I don't understand why there are constant $A,B,C$ s.t. $$\frac{1}{(x-1)(x-2)^2}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-2)^2}.$$
I now how to compute $A,B,C$, but I don't understand how someone though to do this. In what this is natural ?
I don't understand why there are constant $A,B,C$ s.t. $$\frac{1}{(x-1)(x-2)^2}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-2)^2}.$$
I now how to compute $A,B,C$, but I don't understand how someone though to do this. In what this is natural ?
On
Expanding upon request my comment above, I am going to briefly expose the statement that Partial Fractions
naturally proceed from the chain
Residue Theorem $\to$ Laurent Series $\to$ Mittag-Leffler Theorem , and corollaries.
Consider the function $f(z)$ analytic in all the complex plane, except from having a countable number of isolated poles.
If $z_k$ is one of these poles of multiplicity $m_k$, then the Laurent Theorems tells that the function, in an annulus
around and excluding $z_k$, the function can be expressed as (principal and regular part of the expansion)
$$
f(z) = {{a_{k,\; - m_{\,k} } } \over {\left( {z - z_{\,k} } \right)^{m_{\,k} } }}
+ {{a_{k,\; - \left( {m_{\,k} - 1} \right)} } \over {\left( {z - z_{\,k} } \right)^{m_{\,k} - 1} }}
+ \cdots + {{a_{k,\; - 1} } \over {\left( {z - z_{\,k} } \right)}} + g(z)
$$
and where the coefficents $a$ are derived from the Residue theorem.
All other poles of $f(z)$, if any, are contained in $g(z)$.
Thus we can reiterate the above for $g(z)$, and continue untill the remaining regular part of the expansion ($\gamma(z)$)
does not contain any pole in the finite plane and thus it is an entire function.
The behaviour of $f(z)$ and $\gamma(z)$ for $z \to \infty$ is the same.
Finally if the function is meromorphic, i.e the ratio of two entire functions $f(z)=g(z)/h(z)$, the from the above it follows the decomposition into Partial Fractions, which is analyzed and proved under the Mittag-Leffler theorem.
For those interested in a more detailed presentation, I suggest to read the excellent dedicated chapter in " Theory of analytic functions - A.I. Markuševič ".
It's just Bézout's Identity. There are $A,B\in\mathbb R$ s.t. $$A(x-1)+B(x-2)=1.$$ Then $$\frac{1}{(x-1)(x-2)^2}=\frac{A(x-1)+B(x-2)}{(x-1)(x-2)^2}=\frac{A}{(x-2)^2}+\frac{B}{(x-1)(x-2)}=\frac{A}{(x-2)^2}+\frac{B(A(x-1)+B(x-2))}{(x-1)(x-2)}=\frac{A}{(x-2)^2}+\frac{BA}{(x-2)}+\frac{B^2}{(x-1)}.$$
Edit
In fact Bezout says that there are polynomials $P(x)$ and $Q(x)$ s.t. $$P(x)(x-1)+Q(x)(x-2)=1,$$
but this implies that $P(x)$ and $Q(x)$ must has degree $0$.