The beautiful math formula
Good evening english corner and chattering's friends 
.
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.
.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
"Zero87":
My english is awful
(Thanks for corrections!
I don't remember the pronoun you used; boldings are just aimed to emphasize the formula's power.
"Zero87":
I see on you profile that you are 17 years old! It's incredible that you talk of complex analysis at this age, greetings!
Don't expect me to be able to solve a problem in practice...
"Caenorhabditis":
Yes, and therefore the value of any holomorphic function in point 0 is totally determined if we know its behaviour in any loop path including the origin.
My english is awful
(Thanks for corrections!
)I see on you profile that you are 17 years old! It's incredible that you talk of complex analysis at this age, greetings!
Yes, and therefore the value of any holomorphic function in point 0 is totally determined if we know its behaviour in any loop path including the origin.
"Caenorhabditis":
Yes, Eulero's formula is very nice, but this one astounds me every time I see it:
$f(0)=\frac{1}{2 \pi i} \oint\frac{f(z)}{z}dz$
Do you want to explain to us? (in english: this topic is about english improvement
).I suppose it is a theorem of Gauss that says the average value of any holomorphic function on a disc is the value of the same in the centre: it's a thing that doesn't have a corresponding in real analysis.
It's one of the complex analysis theorem that say complex analysis is different (sounds that an "apple advertising") from the real one.
"Caenorhabditis":
Likely it's because I have no idea of how is this possible.
Me too. I am still astonish thinking that $a^z$ could be negative for appropriate complex values $a,z$!
Yes, Eulero's formula is very nice, but this one astounds me every time I see it:
$f(0)=\frac{1}{2 \pi i} \oint\frac{f(z)}{z}dz$
Likely it's because I have no idea of how is this possible.
$f(0)=\frac{1}{2 \pi i} \oint\frac{f(z)}{z}dz$
Likely it's because I have no idea of how is this possible.
"Pianoth":
I wrote them just to repeat a little bit of LaTex
A real kick
"Zero87":
[quote="Stellinelm"]annuntio vobis gaudium magnum , my new love : true table .
"adnuntio", I believe, but my latin is like Lotito's one.
[/quote] "annuntio" is Ok : http://www.vatican.va/holy_father/franc ... dex_it.htm .
"Zero87":
I suppose you was referring to Pianoth's post in your thread "Enigma logico"
yes
"Zero87":
I don't like true tables
I definitely agree!
"Zero87":
I don't like true tables because for complicated formulas they get boring very fast...
I definitely agree! I wrote them just to repeat a little bit of LaTex
"Stellinelm":
annuntio vobis gaudium magnum , my new love : true table .
"adnuntio", I believe, but my latin is like Lotito's one.
I suppose you was referring to Pianoth's post in your thread "Enigma logico": I don't like true tables because for complicated formulas they get boring very fast...
annuntio vobis gaudium magnum , my new love : true table .
"Meringolo":
[quote="Stellinelm"]Hola buddy , very well : chapeau
You've done a mixture of three languages![/quote]
Are you sure? 
"Stellinelm":
Hola buddy , very well : chapeau
You've done a mixture of three languages!
"Zero87":
I remember the course of Phisic I in which the professor couldn't find this expression and for an hour he tried to manipulate various formulas without success... That was comic.
I can't believe
"Meringolo":
TI can write the acceleration as $ddot x$ and so, with some manipulation, becomes the differential equation $ddot x + k/m x =0$
and the solution of this equation is $x=A cos(wt+phi)$
I remember the course of Phisic I in which the professor couldn't find this expression and for an hour he tried to manipulate various formulas without success... That was comic.
How about a simple formula? A basic property of the (real) logarithm
$log(ab)=log(a)+log(b)$, for $a,b>0$ real values.
This is a (very) important property because it make possible transform products in sums:
$log(\prod_(i=1)^n a_i) = \sum_(i=1)^n log(a_i)$.
But take attention: this property holds for real logarithm.
In another way this property make a pair (="fa il paio"
) whit the same of (real) exponential:$x^(a+b)=x^a \cdot x^b$.
Hola buddy , very well : chapeau
The harmonic oscillator's formula describes the motion of a spring, or a pendulum for example.
In the case of spring, we have the negative force that brings the spring at rest, or rather the Hooke's law $-kx$ where k is the spring constant and x is the position.
$F=ma$ but $F$ is Hooke's law $-kx$, so $ma=-kx$
I can write the acceleration as $ddot x$ and so, with some manipulation, becomes the differential equation $ddot x + k/m x =0$
and the solution of this equation is $x=A cos(wt+phi)$
where the angular frequency is $w=sqrt(k/m)$ and the period is $T=2pi sqrt (m/k)$
In the case of spring, we have the negative force that brings the spring at rest, or rather the Hooke's law $-kx$ where k is the spring constant and x is the position.
$F=ma$ but $F$ is Hooke's law $-kx$, so $ma=-kx$
I can write the acceleration as $ddot x$ and so, with some manipulation, becomes the differential equation $ddot x + k/m x =0$
and the solution of this equation is $x=A cos(wt+phi)$
where the angular frequency is $w=sqrt(k/m)$ and the period is $T=2pi sqrt (m/k)$
Under the table : I have no idea ...
"Meringolo":
What do you say about the harmonic oscillator's formula
$x=A cos (wt+\phi)$ ?
Do you want to explain to us this formula?
What do you say about the harmonic oscillator's formula
$x=A cos (wt+\phi)$ ?
$x=A cos (wt+\phi)$ ?
Gugo says ... ok
Gugo says that he wants an "analytic method" without trigonometry: I don't know if your solution solution is suitable 
I found an interesting method (I'm writing it from 9.10 )
I found an interesting method (I'm writing it from 9.10
)
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