Prove that if $f=\frac{1}{(x-a_1)*(x-a_2)...}$, then $f=\frac{k_1}{x-a_1}+\frac{k_2}{x-a_2}+...$ where $k_1=\frac{1}{(a_1-a_2) ... }$ distinct $a_i$

Question

Restatement of Question for Easier Reference

Prove that if $f=\frac{1}{(x-a_1)*(x-a_2)...}$, for distinct $a_i$ then $f=\frac{k_1}{x-a_1}+\frac{k_2}{x-a_2}+...$ where $k_1=\frac{1}{a_1-a_2} ... \tag{Eq. 1}$

Motivation: Steps to Solve $k_1$ and $k_2$ if $f=\frac{k_1}{x-a_1}+\frac{k_2}{x-a_2}+...$ Exists

If $f=\frac{1}{(x-a_1)*(x-a_2)...}=\frac{k_1}{x-a_1}+\frac{k_2}{x-a_2}+...$, distinct $a_i$ , then start as follows:

Multiply left and right sides by $x-a_1$ yielding:

$$f=\frac{x-a_1}{(x-a_1)*(x-a_2)...}=\frac{k_1*(x-a_1)}{x-a_1} +\frac{k_2*(x-a_1)}{x-a_2}+... \tag{Eq. 2}$$

Then take the limit $\underset{x \to {a_1}} \lim f$:

$$\underset{x \to {a_1}} \lim f= \frac{1}{(a_1-a_2)...}=\underset{x \to {a_1}} \lim \frac{k_1*(x-a_1)}{x-a_1} +\frac{k_2*(x-a_1)}{x-a_2}+... =k_1$$ $$\tag{Eq. 3}$$ Then, from Equation 3, arrive at Equation 4: $$\underset{x \to {a_1}} \lim k_1 = \frac{1}{(a_1-a_2)...} \tag{Eq. 4}$$ But how is it possible to show that the Equality Represented by Equation 1 always exists given $f=\frac{1}{(x-a_1)(x-a_2)...}$, distinct $a_i$ ?

I found this one Swarthmore University Reference on "Partial Fraction Expansion (or Decomposition)", but I still do not understand the proof of the concept.

Also, I found the Wolfram Partial-Fraction-Calculator, but I still do not understand how it goes from the multiplied roots in the donominator to the individual roots $\frac{k_1}{x-a_1}...$, the circumstances where this calculation is a proven identity, and the circumstances not provable.

Also I find the Wikipedia reference not so easy for me to understand.

Is there a contained, clear, and concise proof of the equality?

I think that I am understanding the concept of a Polynomial GCD from this reference at Wolfram "PolynomialGCD".

For instance, if one wants to compute the Polynomal GCD of the following polynomials: $$\text{PolynomialGCD( }(x+3)^2*(x-2)*x,(x-2)*(x+3)*(x-1) \text{ ) }\\=(x-2)*(x+3)$$ $$\tag{Eq. 5}$$

I am starting to understand the concepts of the "Fundamental Theorem of Algebra" from "Further linear algebra. Chapter II. Polynomials." by Andrei Yafaev. This ties together the concepts of prime numbers and an irreducible polynomials.

Maybe one way to start towards the answer is to take a simple case.

Say $$f=\frac{k_1}{x-a_1}+\frac{k_2}{x-a_2} \tag{Eq. 6}$$ where $a_1$ and $a_2$ are distinct.

So multiply both sides by $(x-a_1)(x-a_2)$ yielding:

$$f*(x-a_1)(x-a_2)=(k_1)*(x-a_2)+(k_2)*(x-a_1) \tag{Eq. 7}$$

So, to get rid of the $x$ contribution in the numerator, set $k_1=-k_2$. Also, the numerator needs to be $1$; so in this case $-k_1*a_2-k_2*a_1=1$ so $k_1*a_1-k_1*a_2=1$ so then $k_1=\frac{1}{a_1-a_2}$ and $k_2=\frac{1}{a_2-a_1}$.

Dividing Equation 7 again by $(x-a_1)(x-a_2)$ yields (as was to be proven for two variables $a_1$ and $a_2$):

$$f(x)=\frac{1}{(x-a_1)(x-a_2)}$$

So I am looking for an answer similar to this one that extends the concept using simple math. Already the math listed here is very helpful for a separate question that I am in the process of answering, namely:

"Help with the indefinite integral $\int \frac{dx}{2x^4 + 3x^2 + 5}$"

Thank you for your comments and help that assisted me with the thought process to get the result for two distinct variables without advanced math concepts.

If someone could also write-up a self-contained similar answer, I would greatly appreciate that!

Answer 1

Here is a proof by induction. Given $n\in \Bbb Z_+$ and distinct constants $a_1,...,a_n\,$, we need to prove that $$\prod_{j=1}^n\frac1{x-a_j}=\sum_{j=1}^n\frac{c_j}{x-a_j}$$ for some constants $c_1,...,c_n$.

First, note that the result is immediate for $n=1$ (with $c_1=1$). Now suppose the result to have been established for $n=k\,$: $$\prod_{j=1}^k\frac1{x-a_j}=\sum_{j=1}^k\frac{c_j}{x-a_j}$$ for some $c_1,...,c_k\in\Bbb R$. Then $$\begin{align}\prod_{j=1}^{k+1}\frac1{x-a_j}&=\sum_{j=1}^k\frac{c_j}{(x-a_j)(x-a_{k+1})}\\&=\sum_{j=1}^k\frac{c_j}{a_j-a_{k+1}}\left(\frac1{x-a_j}-\frac1{x-a_{k+1}}\right)\\ &=\sum_{j=1}^k\frac{c_j/(a_j-a_{k+1})}{x-a_j}+\frac{\sum_{j=1}^kc_j/(a_{k+1}-a_j)}{x-a_{k+1}}\\ &=\sum_{j=1}^{k+1}\frac{c'_j}{x-a_j},\end{align}$$ where $c'_j:=c_j/(a_j-a_{k+1})\;(j=1,...,k)$ and $c'_{k+1}:=\sum_{j=1}^kc_j/(a_{k+1}-a_j).$ So the result holds also for $n=k+1$. By induction, the result holds for all $n=1,2,...$.

Answer 2

Restatement of Question for Clarity and Mathematical Inductive Approach Towards Its Solution

This answer attempts to follow John Bentin's proof that is well-suited for an advanced audience, but may no be so easy to grasp for those less-versed in induction and the specifics of it as it applies to the question.

Here is a proof by induction. Given $n\in \Bbb Z_+$ and distinct constants $a_1,...,a_n\,$, we need to prove that $$\frac1{\prod_{j=1}^n(x-a_j)}=\sum_{j=1}^n\frac{c_j}{x-a_j} \tag{Eq. 1}$$ for some constants $c_1,...,c_n$.

1. Proof by Induction that $\frac1{\prod_{j=1}^n(x-a_j)}=\sum_{j=1}^n\frac{c_j}{x-a_j}$ for All Distinct $a_j$

First, note that the result is immediate for $n=1$ (with $c_1=1$). Now suppose the result is to be established for $n=k\,$: $$\frac1{\prod_{j=1}^k(x-a_j)}=\sum_{j=1}^k\frac{c_j}{x-a_j} \tag{Eq. 2}$$ for some $c_1,...,c_k\in\Bbb R$. Then $$\begin{align}\frac1{\prod_{j=1}^{k+1}(x-a_j)}&=\sum_{j=1}^k\frac{c_j}{(x-a_j)(x-a_{k+1})} \tag{Eq. 3a}\\ &=\sum_{j=1}^k\frac{c_j}{a_j-a_{k+1}}\left(\frac1{x-a_j}-\frac1{x-a_{k+1}}\right) \tag{Eq. 3b}\\ &=\sum_{j=1}^k\frac{c_j/(a_j-a_{k+1})}{x-a_j}+\frac{\sum_{j=1}^kc_j/(a_{k+1}-a_j)}{x-a_{k+1}} \tag{Eq. 3c}\\ &=\sum_{j=1}^{k+1}\frac{c'_j}{x-a_j}, \tag{Eq. 3d} \end{align}$$ $$\tag{Eqs. 3}$$ where $c'_j:=c_j/(a_j-a_{k+1})\;(j=1,...,k)$ and $c'_{k+1}:=\sum_{j=1}^kc_j/(a_{k+1}-a_j).$ So the result holds also for $n=k+1$. By induction, the result holds for all $n=1,2,...$.

2. Some Simple Examples that $\frac1{\prod_{j=1}^n(x-a_j)}=\sum_{j=1}^n\frac{c_j}{x-a_j}$ for All Distinct $a_j$ Starting With $n=1$

From Equation 1, for $n=1$, $\frac1{\prod_{j=1}^n(x-a_j)}$= $\frac1{\prod_{j=1}^{n=1}(x-a_j)}$=$\frac1{x-a_1}$.

Also, $\sum_{j=1}^n\frac{c_j}{x-a_j} =\sum_{j=1}^{n=1}\frac{c_j}{x-a_j} =\frac{c_1}{x-a_1}$.

Obviously, in this case, if $c_1=1$ then $\frac1{\prod_{j=1}^{n=1}(x-a_j)}$=$\frac1{x-a_1}= \sum_{j=1}^{n=1}\frac{c_j}{x-a_j}$=$\frac1{x-a_2}$.

This result is as to be proven for the case $n=1$.

3. Induction Example for $k=1$, $n=2$ that $\frac1{\prod_{j=1}^n(x-a_j)}=\sum_{j=1}^n\frac{c_j}{x-a_j}$ for All Distinct $a_j$

Now the step of induction needs to be taken for the case that $n=2$, since the case of $n=1$ has already been shown true (for unique $a_1$). The proof by induction proceeds using Equation 2 and Equations 3 (namely Equation 3a, Equation 3b, Equation 3c, and Equation 3d).

4. Proof by Induction that $\frac1{\prod_{j=1}^{n=2}(x-a_j)}=\sum_{j=1}^{n=2}\frac{c_j}{x-a_j}$ for All Distinct $a_j$ for the Inductive Case $n=2$

First, note that the result is immediate for $n=1$, $k=1$, and (with $c_1=1$), as was shown in Section 2, "Some Simple Examples that $\frac1{\prod_{j=1}^n(x-a_j)}=\sum_{j=1}^n\frac{c_j}{x-a_j}$ for All Distinct $a_j$ Starting With $n=1$". Now suppose the result is to be inductively established for $n=2\,$. $k=1\,$ and $k+1=2\,$: $$\frac1{\prod_{j=1}^{n=1}(x-a_j)}= \frac1{x-a_1} =\\= \sum_{j=1}^{n=1}\frac{c_j}{x-a_j}=\frac{c_1}{x-a_1} \tag{From Eq. 2}$$ for some $c_1 \in \Bbb R$, namely here $c_1=1$. Then:

$$\begin{align} \text{Eq. 3a} \underset{implies}\implies \frac1{\prod_{j=1}^{n=1+1}(x-a_j)} &=\sum_{j=1}^{n=1}\frac{c_j}{(x-a_j)(x-a_{k+1})} \\ \underset{implies}\implies \frac1{(x-a_1)}\frac1{(x-a_2)} &=\frac{c_1}{(x-a_1)(x-a_2)} \\ \text{Eq. 3b} \underset{implies}\implies &=\sum_{j=1}^{n=1} \frac{c_j}{a_j-a_{k+1}}\left(\frac1{x-a_j}-\frac1{x-a_{k+1}}\right)\\ &= \frac{c_1}{a_1-a_{2}} \left(\frac1{x-a_1}-\frac1{x-a_{2}}\right)\\ \end{align}$$ $$ \text{Since } \frac1{x-a_1}\frac1{x-a_2} =\frac1{a_1-a_2}\left(\frac1{x-a_1}-\frac1{x-a_2}\right) \\ \text{Because }\frac1{a_1-a_2}\left(\frac1{x-a_1}-\frac1{x-a_2}\right)*\left(x-a_1\right)*\left(x-a_2\right)= \\ =\frac1{a_1-a_2}\left(x-a_2-x+a_1\right)=1 $$ $$ \begin{align} \text{Eq. 3c} \underset{implies}\implies \frac1{x-a_1}\frac1{x-a_2} &=\sum_{j=1}^{n=2} \frac{c_j/(a_j-a_{k+1})}{x-a_j} +\frac{\sum_{j=1}^{n=2} c_j/(a_{k+1}-a_j)}{x-a_{k+1}}\\ \text{Eq. 3d} \underset{implies}\implies \frac1{x-a_1}\frac1{x-a_2}&=\sum_{j=1}^{n=2}\frac{c'_j}{x-a_j}, \\&=\frac1{a_1-a_2}\left(\frac1{x-a_1}-\frac1{x-a_2}\right) \end{align} $$ $$\tag{Implied by Eqs. 3}$$ where $c'_j:=c_j/(a_j-a_{k+1=2})\;(j=1)$ and $c'_{k+1}:=\sum_{j=1}^{n=2} c_j/(a_{k+1}-a_j).$ So the result holds also for $n=2$. By induction, from Section 1, "1. Proof by Induction that $\frac1{\prod_{j=1}^n(x-a_j)}=\sum_{j=1}^n\frac{c_j}{x-a_j}$ for All Distinct $a_j$" this result also holds for all $n=1,2,...$.

Here, in this section, this result is as to be proven for the case n=2.