Lagrange finite-increments formula
Figure 1. Given the chord of the graph of the function $f$ with end points $(a,f(a))$, $(b,f(b))$, then there exists a point $\xi$, $a<\xi<b$, such that the tangent to the graph of the function at the point $(\xi,f(\xi))$ is parallel to the chord.
A formula expressing the increment of a function in terms of the value of its derivative at an intermediate point. If a function $f$ is continuous on an interval $[a,b]$ on the real axis and is differentiable at the interior points of it, then\begin{equation}f(b)-f(a)=f'(\xi)(b-a),\quad a<\xi<b.\end{equation}The finite-increments formula can also be written in the form\begin{equation}f(x+\Delta x)-f(x)=f'(x+\theta\Delta x)\Delta x,\quad 0<\theta<1.\end{equation}The geometric meaning of the finite-increments formula is illustrated in Figure1.
The finite-increments formula can be generalized to functions of several variables: If a function $f$ is differentiable at each point of a convex domain $G$ in an $n$-dimensional Euclidean space, then there exists for each pair of points $x=(x_1,\dots,x_n)\in G$, $x+\Delta x=(x_1+\Delta x_1,\dots,x_n+\Delta x_n)\in G$ a point $\xi=(\xi_1,\ldots,\xi_n)$ lying on the segment joining $x$ and $x+\Delta x$ and such that\begin{equation}f(x+\Delta x)-f(x)=\sum_{i=1}^n\dfrac{\partial f(\xi)}{\partial x_i}\Delta x_i,\quad \xi_i=x_i+\theta\Delta x_i,\quad 0<\theta<1,\quad i=1,\ldots,n.\end{equation}
Comments
This formula is usually called the mean-value theorem (for derivatives). It is a statement for real-valued functions only; consider, e.g., $f(x)=e^{ix}$.
How to Cite This Entry:
Finite-increments formula. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Finite-increments_formula&oldid=38670
This article was adapted from an original article by L.D. Kudryavtsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
FAQs
If a function f is continuous on an interval [a,b] on the real axis and is differentiable at the interior points of it, then f(b)−f(a)=f′(ξ)(b−a),a<ξ<b. The finite-increments formula can also be written in the form f(x+Δx)−f(x)=f′(x+θΔx)Δx,0<θ<1.
What is the formula for the increment in math? ›
The increment in the x or horizontal direction is x2–x1 which is denoted by ∆x=x2–x1. The increment in the y or vertical direction is ∆y=y2–y1.
What is the Lagrange finite increment theorem? ›
Lagrange's theorem in group theory: The order |G| of any finite group G is divisible by the order |H| of any subgroup H of it. The theorem was actually proved by J.L. Lagrange in 1771 in the study of properties of permutations in connection with research on the solvability of algebraic equations in radicals.
What are increments in math? ›
Mathematics. the difference between two values of a variable; a change, positive, negative, or zero, in an independent variable. the increase of a function due to an increase in the independent variable.
What is the increment function? ›
The increment function ϕ is an average of the derivative dy/dt over the time interval ti to ti + h. This average is obtained by evaluating the derivative f(t,y) at several points or “stages” within the time interval.
How do you calculate 5% increment? ›
Divide the number you wish to add 5% to by 100. Multiply this new number by 5. Add the product of the multiplication to your original number.
What is the Lagrange theorem formula? ›
Lagrange's Mean Value Theorem Statement: It states that if f(x) is a function such that: f(x) is continuous on [a, b] f(x) is differentiable on the open interval (a, b) Then there exists at least one real number c ∈ (a, b) such that: f′(c)=f(b)−f(a)b−a.
What is the Lagrange theorem in math? ›
Lagrange theorem is one of the central theorems of abstract algebra. It states that in group theory, for any finite group say G, the order of subgroup H of group G divides the order of G. The order of the group represents the number of elements. This theorem was given by Joseph-Louis Lagrange.
What is the proof of Lagrange's theorem? ›
Proof of Lagrange Statement:
Let H = {h1,h2,…,hn}, then ah1,ah2,…,ahn are the n number of distinct members of aH. This shows that n, the order of H, divides m i.e., is a divisor of m, the order of the finite group G. We also see that the index p is also a divisor of the order of the group.
What is an example of an increment? ›
They increased the dosage of the drug in small increments over a period of several weeks. Fines increase in increments of $10. The volume is adjustable in 10 equal increments.
++ (increment)
Increases the value of an integer variable by 1. Equivalent to the operation i = i + 1. If the value of the variable i is five, then the expression i++ increases the value of i to 6.
What is increment also known as? ›
increment (noun as in small step toward gain) Strong matches. accession accretion accrual addition advancement augmentation enlargement increase profit raise rise supplement.
How do you solve increment and decrement? ›
Increment Operator (++) : Increases the operand's value by one. Decrement Operator (--) : Decreases the operand's value by one. Both of these operators can be used in two forms: Prefix: The operator is placed before the variable.
What is a four-step rule in differentiation? ›
The following is a four-step process to compute f/(x) by definition. Input: a function f(x) Step 1 Write f(x + h) and f(x). Step 2 Compute f(x + h) - f(x). Combine like terms. If h is a common factor of the terms, factor the expression by removing the common factor h.
How do you make an increment? ›
In programming, increment is a common operation used to increase the value of a variable by a fixed amount. It is typically represented by the "++" operator. For example, if you have a variable called "count" with an initial value of 5, you can increment it by 1 using the expression "count++".
What is the ++ a increment operator? ›
In programming (Java, C, C++, JavaScript etc.), the increment operator ++ increases the value of a variable by 1. Similarly, the decrement operator -- decreases the value of a variable by 1. Simple enough till now. However, there is an important difference when these two operators are used as a prefix and a postfix.
