How can partial fractions be used for deductions?

Question

Find partial fractions of the expression,$\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}$ . Hence deduce that; $\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)}+\frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)}+\frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)}=a+b+c+d-p-q-r-s$

My Working

I was able to calculate partial fractions as follows,

$\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}=1+\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)(x-a)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)(x-b)}+..$

But I cannot proceed to deduction part. Highly appreciated if someone can give me a hint to work this out. Thank you!

Answer 1

This is $$\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}+\frac{D}{x-d}$$ The result should be this here $$1+{\frac {{c}^{4}-{c}^{3}p-{c}^{3}q-{c}^{3}r-{c}^{3}s+{c}^{2}pq+{c}^{2 }pr+{c}^{2}ps+{c}^{2}qr+{c}^{2}qs+{c}^{2}rs-cpqr-cpqs-cprs-cqrs+pqrs}{ \left( -c+a \right) \left( -c+b \right) \left( -d+c \right) \left( x-c \right) }}+{\frac {-{d}^{4}+{d}^{3}p+{d}^{3}q+{d}^{3}r+{d} ^{3}s-{d}^{2}pq-{d}^{2}pr-{d}^{2}ps-{d}^{2}qr-{d}^{2}qs-{d}^{2}rs+dpqr +dpqs+dprs+dqrs-pqrs}{ \left( x-d \right) \left( -d+a \right) \left( -d+b \right) \left( -d+c \right) }}+{\frac {-{b}^{4}+{b}^{3}p +{b}^{3}q+{b}^{3}r+{b}^{3}s-{b}^{2}pq-{b}^{2}pr-{b}^{2}ps-{b}^{2}qr-{b }^{2}qs-{b}^{2}rs+bpqr+bpqs+bprs+bqrs-pqrs}{ \left( x-b \right) \left( -b+a \right) \left( -c+b \right) \left( -d+b \right) }}+{ \frac {{a}^{4}-{a}^{3}p-{a}^{3}q-{a}^{3}r-{a}^{3}s+{a}^{2}pq+{a}^{2}pr +{a}^{2}ps+{a}^{2}qr+{a}^{2}qs+{a}^{2}rs-apqr-apqs-aprs-aqrs+pqrs}{ \left( -b+a \right) \left( -c+a \right) \left( -d+a \right) \left( x-a \right) }} $$

Answer 2

You already got the correct partial fraction decomposition $$ \frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)} =1+\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)} \cdot \frac{1}{x-a} \\ + \frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)} \cdot \frac{1}{x-b} \\ + \frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)} \cdot \frac{1}{x-c} \\ + \frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)} \cdot \frac{1}{x-d} \, . $$ With the substitution $x = \frac 1y$ we get $$ \tag{*} \frac{(1-py)(1-qy)(1-ry)(1-sy)}{(1-ay)(1-by)(1-cy)(1-dy)} =1+\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)} \cdot \frac{y}{1-ay} \\ + \frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)} \cdot \frac{y}{1-by} \\ + \frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)} \cdot \frac{y}{1-cy} \\ + \frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)} \cdot \frac{y}{1-dy} \, . $$ For small $y$ the left-hand side has the development $$ \frac{1 - (p+q+r+s)y + O(y^2)}{1-(a+b+c+d)y + O(y^2)} = (1 - (p+q+r+s)y + O(y^2))(1+(a+b+c+d)y + O(y^2) \\ = 1 + (a+b+c+d-p-q-r-s)y + O(y^2) $$ for $y \to 0$. On the right-hand side we have $$ \frac{y}{1-ay} = y(1+ay + O(y^2)) = y + O(y^2) $$ and similarly for the other fractions $\frac{y}{1-by}$, $\frac{y}{1-cy}$ and $\frac{y}{1-dy}$.

Therefore a comparison of the $y$-terms (i.e. the derivates with respect to $y$ at $y = 0$) in equation $(**)$ gives the desired identity $$ a+b+c+d-p-q-r-s = \frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)} \\ +\frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)} \\ +\frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)} \, . $$