Partial Fractions and Interpolations
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Partial Fraction Decompositions
Evaluating partial fraction decomposition is a tedious job. For instance, making partial fractions of $x/((x+1)(x+2)(x+3))$ first demands to set up
\[\frac{x}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3},\]and make a common denominator and compare the numerators:
\begin{equation} \label{eqn:PFD-eg} x = A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2). \end{equation}
However, for simple denominators (i.e., denominators without multiple factors), this task can be detoured using Lagrange Interpolationg Polynomials.
Lagrange Interpolation Polynoimals
The Wikipedia link serves the most general definition of such polynomials, which are not hard to understand through. Thus at this post, we will specifically deal with an example set, ${x_1=-1,x_2=-2,x_3=-3}$.
Think of a polynomial $P(x)$ that has degree $\leq 2$. Then this $P(x)$ is found by the following formula:
\begin{equation} \label{eqn:LIP-elem} P(x) = P(-1)\frac{(x+2)(x+3)}{(-1+2)(-1+3)}+P(-2)\frac{(x+1)(x+3)}{(-2+1)(-2+3)} + P(-3)\frac{(x+1)(x+2)}{(-3+1)(-3+2)}. \end{equation}
In particular, say if $P(x)=\ell_{x_1}(x)$ has $\ell_{x_1}(-1)=1$ while $\ell_{x_1}(-2)=\ell_{x_1}(-3)=0$, we have
\[\ell_{x_1}(x)=\frac{(x+2)(x+3)}{(-1+2)(-1+3)},\]which consists of
- the numerator $(x+2)(x+3)$ multiplying all $(x-x_j)$’s except $(x-x_1)$,
- the denominator $(-1+2)(-1+3)$ that evaluates the numerator at $x=x_1$.
The other two polynomials $\ell_{x_2}(x)$ and $\ell_{x_3}(x)$ are formed with the similar spirit, each respectively designed to have
- degree $\leq 2$, and
- $\ell_{x_i}(x_j)=\delta_{ij}$ (i.e., $=1$ if $i=j$ and $=0$ if $i\neq j$).
By this we have
\begin{equation} \label{eqn:LIP-abs} P(x) = P(x_1)\cdot \ell_{x_1}(x) + P(x_2)\cdot \ell_{x_2}(x)+ P(x_3)\cdot \ell_{x_3}(x). \end{equation}
The polynomials $\ell_{x_1}(x),\ell_{x_2}(x),\ell_{x_3}(x)$ are called Lagrange Interpolation Polynomials. It really depends on the interpolationg points’ set ${x_1,x_2,x_3}$, but we only noted a point for the notational convenience.
Application to Partial Fractions
Having this idea, we apply \eqref{eqn:LIP-abs} as follows.
\[x = (-1)\cdot \ell_{-1}(x) + (-2)\cdot \ell_{-2}(x) + (-3)\cdot \ell_{-3}(x).\]That is,
\[x = -\frac{(x+2)(x+3)}{(-1+2)(-1+3)} - 2\frac{(x+1)(x+3)}{(-2+1)(-2+3)} - 3\frac{(x+1)(x+2)}{(-3+1)(-3+2)},\] \[\therefore x = -\frac12(x+2)(x+3) + 2(x+1)(x+3) -\frac32(x+1)(x+2).\]Compare this with \eqref{eqn:PFD-eg}:
\[x=A(x+2)(x+3) + B(x+1)(x+3)+C(x+1)(x+2).\]This thus tells that $A=-\frac12$, $B=2$, and $C=-\frac32$ should be the case. No equation solving, but obtained all with some magic polynomials!
More Decompositions
The underlying theory is just it. Given a partial fraction decomposition problem, interpret this with Lagrange interpolations, and match the corresponding coefficients.
However this only apparently applies to the denominators of the sort
\[\frac{x^2+2x-1}{x(x-1)(x+1)},\]i.e., with real and distinct roots, and apparently does not apply for the denominators like
\[\frac{v^2-v}{(1+v)(1+v^2)},\quad \frac{t+1}{t^2(t-1)},\]i.e., ones with imaginary roots, or ones with repeating roots.
Imaginary Roots
The Lagrange interpolation polynomials are not limited to reals, and rather it applies for any field (structure with (commutative) addition, multiplication, and division).
Hence for say ${-1,i,-i}$, we still have the formula
\[P(x)=P(-1)\cdot \ell_{-1}(x)+P(i)\cdot \ell_i(x)+P(-i)\cdot \ell_{-i}(x),\]for a polynomial with degree $\leq 2$. It is just the complex number intervening in the computations. Since
\[\ell_{-1}(x)=\frac{(x+i)(x-i)}{(-1+i)(-1-i)}=\frac12(x^2+1),\] \[\ell_i(x)=\frac{(x+1)(x+i)}{(i+1)(i+i)}=-\frac{1+i}{4}(x+1)(x+i),\] \[\ell_{-i}(x)=\frac{(x+1)(x-i)}{(-i+1)(-i-i)}=-\frac{1-i}{4}(x+1)(x-i),\]for say $P(x)=x^2-x$, we have
\[x^2-x=2\cdot\ell_{-1}(x)-(1+i)\cdot\ell_i(x)-(1-i)\ell_{-i}(x),\]which cleans up to
\[x^2-x=(x^2+1)+\frac{i}{2}(x+1)(x+i)-\frac{i}{2}(x+1)(x-i)\] \[=(x^2+1)+(x+1)\left[\frac{i}{2}(x+i)-\frac{i}{2}(x-i)\right]\] \[=(x^2+1)-(x+1),\]so dividing out $(x+1)(x^2+1)$, we have
\[\frac{x^2-x}{(x+1)(x^2+1)}=\frac{1}{x+1}-\frac{1}{x^2+1}.\]Multiple Roots
Algebraically, this one is more tricky. One way is to iterate the partial fraction decompositions:
\[\frac{t+1}{t^2(t-1)}=\frac1t\left[\frac{t+1}{t(t-1)}\right]\] \[=\frac{1}{t}\left[-\frac{1}{t}+\frac{2}{t-1}\right]\] \[=-\frac1{t^2}+\frac{2}{t(t-1)}\] \[=-\frac1{t^2}-\frac{2}{t}+\frac{2}{t-1}.\]Another is to “split” the roots of the denominator $t^2(t-1)$, to say ${0,\epsilon,1}$ (where $\epsilon$ is an infinitesimal). In that case we have the interpolation polynomials
\[\ell_0(t)=\frac{(t-\epsilon)(t-1)}{\epsilon},\] \[\ell_\epsilon(x)=\frac{t(t-1)}{-\epsilon(1-\epsilon)},\] \[\ell_1(x)=\frac{t(t-\epsilon)}{1-\epsilon}.\]Furthermore,
\[t+1=1\cdot\ell_0(t)+(1+\epsilon)\cdot\ell_\epsilon(x)+2\cdot\ell_1(x),\]and cleaning up the formula we obtain,
\[t+1=\frac{1}{\epsilon}(t-\epsilon)(t-1)-\frac{1+\epsilon}{\epsilon(1-\epsilon)}t(t-1)+\frac{2}{1-\epsilon}t(t-\epsilon).\]Clearly the first two terms will be problematic if we let $\epsilon\to 0$. How to remedy this? Probably doing more algebra for the first two terms will help to cancel out $\epsilon$ in the denomninator:
\[(t-1)\left[\frac1\epsilon(t-\epsilon) - \frac{1+\epsilon}{\epsilon(1-\epsilon)}t\right]=(t-1)\left[-1+\frac{t}{\epsilon}\left(1-\frac{1+\epsilon}{1-\epsilon}\right)\right]\] \[=(t-1)\left[-1-\frac{2t}{1-\epsilon}\right].\]Hence we now have a `safe’ version
\[t+1=-(t-1)-\frac{2}{1-\epsilon}t(t-1)+\frac{2}{1-\epsilon}t(t-\epsilon),\]so sending $\epsilon\to 0$ we have
\[t+1=-(t-1)-2t(t-1)+2t^2,\]and dividing out $t^2(t-1)$ gives
\[\frac{t+1}{t^2(t-1)}=-\frac1{t^2}-\frac2t+\frac2{t-1},\]recovering the same partial fraction decomposition as above.
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- 211208: Created