagoras

 

Supercalifragilisticexpialidocious

He traveled all around the world and everywhere he went
He'd use his word and all would say there goes a clever gent
When dukes or Maharajas pass the time of day with me
I say me special word and then they ask me out to tea (woo)

But if you say HYPERBOLICfragilisticexpialidocious, then you might be even smarter!

e^ix=cos(x)+isin(x), where cosh(x) is the hyperbolic version of the trigonometric cosine, cos(x), is Euler’s Formula, where i is  the imaginary number, √-1, which is called j by electrical engineers.  This can be restated as y=f(x)=ln(cos(x)+i*sin(x))/i, and which implies cos(x)=e^ix - isin(x). If f(x) is coordinate transformed by rotating by 90° it becomes x=f(y)=ln(cos(y)+i*sin(y))/i. This can restated as y=g(x) where g(x)=sin(90°)f(x).  But this also means that cosh(ix)= cos(y) + isin(y). Any cosine can also restated as √(1-sin²). This means that cosh(ix)=cos(x)+isin(x) therefore becomes cosh(ix)=√(1 - sin²(x))+isin(x) and also cosh(ix)=√(1-sin²(y)+2isin(y)). If only the real components of the complex numbers are used, this means that the rotation of Euler’s Formula by 90° is true only if y and x are identical and zero.

Pythagoras’ Theorem, c=√(a²+b²), is true because cos²(x)+sin²(x)=1, which is also the formula for a circle.  By contrast, cosh²(x)-sinh²(x)=1 is the formula for a hyperbola.  Pythagoras’ Theorem is only true for a flat surface.  If the surface is spherical, whose sphere has a radius R, then the formula for the hypotenuse, c, of a right triangle whose other sides are a and b is cos(c/R)=cos(a/R)cos(b/R).  If this is expressed as a series, then as R approaches infinity, the series becomes c=√(a²+b²). For a hyperbolic surface, the formula for a hypotenuse is cosh(c)=cosh(a)cosh(b), but if the coefficient of the imaginary number is zero, or only the real portion is used, then this also becomes Pythagoras’ Theorem. 

For a finite spherical surface, as R, the radius of the sphere, becomes very large compared to a or b, this becomes Pythagoras’ Theorem.  On a infinte surface, it also becomes Pythagoras’ Theorem but we appaer t only use the ral portionof the solution. Using only the real portion of a complex solution is precisely how electrical engineers treat the imaginary portion when their solution is a complex number. And now you know the rest of the story. If you use only Pythagoras' Theorem, then apparently you have no imagination!  If you have an imagination then you are HYPERBOLICfragilisticexpialidocious!