The three-levels formula

1) Linear forms on \(\mathbb{R}\left[ X \right].\)

Let \(a\) be a real. We note \({\varphi _a}\) the application from \(\mathbb{R}\left[ X \right]\) in \(\mathbb{R}\) which to a polynomial \(P\) associates \(P\left( a \right)\).
The application \({\varphi _a}\) is called the evaluation in \(a.\)
For all real \(a,\) \({\varphi _a}\) is a linear form on \(\mathbb{R}\left[ X \right]\). In fact, if \(P\) and \(Q\) are two polynomials and \(\lambda \) and \(\mu \) two reals, we have
\({\varphi _a}\left( {\lambda P + \mu Q} \right) = \left( {\lambda P + \mu Q} \right)\left( a \right) = \lambda P\left( a \right) + \mu Q\left( a \right) = \lambda {\varphi _a}\left( P \right) + \mu {\varphi _a}\left( Q \right)\) .

Let \(a\) be a real and \({\varphi _a}:\) \(\begin{array}{c}\mathbb{R}\left[ X \right] \to \mathbb{R}\\{\rm{ }}P \mapsto P\left( a \right)\end{array}\). \({\varphi _a}\) is a linar form on \(\mathbb{R}\left[ X \right].\)

On the other hand, Let \(a\) and \(b\) be two reals. We note furthermore \(\psi \) the application from \(\mathbb{R}\left[ X \right]\)in \(\mathbb{R}\)which to a polynomial \(P\) associates \(\int_a^b {P\left( t \right)dt} \). We know \(\psi \) is a linear form on \(\mathbb{R}\left[ X \right].\)

2) Linear independency of the set $\left( {{\varphi }_{a}},~{{\varphi }_{b}},~{{\varphi }_{c}} \right)$ in the dual of \({\mathbb{R}_3}\left[ X \right].\)

We take place from now on \({\mathbb{R}_3}\left[ X \right]\). We give ourselves three reals distincts two, two \(a,\) \(b\) et \(c\).
Let’s demonstrate that the set $\left( {{\varphi }_{a}},~{{\varphi }_{b}},~{{\varphi }_{c}} \right)$ is a linearly independent set in the dual of \({R_3}\left[ X \right].\)
Let $\left( \lambda ,~\mu ,\nu \right)\in {{\mathbb{R}}^{3}}.$
$\lambda {{\varphi }_{a}}+\mu {{\varphi }_{b}}+\nu {{\varphi }_{c}}=0\Rightarrow \forall P\in {{\mathbb{R}}_{3}}\left[ X \right],~\left( \lambda {{\varphi }_{a}}+\mu {{\varphi }_{b}}+\nu {{\varphi }_{c}} \right)\left( P \right)=0 \\ \Rightarrow \forall P\in {{\mathbb{R}}_{3}}\left[ X \right],~\lambda P\left( a \right)+\mu P\left( b \right)+vP\left( c \right)=0.$
We then apply the previous equality, valid for any polynomial of degree at most 3, successively to the three polynomials.
\(\displaystyle {P_a} = \frac{{\left( {X - b} \right)\left( {X - c} \right)}}{{\left( {a - b} \right)\left( {a - c} \right)}},\) \(\displaystyle {P_b} = \frac{{\left( {X - a} \right)\left( {X - c} \right)}}{{\left( {b - a} \right)\left( {b - c} \right)}}\) and \(\displaystyle {P_c} = \frac{{\left( {X - a} \right)\left( {X - b} \right)}}{{\left( {c - a} \right)\left( {c - b} \right)}}\). Since\({P_a}\left( a \right) = {P_b}\left( b \right) = {P_c}\left( c \right) = 1\) and\({P_a}\left( b \right) = {P_a}\left( c \right) = {P_b}\left( a \right) = {P_b}\left( c \right) = {P_c}\left( a \right) = {P_c}\left( b \right) = 0\), we obtain \(\lambda = \mu = \gamma = 0\). We have demonstrate that

If \(a,\) \(b,\) \(c\) are three reals two, two distinct, the set $\left( {{\varphi }_{a}},~{{\varphi }_{b}},~{{\varphi }_{c}} \right)$ is a linearly independent set in the dual of \({\mathbb{R}_3}\left[ X \right].\)

3) Independance of the linear forms \({\varphi _a},\) \({\varphi _b},\) \({\varphi _c}\) and \(\psi \) in the dual of \({\mathbb{R}_3}\left[ X \right].\)

\(a,\) \(b\) and \(c\) designate three reals distinct two, two. Remember that the dual of \({\mathbb{R}_3}\left[ X \right]\) has same dimension than \({\mathbb{R}_3}\left[ X \right]\), namely 4. According 2), the set $\left( {{\varphi }_{a}},~{{\varphi }_{b}},~{{\varphi }_{c}} \right)$ is linearly independnet and so two cases arise for the set $\left( {{\varphi }_{a}},~{{\varphi }_{b}},~{{\varphi }_{c}},~\psi \right)$ namely :

• l^st case : the set $\left( {{\varphi }_{a}},~{{\varphi }_{b}},~{{\varphi }_{c}},~\psi \right)$ is linearly independnet and so a base in the dual of \({\mathbb{R}_3}\left[ X \right],\)
• 2^nd case : the set is linearly dependent which equals, since the set $\left( {{\varphi }_{a}},~{{\varphi }_{b}},~{{\varphi }_{c}} \right)$ linearly independent, the fact that \(\psi \) is linear combinaison of \({\varphi _a},\) \({\varphi _b}\) and \({\varphi _c}.\)

Let $F=Vect\left( {{\varphi }_{a}},~{{\varphi }_{b}},~{{\varphi }_{c}} \right)$ . We note \({F^o}\) the set of polynomials \(P\) for which cancels out all linear form member of F.
\(F\) is a subspace of the dual of \({\mathbb{R}_3}\left[ X \right]\) with dimension 3. We know that \({F^o}\) is a vector subspace of \({\mathbb{R}_3}\left[ X \right]\) with dimension \(4 - 3 = 1\). Now, the polynomial \({P_0} = \left( {X - a} \right)\left( {X - b} \right)\left( {X - c} \right)\) is a non-zero polynomial such as \({\varphi _a}\left( {{P_0}} \right) = {\varphi _b}\left( {{P_0}} \right) = {\varphi _c}\left( {{P_0}} \right)\). But then, all element of \(F\) cancels out in \(P\) and then

\[{F^o} = Vect\left( P \right) \ où \ P = \left( {X - a} \right)\left( {X - b} \right)\left( {X - c} \right)\] .

Inversly, we now that the set of linear forms that cancel out in \({P_0}\) is a subspace of the dual of \({\mathbb{R}_3}\left[ X \right]\) with dimension \(4 - 1 = 3\) and is then $Vect\left( {{\varphi }_{a}},~{{\varphi }_{b}},~{{\varphi }_{c}} \right)$ .
We deduce $\psi \in Vect\left( {{\varphi }_{a}},~{{\varphi }_{b}},~{{\varphi }_{c}} \right)\Leftrightarrow \psi \left( {{P}_{0}} \right)=0\Leftrightarrow \int_{a}^{b}{(t-a)\left( t-b \right)\left( t-c \right)dt}=0$. Yet,
\[ \begin{array}{l}\displaystyle \int_a^b {(t - a)\left( {t - b} \right)\left( {t - c} \right)dt} =\displaystyle  \int_a^b {({t^3} - \left( {a + b + c} \right){t^2} + \left( {ab + ac + bc} \right)t - abc)dt} \\ =\displaystyle \frac{1}{4}\left( {{b^4} - {a^4}} \right) - \frac{1}{3}\left( {a + b + c} \right)\left( {{b^3} - {a^3}} \right) + \frac{1}{2}\left( {ab + ac + bc} \right)\left( {{b^2} - {a^2}} \right) - abc\left( {b - a} \right)\end{array}\]
And so,
\[ \begin{array}{l}\displaystyle \int_a^b {(t - a)\left( {t - b} \right)\left( {t - c} \right)dt}\\ =\displaystyle \frac{{b - a}}{{12}}\left( {3\left( {{a^3} + {a^2}b + a{b^2} + {b^3}} \right) - 4\left( {a + b + c} \right)\left( {{a^2} + ab + {b^2}} \right) + 6\left( {ab + ac + bc} \right)\left( {a + b} \right) - 12abc} \right)\\ =\displaystyle \frac{{b - a}}{{12}}\left( { - {a^3} - {b^3} + {a^2}b + a{b^2} + c\left( {2{a^2} + 2{b^2} - 4ab} \right)} \right)\\ =\displaystyle \frac{{b - a}}{{12}}\left( {\left( {{a^2} - {b^2}} \right)\left( {b - a} \right) + 2c{{\left( {b - a} \right)}^2}} \right)\\ =\displaystyle \frac{{{{\left( {b - a} \right)}^3}}}{{12}}\left( {2c - \left( {a + b} \right)} \right)\end{array}\]
Finally, since \(a\) and \(b\) are distincts

\[\psi \in Vect\left( {{\varphi _a},{\varphi _b},{\varphi _c}} \right) \Leftrightarrow c = \frac{{a + b}}{2}.\]

In the case where \(c = \frac{{a + b}}{2},\) \(\psi \) is a linear combinaison of \({\varphi _a},{\varphi _b},{\varphi _c}\). Then, it exist three reals \(\lambda ,\) \(\mu \) and \(\gamma \), only defined since $\left( {{\varphi }_{a}},~{{\varphi }_{b}},~{{\varphi }_{c}} \right)$ is linearly independent, such as \(\psi = \lambda {\varphi _a} + \mu {\varphi _b} + v{\varphi _c}\). This last equality is still written

\[\exists \left( \lambda ,~\mu ,\nu \right)\in {{\mathbb{R}}^{3}}/\forall P\in {{\mathbb{R}}_{3}}\left[ X \right],\\ ~\int_{a}^{b}{P\left( t \right)dt}=\lambda P\left( a \right)+\mu P\left( \frac{a+b}{2} \right)+vP\left( b \right)\] .

4) The three-levels formula.

Let's explicitly determine the three reals \(\lambda ,\) \(\mu \) et \(\nu \) of 3). For this, we apply the previous equality, valid for any polynomial \(P\) with degree at most 3, to polynomials \({P_a},\) \({P_b}\) et \({P_c}\), defined at the end of 2), with \(c = \frac{{a + b}}{2}\). We obtain
\[\begin{array}{l}\lambda = \lambda \times 1 + \times 0 + \nu \times 0 = \lambda {P_a}\left( a \right) + \mu {P_a}\left( c \right) + v{P_a}\left( b \right) =\displaystyle \int_a^b {{P_a}\left( t \right)dt} \\ =\displaystyle \int_a^b {\frac{{\left( {t - a} \right)\left( {t - b} \right)}}{{\left( {a - b} \right)\left( {a - c} \right)}}dt} =\displaystyle \frac{1}{{\left( {a - b} \right)\left( {a - c} \right)}}\left( {\frac{1}{3}\left( {{b^3} - {a^3}} \right) - \frac{1}{2}\left( {{b^2} - {a^2}} \right)\left( {b + c} \right) + bc\left( {b - a} \right)} \right)\\ =\displaystyle \frac{1}{{6\left( {c - a} \right)}}\left( {2\left( {{a^2} + ab + {b^2}} \right) - 3\left( {a + b} \right)\left( {b + c} \right) + 6bc} \right) \\ =\displaystyle \frac{1}{{6\left( {c - a} \right)}}\left( {2{a^2} - {b^2} - ab + c\left( { - 3a + 3b} \right)} \right)\\ =\displaystyle \frac{1}{{3\left( {b - a} \right)}}\left( {2{a^2} - {b^2} - ab + \frac{{a + b}}{2}\left( { - 3a + 3b} \right)} \right) \\= \displaystyle \frac{1}{{3\left( {b - a} \right)}}\left( {b - a} \right)\left( {3\frac{{a + b}}{2} - \left( {2a + b} \right)} \right)\\ =\displaystyle \frac{{b - a}}{6}\end{array}\]
Then exchanging roles between \(a\) and \(b\), \(v = \frac{{b - a}}{6}\). Finally,
\[\begin{array}{l}\mu =\displaystyle \int_{a}^{b}{{{P}_{c}}\left( t \right)dt}=\displaystyle \int_{a}^{b}{\frac{\left( t-a \right)\left( t-b \right)}{\left( c-a \right)\left( c-b \right)}dt} \\ =\displaystyle\frac{1}{\left( c-a \right)\left( c-b \right)}~\left( \frac{1}{3}\left( {{b}^{3}}-{{a}^{3}} \right)-\frac{1}{2}\left( {{b}^{2}}-{{a}^{2}} \right)\left( a+b \right)+ab\left( b-a \right) \right) \\ =-\displaystyle\frac{4}{6{{(b-a)}^{2}}}\left( b-a \right)\left( 2\left( {{a}^{2}}+ab+{{b}^{2}} \right)-3\left( a+b \right)\left( a+b \right)+6ab \right) \\ =-\displaystyle\frac{4}{6\left( b-a \right)}\left( -{{a}^{2}}-{{b}^{2}}+2ab \right) \\ =\displaystyle\frac{4\left( b-a \right)}{6} \end{array}\]
We obtained the three-level formula making it possible to calculate the value of the integral of a polynomial with degree lower or equal to 3 on a segment knowing the values of this polynomial at the beginning, in the middle and at the end of this segment:

Let \(a\) and \(b\) be two distinct real. \(\forall P \in {\mathbb{R}_3}\left[ X \right],\) \(\displaystyle \int_{a}^{b} {P\left( t \right)dt} = \frac{{b - a}}{6}\left( {P\left( a \right) + 4P\left( {\frac{{a + b}}{2}} \right) + P\left( b \right)} \right)\) .