- #1
SanjeevGupta
f(x) is a polynomial. A product of n distinct factors
[tex](x-a_{i}).[/tex]
Prove that
[tex]\frac{1}{f(x)}=\sum\frac{1}{f'(a_{i})}.\frac{1}{(x-a_{i})}[/tex]
This I can do by writing f(x)=(x-a)g(x) where g(a)<>0. Then splitting
[tex]\frac{1}{f(x)}[/tex]
into
[tex]\frac{A}{(x-a)}+\frac{h(x)}{g(x)}[/tex]
for some h(x) so
[tex]A=\frac{1}{g(a)}[/tex]
and differentiating f(x), f'(a)=g(a) so
[tex]A=\frac{1}{f'(a)}[/tex]
and I repeat that for each factor so I get the sum required.
The next question I can't answer:
Show that
[tex]\sum\frac{(a_{i})^r}{f'(a_{i})}[/tex]
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?
[tex](x-a_{i}).[/tex]
Prove that
[tex]\frac{1}{f(x)}=\sum\frac{1}{f'(a_{i})}.\frac{1}{(x-a_{i})}[/tex]
This I can do by writing f(x)=(x-a)g(x) where g(a)<>0. Then splitting
[tex]\frac{1}{f(x)}[/tex]
into
[tex]\frac{A}{(x-a)}+\frac{h(x)}{g(x)}[/tex]
for some h(x) so
[tex]A=\frac{1}{g(a)}[/tex]
and differentiating f(x), f'(a)=g(a) so
[tex]A=\frac{1}{f'(a)}[/tex]
and I repeat that for each factor so I get the sum required.
The next question I can't answer:
Show that
[tex]\sum\frac{(a_{i})^r}{f'(a_{i})}[/tex]
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?
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