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D.D. Stancu
Institutul de Calcul
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D.D. Stancu, Sur la formule d’interpolation d’Hermite et quelques applications de celle-ci. (Romanian) Acad. R. P. Romane. Fil. Cluj. Stud. Cerc. Mat. 8 1957 339-355.
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Studii si Cercetari Matematice
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ON HERMITE'S INTERPOLATION FORMULA AND SOME APPLICATIONS OF IT*
- Ch. Hermite [1] formulated the following general interpolation problem:
To construct a polynomial, of minimal degree, which together with its successive derivatives - up to and including certain ordersr_(1)-1,dots,r_(s)-1r_{1}-1, \ldots, r_{s}-1, to take on distinct points
{:(1)x_(1)","x_(2)","dots","x_(s):}\begin{equation*} x_{1}, x_{2}, \ldots, x_{s} \tag{1} \end{equation*}
values ​​given in advance.
It is known^(1){ }^{1}that this polynomial exists and is unique, and its degree isnn, wheren=r_(1)+dots+r_(s)-1n=r_{1}+\ldots+r_{s}-1.
It is known^(1){ }^{1}that this polynomial exists and is unique, and its degree isnn, wheren=r_(1)+dots+r_(s)-1n=r_{1}+\ldots+r_{s}-1.
Let us denote this polynomial byH_(n)(x)H_{n}(x)and suppose that the values ​​given before are the values ​​taken, on the pointsx_(1),x_(2),dots,x_(s)x_{1}, x_{2}, \ldots, x_{s}, of a functionf(x)f(x)and its derivatives. Then the Hermite interpolation polynomial relative to the functionf(x)f(x)and points (1), to which are attached the multiplicity orders respectivelyr_(1),r_(2),dots,r_(s)r_{1}, r_{2}, \ldots, r_{s},
{:(2)H_(n)(x)=H_(n)(ubrace(x_(1),dots,x_(1))_(r_(1))","ubrace(x_(2),dots,x_(2))_(r_(2))","dots","ubrace(x_(s),dots,x_(s))_(r_(s));f∣x):}\begin{equation*}
H_{n}(x)=H_{n}(\underbrace{x_{1}, \ldots, x_{1}}_{r_{1}}, \underbrace{x_{2}, \ldots, x_{2}}_{r_{2}}, \ldots, \underbrace{x_{s}, \ldots, x_{s}}_{r_{s}} ; f \mid x) \tag{2}
\end{equation*}
will be the polynomial that verifies the conditions
{:(3)H_(n)^((l))(x_(k))=f_(n)^((l))(x_(k))((l=0,1,dots,r_(k)-1)/(k=1,2,dots,s)):}\begin{equation*}
H_{n}^{(l)}\left(x_{k}\right)=f_{n}^{(l)}\left(x_{k}\right)\binom{l=0,1, \ldots, r_{k}-1}{k=1,2, \ldots, s} \tag{3}
\end{equation*}
- If
{:(4)H_(n)(x)=sum_(j=0)^(n)a_(j)x^(j)",":}\begin{equation*}
H_{n}(x)=\sum_{j=0}^{n} a_{j} x^{j}, \tag{4}
\end{equation*}
the previous conditions lead us to the system ofn+1n+1equations withn+1n+1unknown:a_(0),a_(1),dots,a_(n)a_{0}, a_{1}, \ldots, a_{n}
{:[(5)sum_(j=i)^(n)j(j-1)dots(j-l+1)a_(j)x_(k)^(j-l)=f^((l))(x_(k))],[(l=0,1,dots,r_(k)-1;k=1,2,dots,s).]:}\begin{gather*} \sum_{j=i}^{n} j(j-1) \ldots(j-l+1) a_{j} x_{k}^{j-l}=f^{(l)}\left(x_{k}\right) \tag{5}\\ \left(l=0,1, \ldots, r_{k}-1 ; k=1,2, \ldots, s\right) . \end{gather*}
{:[(5)sum_(j=i)^(n)j(j-1)dots(j-l+1)a_(j)x_(k)^(j-l)=f^((l))(x_(k))],[(l=0,1,dots,r_(k)-1;k=1,2,dots,s).]:}\begin{gather*} \sum_{j=i}^{n} j(j-1) \ldots(j-l+1) a_{j} x_{k}^{j-l}=f^{(l)}\left(x_{k}\right) \tag{5}\\ \left(l=0,1, \ldots, r_{k}-1 ; k=1,2, \ldots, s\right) . \end{gather*}
The determinant of this system - as is known - is different from zero, because the points (1) are distinct.
To find the expression of the polynomialH_(n)(x)H_{n}(x)we could eliminate the unknownsa_(0),a_(1),dots,a_(n)a_{0}, a_{1}, \ldots, a_{n}between equations (4) and (5). This path is, however, very difficult. G. Z emp 1 en [3] nevertheless tried to find the explicit expression of the polynomialH_(n)(x)H_{n}(x). We note, however, that the results obtained by this author are inaccurate; the formula he gave does not correspond to the solution of the problem ifs >= 2s \geqq 2and the numbersr_(i)r_{i}are greater than 2 . Of course, the results he gave in the second part of his work, relating to the decomposition of a rational function into simple fractions, are also erroneous. In the review made by We1tzien Zeh1endorf [4] and in the memoir of E. Netto in [5], in which this work is mentioned, no observations were made on the results obtained by G. Zemplen.
3. From the above-mentioned elimination of the unknownsa_(0),a_(1),dots,a_(n)a_{0}, a_{1}, \ldots, a_{n}and from the formal solution in relation toH_(n)(x)H_{n}(x)of the equation that was obtained, we can realize that the polynomial sought is of the form
3. From the above-mentioned elimination of the unknownsa_(0),a_(1),dots,a_(n)a_{0}, a_{1}, \ldots, a_{n}and from the formal solution in relation toH_(n)(x)H_{n}(x)of the equation that was obtained, we can realize that the polynomial sought is of the form
{:(6)H_(n)(x)=sum_(i=1)^(s)sum_(k=0)^(r_(i)-1)l_(i,k)(x)f^((k))(x_(i))",":}\begin{equation*}
H_{n}(x)=\sum_{i=1}^{s} \sum_{k=0}^{r_{i}-1} l_{i, k}(x) f^{(k)}\left(x_{i}\right), \tag{6}
\end{equation*}
wherel_(l,k)(x)l_{l, k}(x)are polynomials of degreennFrom the formal calculation above it is immediately observed that the polynomialsl_(l,k)(x)l_{l, k}(x)are independent off(x)f(x)and that, taking into account (3), they must verify the following conditions:
{:[(7)l_(i,k)^((p))(x_(j))=0(j!=i;p=0,1,dots,r_(j)-1)],[(8)l_(i,k)^((p))(x_(i))={[0","" pentru "p!=k],[1","" pentru "p=k]quad(p=0,1,dots,r_(i)-1):}]:}\begin{gather*}
l_{i, k}^{(p)}\left(x_{j}\right)=0\left(j \neq i ; p=0,1, \ldots, r_{j}-1\right) \tag{7}\\
l_{i, k}^{(p)}\left(x_{i}\right)=\left\{\begin{array}{l}
0, \text { pentru } p \neq k \\
1, \text { pentru } p=k
\end{array} \quad\left(p=0,1, \ldots, r_{i}-1\right)\right. \tag{8}
\end{gather*}
These conditions will completely determine our fundamental interpolation polynomialsl_(i,k)(x)l_{i, k}(x)Indeed, based on (7) it is observed thatl_(i,k)(x)l_{i, k}(x)contains as a factor the product
{:(9)g_(i)(x)=(x-x_(1))^(gamma_(1))dots(x-x_(i-1))^(gamma_(i-1))(x-x_(i+1))^(gamma_(i+1))dots(x-x_(s))^(gamma_(s)).:}\begin{equation*}
g_{i}(x)=\left(x-x_{1}\right)^{\gamma_{1}} \ldots\left(x-x_{i-1}\right)^{\gamma_{i-1}}\left(x-x_{i+1}\right)^{\gamma_{i+1}} \ldots\left(x-x_{s}\right)^{\gamma_{s}} . \tag{9}
\end{equation*}
Taking into account conditions (8), we conclude that(x-x_(i))^(k)\left(x-x_{i}\right)^{k}is also a factor ofl_(l,k)(x)l_{l, k}(x)It follows that this is of the form
{:(10)l_(i,k)(x)=g_(i)(x)(x-x_(i))^(k)h_(i,k)(x)",":}\begin{equation*}
l_{i, k}(x)=g_{i}(x)\left(x-x_{i}\right)^{k} h_{i, k}(x), \tag{10}
\end{equation*}
whereh_(i,k)(x)h_{i, k}(x)is a polynomial of degreer_(i)-k-1r_{i}-k-1. It remains to determine this last polynomial. If we expand it according to Taylor's formula in the neighborhood ofx=x_(i)x=x_{i}, it is obtained
{:(11)h_(i,k)(x)=sum_(m=0)^(r_(i)-k-1)((x-x_(i))^(m))/(m!)h_(i,k)^((m))(x_(i)).:}\begin{equation*}
h_{i, k}(x)=\sum_{m=0}^{r_{i}-k-1} \frac{\left(x-x_{i}\right)^{m}}{m!} h_{i, k}^{(m)}\left(x_{i}\right) . \tag{11}
\end{equation*}
From (10) we have
(x-x_(i))^(k)h_(i,k)(x)=l_(i,k)(x)(1)/(g_(i)(x))\left(x-x_{i}\right)^{k} h_{i, k}(x)=l_{i, k}(x) \frac{1}{g_{i}(x)}
If we calculate, according to Leibniz's formula, the derivative of the orderqqof both members of this equality, we obtain
sum_(j=0)^(q)((q)/(j))[(x-x_(i))^(k)]^((q-j))h_(i,k)^((j))(x)=sum_(j=0)^(q)((q)/(j))l_(i,k)^((q-j))(x)((1)/(g_(i)(x)))^((j))\sum_{j=0}^{q}\binom{q}{j}\left[\left(x-x_{i}\right)^{k}\right]^{(q-j)} h_{i, k}^{(j)}(x)=\sum_{j=0}^{q}\binom{q}{j} l_{i, k}^{(q-j)}(x)\left(\frac{1}{g_{i}(x)}\right)^{(j)}
If it is takenq=k+mq=k+m, it is donex=x_(i)x=x_{i}and taking into account (7) and (8), we obtain the relations
{:[k!h_(i,k)^((m))(x_(i))=((1)/(g_(i)(x)))_(x=x_(i))^((m))],[(m=0,1,dots,r_(i)-k-1)]:}\begin{aligned}
& k!h_{i, k}^{(m)}\left(x_{i}\right)=\left(\frac{1}{g_{i}(x)}\right)_{x=x_{i}}^{(m)} \\
& \left(m=0,1, \ldots, r_{i}-k-1\right)
\end{aligned}
In this way we arrive at the following expressions for the fundamental interpolation polynomials
{:(12)l_(i,k)(x)=sum_(r=0)^(r_(i)-k-1)[((x-x_(i))^(r))/(r!)((1)/(g_(i)(x)))_(x_(i))^((r))]g_(i)(x)",":}\begin{equation*}
l_{i, k}(x)=\sum_{r=0}^{r_{i}-k-1}\left[\frac{\left(x-x_{i}\right)^{r}}{r!}\left(\frac{1}{g_{i}(x)}\right)_{x_{i}}^{(r)}\right] g_{i}(x), \tag{12}
\end{equation*}
and the polynomial (6) becomes
{:(13)H_(n)(x)=sum_(i=1)^(s)sum_(k=0)^(r_(i)-1)sum_(r=0)^(r_(i)-k-2)((x-x_(i))^(k))/(k!)[((x-x_(i))^(r))/(r!)((1)/(g_(i)(x)))_(x_(i))^((r))]g_(i)(x)f^((k))(x_(i)).:}\begin{equation*}
H_{n}(x)=\sum_{i=1}^{s} \sum_{k=0}^{r_{i}-1} \sum_{r=0}^{r_{i}-k-2} \frac{\left(x-x_{i}\right)^{k}}{k!}\left[\frac{\left(x-x_{i}\right)^{r}}{r!}\left(\frac{1}{g_{i}(x)}\right)_{x_{i}}^{(r)}\right] g_{i}(x) f^{(k)}\left(x_{i}\right) . \tag{13}
\end{equation*}
To this expression of the polynomialH_(n)(x)H_{n}(x)VL Gonciarov arrived at this, in a slightly different way [6].
4. If the derivative of the order is calculatedjj, at the pointx_(i)x_{i}, his(1)/(g_(i)(x))\frac{1}{g_{i}(x)}, we obtain
4. If the derivative of the order is calculatedjj, at the pointx_(i)x_{i}, his(1)/(g_(i)(x))\frac{1}{g_{i}(x)}, we obtain
{:[((1)/(g_(i)(x)))_(x_(i))^((j))=(-1)^(j)sum(j!)/(alpha_(1)!dots//dotsalpha_(s)!)xx],[xx(r_(1)(r_(1)+1)dots(r_(1)+alpha_(1)-1)dots//dotsr_(s)(r_(s)+1)dots(r_(s)+alpha_(s)-1))/((x_(i)-x_(1))^(r_(1)+alpha_(1))dots//dots(x_(i)-x_(s))^(r_(s)+alpha_(s)))","]:}\begin{gathered}
\left(\frac{1}{g_{i}(x)}\right)_{x_{i}}^{(j)}=(-1)^{j} \sum \frac{j!}{\alpha_{1}!\ldots / \ldots \alpha_{s}!} \times \\
\times \frac{r_{1}\left(r_{1}+1\right) \ldots\left(r_{1}+\alpha_{1}-1\right) \ldots / \ldots r_{s}\left(r_{s}+1\right) \ldots\left(r_{s}+\alpha_{s}-1\right)}{\left(x_{i}-x_{1}\right)^{r_{1}+\alpha_{1}} \ldots / \ldots\left(x_{i}-x_{s}\right)^{r_{s}+\alpha_{s}}},
\end{gathered}
dash indicating the absence of factors with all indicesiiHere the sum extends to all systems of non-negative integers that verify the relation
alpha_(1)+alpha_(2)+dots+alpha_(i-1)+alpha_(i+1)+dots+alpha_(s)=j\alpha_{1}+\alpha_{2}+\ldots+\alpha_{i-1}+\alpha_{i+1}+\ldots+\alpha_{s}=j
Taking this into account, the fundamental interpolation polynomiall_(i,k)(x)l_{i, k}(x)can be put in the form
{:[(14)l_(i,k)(x)=(g_(i)(x))/(g_(i)(x_(i))){((x-x_(i))^(k))/(k!)sum_(j=0)^(r_(i)-k-1)(x-x_(i))^(j)xx:}],[{: xxsum_(a_(1)+dots+a_(i-1)+alpha_(i+1)+dots+alpha_(s)=j)(((r_(1)+alpha_(1)-1)/(alpha_(1)))dots//dots((r_(s)+alpha_(s)-1)/(alpha_(s))))/((x_(1)-x_(i))^(alpha_(1))dots//dots(x_(s)-x_(i))^(alpha_(s)))}]:}\begin{gather*}
l_{i, k}(x)=\frac{g_{i}(x)}{g_{i}\left(x_{i}\right)}\left\{\frac{\left(x-x_{i}\right)^{k}}{k!} \sum_{j=0}^{r_{i}-k-1}\left(x-x_{i}\right)^{j} \times\right. \tag{14}\\
\left.\times \sum_{a_{1}+\ldots+a_{i-1}+\alpha_{i+1}+\ldots+\alpha_{s}=j} \frac{\binom{r_{1}+\alpha_{1}-1}{\alpha_{1}} \ldots / \ldots\binom{r_{s}+\alpha_{s}-1}{\alpha_{s}}}{\left(x_{1}-x_{i}\right)^{\alpha_{1}} \ldots / \ldots\left(x_{s}-x_{i}\right)^{\alpha_{s}}}\right\}
\end{gather*}
In this way we arrived at the definitive expression of Hermite's interpolation polynomial (2)
{:[H_(n)(x)=],[(15)=sum_(i=1)^(s)(g_(i)(x))/(g_(i)(x_(i))){sum_(k=0)^(r_(i)-1)((x-x_(i))^(k))/(k!)[sum_(j=0)^(r_(i)-k-1)A_(alpha_(1),dots,alpha_(i-1),alpha_(i+1),dots,alpha_(s))^((j))(x-x_(i))^(j)]f^((k))(x_(i))}]:}\begin{align*}
& H_{n}(x)= \\
& =\sum_{i=1}^{s} \frac{g_{i}(x)}{g_{i}\left(x_{i}\right)}\left\{\sum_{k=0}^{r_{i}-1} \frac{\left(x-x_{i}\right)^{k}}{k!}\left[\sum_{j=0}^{r_{i}-k-1} A_{\alpha_{1}, \ldots, \alpha_{i-1}, \alpha_{i+1}, \ldots, \alpha_{s}}^{(j)}\left(x-x_{i}\right)^{j}\right] f^{(k)}\left(x_{i}\right)\right\} \tag{15}
\end{align*}
where
{:[A_(alpha_(1),dots,alpha_(i-1),alpha_(i+1),dots,alpha_(s))^((j))=],[(16)=sum_(alpha_(1)+dots+alpha_(i-1)+alpha_(i-1)+dots+alpha_(s)=j)(((gamma_(1)+alpha_(1)-1)/(alpha_(1)))dots//dots((gamma_(s)+alpha_(s)-1)/(alpha_(s))))/((x_(1)-x_(i))^(alpha_(1))dots//dots(x_(s)-x_(i))^(alpha_(s))).]:}\begin{gather*}
A_{\alpha_{1}, \ldots, \alpha_{i-1}, \alpha_{i+1}, \ldots, \alpha_{s}}^{(j)}= \\
=\sum_{\alpha_{1}+\ldots+\alpha_{i-1}+\alpha_{i-1}+\ldots+\alpha_{s}=j} \frac{\binom{\gamma_{1}+\alpha_{1}-1}{\alpha_{1}} \ldots / \ldots\binom{\gamma_{s}+\alpha_{s}-1}{\alpha_{s}}}{\left(x_{1}-x_{i}\right)^{\alpha_{1}} \ldots / \ldots\left(x_{s}-x_{i}\right)^{\alpha_{s}}} . \tag{16}
\end{gather*}
- If we consider Hermite's interpolation formula
{:(17)f(x)=H_(n)(x)+R_(n+1)(x)",":}\begin{equation*}
f(x)=H_{n}(x)+R_{n+1}(x), \tag{17}
\end{equation*}
RESTR_(n+1)(x)R_{n+1}(x)has, as is known, the expression
R_(n+1)(x)=(x-x_(1))^(r_(1))dots(x-x_(s))^(r_(s))[x,ubrace(x_(1),dots,x_(1))_(r_(1)),ubrace(x_(2),dots,x_(2))_(r_(2)),dots,ubrace(x_(s),dots,x_(s))_(r_(s));f]", "R_{n+1}(x)=\left(x-x_{1}\right)^{r_{1}} \ldots\left(x-x_{s}\right)^{r_{s}}[x, \underbrace{x_{1}, \ldots, x_{1}}_{r_{1}}, \underbrace{x_{2}, \ldots, x_{2}}_{r_{2}}, \ldots, \underbrace{x_{s}, \ldots, x_{s}}_{r_{s}} ; f] \text {, }
where
[alpha_(1),alpha_(2),dots,alpha_(m+1);f]\left[\alpha_{1}, \alpha_{2}, \ldots, \alpha_{m+1} ; f\right]
is the difference divided by the ordermmof the functionf(x)f(x)on the pointsalpha_(1),alpha_(2),dots,alpha_(m+1)\alpha_{1}, \alpha_{2}, \ldots, \alpha_{m+1}.
Assuming thatf(x)f(x)has a derivative of the ordern+1n+1in the smallest interval containing the valuesx_(1),dots,x_(s),xx_{1}, \ldots, x_{s}, x, the rest can be expressed, as is known, by the formula
Assuming thatf(x)f(x)has a derivative of the ordern+1n+1in the smallest interval containing the valuesx_(1),dots,x_(s),xx_{1}, \ldots, x_{s}, x, the rest can be expressed, as is known, by the formula
{:(18)R_(n+1)(x)=((x-x_(1))^(r_(1))(x-x_(2))^(r_(2))dots(x-x_(s))^(r_(s)))/((n+1)!)f^((n+1))(xi)",":}\begin{equation*}
R_{n+1}(x)=\frac{\left(x-x_{1}\right)^{r_{1}}\left(x-x_{2}\right)^{r_{2}} \ldots\left(x-x_{s}\right)^{r_{s}}}{(n+1)!} f^{(n+1)}(\xi), \tag{18}
\end{equation*}
wherexi\xibelongs to the smallest interval containing the valuesx_(i)x_{i}andxx.
6. Example. Find the interpolation polynomial of minimum degree relative to the functionf(x)f(x)and at the nodesx_(1)=x_(2)=x_(3)=-1,x_(4)=0x_{1}=x_{2}=x_{3}=-1, x_{4}=0,x_(5)=x_(6)=x_(7)=1x_{5}=x_{6}=x_{7}=1.
6. Example. Find the interpolation polynomial of minimum degree relative to the functionf(x)f(x)and at the nodesx_(1)=x_(2)=x_(3)=-1,x_(4)=0x_{1}=x_{2}=x_{3}=-1, x_{4}=0,x_(5)=x_(6)=x_(7)=1x_{5}=x_{6}=x_{7}=1.
Based on the previous formula we find
{:[H_(6)(x)=H_(6)(-1","-1","-1","0","1","1","1;f∣x)=(1-x^(2))^(3)f(0)+],[+x(x-1)^(3)[(1)/(8)+(5(x+1))/(16)+((x+1)^(2))/(2)]f(-1)+x(x-1)^(3)(x+1)[(1)/(8)+(5(x+1))/(16)]f^(')(-1)+],[+(x(x-1)^(3)(x+1)^(2))/(16)f^('')(-1)+x(x+1)^(3)[(1)/(8)-(5)/(16)(x-1)+((x-1)^(2))/(2)]f(1)+],[(19)+x(x+1)^(3)(x-1)[(1)/(8)-(5(x-1))/(16)]f^(')(1)+(x(x+1)^(3)(x-1)^(2))/(16)f^('')(1)]:}\begin{gather*}
H_{6}(x)=H_{6}(-1,-1,-1,0,1,1,1 ; f \mid x)=\left(1-x^{2}\right)^{3} f(0)+ \\
+x(x-1)^{3}\left[\frac{1}{8}+\frac{5(x+1)}{16}+\frac{(x+1)^{2}}{2}\right] f(-1)+x(x-1)^{3}(x+1)\left[\frac{1}{8}+\frac{5(x+1)}{16}\right] f^{\prime}(-1)+ \\
+\frac{x(x-1)^{3}(x+1)^{2}}{16} f^{\prime \prime}(-1)+x(x+1)^{3}\left[\frac{1}{8}-\frac{5}{16}(x-1)+\frac{(x-1)^{2}}{2}\right] f(1)+ \\
+x(x+1)^{3}(x-1)\left[\frac{1}{8}-\frac{5(x-1)}{16}\right] f^{\prime}(1)+\frac{x(x+1)^{3}(x-1)^{2}}{16} f^{\prime \prime}(1) \tag{19}
\end{gather*}
7. Particular cases.
- Ifr_(1)=r_(2)=dots=r_(3)=1r_{1}=r_{2}=\ldots=r_{3}=1, (15) reduces to the Lagrange interpolation polynomial
H_(s-1)(x)=sum_(i=1)^(s)(g_(i)(x))/(g_(i)(x_(i)))f(x_(i)).H_{s-1}(x)=\sum_{i=1}^{s} \frac{g_{i}(x)}{g_{i}\left(x_{i}\right)} f\left(x_{i}\right) .
2^(@)2^{\circ}Ifr_(1)=r_(2)=dots=r_(s)=2r_{1}=r_{2}=\ldots=r_{s}=2, from (13) we obtain
H_(2s-1)(x)=sum_(i=1)^(s)(g_(i)(x))/(g_(i)(x_(i)))[1-(x-x_(i))(g_(i)^(')(x_(i)))/(g_(i)(x_(i)))]f(x_(i))+sum_(i=1)^(s)(x-x_(i))(g_(i)(x))/(g_(i)(x_(i)))f^(')(x_(i)),H_{2 s-1}(x)=\sum_{i=1}^{s} \frac{g_{i}(x)}{g_{i}\left(x_{i}\right)}\left[1-\left(x-x_{i}\right) \frac{g_{i}^{\prime}\left(x_{i}\right)}{g_{i}\left(x_{i}\right)}\right] f\left(x_{i}\right)+\sum_{i=1}^{s}\left(x-x_{i}\right) \frac{g_{i}(x)}{g_{i}\left(x_{i}\right)} f^{\prime}\left(x_{i}\right),
and from (14) it follows
{:[H_(2s-1)(x)=sum_(i=1)^(s)(g_(i)(x))/(g_(i)(x_(i)))(x-x_(i))[(1)/(x-x_(i))+(1)/(x_(1)-x_(i))+dots+(1)/(x_(s)-x_(i))]+],[+sum_(i=1)^(s)(g_(i)(x))/(g_(l)(x_(i)))(x-x_(i))f^(')(x_(i)).]:}\begin{gathered}
H_{2 s-1}(x)=\sum_{i=1}^{s} \frac{g_{i}(x)}{g_{i}\left(x_{i}\right)}\left(x-x_{i}\right)\left[\frac{1}{x-x_{i}}+\frac{1}{x_{1}-x_{i}}+\ldots+\frac{1}{x_{s}-x_{i}}\right]+ \\
+\sum_{i=1}^{s} \frac{g_{i}(x)}{g_{l}\left(x_{i}\right)}\left(x-x_{i}\right) f^{\prime}\left(x_{i}\right) .
\end{gathered}
This formula was also found by A. Markoff [7].
3^(@)3^{\circ}In the case ofr_(1)=r_(2)=dots=r_(s)=3r_{1}=r_{2}=\ldots=r_{s}=3, we have
3^(@)3^{\circ}In the case ofr_(1)=r_(2)=dots=r_(s)=3r_{1}=r_{2}=\ldots=r_{s}=3, we have
H_(3s-1)(x)=sum_(i=1)^(s)[l_(i,0)(x)f(x_(i))+l_(i,1)(x)f^(')(x_(i))+l_(i,2)(x)f^('')(x_(i))]H_{3 s-1}(x)=\sum_{i=1}^{s}\left[l_{i, 0}(x) f\left(x_{i}\right)+l_{i, 1}(x) f^{\prime}\left(x_{i}\right)+l_{i, 2}(x) f^{\prime \prime}\left(x_{i}\right)\right]
where
l_(i,0)(x)=(g_(i)(x))/(g^(i)(x_(i)))[1-(x-x_(i))(g_(i)^(')(x_(i)))/(g_(i)(x_(i)))-(x-x_(i))^(2)(g_(i)^('')(x_(i))g_(i)(x_(i))-2g_(i)^('2)(x_(i)))/(2g_(i)^(2)(x_(i)))]l_{i, 0}(x)=\frac{g_{i}(x)}{g^{i}\left(x_{i}\right)}\left[1-\left(x-x_{i}\right) \frac{g_{i}^{\prime}\left(x_{i}\right)}{g_{i}\left(x_{i}\right)}-\left(x-x_{i}\right)^{2} \frac{g_{i}^{\prime \prime}\left(x_{i}\right) g_{i}\left(x_{i}\right)-2 g_{i}^{\prime 2}\left(x_{i}\right)}{2 g_{i}^{2}\left(x_{i}\right)}\right]
{:[l_(i,1)(x)=(x-x_(i))(g_(i)(x))/(g_(i)(x_(i)))[1-(x-x_(i))(g_(i)^(')(x_(i)))/(g_(i)(x_(i)))]],[l_(i,2)(x)=((x-x_(i))^(2))/(2)(g_(i)(x))/(g_(i)(x_(i))).]:}\begin{aligned}
& l_{i, 1}(x)=\left(x-x_{i}\right) \frac{g_{i}(x)}{g_{i}\left(x_{i}\right)}\left[1-\left(x-x_{i}\right) \frac{g_{i}^{\prime}\left(x_{i}\right)}{g_{i}\left(x_{i}\right)}\right] \\
& l_{i, 2}(x)=\frac{\left(x-x_{i}\right)^{2}}{2} \frac{g_{i}(x)}{g_{i}\left(x_{i}\right)} .
\end{aligned}
- Let's consider the particular cases=2s=2and let's notex_(1)=a,x_(2)=b,r_(1)=mx_{1}=a, x_{2}=b, r_{1}=m,r_(2)=nr_{2}=nThe interpolation formula (15) reduces in this case to
{:[H_(m+n-1)(x)=H_(m+n-1)(ubrace(a,dots,a)_(m)","ubrace(b,dots,b)_(n);f(x)=],[(22)=((x-b)/(a-b))^(n)sum_(k=0)^(m-1)((x-a)^(k))/(k!)[sum_(i=0)^(m-k-1)((n+i-1)/(i))((x-a)/(b-a))^(i)]f^((k))(a)+],[(20)+((x-a)/(b-a))^(m)sum_(r=0)^(n-1)((x-b)^(r))/(r!)[sum_(j=0)^(n-r-1)((m+j-1)/(j))((x-b)/(a-b))^(j)]f^((r))(b).]:}\begin{gather*}
H_{m+n-1}(x)=H_{m+n-1}(\underbrace{a, \ldots, a}_{m}, \underbrace{b, \ldots, b}_{n} ; f(x)= \\
=\left(\frac{x-b}{a-b}\right)^{n} \sum_{k=0}^{m-1} \frac{(x-a)^{k}}{k!}\left[\sum_{i=0}^{m-k-1}\binom{n+i-1}{i}\left(\frac{x-a}{b-a}\right)^{i}\right] f^{(k)}(a)+ \tag{22}\\
+\left(\frac{x-a}{b-a}\right)^{m} \sum_{r=0}^{n-1} \frac{(x-b)^{r}}{r!}\left[\sum_{j=0}^{n-r-1}\binom{m+j-1}{j}\left(\frac{x-b}{a-b}\right)^{j}\right] f^{(r)}(b) . \tag{20}
\end{gather*}
In the corresponding interpolation formula
{:(21)f(x)=H_(m+n-1)(x)+R_(m+n)(x):}\begin{equation*}
f(x)=H_{m+n-1}(x)+R_{m+n}(x) \tag{21}
\end{equation*}
the rest has the expression
{:[R_(m+n)(x)=(x-a)^(m)(x-b)^(n)[x","a","dots","a","b","dots","b;f]=],[quad=((x-a)^(m)(x-b)^(n))/((m+n)!)f^((m+n))(xi)","(a < xi < b)]:}\begin{gathered}
R_{m+n}(x)=(x-a)^{m}(x-b)^{n}[x, a, \ldots, a, b, \ldots, b ; f]= \\
\quad=\frac{(x-a)^{m}(x-b)^{n}}{(m+n)!} f^{(m+n)}(\xi),(a<\xi<b)
\end{gathered}
- An important application of the interpolation formula (15) is that concerning the decomposition of a rational function into simple fractions, in the general case when the denominator of the rational function has multiple roots. Moreover, there is an equivalence between these two problems, as can be easily seen.
Suppose we are given the rational function
R(x)=(f(x))/(omega(x))R(x)=\frac{f(x)}{\omega(x)}
where
omega(x)=prod_(p=1)^(s)(x-x_(p))^(r)_(p)=g_(i)(x)(x-x_(i))^(r_(i))\omega(x)=\prod_{p=1}^{s}\left(x-x_{p}\right)^{r}{ }_{p}=g_{i}(x)\left(x-x_{i}\right)^{r_{i}}
andf(x)f(x)is a polynomial of degreem < nm<n(a case that is always of interest), where we noted as beforen=r_(1)+r_(2)+dots+r_(s)-1n=r_{1}+r_{2}+\ldots+r_{s}-1.
It is known thatR(x)R(x)can be represented inmod\bmodunique in form
{:(23)R(x)=sum_(i=1)^(s)sum_(p=0)^(r_(i)-1)(A_(ip))/((x-x_(i))^(r_(i)-p))",":}\begin{equation*}
R(x)=\sum_{i=1}^{s} \sum_{p=0}^{r_{i}-1} \frac{A_{i p}}{\left(x-x_{i}\right)^{r_{i}-p}}, \tag{23}
\end{equation*}
whereA_(ip)A_{i p}are constants.
If the interpolation polynomial is written relative to the polynomialf(x)f(x)and at the nodesx_(1),x_(2),dots,x_(s)-x_{1}, x_{2}, \ldots, x_{s}-of orders of multiplicity equal respectively tor_(1),r_(2),dots,x_(s)r_{1}, r_{2}, \ldots, x_{s}- the polynomial is obtainedH_(n)(x)H_{n}(x)from (13) or (15). Considering thatf(x)f(x)is a polynomial of degreem < nm<n, we have
If the interpolation polynomial is written relative to the polynomialf(x)f(x)and at the nodesx_(1),x_(2),dots,x_(s)-x_{1}, x_{2}, \ldots, x_{s}-of orders of multiplicity equal respectively tor_(1),r_(2),dots,x_(s)r_{1}, r_{2}, \ldots, x_{s}- the polynomial is obtainedH_(n)(x)H_{n}(x)from (13) or (15). Considering thatf(x)f(x)is a polynomial of degreem < nm<n, we have
H_(n)(x)-=f(x),H_{n}(x) \equiv f(x),
so we will have
(f(x))/(omega(x))-=(H_(n)(x))/(omega(x))=sum_(i=1)^(s)sum_(p=0)^(r_(i)-1)(A_(ip))/((x-x_(i))^(r_(i)-p))\frac{f(x)}{\omega(x)} \equiv \frac{H_{n}(x)}{\omega(x)}=\sum_{i=1}^{s} \sum_{p=0}^{r_{i}-1} \frac{A_{i p}}{\left(x-x_{i}\right)^{r_{i}-p}}
where, based on formula (13)
{:[(21))A_(ip)=sum_(k=0)^(p)(f^((k))(x_(i)))/(k!(p-k)!)((1)/(g_(i)(x)))_(x=x_(i))^((p-k))=(1)/(p!)((f(x))/(g_(i)(x)))_(x=x_(i))^((p))],[(p=0,1,dots,r_(i)-1;i=1,2,dots,s).]:}\begin{gather*}
A_{i p}=\sum_{k=0}^{p} \frac{f^{(k)}\left(x_{i}\right)}{k!(p-k)!}\left(\frac{1}{g_{i}(x)}\right)_{x=x_{i}}^{(p-k)}=\frac{1}{p!}\left(\frac{f(x)}{g_{i}(x)}\right)_{x=x_{i}}^{(p)} \tag{21)}\\
\left(p=0,1, \ldots, r_{i}-1 ; i=1,2, \ldots, s\right) .
\end{gather*}
Thus we have
againl_(2,k)(x),l_(3,k)(x)l_{2, k}(x), l_{3, k}(x)are obtained from here by circular permutations of the indices1,2,31,2,3.
assuming of course that in the interval (a,ba, b) functionf(x)f(x)has a derivative of the orderm+nm+n.
9. In the case ofs=3s=3, (15) reduces to 1a
where
againl_(2,k)(x),l_(3,k)(x)l_{2, k}(x), l_{3, k}(x)are obtained from here by circular permutations of the indices1,2,31,2,3.
assuming of course that in the interval (a,ba, b) functionf(x)f(x)has a derivative of the orderm+nm+n.
9. In the case ofs=3s=3, (15) reduces to 1a
where
{:[H_(r_(1)+r_(2)+r_(3)-1)(x)=H_(r_(1)+r_(2)+r_(3)-1)(ubrace(x_(1),dots,x_(1))_(r_(1))","ubrace(x_(2),dots,x_(2))_(r_(2))","ubrace(x_(3),dots,x_(3))_(r_(3));f∣x)=],[quad=sum_(k=0)^(r_(1)-1)l_(1,k)(x)f^((k))(x_(1))+sum_(k=0)^(r_(2)-1)l_(2,k)(x)f^((k))(x_(2))+sum_(k=0)^(r_(3)-1)l_(3,k)(x)f^((k))(x_(3))]:}\begin{aligned}
& H_{r_{1}+r_{2}+r_{3}-1}(x)=H_{r_{1}+r_{2}+r_{3}-1}(\underbrace{x_{1}, \ldots, x_{1}}_{r_{1}}, \underbrace{x_{2}, \ldots, x_{2}}_{r_{2}}, \underbrace{x_{3}, \ldots, x_{3}}_{r_{3}} ; f \mid x)= \\
& \quad=\sum_{k=0}^{r_{1}-1} l_{1, k}(x) f^{(k)}\left(x_{1}\right)+\sum_{k=0}^{r_{2}-1} l_{2, k}(x) f^{(k)}\left(x_{2}\right)+\sum_{k=0}^{r_{3}-1} l_{3, k}(x) f^{(k)}\left(x_{3}\right)
\end{aligned}
{:[l_(1,k)(x)=((x-x_(2))/(x_(1)-x_(2)))^(r_(2))((x-x_(3))/(x_(1)-x_(3)))^(r_(3))xx],[xx{((x-x_(1))^(k))/(k!)sum_(alpha=0)^(r_(1)-k-1)((x-x_(1))/(x_(2)-x_(1)))^(alpha)[sum_(j=0)^(alpha)((r_(2)+alpha-j-1)/(r_(2)-1))((r_(3)+j-1)/(j))((x_(1)-x_(2))/(x_(1)-x_(3)))^(j)]}","]:}\begin{gathered}
l_{1, k}(x)=\left(\frac{x-x_{2}}{x_{1}-x_{2}}\right)^{r_{2}}\left(\frac{x-x_{3}}{x_{1}-x_{3}}\right)^{r_{3}} \times \\
\times\left\{\frac{\left(x-x_{1}\right)^{k}}{k!} \sum_{\alpha=0}^{r_{1}-k-1}\left(\frac{x-x_{1}}{x_{2}-x_{1}}\right)^{\alpha}\left[\sum_{j=0}^{\alpha}\binom{r_{2}+\alpha-j-1}{r_{2}-1}\binom{r_{3}+j-1}{j}\left(\frac{x_{1}-x_{2}}{x_{1}-x_{3}}\right)^{j}\right]\right\},
\end{gathered}
If formula (15) is used, the explicit formulas are obtained
{:(25)A_(ip)=(1)/(g_(i)(x_(i)))sum_(k=0)^(k)[sum_(alpha_(1)+dots//dots+alpha_(s)=p-k)(((r_(1)+alpha_(1)-1)/(a_(1)))cdots//cdots((r_(s)+alpha_(s)-1)/(alpha_(s))))/((x_(1)-x_(i))^(alpha_(1))dots//cdots(x_(s)-x_(i))^(u_(s)))](f^((k))(x_(i)))/(k!):}\begin{equation*}
A_{i p}=\frac{1}{g_{i}\left(x_{i}\right)} \sum_{k=0}^{k}\left[\sum_{\alpha_{1}+\ldots / \ldots+\alpha_{s}=p-k} \frac{\binom{r_{1}+\alpha_{1}-1}{a_{1}} \cdots / \cdots\binom{r_{s}+\alpha_{s}-1}{\alpha_{s}}}{\left(x_{1}-x_{i}\right)^{\alpha_{1}} \ldots / \cdots\left(x_{s}-x_{i}\right)^{u_{s}}}\right] \frac{f^{(k)}\left(x_{i}\right)}{k!} \tag{25}
\end{equation*}
- In the particular case
{:(26)R(x)=(f(x))/((x-a)^(m)(x-b)^(n)):}\begin{equation*}
R(x)=\frac{f(x)}{(x-a)^{m}(x-b)^{n}} \tag{26}
\end{equation*}
the decomposition into simple fractions is of the form
R(x)=sum_(p=0)^(m-1)(A_(p))/((x-a)^(p))+sum_(q=0)^(n-1)(B_(q))/((x-b)^(q))R(x)=\sum_{p=0}^{m-1} \frac{A_{p}}{(x-a)^{p}}+\sum_{q=0}^{n-1} \frac{B_{q}}{(x-b)^{q}}
Based on the previous formulas, we have
{:[(27)A_(p)=(1)/((a-b)^(n))sum_(i=0)^(p)(-1)^(p-i)(((n+p-i-1)/(n-1)))/((a-b)^(p-i))(f^((i))(a))/(i!)],[B_(q)=(1)/((b-a)^(m))sum_(j=0)^(q)(-1)^(q-j)(((m+q-j-1)/(m-1)))/((b-a)^(q-j))(f^((j))(b))/(j!)]:}\begin{align*}
& A_{p}=\frac{1}{(a-b)^{n}} \sum_{i=0}^{p}(-1)^{p-i} \frac{\binom{n+p-i-1}{n-1}}{(a-b)^{p-i}} \frac{f^{(i)}(a)}{i!} \tag{27}\\
& B_{q}=\frac{1}{(b-a)^{m}} \sum_{j=0}^{q}(-1)^{q-j} \frac{\binom{m+q-j-1}{m-1}}{(b-a)^{q-j}} \frac{f^{(j)}(b)}{j!}
\end{align*}
- Let us now assume that we have to calculate the definite integral
I=int_(a)^(b)f(x)dxI=\int_{a}^{b} f(x) d x
If the interpolation formula (17) is used, the quadrature formula is obtained
{:(28)int_(a)^(b)f(x)dx=sum_(i=1)^(s)sum_(k=0)^(r_(i)-1)A_(l,k)f^((k))(x_(i))+rho(f):}\begin{equation*}
\int_{a}^{b} f(x) d x=\sum_{i=1}^{s} \sum_{k=0}^{r_{i}-1} A_{l, k} f^{(k)}\left(x_{i}\right)+\rho(f) \tag{28}
\end{equation*}
where
again
again
A_(l,k)=int_(a)^(b)l_(l,k)(x)dxA_{l, k}=\int_{a}^{b} l_{l, k}(x) d x
rho(f)=int_(a)^(b)R(x)dx\rho(f)=\int_{a}^{b} R(x) d x
The quadrature formula (28) generally has the degree of accuracynn, that is, the remainder is zero iff(x)f(x)is a polynomial of degree at mostnnHowever, it may happen that through a convenient choice of nodes, the degree of accuracy may be higher.
Examples.1^(@)1^{\circ}. Choosing the nodesx_(1)=x_(2)=x_(3)=-1,x_(4)=0,x_(5)=x_(6)==x_(7)=1x_{1}=x_{2}=x_{3}=-1, x_{4}=0, x_{5}=x_{6}= =x_{7}=1and writing the corresponding interpolation formula, we have
f(x)=H_(6)(x)+R_(7)(x)f(x)=H_{6}(x)+R_{7}(x)
whereH_(6)(x)H_{6}(x)is the interpolation polynomial (19) and
R_(eta)(x)=x(x^(2)-1)^(3)[x,-1,-1,-1,0,1,1,1;f]R_{\eta}(x)=x\left(x^{2}-1\right)^{3}[x,-1,-1,-1,0,1,1,1 ; f]
Integrating from -1 to +1, we arrive at the quadrature formula of degree of accuracy 7:
{:[(29)int_(-1)^(+1)f(x)dx=(1)/(105)[57 f(-1)+12f^(')(-1)+f^('')(-1)+96(0)+f^('')(1)-:}],[{:-12f^(')(1)+57 f(1)]+rho(f)]:}\begin{gather*}
\int_{-1}^{+1} f(x) d x=\frac{1}{105}\left[57 f(-1)+12 f^{\prime}(-1)+f^{\prime \prime}(-1)+96(0)+f^{\prime \prime}(1)-\right. \tag{29}\\
\left.-12 f^{\prime}(1)+57 f(1)\right]+\rho(f)
\end{gather*}
For the restrho(f)\rho(f)the expression has been established
rho(f)=(-1)/(396900)f^((8))(xi),quad-1 < xi < +1\rho(f)=\frac{-1}{396900} f^{(8)}(\xi), \quad-1<\xi<+1
2^(@)2^{\circ}If nodes are usedx_(1)=x_(2)=x_(3)=x_(4)=x_(5)=0x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0and-x_(6)=x_(7)=sqrt((7)/(8))-x_{6}=x_{7}=\sqrt{\frac{7}{8}}, write the corresponding Hermite interpolation formula, multiply byp(x)=(1-x^(2))^(-(1)/(2))p(x)=\left(1-x^{2}\right)^{-\frac{1}{2}}and integrating from -1 to +1, we obtain the quadrature formula with accuracy degree 9
{:[int_(-1)^(+1)(f(x))/(sqrt(1-x^(2)))dx=(pi)/(65856){35136 f(0)+3024f^('')(0)+49f^((IV))(0)+:}],[('")"{: quad+15360[f(-sqrt((7)/(8)))+f(|sqrt((7)/(8)))]}+(pi)/(530841600)f^((10))(xi)]:}\begin{align*}
& \int_{-1}^{+1} \frac{f(x)}{\sqrt{1-x^{2}}} d x=\frac{\pi}{65856}\left\{35136 f(0)+3024 f^{\prime \prime}(0)+49 f^{(I V)}(0)+\right. \\
& \left.\quad+15360\left[f\left(-\sqrt{\frac{7}{8}}\right)+f\left(\left\lvert\, \sqrt{\frac{7}{8}}\right.\right)\right]\right\}+\frac{\pi}{530841600} f^{(10)}(\xi) \tag{$\prime$}
\end{align*}
Observation. The quadrature formulas (with remainder) from (29) and (29́) were arrived at by doubling the node in the first casex=0x=0, and in the second case doubling the nodesx_(6)x_{6}andx_(7)x_{7}and taking thex=0x=0multiple of order 6.
13. Let's see what the quadrature formula (28) becomes in the particular case from no. 8. We will use the interpolation polynomial (20).
13. Let's see what the quadrature formula (28) becomes in the particular case from no. 8. We will use the interpolation polynomial (20).
Using the generalized integration by parts formula, after systematically performing the calculations, we obtain
{:[int_(a)^(b)H_(m+n=1)(x)dx=nsum_(k=0)^(m-1)((b-a)^(k+1))/((k+2)!)[sum_(i=0)^(m-k-1)(((k+i)/(i)))/(((n+k+i+1)/(n+i-1)))]f^((k))(a)+],[+msum_(r=0)^(n-1)((a-b)^(k))/((r+2)!)[sum_(r=0)^(n-r-1)(((r+j)/(i)))/(((m+r+j+1)/(r+2)))]f^((j))(b)]:}\begin{gathered}
\int_{a}^{b} H_{m+n=1}(x) d x=n \sum_{k=0}^{m-1} \frac{(b-a)^{k+1}}{(k+2)!}\left[\sum_{i=0}^{m-k-1} \frac{\binom{k+i}{i}}{\binom{n+k+i+1}{n+i-1}}\right] f^{(k)}(a)+ \\
+m \sum_{r=0}^{n-1} \frac{(a-b)^{k}}{(r+2)!}\left[\sum_{r=0}^{n-r-1} \frac{\binom{r+j}{i}}{\binom{m+r+j+1}{r+2}}\right] f^{(j)}(b)
\end{gathered}
To obtain a simpler expression of the coefficients of this formula, we will use the identity
(n)/(k+3)sum_(i=0)^(m-k-1)(((k+i)/(i)))/(((n+k+i+1)/(k+2)))=(((m)/(k+1)))/(((m+n)/(k+1)))\frac{n}{k+3} \sum_{i=0}^{m-k-1} \frac{\binom{k+i}{i}}{\binom{n+k+i+1}{k+2}}=\frac{\binom{m}{k+1}}{\binom{m+n}{k+1}}
which is proven without difficulty.
In this way we arrive at the quadrature formula of degree of accuracym+n-1m+n-1:
In this way we arrive at the quadrature formula of degree of accuracym+n-1m+n-1:
{:[(30)int_(a)^(b)f(x)dx=sum_(k=0)^(m-1)((b-a)^(k+1))/((k+1)!)*(((m)/(k+1)))/(((m+n)/(k+1)))f^((k))(a)-],[-sum_(r=0)^(n-1)((a-b)^(r+1))/((r+1)!)*(((n)/(r+1)))/(((m+n)/(r+1)))f^((r))(b)+rho_(m+n)(f)]:}\begin{align*}
& \int_{a}^{b} f(x) d x=\sum_{k=0}^{m-1} \frac{(b-a)^{k+1}}{(k+1)!} \cdot \frac{\binom{m}{k+1}}{\binom{m+n}{k+1}} f^{(k)}(a)- \tag{30}\\
& -\sum_{r=0}^{n-1} \frac{(a-b)^{r+1}}{(r+1)!} \cdot \frac{\binom{n}{r+1}}{\binom{m+n}{r+1}} f^{(r)}(b)+\rho_{m+n}(f)
\end{align*}
where
rho_(m+n)(f)=int_(a)^(b)(x-a)^(m)(x-b)^(n)[x,a,dots,a,b,dots,b;f]dx\rho_{m+n}(f)=\int_{a}^{b}(x-a)^{m}(x-b)^{n}[x, a, \ldots, a, b, \ldots, b ; f] d x
or
{:(31)rho_(m+n)(f)=((-1)^(n)(b-a)^(m+n+1))/((m+n+1)!((m+n)/(n)))f^((m+n))(xi)","quad a < xi < b:}\begin{equation*}
\rho_{m+n}(f)=\frac{(-1)^{n}(b-a)^{m+n+1}}{(m+n+1)!\binom{m+n}{n}} f^{(m+n)}(\xi), \quad a<\xi<b \tag{31}
\end{equation*}
makingm=nm=n, this reduces to the formula
{:(32)int_(a)^(b)f(x)dx=sum_(k=0)^(m-1)((b-a)^(k+1))/((k+1)!)(((m)/(k+1)))/(((2m)/(k+1)))[f^((k))(a)+(-1)^(k)f^((k))(b)]+rho_(2m)(f):}\begin{equation*}
\int_{a}^{b} f(x) d x=\sum_{k=0}^{m-1} \frac{(b-a)^{k+1}}{(k+1)!} \frac{\binom{m}{k+1}}{\binom{2 m}{k+1}}\left[f^{(k)}(a)+(-1)^{k} f^{(k)}(b)\right]+\rho_{2 m}(f) \tag{32}
\end{equation*}
where
rho_(2m)(f)=(-1)^(n)((b-a)^(2m+1))/((2m+1)!((2m)/(m)))f^((2m))(xi)\rho_{2 m}(f)=(-1)^{n} \frac{(b-a)^{2 m+1}}{(2 m+1)!\binom{2 m}{m}} f^{(2 m)}(\xi)
Formulas (20) and (32) are due to^(2){ }^{2}Hermite [1]. He deduced them with a special method, without using the interpolation formula (20), which he gave up establishing because it seemed to him too complicated.
These formulas have recently been generalized, in a certain sense, by Prof. DV Ionescu [12].
14. Using the interpolation formula (15), one can also construct formulas for the numerical calculation of derivatives of different orders of a functionf(x)f(x)at a pointx_(0)x_{0}.
Now we will deal with a concrete case that seems important to us. Namely, we will use the interpolation formula (20) to establish formulas for calculating derivativesf^(')(a),f^('')(a),dots,f^((m+p))f^{\prime}(a), f^{\prime \prime}(a), \ldots, f^{(m+p)}, whereppis a natural number smaller thannn.
For this, we will seek to calculate
{:(33)(d^(m+p))/(dx^(m+p))H_(m+n-1)(x)|_(x=a):}\begin{equation*}
\left.\frac{d^{m+p}}{d x^{m+p}} H_{m+n-1}(x)\right|_{x=a} \tag{33}
\end{equation*}
Considering formula (20), the coefficient off^((k))(a)f^{(k)}(a)from (33) it is found that
{:(34)sum_(i=0)^(m-k-1)((-1)^(i))/((a-b)^(n+i))*(1)/(k!)((n+i-1)/(i))(d^(m+p))/(dx^(m+p))[(x-b)^(n)(x-a)^(k+i)]:}\begin{equation*}
\sum_{i=0}^{m-k-1} \frac{(-1)^{i}}{(a-b)^{n+i}} \cdot \frac{1}{k!}\binom{n+i-1}{i} \frac{d^{m+p}}{d x^{m+p}}\left[(x-b)^{n}(x-a)^{k+i}\right] \tag{34}
\end{equation*}
Since based on Leibniz's formula it is found that
{:[[(x-b)^(n)(x-a)^(k)]_(x=a)^((m+p))=],[=n(n-1)dots(n-m-p+k+1)!k!((m+p)/(k))(a-b)^(n-m-p+k)]:}\begin{gathered}
{\left[(x-b)^{n}(x-a)^{k}\right]_{x=a}^{(m+p)}=} \\
=n(n-1) \ldots(n-m-p+k+1)!k!\binom{m+p}{k}(a-b)^{n-m-p+k}
\end{gathered}
expression (34) can be written
((m+p)!)/(k!(a-b)^(m+p-k))sum_(i=0)^(m-k-1)(-1)^(i)((n+i-1)/(i))((n)/(m+p-k-i))\frac{(m+p)!}{k!(a-b)^{m+p-k}} \sum_{i=0}^{m-k-1}(-1)^{i}\binom{n+i-1}{i}\binom{n}{m+p-k-i}
If we use the identity
sum_(alpha=0)^(k)(-1)^(alpha)((p+alpha)/(alpha))((p+1)/(j-alpha))=(-1)^(k)((j-1)/(k))((p+1+k)/(j))\sum_{\alpha=0}^{k}(-1)^{\alpha}\binom{p+\alpha}{\alpha}\binom{p+1}{j-\alpha}=(-1)^{k}\binom{j-1}{k}\binom{p+1+k}{j}
which is easily proven, is ultimately found for the coefficient off^((k))(a)f^{(k)}(a)from (33) the following expression
{:(35)(-1)^(p+1)((m+p)!)/(k!(b-a)^(m+p-k))((m+p-k-1)/(p))((m+n-k-1)/(n-p-1)):}\begin{equation*}
(-1)^{p+1} \frac{(m+p)!}{k!(b-a)^{m+p-k}}\binom{m+p-k-1}{p}\binom{m+n-k-1}{n-p-1} \tag{35}
\end{equation*}
- Let's now find the coefficient off(r)(b)f(r)(b)from (34).
This is
{:[(36)(1)/((b-a)^(m))*(1)/(r!)sum_(j=0)^(n-r-1)((m+j-1)/(j))(1)/((b-a)^(j))[(x-a)^(m)(x-b)^(r+j)]_(x=a)^((m+p))=],[quad=(-1)^(r-p)((m+p)!)/(r!)(1)/((b-a)^(m+p-r))sum_(j=0)^(n-r-1)((r+j)/(p))((m+j-1)/(j))]:}\begin{gather*}
\frac{1}{(b-a)^{m}} \cdot \frac{1}{r!} \sum_{j=0}^{n-r-1}\binom{m+j-1}{j} \frac{1}{(b-a)^{j}}\left[(x-a)^{m}(x-b)^{r+j}\right]_{x=a}^{(m+p)}= \tag{36}\\
\quad=(-1)^{r-p} \frac{(m+p)!}{r!} \frac{1}{(b-a)^{m+p-r}} \sum_{j=0}^{n-r-1}\binom{r+j}{p}\binom{m+j-1}{j}
\end{gather*}
Let's then deal with the evaluation of the amount
{:(37)C_(r)=sum_(j=0)^(n-r-1)((m-1+j)/(j))((r+j)/(p)):}\begin{equation*}
C_{r}=\sum_{j=0}^{n-r-1}\binom{m-1+j}{j}\binom{r+j}{p} \tag{37}
\end{equation*}
Here some terms at the beginning are null, because in the development according to Leibniz's formula, which we used above, it must be assumedr+j >= pr+j \geqq p, so thatC_(r)C_{r}is reduced to 1
{:(38)C_(r)=sum_(j=p-r)^(n-r-1)((m-1+j)/(j))((r+j)/(r-p+j)):}\begin{equation*}
C_{r}=\sum_{j=p-r}^{n-r-1}\binom{m-1+j}{j}\binom{r+j}{r-p+j} \tag{38}
\end{equation*}
This can also be put in the form
{:(39)C_(r)=sum_(i=0)^(n-p-1)((m+p-r-1+i)/(p-r+i))((p+i)/(i)):}\begin{equation*}
C_{r}=\sum_{i=0}^{n-p-1}\binom{m+p-r-1+i}{p-r+i}\binom{p+i}{i} \tag{39}
\end{equation*}
or
{:(40)C_(r)=((m+p-r-1)/(m-1))sum_(i=0)^(n-p-1)((p+i)/(i))(((m+p-r+i)/(i)))/(((p-r+i)/(i))):}\begin{equation*}
C_{r}=\binom{m+p-r-1}{m-1} \sum_{i=0}^{n-p-1}\binom{p+i}{i} \frac{\binom{m+p-r+i}{i}}{\binom{p-r+i}{i}} \tag{40}
\end{equation*}
Applying the formula
((a)/(b))=sum_(j=0)^(l)((l)/(j))((a-l)/(b-j))\binom{a}{b}=\sum_{j=0}^{l}\binom{l}{j}\binom{a-l}{b-j}
get
((p+i)/(i))=sum_(j=0)^(r)((r)/(j))((p+i-v)/(i-j)).\binom{p+i}{i}=\sum_{j=0}^{r}\binom{r}{j}\binom{p+i-v}{i-j} .
With this
{:[C_(r)=sum_(i=0)^(n-p-1)sum_(j=0)^(r)((r)/(j))((p+i-r)/(i-j))((m+p-r-1+i)/(p-r+i))=],[=sum_(j=0)^(r)((r)/(j))sum_(i=j)^(n-p-1)((p+i-r)/(i-j))((m+p-r-1+i)/(p-r+i))","]:}\begin{aligned}
C_{r} & =\sum_{i=0}^{n-p-1} \sum_{j=0}^{r}\binom{r}{j}\binom{p+i-r}{i-j}\binom{m+p-r-1+i}{p-r+i}= \\
& =\sum_{j=0}^{r}\binom{r}{j} \sum_{i=j}^{n-p-1}\binom{p+i-r}{i-j}\binom{m+p-r-1+i}{p-r+i},
\end{aligned}
based on the usual convention that((m)/(n))\binom{m}{n}be null ifmmornnwould be negative.
Noting
Noting
beta=i-j,quad m=p+j-r,quad n=m+p+j-r-1\beta=i-j, \quad m=p+j-r, \quad n=m+p+j-r-1
HAVE
C_(r)=sum_(j=0)^(r)((r)/(j))^(n-p-j-1)sum_(beta=0)^(-1)((m+beta)/(beta))((n+beta)/(m+beta)).C_{r}=\sum_{j=0}^{r}\binom{r}{j}^{n-p-j-1} \sum_{\beta=0}^{-1}\binom{m+\beta}{\beta}\binom{n+\beta}{m+\beta} .
Applying the identity
sum_(p=0)^(k)((m+beta)/(beta))((n+beta)/(m+beta))=((n+k+1)/(k))((n)/(m))\sum_{p=0}^{k}\binom{m+\beta}{\beta}\binom{n+\beta}{m+\beta}=\binom{n+k+1}{k}\binom{n}{m}
we finally get
{:(41)C_(r)=sum_(j=0)^(r)((r)/(j))((m+n-r-1)/(n-p-j-1))((m+p+j-r-1)/(p+j-r)):}\begin{equation*}
C_{r}=\sum_{j=0}^{r}\binom{r}{j}\binom{m+n-r-1}{n-p-j-1}\binom{m+p+j-r-1}{p+j-r} \tag{41}
\end{equation*}
For example, fork=0,1,2k=0,1,2HAVE
{:[C_(0)=((m+p-1)/(m-1))((m+n-1)/(m+p))","],[C_(1)=((m+n-2)/(m+p-1))((m+p-2)/(m-1))+((m+n-2)/(m+p))((m+p-1)/(m-1))","],[C_(2)=((m+n-3)/(n-p-1))((m+p-3)/(p-2))+2((m+n-3)/(n-p-2))((m+p-2)/(p-1))+((m+n-3)/(n-p-3))((m+p-1)/(p))]:}\begin{gathered}
C_{0}=\binom{m+p-1}{m-1}\binom{m+n-1}{m+p}, \\
C_{1}=\binom{m+n-2}{m+p-1}\binom{m+p-2}{m-1}+\binom{m+n-2}{m+p}\binom{m+p-1}{m-1}, \\
C_{2}=\binom{m+n-3}{n-p-1}\binom{m+p-3}{p-2}+2\binom{m+n-3}{n-p-2}\binom{m+p-2}{p-1}+\binom{m+n-3}{n-p-3}\binom{m+p-1}{p}
\end{gathered}
- T, Taking into account (35), (36), (37), (41), it is seen that we were able to construct the following numerical derivation formula of degree of accuracym+n-1m+n-1
{:(42)f^((m+p))(a)=sum_(k=0)^(m-1)A_(k)f^((k))(a)+sum_(r=0)^(n-1)B_(r)f^((r))(b)+rho(f):}\begin{equation*}
f^{(m+p)}(a)=\sum_{k=0}^{m-1} A_{k} f^{(k)}(a)+\sum_{r=0}^{n-1} B_{r} f^{(r)}(b)+\rho(f) \tag{42}
\end{equation*}
where
{:(43)A_(k)=(-1)^(p+1)((m+p)!)/(k!(b-a)^(m+p-k))((m+p-k-1)/(p))((n+m-k-1)/(n-p-1)):}\begin{equation*}
A_{k}=(-1)^{p+1} \frac{(m+p)!}{k!(b-a)^{m+p-k}}\binom{m+p-k-1}{p}\binom{n+m-k-1}{n-p-1} \tag{43}
\end{equation*}
and
{:(44)B_(r)=(-1)^(r+p)((m+p)!)/(r!(b-a)^(m+p-r))C_(r):}\begin{equation*}
B_{r}=(-1)^{r+p} \frac{(m+p)!}{r!(b-a)^{m+p-r}} C_{r} \tag{44}
\end{equation*}
and
{:(45)C_(r)=sum_(j-0)^(r)((r)/(j))((m+n-r-1)/(n-p-j-1))((m+p+j-r-1)/(p+j-r)):}\begin{equation*}
C_{r}=\sum_{j-0}^{r}\binom{r}{j}\binom{m+n-r-1}{n-p-j-1}\binom{m+p+j-r-1}{p+j-r} \tag{45}
\end{equation*}
or withC_(r)C_{r}given by (38), (39) or (40).
For the rest of this formula, the expression was obtained
For the rest of this formula, the expression was obtained
{:(46)rho(f)=((m+p)!)/((m+n)!)((n)/(p))(a-b)^(n-p)f^((m+n))(xi)","a < xi < b.:}\begin{equation*}
\rho(f)=\frac{(m+p)!}{(m+n)!}\binom{n}{p}(a-b)^{n-p} f^{(m+n)}(\xi), a<\xi<b . \tag{46}
\end{equation*}
- Particular cases of the numerical derivation formula (42):
1^(@).m=n=2,p=11^{\circ} . m=n=2, p=1:
f^(''')(a)=-(12)/((b-a)^(3))[f(a)-f(b)]+(6)/((b-a)^(2))[f^(')(a)+f^(')(b)]+(a-b)/(2)f^((4))(xi).f^{\prime \prime \prime}(a)=-\frac{12}{(b-a)^{3}}[f(a)-f(b)]+\frac{6}{(b-a)^{2}}\left[f^{\prime}(a)+f^{\prime}(b)\right]+\frac{a-b}{2} f^{(4)}(\xi) .
2^(@).m=3,n=4,p=02^{\circ} . m=3, n=4, p=0:
{:[f^(''')(a)=(120)/((b-a)^(3))[f(b)-f(a)]-(60)/((b-a)^(2))[f^(')(b)+f^(')(a)]+],[quad+(12)/(b-a)[f^('')(b)-f^('')(a)]-f^(''')(b)+((b-a)^(4))/(840)f^((7))(xi)]:}\begin{aligned}
& f^{\prime \prime \prime}(a)=\frac{120}{(b-a)^{3}}[f(b)-f(a)]-\frac{60}{(b-a)^{2}}\left[f^{\prime}(b)+f^{\prime}(a)\right]+ \\
& \quad+\frac{12}{b-a}\left[f^{\prime \prime}(b)-f^{\prime \prime}(a)\right]-f^{\prime \prime \prime}(b)+\frac{(b-a)^{4}}{840} f^{(7)}(\xi)
\end{aligned}
3^(@).m=3,n=4,p=13^{\circ} . m=3, n=4, p=1:
{:[f^((IV))(a)=(1080)/((b-a)^(4))f(a)+(480)/((b-a)^(3))f^(')(a)+(72)/((b-a)^(2))f^('')(a)-(1080)/((b-a)^(4))f(b)+],[quad+(600)/((b-a)^(3))f^(')(b)-(132)/((b-a)^(2))f^('')(b)+(12)/(b-a)f^(''')(b)-(2(b-a)^(3))/(105)f^((7))(xi)]:}\begin{aligned}
& f^{(I V)}(a)=\frac{1080}{(b-a)^{4}} f(a)+\frac{480}{(b-a)^{3}} f^{\prime}(a)+\frac{72}{(b-a)^{2}} f^{\prime \prime}(a)-\frac{1080}{(b-a)^{4}} f(b)+ \\
& \quad+\frac{600}{(b-a)^{3}} f^{\prime}(b)-\frac{132}{(b-a)^{2}} f^{\prime \prime}(b)+\frac{12}{b-a} f^{\prime \prime \prime}(b)-\frac{2(b-a)^{3}}{105} f^{(7)}(\xi)
\end{aligned}
4^(@).m==3,n=4,p=24^{\circ} . m==3, n=4, p=2:
{:[f^((V))(a)=(-4320)/((b-a)^(5))f(a)-(1800)/((b-a)^(4))f^(')(a)-(240)/((b-a)^(3))f^('')(a)+(4320)/((b-a)^(5))f(b)-],[quad-(2520)/((b-a)^(4))f^(')(b)+(600)/((b-a)^(3))f^('')(b)-(60)/((b-a)^(2))f^(''')(b)+(1)/(7)(b-a)^(2)f^((7))(xi)]:}\begin{aligned}
& f^{(V)}(a)=\frac{-4320}{(b-a)^{5}} f(a)-\frac{1800}{(b-a)^{4}} f^{\prime}(a)-\frac{240}{(b-a)^{3}} f^{\prime \prime}(a)+\frac{4320}{(b-a)^{5}} f(b)- \\
& \quad-\frac{2520}{(b-a)^{4}} f^{\prime}(b)+\frac{600}{(b-a)^{3}} f^{\prime \prime}(b)-\frac{60}{(b-a)^{2}} f^{\prime \prime \prime}(b)+\frac{1}{7}(b-a)^{2} f^{(7)}(\xi)
\end{aligned}
5^(@).m=3,n=4,p=35^{\circ} . m=3, n=4, p=3:
{:[f^((VI))(a)=(7200)/((b-a)^(6))f(a)+(2880)/((b-a)^(5))f^(')(a)+(360)/((b-a)^(4))f^('')(a)-(7200)/((b-a)^(6))f(b)+],[+(4320)/((b-a)^(5))f^(')(b)-(1080)/((b-a)^(4))f^('')(b)+(120)/((b-a)^(4))f^(''')(b)-(4(b-a))/(7)f^((7))(xi)]:}\begin{aligned}
& f^{(V I)}(a)=\frac{7200}{(b-a)^{6}} f(a)+\frac{2880}{(b-a)^{5}} f^{\prime}(a)+\frac{360}{(b-a)^{4}} f^{\prime \prime}(a)-\frac{7200}{(b-a)^{6}} f(b)+ \\
& +\frac{4320}{(b-a)^{5}} f^{\prime}(b)-\frac{1080}{(b-a)^{4}} f^{\prime \prime}(b)+\frac{120}{(b-a)^{4}} f^{\prime \prime \prime}(b)-\frac{4(b-a)}{7} f^{(7)}(\xi)
\end{aligned}
6^(@).m=1,n=6,p=06^{\circ} . m=1, n=6, p=0:
{:[f^(')(a)=(6)/(b-a)[f(b)-f(a)]-5f^(')(b)+2(b-a)f^('')(b)-(1)/(2)f^(''')(b)+],[quad+(1)/(12)(b-a)^(3)f^((IV))(b)-(1)/(120)(b-a)^(4)f^((V))(b)+((b-a)^(6))/(5040)f^((7))(xi)]:}\begin{aligned}
& f^{\prime}(a)=\frac{6}{b-a}[f(b)-f(a)]-5 f^{\prime}(b)+2(b-a) f^{\prime \prime}(b)-\frac{1}{2} f^{\prime \prime \prime}(b)+ \\
& \quad+\frac{1}{12}(b-a)^{3} f^{(I V)}(b)-\frac{1}{120}(b-a)^{4} f^{(V)}(b)+\frac{(b-a)^{6}}{5040} f^{(7)}(\xi)
\end{aligned}
7^(@).m=1,n=6,p=17^{\circ} . m=1, n=6, p=1:
{:[f^('')(a)=(30)/((b-a)^(2))[f(a)-f(b)]+(30)/(b-a)f^(')(b)-14f^('')(b)+4(b-a)f^(''')(b)-],[quad-(3)/(4)(b-a)^(2)f^((IV))(b)+(1)/(12)(b-a)^(3)f^((V))(b)-((b-a)^(5))/(420)f^((7))(xi)]:}\begin{aligned}
& f^{\prime \prime}(a)=\frac{30}{(b-a)^{2}}[f(a)-f(b)]+\frac{30}{b-a} f^{\prime}(b)-14 f^{\prime \prime}(b)+4(b-a) f^{\prime \prime \prime}(b)- \\
& \quad-\frac{3}{4}(b-a)^{2} f^{(I V)}(b)+\frac{1}{12}(b-a)^{3} f^{(V)}(b)-\frac{(b-a)^{5}}{420} f^{(7)}(\xi)
\end{aligned}
8^(@).m=1,n=6,p=28^{\circ} . m=1, n=6, p=2:
{:[f^(''')(a)=(120)/((b-a)^(3))[f(b)-f(a)]-(120)/((b-a)^(2))f^(')(b)+(60)/(b-a)f^('')(b)-],[-19f^(''')(b)+4(b-a)f^((IV))(b)-(1)/(2)(b-a)^(2)f^((V))(b)+((b-a)^(4))/(56)f^((7))(xi)]:}\begin{aligned}
& f^{\prime \prime \prime}(a)=\frac{120}{(b-a)^{3}}[f(b)-f(a)]-\frac{120}{(b-a)^{2}} f^{\prime}(b)+\frac{60}{b-a} f^{\prime \prime}(b)- \\
& -19 f^{\prime \prime \prime}(b)+4(b-a) f^{(I V)}(b)-\frac{1}{2}(b-a)^{2} f^{(V)}(b)+\frac{(b-a)^{4}}{56} f^{(7)}(\xi)
\end{aligned}
9^(@).m=1,n=6,p=39^{\circ} . m=1, n=6, p=3:
{:[f^((IV))(a)=(360)/((b-a)^(4))[f(a)-f(b)]+(360)/((b-a)^(3))f^(')(b)-(180)/((b-a)^(2))f^('')(b)+],[quad+(60)/(b-a)f^(''')(b)-14f^((IV))(b)+2(b-a)f^((V))(b)-(2(b-a)^(3))/(21)f^((7))(xi)]:}\begin{aligned}
& f^{(I V)}(a)=\frac{360}{(b-a)^{4}}[f(a)-f(b)]+\frac{360}{(b-a)^{3}} f^{\prime}(b)-\frac{180}{(b-a)^{2}} f^{\prime \prime}(b)+ \\
& \quad+\frac{60}{b-a} f^{\prime \prime \prime}(b)-14 f^{(I V)}(b)+2(b-a) f^{(V)}(b)-\frac{2(b-a)^{3}}{21} f^{(7)}(\xi)
\end{aligned}
10^(@).m=1,n=6,p=410^{\circ} . m=1, n=6, p=4:
{:[f^((V))(a)=(720)/((b-a)^(5))[f(b)-f(a)]-(720)/((b-a)^(4))f^(')(b)+(360)/((b-a)^(3))f^('')(b)-],[-(120)/((b-a)^(2))f^(''')(b)+(30)/(b-a)f^((IV))(b)-5f^((V))(b)+(5(b-a)^(2))/(14)f^((7))(xi)]:}\begin{aligned}
& f^{(V)}(a)=\frac{720}{(b-a)^{5}}[f(b)-f(a)]-\frac{720}{(b-a)^{4}} f^{\prime}(b)+\frac{360}{(b-a)^{3}} f^{\prime \prime}(b)- \\
& -\frac{120}{(b-a)^{2}} f^{\prime \prime \prime}(b)+\frac{30}{b-a} f^{(I V)}(b)-5 f^{(V)}(b)+\frac{5(b-a)^{2}}{14} f^{(7)}(\xi)
\end{aligned}
11^(@).m=1,n=6,p=511^{\circ} . m=1, n=6, p=5:
{:[f^((VI))(a)=(720)/((b-a)^(6))[f(a)-f(b)]+(720)/((b-a)^(5))f^(')(b)-(360)/((b-a)^(4))f^('')(b)+],[quad+(120)/((b-a)^(3))f^(''')(b)-(30)/((b-a)^(2))f^((IV))(b)+(6)/(b-a)f^((V))(b)-(6(b-a))/(7)f^((7))(xi)]:}\begin{aligned}
& f^{(V I)}(a)=\frac{720}{(b-a)^{6}}[f(a)-f(b)]+\frac{720}{(b-a)^{5}} f^{\prime}(b)-\frac{360}{(b-a)^{4}} f^{\prime \prime}(b)+ \\
& \quad+\frac{120}{(b-a)^{3}} f^{\prime \prime \prime}(b)-\frac{30}{(b-a)^{2}} f^{(I V)}(b)+\frac{6}{b-a} f^{(V)}(b)-\frac{6(b-a)}{7} f^{(7)}(\xi)
\end{aligned}
- We mention that the first 22 numerical derivation formulas - explicitly constructed, as examples, by Prof. T. Popoviciu in his important memoir [13] - can be obtained immediately from the more general formula (42). Above we gave other formulas than those found in the aforementioned work.
BIBLIOGRAPHY
- Ch. Hermite, Sur to Lagrange's interpolation formulas. Journ. fd reign n. angew. Math., 1878, vol. LXXXIV, p. 70-79.
- IP Natanson, Konstruktivnaia leoria functii. Moscow-Leningrad, 1949, p. 501.
- Gy. Zemp1en, Étude sur l'interpolation et la décomposition des fonctions rationalelles en fractions partielles. Archiv der Mathematik und Physik, 1905 (3) 8, pp. 214-226.
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- Jahrbuch über die Fortschritte der Mathematik, 1905, vol. XXXV, pp. 283-284.
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- E. Netto, Les fonctions rationalelles. Encyclopédie des Sci. Math., 1910, vol. I, vol. II. i do 2, p. 62.
- VL Gonciarov, Teoria interpolirovania i priblijenia tunctii. Moscow, 1954, p. 66.
- A. Markoff, Sur la méthode de Gauss pour le calcul approché des intégrales. Math. Annalen, 1885, vol. XXV, p. 427.
- K. Petr, O jedene formulas pro numerický výpocet certain integrals. Čaopis pro péstováni Mathematiky a Fysiky, 1915, tom. XLIV, pp. 454-455.
- N. Obreschkoff, Neue Quadraturformeln. Abhandl. d. preuss. Akkadian d. Wis., Math. - naturwiss. Kl., 1940, no. 4, pp. 1-20.
- Sh. E. Mikeladze, Cislennîe metodi matematiceskogo analysis. Moscow, 1953, p. 394.
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- Cislennoe integrirovanie. Good luck Matt. Nauk, t. III, fasc. 6, 1948.
- DV Ionescu, Generalization of the quadrature theorem of N. Obreschkoff. Studies and scientific research, -Acad. RPR Cluj Branch, 1952, no. 3-4, p. 1-9.
- T. Popoviciu, On the remainder in some numerical derivation formulas. Studies and research. mat. - Acad. RPR, tom. III, no. 1-2, p. 103.
About the interpolation formula of Hermit and about some of their applications
(Краткое онлайн)
After the fact that the obvious expression given by G. is noted. Земпленом [3], for the Hermite interpolation polynomial [2] is not corrects >= 2s \geqq 2andr > 2r>2, is established, a few more simple way than in [6], formula (13); then a more obvious formula (15) is given for the Hermite interpolation polynomial satisfying conditions (3). In (20) the effective expression for the interpolation polynomial is given:
After the fact that the obvious expression given by G. is noted. Земпленом [3], for the Hermite interpolation polynomial [2] is not corrects >= 2s \geqq 2andr > 2r>2, is established, a few more simple way than in [6], formula (13); then a more obvious formula (15) is given for the Hermite interpolation polynomial satisfying conditions (3). In (20) the effective expression for the interpolation polynomial is given:
H_(m+n-1)ubrace((a,dots,a)_(m),ubrace(b,dots,b)_(n);f∣x).H_{m+n-1} \underbrace{(a, \ldots, a}_{m}, \underbrace{b, \ldots, b}_{n} ; f \mid x) .
В №. 10-11 the previous formulas are used to decompose the rational function into simple fractions. Thus, formulas (24) were established for the expansion coefficients (23). In (27) effective expressions are given for the coefficients of the expansion of the private rational function (26).
В №. 12-13 give some applications to the numerical integration of functions. As examples, constructed in (29) and (29') are two high-accuracy quadrature formulas obtained with appropriate selection of nodes. Using the interpolation formula (20), the quadrature formula (30) was established. In connection with this, it is noted that both formula (30) and partial formula (32) were given by Hermit [1].
In (42) the formula of numerical differentiation is established, analogous to Hermit's quadrature formula (30).
В №. 17 days 11 examples of such formulas.
Sur la formulas d'interpolation d'Hermite et quelques applications de celle-ci
Après avoir mentioned that the expression given by G. Zemp1en [3] pour le polynôme d'interpolation d' Hermite (2) is inaccurate ands >= 2s \geqq 2and if the numbersr_(i)r_{i}sont plus grands que 2 , on établit, par une voie plus simple que celle utilisateur dans [6], to formulas (13); puis on donne une formulae more explicite (15) pour le polynôme d'interpolation d'Hermite qui verifie les conditions (3). A (20) on donne l'effective expression pour le polynôme d'interpolationH_(m+n-1)ubrace((a,dots,a)_(m),ubrace(b,dots,b)_(n);f∣x)H_{m+n-1} \underbrace{(a, \ldots, a}_{m}, \underbrace{b, \ldots, b}_{n} ; f \mid x).
Aux nos. 10-11 on applique les formules précédentes à la décomposition d'une fonction rationelle en simple fractions. Ainsi, pour les coefficients de la décomposition (23) on a établi les formulas (24). A (27) on donne les expressions effecientes des coefficients de la décomposition de la fonction rationalnelle particulière (26).
Aux nos. 12-13 on fait quelques applications à l'intégration numérique des fonctions. As an example, two quadrature formulas with a high degree of accuracy were constructed in (29) and (29'). Using the interpolation formulas (20) we established the quadrature formulas (30). Ici on fait la remarque que la formule (30) ainsi que la formulale particulier (32) ont été données pour 1a première fois par Hermite [1].
In (42) we established a numerical derivation formula analogous to Hermite's quadrature formula (30). They have no. 17 on donne 11 examples of pareilles formulas.
- Presented at the communications meeting of December 1, 1956, of the Cluj Branch of the Society of Mathematics and Physics.
^(1){ }^{1}See for example: IP Natanson [2]
- Presented at the communications meeting of December 1, 1956, of the Cluj Branch of the Society of Mathematics and Physics.
- ^(2){ }^{2}Formulas (30) and (32) are found in a famous memoir by Hermite [1].