Toronto Math Forum

Find the first four terms in power-series expansion about the given point for the given function; find the largest disc in which the series is valid:
$$[Log(1-z)]^2 \ about \ z_0=0$$

Answer:
$$[Log(1-z)]' = -\frac{1}{1-z} = -\sum_{n=0}^{\infty}z^n, \text{   valid at }  |z| <1$$
$$Log(1-z) =- \int\sum_{n=0}^{\infty}z^n dz =-\sum_{n=0}^{\infty}\int z^n dz = -\sum_{n=0}^{\infty}\frac{z^n+1}{n+1} =- \sum_{m=1}^{\infty}\frac{z^m}{m}$$

$\lim \limits_{z_{0} \to i}\ f(x) = \lim \limits_{z_{0} \to i}\ |1-z|^2 = |1-i|^2 = 1^2+-(1)^2 = 2$

show that the given equation is not exact but becomes exact
when multiplied by the given integrating factor. Then solve the equation.
$$x^2y^3+x(1+y^2)y'=0, u(x.y)=\frac{1}{xy^3}$$

Solution:
This is not exact because
$$\frac{\partial}{\partial y}(x^2y^3)=3x^2y^2 \neq \frac{\partial}{\partial x}[x(1+y^2)]=q=y^2$$
However, multiplyin by $\mu$, we get:     $$x+\frac{1+y^2}{y^3}y'=0$$
it is exact since $$\frac{\partial}{\partial y}(x) = \frac{\partial}{\partial x}(\frac{1+y^2}{y^3})=0$$
Next, I will find the function $\phi$.
$$\frac{\partial \phi}{\partial x}= x$$
$$\phi = \int x dx = \frac{x^2}{2}+h(y)$$
$$\frac{\partial \phi}{\partial y} = h'(y) = \frac{1+y^2}{y^3}$$
$$h(y) = \int  \frac{1+y^2}{y^3} dy = -\frac{1}{2y^2} + ln|y|$$
As a result, the function is $$\phi (x,y) = \frac{x^2}{2}-\frac{1}{2y^2}+ln|y|$$