ln(2+x)−tx=0 la ecuación
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
/ / -2*t\\ / / -2*t\\
|W\-t*e /| |W\-t*e /|
x1 = -2 - re|-----------| - I*im|-----------|
\ t / \ t /
$$x_{1} = - \operatorname{re}{\left(\frac{W\left(- t e^{- 2 t}\right)}{t}\right)} - i \operatorname{im}{\left(\frac{W\left(- t e^{- 2 t}\right)}{t}\right)} - 2$$
x1 = -re(LambertW(-t*exp(-2*t))/t) - i*im(LambertW(-t*exp(-2*t))/t) - 2
Suma y producto de raíces
[src]
/ / -2*t\\ / / -2*t\\
|W\-t*e /| |W\-t*e /|
-2 - re|-----------| - I*im|-----------|
\ t / \ t /
$$- \operatorname{re}{\left(\frac{W\left(- t e^{- 2 t}\right)}{t}\right)} - i \operatorname{im}{\left(\frac{W\left(- t e^{- 2 t}\right)}{t}\right)} - 2$$
/ / -2*t\\ / / -2*t\\
|W\-t*e /| |W\-t*e /|
-2 - re|-----------| - I*im|-----------|
\ t / \ t /
$$- \operatorname{re}{\left(\frac{W\left(- t e^{- 2 t}\right)}{t}\right)} - i \operatorname{im}{\left(\frac{W\left(- t e^{- 2 t}\right)}{t}\right)} - 2$$
/ / -2*t\\ / / -2*t\\
|W\-t*e /| |W\-t*e /|
-2 - re|-----------| - I*im|-----------|
\ t / \ t /
$$- \operatorname{re}{\left(\frac{W\left(- t e^{- 2 t}\right)}{t}\right)} - i \operatorname{im}{\left(\frac{W\left(- t e^{- 2 t}\right)}{t}\right)} - 2$$
/ / -2*t\\ / / -2*t\\
|W\-t*e /| |W\-t*e /|
-2 - re|-----------| - I*im|-----------|
\ t / \ t /
$$- \operatorname{re}{\left(\frac{W\left(- t e^{- 2 t}\right)}{t}\right)} - i \operatorname{im}{\left(\frac{W\left(- t e^{- 2 t}\right)}{t}\right)} - 2$$
-2 - re(LambertW(-t*exp(-2*t))/t) - i*im(LambertW(-t*exp(-2*t))/t)