# Partial Fractions Question

1. Sep 6, 2007

### SanjeevGupta

f(x) is a polynomial. A product of n distinct factors

$$(x-a_{i}).$$

Prove that

$$\frac{1}{f(x)}=\sum\frac{1}{f'(a_{i})}.\frac{1}{(x-a_{i})}$$

This I can do by writing f(x)=(x-a)g(x) where g(a)<>0. Then splitting

$$\frac{1}{f(x)}$$

into

$$\frac{A}{(x-a)}+\frac{h(x)}{g(x)}$$

for some h(x) so

$$A=\frac{1}{g(a)}$$

and differentiating f(x), f'(a)=g(a) so

$$A=\frac{1}{f'(a)}$$

and I repeat that for each factor so I get the sum required.

The next question I can't answer:

Show that

$$\sum\frac{(a_{i})^r}{f'(a_{i})}$$

is 0 when r=0,1,...,n-2 and is 1 when r=n-1

I've tried expanding f(x), differentiating and putting x equal to the roots and summing over all the roots so I see the result works but I'm having no success in proving it.
Could someone help with a hint or two?

Last edited by a moderator: Sep 6, 2007
2. Sep 6, 2007

### EnumaElish

What do you get when n = 2, say?

3. Sep 6, 2007

### HallsofIvy

Staff Emeritus
You might consider using logarithmic differentiation.

If $f(x)= a(x-x_1)(x- x_2)\cdot\cdot\cdot(x- x_n)$
then $ln(f(x)) ln a+ ln(x-x_1)+ \cdot\cdot\cdot+ ln(x-x_n)$
what do you get when you differentiate that?

4. Sep 7, 2007

### SanjeevGupta

Yes I see that the result works for n=2 and even 3. But I do not see how that helps starting off a proof by Induction.

5. Sep 7, 2007

### SanjeevGupta

Yes I tried this too but I still cannot see how I get the $$(a_{i})^r$$ terms unless I do something like what I tried in my first post i.e. use an expanded form of f(x) and differentiate as the highest order term is of order n-1.

6. Sep 7, 2007

### learningphysics

How does 1/f(x) look if you use partial fractions... get a few terms... you'll see a pattern...

Then examine what f'(ai) looks like.

Relate the above 2 ideas...

7. Sep 8, 2007

### SanjeevGupta

Thanks very much for the help.
I see it now: Consider a1^r/f'(a1) as the polynomial x^r/[(x-a2)(x-a3)...(x-an)] and split this into partial fractions just like 1/f(x) above and you find a1^r/f'(a1) is just he negative of the sum of all other such terms with the other n-1 factors of f(x).
This works for r<n-1. For the r=n-1 case I differentiated the expanded f(x), divided by f'(x), and then substituted in a1,a2,...,an and added. From above all terms with r<n-1 are zero so we are left with n=n*Sum(a^(n-1)/f'(a)).

Thanks again for the hints.