Demostrar $\exp(x+y) = \exp(x) \exp(y)$

Question

Estoy tratando de demostrar $\exp(x+y) = \exp(x) \exp(y)$.

Puedo usar ese $$\exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}$$ Además yo sé cómo multiplicar dos de alimentación de la serie en un punto, es decir, si $f(x) = \sum_{n=0}^\infty c_n(x-a)^n$ $g(x) = \sum_{k=0}^\infty d_n(x-a)^n$ $$ f(x)g(x) = \sum_{n=0}^\infty e_n(x-a)^n $$ con $$ e_n = \sum_{m=0}^n c_md_{n-m} $$

Answer 1

\begin{align} \exp(x+y)&=\sum_n\frac{(x+y)^n}{n!} \\\\ &=\sum_{n}\frac{1}{n!}\sum_{a+b=n} {n \choose a} x^ay^b \\\\ &= \sum_{n}\frac{n!}{n!}\sum_{a+b=n}\frac{x^a}{a!}\frac{y^b}{b!} \\\\ &= \sum_{a,b} \frac{x^a}{a!}\frac{y^b}{b!} \\\\ &= \exp(x)\cdot\exp(y) \end{align}

Answer 2

Esto puede ser hecho sin escribir una sola suma. Considere la función $$ f(x, y) = \frac{e^x e^y}{e^{x+y}}. $$ Observe that $$ \frac{\partial f}{\partial x} = \frac{e^x e^y e^{x+y} - e^x e^y e^{x+y}}{(e^{x+y})^2} = 0. $$ Del mismo modo, $$ \frac{\partial f}{\partial y} = \frac{e^x e^y e^{x+y} - e^x e^y e^{x+y}}{(e^{x+y})^2} = 0. $$ This shows that $f$ is a constant function. Now, we need only to use the series definition to show $f(0, 0) = 1$. Then, by rearrangement, we have the desired result: $$ e^{x+y} = e^x e^y. $$

Answer 3

Mi solución

Deje $x,y \in \mathbb R$ y

$f(z) := \sum_{n=0}^\infty \left(\frac {x^n}{n!} \right )z^n$ $g(z) := \sum_{n=0}^\infty \left(\frac {y^n}{n!} \right )z^n$ . A continuación,$\exp(x) \exp(y) = f(1)g(1)$. Que es $$ f(z)g(z) = \sum_{n=0}^\infty \left( \sum_{k=0}^m \frac {x^m^{n-m}}{m! (n-m)!} \right)z^n $$ $$ = \sum_{n=0}^\infty \frac 1 {n!} (x+y)^n z^n $$ thus $f(1)g(1) = \exp(x+y)$.

Answer 4

$A(t)=\exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}t^n$
$B(t)=\exp(y) = \sum_{n=0}^\infty \frac {y^n}{n!}t^n$
$C(t) = A(t)*B(t)=\sum_{n=0}^\infty (\sum_{k+z=n}^\ \frac {x^k}{k!}*\frac {y^z}{z!})t^n=\sum_{n=0}^\infty \frac {(x+y)^n}{n!}t^n=exp(x+y)$

y el uso de $t=1$

sry yo era demasiado tarde^^

Answer 5

Fórmula de Euler dice que (forma exponencial de un número complejo) $$e^{i\theta}=\cos\theta+i\sin\theta.$$ Therefor $$e^{x+y}=\cos(-i(x+y))+i\sin(-i(x+y))\\=\cos(xi+yi)-i\sin(xi+yi)\\=(\cos ix\cos iy-\sin ix\sin iy)+i(\sin ix\cos iy+\cos ix\sin iy)\\=(\cos ix+i\sin iy)(\cos iy+i\sin iy)\\=e^xe^y.$$