| > Let R be a rational function given by R(x) = P(x)/x^n, where P is a
> polynomial with real coefficients and P(0) != 0. Find P(x) if
> R'(1/x) = -x^2 R'(x) for all x != 0.
d R(1/x) d R(x)
=> -------- = ------ for all x <> 0
dx dx
=> R(1/x) = R(x) + C for all x <> 0
Consideration at x = 1 shows that C is 0, so:
=> R(1/x) = R(x).
The relationship will hold for those P(x) which are linear sums of
the polynomials Q_k(x) = x^(n+k) + x^(n-k). (& Q_n(x) must be present in the
sum so that P(0) <> 0.)
Andrew.
|
| >> Let R be a rational function given by R(x) = P(x)/x^n, where P is a
>> polynomial with real coefficients and P(0) != 0. Find P(x) if
>> R'(1/x) = -x^2 R'(x) for all x != 0.
> d R(1/x) d R(x)
>=> -------- = ------ for all x <> 0
> dx dx
who does that follow?
we were told that R'(1/x) = -x^2 R'(x), why are you writing
R'(1/x) = R'(x) ??
is there something obviouse iam missing? (please note i still had not
had my morning coffee).
thanks!
\nasser
|
| >>> Let R be a rational function given by R(x) = P(x)/x^n, where P is a
>>> polynomial with real coefficients and P(0) != 0. Find P(x) if
>>> R'(1/x) = -x^2 R'(x) for all x != 0.
>
>> d R(1/x) d R(x)
>>=> -------- = ------ for all x <> 0
>> dx dx
>
> who does that follow?
>
> we were told that R'(1/x) = -x^2 R'(x), why are you writing
> R'(1/x) = R'(x) ??
I guess it's a question of how you interpret R'(1/x).
I take it to be:
d R(1/x)
--------
d (1/x)
ie: the formal derivative of R(y), dR(y)/dy, with 1/x then substituted for y.
I think this is the most common interpretation, although there was some
flaming in sci.math over the issue recently, I believe.
Andrew.
|