The beautiful math formula
Good evening english corner and chattering's friends 
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I open this thread to talk (in english obviously) about a beautiful - in our opinion - math formula and describe it in some words: we are in a math forum
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I begin with the Euler formula
$e^(iy) = cos(y)+i sin(y)$.
Apply it with the basic property of an exponential
$e^z = e^(x+iy) = e^x \cdot e^(iy) = e^x(cos(y)+i sin(y))$
forall $z=x+iy \in \CC$.
This is a great math law that, in a first look, seems to break out our high school knowledge. We grew with the axioms "exponential is one-to-one" and similar properties of the real exponential.
In complex analysis we learn - especially based on this formula - that for purely imaginary values complex exponential is limited and periodic:
$e^(iy) = cos(y)+i sin(y)$.
A first consequence of this is that the modulus of a complex exponential is equal to the modulus of the corresponding real exponential obtained elevating the base to the real part of the complex number:
$|e^z|= |e^(x+iy)| = |e^x| |e^(iy)| = |e^x| = e^x$ (because $x=Re(z)\in \RR$).
We can note that, reverting this, complex sine and cosine are not limited for purely imaginary values.
.I open this thread to talk (in english obviously) about a beautiful - in our opinion - math formula and describe it in some words: we are in a math forum
.I begin with the Euler formula
$e^(iy) = cos(y)+i sin(y)$.
Apply it with the basic property of an exponential
$e^z = e^(x+iy) = e^x \cdot e^(iy) = e^x(cos(y)+i sin(y))$
forall $z=x+iy \in \CC$.
This is a great math law that, in a first look, seems to break out our high school knowledge. We grew with the axioms "exponential is one-to-one" and similar properties of the real exponential.
In complex analysis we learn - especially based on this formula - that for purely imaginary values complex exponential is limited and periodic:
$e^(iy) = cos(y)+i sin(y)$.
A first consequence of this is that the modulus of a complex exponential is equal to the modulus of the corresponding real exponential obtained elevating the base to the real part of the complex number:
$|e^z|= |e^(x+iy)| = |e^x| |e^(iy)| = |e^x| = e^x$ (because $x=Re(z)\in \RR$).
We can note that, reverting this, complex sine and cosine are not limited for purely imaginary values.
Risposte
Again ...
$arctan(x) + arctan(y) = arctan((x + y)/(1 - x*y))$
Let A = arctan(x) and tan(A) = x.
Let B = arctan(y) and tan(B) = y.
tan(A + B) = (tan(A) + tan(B))/(1 - tan(A)*tan(B)).
tan(A + B) = (x + y)/(1 - x*y)
Hence A + B = arctan((x + y)/(1 - x*y))
$arctan(x) + arctan(y) = arctan((x + y)/(1 - x*y))$
Let A = arctan(x) and tan(A) = x.
Let B = arctan(y) and tan(B) = y.
tan(A + B) = (tan(A) + tan(B))/(1 - tan(A)*tan(B)).
tan(A + B) = (x + y)/(1 - x*y)
Hence A + B = arctan((x + y)/(1 - x*y))
hey friends , about viewtopic.php?f=36&t=114200
uhm not sure but I'll give it a go
$x = tan(a)$ , $y = tan(b)$
then
$tan(a + b) = ( tan(a) + tan(b) ) / ( 1 - tan(a)tan(b) )$
$tan(a + b) = ( x + y)/( 1 - xy)$
$a + b = arctan( ( x + y)/(1 - xy) )$
or
$arctan(x) + arctan(y) = arctan ( ( x + y)/( 1 - xy) )
$
similarly
$arctan(x) - arctan(y) = arctan( ( x - y)/(1 + xy) )$
Note sure you would consider this a formidable substitution! but its a start,
Hope this is ok , what do you think about
uhm not sure but I'll give it a go
$x = tan(a)$ , $y = tan(b)$
then
$tan(a + b) = ( tan(a) + tan(b) ) / ( 1 - tan(a)tan(b) )$
$tan(a + b) = ( x + y)/( 1 - xy)$
$a + b = arctan( ( x + y)/(1 - xy) )$
or
$arctan(x) + arctan(y) = arctan ( ( x + y)/( 1 - xy) )
$
similarly
$arctan(x) - arctan(y) = arctan( ( x - y)/(1 + xy) )$
Note sure you would consider this a formidable substitution! but its a start,
Hope this is ok , what do you think about
In fact, I was mistaken too, I don't know what convenient is write at that hour
$i=+-sqrt-1$ , ok, but if we look for the most beautiful formula, I think $i=sqrt-1$ is more romantic
$i=+-sqrt-1$ , ok, but if we look for the most beautiful formula, I think $i=sqrt-1$ is more romantic
The problem with $i = sqrt(-1)$ is that the notation $sqrt(-1)$ is reserved for the principal square root function, which returns only the positive value. This can produce false results:
$-1 = i^2 = i*i = sqrt(-1)*sqrt(-1)=sqrt((-1)*(-1))=sqrt(1)=1$
So, if you want to be correct you must write $i = pm sqrt(-1)$.
I do think that what I wrote in the previous post is actually wrong as well now
$-1 = i^2 = i*i = sqrt(-1)*sqrt(-1)=sqrt((-1)*(-1))=sqrt(1)=1$
So, if you want to be correct you must write $i = pm sqrt(-1)$.
"myself":
$sqrt(-1) = pm i$
I do think that what I wrote in the previous post is actually wrong as well now
"Pianoth":
$E = mc$? Are you sure?
Ehm... $E/c=mc$
"Pianoth":
Actually, the definition is $i^2 = -1 => sqrt(-1) = pm i$.
You right, but (I think) isn't wrong say $i=sqrt(-1) $, because anyway $+i=sqrt(-1)$ is right, do you agree?
"Meringolo":
$E=mc$
$E = mc$? Are you sure?
"Meringolo":
$i=sqrt(-1)$
Actually, the definition is $i^2 = -1 => sqrt(-1) = pm i$.
Anyway, the $tau$ example was just for fun... The $e^(ipi)+1=0$ is definitely more useful than $e^(itau)-1=0$.
Well Zero,
I think that $E=mc$ is a formula that makes us reflect on the philosophical meaning of the things that surround us.
And I think most beautiful formulas are simple ones, for example $i=sqrt(-1)$
But I don't mean that the Einstein's field equation (the tensor equation that describes the geometry of space-time as a function of various parameters) is rough, on the contrary...
$R_(muv)-1/2g_(muv)R+\Lambdag_(muv)=-(8piG)/(c^4)T_(muv)$
I think that $E=mc$ is a formula that makes us reflect on the philosophical meaning of the things that surround us.
And I think most beautiful formulas are simple ones, for example $i=sqrt(-1)$
But I don't mean that the Einstein's field equation (the tensor equation that describes the geometry of space-time as a function of various parameters) is rough, on the contrary...
$R_(muv)-1/2g_(muv)R+\Lambdag_(muv)=-(8piG)/(c^4)T_(muv)$
It's wonderful in all over the form.
But, if you like, I opened this thread to talk about various formulas (whatever you like), just in order to improve our math english.
But, if you like, I opened this thread to talk about various formulas (whatever you like), just in order to improve our math english.
wonderful 
$1=-e^(ipi)$
It has something to charm even so. Or not?
It has something to charm even so. Or not?
yes of course ; ... i prefer $pi$ !
but now i need a hero for this questions
viewtopic.php?f=3&t=114172
Save me , please ...
I can't face this question alone
but now i need a hero for this questions
viewtopic.php?f=3&t=114172
Save me , please ...
I can't face this question alone
"Stellinelm":
[quote="Pianoth"]Well, $ e^(i tau) = cos(tau) + i sin(tau) = 1 $
I didn't know
[/quote]Recently there was a discussion in the section "generale" about the notation $\tau= 2\pi$.
Do you remember? I remember you took part on it (it was "$\tau$ or $\pi$?" or something similar).
"Zero87":
[quote="Stellinelm"]The most beautiful $e^(iπ) + 1 = 0$
I completely agree!
But - for Euler formula explained some posts above - $e^(i\pi)= cos(\pi)+ i sin(\pi)= -1$
Do you have some formula you like to talk about for interest or other?
[/quote]just because it is easy to remember
My love is TdN
"Pianoth":
Well, $ e^(i tau) = cos(tau) + i sin(tau) = 1 $
I didn't know
I know, that was just an example 
Anyway, since we are talking about complex numbers and $e$, I also like the fact that $sinh(x) = -i sin(ix)$, $cosh(x) = cos(ix)$ and $tanh(x) = -i tan(ix)$.
Anyway, since we are talking about complex numbers and $e$, I also like the fact that $sinh(x) = -i sin(ix)$, $cosh(x) = cos(ix)$ and $tanh(x) = -i tan(ix)$.
"Pianoth":
Well, $e^(i tau) = cos(tau) + i sin(tau) = 1$
But, with the use of $\tau$ we have $e^(i \tau) - 1 =0$ that is not romantic as $e^(i \pi)+1=0$.
Well, $e^(i tau) = cos(tau) + i sin(tau) = 1$
"Stellinelm":
The most beautiful $e^(iπ) + 1 = 0$
I completely agree!
But - for Euler formula explained some posts above - $e^(i\pi)= cos(\pi)+ i sin(\pi)= -1$
Do you have some formula you like to talk about for interest or other?
The most beautiful $e^(iπ) + 1 = 0$
"gio73":
My knowledges about complex analysis are next to zero, but I think this subject very interesting.
Don't worry: I made this thread to talk about a formula we love (or simply we know). I began with Euler formula, but the object is "everything" (I meaning math).
"gio73":
Good night all the forumists, I'm going to sleep
Me too! Good night forumists.
Hi dottor Zero, nice to read you again on the English corner!
My knowledges about complex analysis are next to zero, but I think this subject very interesting.
Good night all the forumists, I'm going to sleep
My knowledges about complex analysis are next to zero, but I think this subject very interesting.
Good night all the forumists, I'm going to sleep
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