constants of polynomial

bluemath

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Sep 20, 2015
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Hello,

I'm trying to find constants a,b,c of polynomial P(x) who verifies for degree 0,1,2 and 3 :

Integral -1 to 1 P(x)dx = aP(-1) + bP(0) + cP(1)

About degree 0, I took P(x) = k (constant)

I found 2k = ak + bk + ck , then a + b + c = 2

About degree 1, I took P(x) = mx + s

I found a(m + s) + b(-m + s) + cs = 2s

By the way, all that sounds very strange and my final system will be very difficult with the continuation. Am I in the right way please or it's possible to do that more easily ?

Thanks in advance
 
Hello,

I'm trying to find constants a,b,c of polynomial P(x) who verifies for degree 0,1,2 and 3 :

Integral -1 to 1 P(x)dx = aP(-1) + bP(0) + cP(1)

About degree 0, I took P(x) = k (constant)

I found 2k = ak + bk + ck , then a + b + c = 2

About degree 1, I took P(x) = mx + s

I found a(m + s) + b(-m + s) + cs = 2s

By the way, all that sounds very strange and my final system will be very difficult with the continuation. Am I in the right way please or it's possible to do that more easily ?

Thanks in advance

Let Q be the general integral of P
\(\displaystyle \int\, P(x)\)
That is Q'(x) = P(x). Your integral value is Q(1) minus Q(-1)
\(\displaystyle \int_{-1}^1\, P(x)\) = Q(1) - Q(-1)

So, assuming P(1) is not zero, choose a and b and let
c = \(\displaystyle \frac{Q(1)\, -\, [ Q(-1)\, +\, a\, P(-1)\, +\, b\, P(0)]}{P(1)}\)
 
Thank you for this very elegant answer :)

It means that there is no value fixed for a, b and c but simply an infinity of solutions ?
 
Thank you for this very elegant answer :)

It means that there is no value fixed for a, b and c but simply an infinity of solutions ?

That is correct, there is an infinity of solutions [except for certain esoteric cases, i.e. when two of P(-1), P(0), and P(1) are zero].
 
Last edited:
Hello,

I'm back again.

The solution seen here is for all degrees actually.

If we work only with degree 0,1,2 or 3 , is it possible to get specific constant for a, b and c ?
 
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