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T. Popoviciu, Sur la formule des accroissements finis, Mathematica, 23 (1947-1948), pp. 123-126 (in French).
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MATHEMATICA
TIBERIU POPOVICIU:
ON THE FORMULA FOR FINITE INCREASES
ON THE FORMULA FOR FINITE INCREASES
BY
TIBERIU POPOVICIU
-
- Eitherf(x)f(x)a continuous function in the bounded and closed interval[a,b][a, b]. We will designate by X the abscissa pointxxof the real axis and byX^(')X^{\prime}the coordinate point(x,f(x))(x, f(x)). EspeciallyA,A^('),B,B^(')A, A^{\prime}, B, B^{\prime}are the points(a,0),(a,f(a),(b,0),(b,f(b))(a, 0),(a, f(a),(b, 0),(b, f(b)). We will also designate by
[x_(1),x_(2);f]=(f(x_(2))-f(x_(1)))/(x_(2)-x_(1))\left[x_{1}, x_{2}; f\right]=\frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}
the divided difference of the functionf(x)f(x)on the pointsx_(1),x_(2)x_{1}, x_{2}and by
[x_(1),x_(2),x_(3);f]=([x_(2),x_(3);f]-[x_(1),x_(2);f])/(x_(3)-x_(1))\left[x_{1}, x_{2}, x_{3}; f\right]=\frac{\left[x_{2}, x_{3}; f\right]-\left[x_{1}, x_{2}; f\right]}{x_{3}-x_{1}}
the divided difference of this same function on the pointsx_(1),x_(2),x_(8)x_{1}, x_{2}, x_{8}.
The area of ​​the trapezoidXX^(')Y^(')Y(x <= y)\mathrm{X} \mathrm{X}^{\prime} \mathrm{Y}^{\prime} \mathrm{Y}(x \leqq y)is then, by definition, equal to
The area of ​​the trapezoidXX^(')Y^(')Y(x <= y)\mathrm{X} \mathrm{X}^{\prime} \mathrm{Y}^{\prime} \mathrm{Y}(x \leqq y)is then, by definition, equal to
varphi(x,y)=(1)/(2)[f(x)+f(y)](yx)\varphi(x, y)=\frac{1}{2}[f(x)+f(y)](yx)
The formula for finite increments results from the observation that at least one of the absolute extrema (maximum or minimum) of the sum of the areas of the trapezoidsAA^(')X^(')X,XX^(')B^(')BAA^{\prime} X^{\prime}is necessarily reached at least at one point of the open interval(a,b)(a, b). In fact, this sum is equal to
F_(1)(x)=(1)/(2)[f(a)+f(x)](xa)+(1)/(2)[f(x)+f(b)](bx)\mathrm{F}_{1}(x)=\frac{1}{2}[f(a)+f(x)](xa)+\frac{1}{2}[f(x)+f(b)](bx)
AndF_(1)(x)\mathrm{F}_{1}(x)is a continuous function ofxxIn[a,b][a, b], taking the same value (1)
F_(0)=varphi(a,b)\mathrm{F}_{0}=\varphi(a, b)
at the endsa,ba, bof this interval. If the derivativef^(')(x)f^{\prime}(x)exists in the open interval (a,ba, b), we conclude the existence of at least onex in(a,b)x \in(a, b)such that we haveF_(1)^(')(x)=0F_{1}^{\prime}(x)=0, SO
f^(')(x)=[a,b;f]=(f(b)-f(a))/(ba)f^{\prime}(x)=[a, b ; f]=\frac{f(b)-f(a)}{ba}
-
- Let us considernnpointsx_(1),x_(2),dots,x_(n)x_{1}, x_{2}, \ldots, x_{n}in the meantime[a,b][a, b]and let's still assume thata <= x_(1) <= x_(2) <= cdots <= x_(n) <= ba \leqq x_{1} \leqq x_{2} \leqq \cdots \leqq x_{n} \leqq b. The sum of the areas of the trapezoidsAA^(')X_(1)^(')X_(1),X_(1)X_(1)^(')X_(2)^(')X_(2),dots,X_(n-1)X_(n-1)^(')X_(n)^(')X_(n),X_(n)X_(n)^(')B^(')BA A^{\prime} X_{1}^{\prime} X_{1}, X_{1} X_{1}^{\prime} X_{2}^{\prime} X_{2}, \ldots, X_{n-1} X_{n-1}^{\prime} X_{n}^{\prime} X_{n}, X_{n} X_{n}^{\prime} B^{\prime} Bis equal to
{:[(2)F_(n)(M)=F_(n)(x_(1),x_(2),dots,x_(n))=],[=(1)/(2)sum_(i=0)^(n)[f(x_(i))+f(x_(i+1))](x_(i+1)-x_(i))=sum_(i=0)^(n)varphi(x_(i),x_(i+1))]:}\begin{align*}
\mathrm{F}_{n}(\mathrm{M}) & =\mathrm{F}_{n}\left(x_{1}, x_{2}, \ldots, x_{n}\right)= \tag{2}\\
& =\frac{1}{2} \sum_{i=0}^{n}\left[f\left(x_{i}\right)+f\left(x_{i+1}\right)\right]\left(x_{i+1}-x_{i}\right)=\sum_{i=0}^{n} \varphi\left(x_{i}, x_{i+1}\right)
\end{align*}
by posingx_(0)=a,x_(n+1)=bx_{0}=a, x_{n+1}=band designating by M the point with coordinates(x_(1),x_(2),dots,x_(n))\left(x_{1}, x_{2}, \ldots, x_{n}\right)in ordinary space atnndimensions.F_(n)(x_(1),x_(2),dots,x_(n))\mathrm{F}_{n}\left(x_{1}, x_{2}, \ldots, x_{n}\right)is then a continuous function ofx_(1),x_(2),dots,x_(n)x_{1}, x_{2}, \ldots, x_{n}in a limited and closed areaDD. This domainDDis formed by a simplexM_(0)M_(1)dotsM_(n)M_{0} M_{1} \ldots M_{n}whose vertices are the points
{:[M_(i)ubrace((a,a,dots,aubrace)_(i)","ubrace(b,b,dots,bubrace)_(n-i))","quad i=0","1","dots","n],[(M_(0)(b,b,dots,b),quadM_(n)(a,a,dots,a)).]:}\begin{aligned}
& \mathrm{M}_{i} \underbrace{(a, a, \ldots, a}_{i},\underbrace{b, b, \ldots, b}_{n-i}), \quad i=0,1, \ldots, n \\
&\left(\mathrm{M}_{0}(b, b, \ldots, b), \quad \mathrm{M}_{n}(a, a, \ldots, a)\right) .
\end{aligned}
The interior pointsM(x_(1),x_(2),dots,x_(n))M\left(x_{1}, x_{2}, \ldots, x_{n}\right)ofDDare then characterized by inequalitiesa < x_(1) < x_(2) < cdots < x_(n) < ba<x_{1}<x_{2}<\cdots<x_{n}<b. Point M belongs to the faceD_(i)D_{i}opposite the summitM_(i)M_{i}ofDDifx_(i)=x_(i+1)x_{i}=x_{i+1}and this pointMMis an interior point of this face ifa < x_(1) < cdots < x_(i)=x_(i+1) < x_(i+2)<<cdots < x_(n) < ba<x_{1}<\cdots<x_{i}=x_{i+1}<x_{i+2}< <\cdots<x_{n}<b.
From formula (2) it immediately follows that ifM(x_(1),x_(2),dots,x_(n))M\left(x_{1}, x_{2}, \ldots, x_{n}\right)is on the faceD_(i)D_{i}we have
F_(n)(M)=F_(n-1)(x_(1),x_(2),dots,x_(i),x_(i+2),x_(i+3),dots,x_(n))\mathrm{F}_{n}(\mathrm{M})=\mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{i}, x_{i+2}, x_{i+3}, \ldots, x_{n}\right)
In particular we have
(3)
(3)
{:[F_(n)(a,x_(1),x_(2),dots,x_(n-1))=F_(n)(x_(1),x_(1),x_(2),dots,x_(n-1))=],[=F_(n)(x_(1),x_(2),x_(2),x_(3),dots,x_(n-1))=dots=F(x_(1),x_(2),dots,x_(n-2),x_(n-1),x_(n-1))=],[=F_(n)(x_(1),x_(2),dots,x_(n-1),b)=F_(n-1)(x_(1),x_(2),dots,x_(n-1))","]:}\begin{aligned}
& \mathrm{F}_{n}\left(a, x_{1}, x_{2}, \ldots, x_{n-1}\right)=\mathrm{F}_{n}\left(x_{1}, x_{1}, x_{2}, \ldots, x_{n-1}\right)= \\
= & \mathrm{F}_{n}\left(x_{1}, x_{2}, x_{2}, x_{3}, \ldots, x_{n-1}\right)=\ldots=\mathrm{F}\left(x_{1}, x_{2}, \ldots, x_{n-2}, x_{n-1}, x_{n-1}\right)= \\
= & \mathrm{F}_{n}\left(x_{1}, x_{2}, \ldots, x_{n-1}, b\right)=\mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{n-1}\right),
\end{aligned}
which shows us thatF_(n)(M)F_{n}(M)takes the same values ​​on the facesD_(i)D_{i}'i=0,1,dots,ni=0,1, \ldots, nThe geometric interpretation of this fact is, moreover, very simple.
Finally we can also note that
(4)
(4)
F_(n)(M_(i))=F_(0),quad i=0,1,dots,n,\mathrm{F}_{n}\left(\mathrm{M}_{i}\right)=\mathrm{F}_{0}, \quad i=0,1, \ldots, n,
F_(0)\mathrm{F}_{0}being defined by (1).
3. - This being so, we can demonstrate the
3. - This being so, we can demonstrate the
Theorem 1. - If the functionf(x)f(x)is continuous in the closed interval[a,b][a, b]and is differentiable in the open interval(a,b)(a, b), the function
F_(n)(x_(1),x_(2),dots,x_(n))\mathrm{F}_{n}\left(x_{1}, x_{2}, \ldots, x_{n}\right)reaches at least one of its absolute extrema at at least one interior point of the domain D .
F_(n)(x_(1),x_(2),dots,x_(n))\mathrm{F}_{n}\left(x_{1}, x_{2}, \ldots, x_{n}\right)reaches at least one of its absolute extrema at at least one interior point of the domain D .
The functionF_(n)(M)\mathrm{F}_{n}(\mathrm{M})takes the valueF_(0)\mathrm{F}_{0}. If this function reduces to a constant the theorem is demonstrated. Otherwise, and to clarify the ideas, let us suppose thatF_(n)(M)F_{n}(M)take values> F_(0)>F_{0}. It is then sufficient to demonstrate that the maximum ofF_(n)(M)F_{n}(M)is reached at an interior point of D.
We can now prove the theorem by complete induction onnn. Suppose the theorem is true for the functionF_(n-1)(x_(1),x_(2),dots,x_(n-1))\mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{n-1}\right). We will demonstrate that the maximum ofF_(n)(M)\mathrm{F}_{n}(\mathrm{M})can only be reached on the facesD_(i)D_{i}of the domainDDOtherwise this maximum would be reached at an interior point of the face.D_(i)D_{i}, as a result of equalities (3), (4) and as a result of the hypothesis that the theorem is verified for the functionF_(n-1)\mathrm{F}_{n-1}. Either(x_(1),x_(2),dots,x_(i),x_(i),x_(i+1),dots,x_(n-1))\left(x_{1}, x_{2}, \ldots, x_{i}, x_{i}, x_{i+1}, \ldots, x_{n-1}\right)an interior point ofD_(i)D_{i}where the maximum is reached. We havea < x_(1)<<x_(2) < dots < x_(n-1) < ba<x_{1}< <x_{2}<\ldots<x_{n-1}<band the fact that insideDF_(n)(M)\mathrm{D} \mathrm{F}_{n}(\mathrm{M})takes smaller values ​​it results that
F_(n)(x_(1),x_(2),dots,x_(i),t,x_(i+1),dots,x_(n-1)) < F_(n-1)(x_(1),x_(2),dots,x_(n-1)),x_(i) < t < x_(i+1)\mathrm{F}_{n}\left(x_{1}, x_{2}, \ldots, x_{i}, t, x_{i+1}, \ldots, x_{n-1}\right)<\mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{n-1}\right), x_{i}<t<x_{i+1},
F_(n)(x_(1),x_(2),dots,x_(i),t,x_(i+1),dots,x_(n-1)) < F_(n-1)(x_(1),x_(2),dots,x_(n-1)),x_(i) < t < x_(i+1)\mathrm{F}_{n}\left(x_{1}, x_{2}, \ldots, x_{i}, t, x_{i+1}, \ldots, x_{n-1}\right)<\mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{n-1}\right), x_{i}<t<x_{i+1},
i=0,1,dots,n-1i=0,1, \ldots, n-1
Or
varphi(x_(i),t)+varphi(t,x_(i+1)) < varphi(x_(i),x_(i+1)),x_(i) < t < x_(i+1),quad i=0,1,dots,1-n\varphi\left(x_{i}, t\right)+\varphi\left(t, x_{i+1}\right)<\varphi\left(x_{i}, x_{i+1}\right), x_{i}<t<x_{i+1}, \quad i=0,1, \ldots, 1-n. A simple calculation gives us
varphi(x_(i),t)+varphi(t,x_(i+1)) < varphi(x_(i),x_(i+1)),x_(i) < t < x_(i+1),quad i=0,1,dots,1-n\varphi\left(x_{i}, t\right)+\varphi\left(t, x_{i+1}\right)<\varphi\left(x_{i}, x_{i+1}\right), x_{i}<t<x_{i+1}, \quad i=0,1, \ldots, 1-n. A simple calculation gives us
{:[{:x_(i+1))-varphi(x_(i),t)-varphi(t,x_(i+1))=],[=(1)/(2)(x_(i+1)-x_(i))(t-x_(i))(x_(i+1)-t)[x_(i),t,x_(i+1);f]=],[=(1)/(2)(x_(i+1)-x_(i))(t-x_(i)){[x_(i),x_(i+1);f]-[x_(i),t;f]}=],[=(1)/(2)(x_(i+1)-x_(i))(t-x_(i+1)){[x_(i),x_(i+1);f]-[x_(i+1),t;f]}]:}\begin{aligned}
& \left.x_{i+1}\right)-\varphi\left(x_{i}, t\right)-\varphi\left(t, x_{i+1}\right)= \\
& =\frac{1}{2}\left(x_{i+1}-x_{i}\right)\left(t-x_{i}\right)\left(x_{i+1}-t\right)\left[x_{i}, t, x_{i+1} ; f\right]= \\
& =\frac{1}{2}\left(x_{i+1}-x_{i}\right)\left(t-x_{i}\right)\left\{\left[x_{i}, x_{i+1} ; f\right]-\left[x_{i}, t ; f\right]\right\}= \\
& =\frac{1}{2}\left(x_{i+1}-x_{i}\right)\left(t-x_{i+1}\right)\left\{\left[x_{i}, x_{i+1} ; f\right]-\left[x_{i+1}, t ; f\right]\right\}
\end{aligned}
and inequality (5) allows us to write
[x_(i),t;f] < [x_(i),x_(i+1);f] < [x_(i+1),t;f],quadx_(i) < t < x_(i+1),quad i=0,1,dots,n-1\left[x_{i}, t ; f\right]<\left[x_{i}, x_{i+1} ; f\right]<\left[x_{i+1}, t ; f\right], \quad x_{i}<t<x_{i+1}, \quad i=0,1, \ldots, n-1.
Making it tendertttowardsx_(i)x_{i}then towardsx_(i+1)x_{i+1}, we deduce
from where
[x_(i),t;f] < [x_(i),x_(i+1);f] < [x_(i+1),t;f],quadx_(i) < t < x_(i+1),quad i=0,1,dots,n-1\left[x_{i}, t ; f\right]<\left[x_{i}, x_{i+1} ; f\right]<\left[x_{i+1}, t ; f\right], \quad x_{i}<t<x_{i+1}, \quad i=0,1, \ldots, n-1.
Making it tendertttowardsx_(i)x_{i}then towardsx_(i+1)x_{i+1}, we deduce
from where
{:[f^(')(x_(i)) <= [x_(i),x_(i+1);f]",",i=1","2","dots","n-1],[[x_(i),x_(i+1);f] <= f^(')(x_(i+1))",",i=0","1","dots","n-2]:}\begin{array}{ll}
f^{\prime}\left(x_{i}\right) \leqq\left[x_{i}, x_{i+1} ; f\right], & i=1,2, \ldots, n-1 \\
{\left[x_{i}, x_{i+1} ; f\right] \leqq f^{\prime}\left(x_{i+1}\right),} & i=0,1, \ldots, n-2
\end{array}
or else
[a,x_(1);f] <= [x_(1),x_(2),;f] <= [x_(2),x_(3);f] <= dots <= [x_(n-1),b;f]\left[a, x_{1} ; f\right] \leqq\left[x_{1}, x_{2}, ; f\right] \leqq\left[x_{2}, x_{3} ; f\right] \leqq \ldots \leqq\left[x_{n-1}, b ; f\right]
(7)
[x_(i),x_(i+1),x_(i+2);f] >= 0,quad i==0,1,dots,n-2.\left[x_{i}, x_{i+1}, x_{i+2} ; f\right] \geqq 0, \quad i==0,1, \ldots, n-2 .
The functionf(x)f(x)is therefore non-concave (of order 1) on the pointsaa,x_(1),dots,x_(n-1),bx_{1}, \ldots, x_{n-1}, b. We then know that we also have
(8)
(8)
[a,x_(i),b;f] >= 0,quad i=1,2,dots,n-1\left[a, x_{i}, b ; f\right] \geqq 0, \quad i=1,2, \ldots, n-1
But, I say that these inequalities are in contradiction with
{:(9)F_(n-1)(x_(1),x_(2),dots,x_(n-1)) > F_(0):}\begin{equation*}
\mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{n-1}\right)>\mathrm{F}_{0} \tag{9}
\end{equation*}
which results from the hypothesis made on the pointsx_(1),x_(2),dots,x_(n-1)x_{1}, x_{2}, \ldots, x_{n-1}. Indeed, a simple calculation gives us
sum_(i=1)^(n-1)[varphi(a,x_(i))+( bar(varphi))(x_(i),b)-varphi(a,b)]=(b-a)[F_(n-1)(x_(1),x_(2),dots,x_(n-1))-F_(0)]\sum_{i=1}^{n-1}\left[\varphi\left(a, x_{i}\right)+\bar{\varphi}\left(x_{i}, b\right)-\varphi(a, b)\right]=(b-a)\left[\mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{n-1}\right)-\mathrm{F}_{0}\right]
and, taking into account (6),
-(1)/(2)sum_(i=1)^(n-1)(x_(i)-a)(b-x_(i))[a,x_(i),b;f]=F_(n-1)(x_(1),x_(2),dots,x_(n-1))-F_(0)-\frac{1}{2} \sum_{i=1}^{n-1}\left(x_{i}-a\right)\left(b-x_{i}\right)\left[a, x_{i}, b ; f\right]=\mathrm{F}_{n-1}\left(x_{1}, x_{2}, \ldots, x_{n-1}\right)-\mathrm{F}_{0}
which proves our assertion.
This contradiction demonstrates Theorem 1 since forn=1n=1the property results easily.
This contradiction demonstrates Theorem 1 since forn=1n=1the property results easily.
The example of the function
f(x)={[(x-a)^(2)",",x in[a,(a+b)/(2)]],[(x-b)^(2)",",x in[(a+b)/(2),b]]:}f(x)= \begin{cases}(x-a)^{2}, & x \in\left[a, \frac{a+b}{2}\right] \\ (x-b)^{2}, & x \in\left[\frac{a+b}{2}, b\right]\end{cases}
shows us that the property expressed by Theorem 1 is no longer true in general if the functionffis not differentiable at every point in the interval(a,b)(a, b).
4. - We have
4. - We have
{:[(delF_(n)(x_(1),x_(2),dots,x_(n)))/(delx_(i))=(1)/(2)[f^(')(x_(i))(x_(i+1)-x_(i-1))-f(x_(i+1))+f(x_(i-1))]],[i=1","2","dots","n]:}\begin{gathered}
\frac{\partial F_{n}\left(x_{1}, x_{2}, \ldots, x_{n}\right)}{\partial x_{i}}=\frac{1}{2}\left[f^{\prime}\left(x_{i}\right)\left(x_{i+1}-x_{i-1}\right)-f\left(x_{i+1}\right)+f\left(x_{i-1}\right)\right] \\
i=1,2, \ldots, n
\end{gathered}
and we deduce
Theorem 2. - If the functionf(x)f(x)is continuous in the closed interval[a,b][a, b]and is differentiable in the open interval(a,b)(a, b)we can findnndistinct pointsx_(1),x_(2),dots,x_(n),x_(1) < x_(2) < dots < x_(n)x_{1}, x_{2}, \ldots, x_{n}, x_{1}<x_{2}<\ldots<x_{n}of(a,b)(a, b)such that we have
Theorem 2. - If the functionf(x)f(x)is continuous in the closed interval[a,b][a, b]and is differentiable in the open interval(a,b)(a, b)we can findnndistinct pointsx_(1),x_(2),dots,x_(n),x_(1) < x_(2) < dots < x_(n)x_{1}, x_{2}, \ldots, x_{n}, x_{1}<x_{2}<\ldots<x_{n}of(a,b)(a, b)such that we have
{:[f^(')(x_(i))=[x_(i-1),x_(i+1);f]=(f(x_(i+1))-f(x_(i-1)))/(x_(i+1)-x_(i-1))],[i=1","2","dots","n quad(x_(0)=a,x_(n+1)=b)]:}\begin{gathered}
f^{\prime}\left(x_{i}\right)=\left[x_{i-1}, x_{i+1} ; f\right]=\frac{f\left(x_{i+1}\right)-f\left(x_{i-1}\right)}{x_{i+1}-x_{i-1}} \\
i=1,2, \ldots, n \quad\left(x_{0}=a, x_{n+1}=b\right)
\end{gathered}
The geometric interpretation is simple.