This article may require cleanup to meet Wikipedia's quality standards. Please improve this article if you can. (October 2007)

In the branch of mathematics called multivariable calculus, the implicit function theorem is a tool which allows relations to be converted to functions. It does this by representing the relation as the graph of a function. There may not be a single function whose graph is the entire relation, but there may be such a function on a restriction of the domain of the relation. The implicit function theorem gives a sufficient condition to ensure that there is such a function. Image File history File links Broom_icon. ... For other meanings of mathematics or uses of math and maths, see Mathematics (disambiguation) and Math (disambiguation). ... Multivariable calculus is the extension of calculus in one variable to calculus in several variables: the functions which are differentiated and integrated involve several variables rather than one variable. ... In mathematics, the concept of a relation is a generalization of 2-place relations, such as the relation of equality, denoted by the sign = in a statement like 5 + 7 = 12, or the relation of order, denoted by the sign < in a statement like 5 < 12. Relations that involve two... Graph of example function, The mathematical concept of a function expresses the intuitive idea of deterministic dependence between two quantities, one of which is viewed as primary (the independent variable, argument of the function, or its input) and the other as secondary (the value of the function, or output). A... relation graph theory In mathematics, the graph of a function f is the collection of all ordered pairs (x,f(x)). In particular, graph means the graphical representation of this collection, in the form of a curve or surface, together with axes, etc. ... In mathematics, the domain of a function is the set of all input values to the function. ...

The theorem states that if the equation R(x, y) = 0 (an Implicit function) satisfies some mild conditions on its partial derivatives, then one can in principle solve this equation for y, at least over some small interval. Geometrically, the graph defined by R(x,y) = 0 will overlap locally with the graph of a function y = f(x). In mathematics, an implicit function is a generalization for the concept of a function in which the dependent variable may not be given explicitly in terms of the independent variable. ... In mathematics, a partial derivative of a function of several variables is its derivative with respect to one of those variables with the others held constant (as opposed to the total derivative, in which all variables are allowed to vary). ... In mathematics, interval is a concept relating to the sequence and set-membership of one or more numbers. ... In mathematics, a phenomenon is sometimes said to occur locally if, roughly speaking, it occurs on sufficiently small or arbitrarily small neighborhoods of points. ...

Example

A circle specified by the implicit function f(x,y) = x² + y² − 1. Around point A y can be expressed as a function of x, specifically . No such function exists for point B.

Consider the unit circle. If we define the function f as f(x,y) = x² + y² − 1, then the relation f(x,y) = 1 cuts out the unit circle. Explicitly, the unit circle is the set {(x,y) | f(x,y) = 1}. There is no way to represent the unit circle as the graph of a function y = g(x) because for each choice of , there are two choices of y. Namely, and . Image File history File links Implicit_circle. ... Image File history File links Implicit_circle. ... Illustration of a unit circle. ...

However, it is possible to represent part of the circle as a function. If we let for − 1 < x < 1, then the graph of y = g₁(x) provides the upper half of the circle. Similarly, if , then the graph of y = g₂(x) gives the lower half of the circle.

It is not possible to find a function which will cut out a neighbourhood of (1,0) or ( − 1,0). Any neighbourhood of (1,0) or ( − 1,0) contains both the upper and lower halves of the circle. Because functions must be single-valued, there is no way of writing both the upper and lower halves using one function y = g(x). Consequently, there is no function whose graph looks like a neighbourhood of (1,0) or ( − 1,0). In these two cases, the conclusion of the implicit function theorem fails.

The purpose of the implicit function theorem is to tell us the existence of functions like g₁(x) and g₂(x) in situations where we cannot write down explicit formulas. It guarantees that g₁(x) and g₂(x) are differentiable, and it even works in situations where we do not have a formula for f(x,y).

Statement of the theorem

Let f : R^n+m → R^m be a continuously differentiable function. We think of R^n+m as the cartesian product Rⁿ × R^m, and we write a point of this product as (x₁, ..., x_n, y₁, ..., y_m). f is the given relation. Our goal is to construct a function g : Rⁿ → R^m whose graph (x₁, ..., x_n, g(x₁, ..., x_n)) is precisely the set of all (x₁, ..., x_n, y₁, ..., y_m) such that f(x₁, ..., x_n, y₁, ..., y_m) = 0. In mathematics, a smooth function is one that is infinitely differentiable, i. ... In mathematics, the Cartesian product is a direct product of sets. ...

As noted above, this may not always be possible. As such, we will fix a point (a₁, ..., a_n, b₁, ..., b_m) which satisfies f(a₁, ..., a_n, b₁, ..., b_m) = 0, and we will ask for a g that works near the point (a₁, ..., a_n, b₁, ..., b_m). In other words, we want an open set U of Rⁿ, an open set V of R^m, and a function g : U → V such that the graph of g equals the relation f = 0 on U × V. In symbols,

To state the implicit function theorem, we need the Jacobian, also called the differential or total derivative, of f. This is the matrix of partial derivatives of f. Abbreviating to (a,b), the Jacobian matrix is In vector calculus, the Jacobian is shorthand for either the Jacobian matrix or its determinant, the Jacobian determinant. ... In mathematics, a partial derivative of a function of several variables is its derivative with respect to one of those variables with the others held constant (as opposed to the total derivative, in which all variables are allowed to vary). ...

where X is the matrix of partial derivatives in the x's and Y is the matrix of partial derivatives in the y's. The implicit function theorem says that if Y is an invertible matrix, then there are U, V, and g as desired. Writing all the hypotheses together gives the following statement.

Let f : R^n+m → R^m be a continuously differentiable function, and let R^n+m have coordinates (x₁,...,x_n, y₁, ..., y_m). Fix a point (a₁,...,a_n,b₁,...,b_m) = (a,b) with f(a,b)=c, where c∈ R^m. If the matrix [(∂f_i/∂y_j)(a,b)] is invertible, then there exists an open set U containing (a₁,..., a_n), an open set V containing (b₁,...,b_m), and a differentiable function g:U → V such that

In mathematics, a smooth function is one that is infinitely differentiable, i. ... In mathematics, the idea of inverse element generalises both of the concepts of negation, in relation to addition (see additive inverse), and reciprocal, in relation to multiplication. ...

Example

Lets go back to the example of the unit circle. In this case n = m = 1 and f(x,y) = x² + y² − 1. The matrix of partial derivatives is just a -matrix, given by Illustration of a unit circle. ...

Thus, here, Y is just a number; the linear map defined by it is invertible iff . By the implicit function theorem we see that we can write the circle in the form y = g(x) for all points where . For ( − 1,0) and (1,0) we run into trouble, as noted before.

Example

Suppose we have an m-dimensional space, parametrised by a set of coordinates . We can introduce a new coordinate system by giving m functions . These functions allow to calculate the new coordinates of a point, given the old coordinates . One might want to verify if the opposite is possible: given coordinates , can we 'go back' and calculate ? The implicit function theorem will provide an answer to this question. The (new and old) coordinates are related by f = 0, with

Now the Jacobian matrix of f at a certain point (a,b) is given by

Where 1_m denotes the identity matrix, and J is the matrix of partial derivatives, evaluated at (a,b). (In the above, these blocks were denoted by X and Y.) The implicit function theorem now states that we can locally express as a function of if J is invertible. Demanding J is invertible is equivalent to , thus we see that we can go back from the primed to the unprimed coordinates if the determinant of the Jacobian J is non-zero. This statement is also known as the inverse function theorem. In linear algebra, the identity matrix of size n is the n-by-n square matrix with ones on the main diagonal and zeros elsewhere. ... In mathematics, the inverse function theorem gives sufficient conditions for a vector-valued function to be invertible on an open region containing a point in its domain. ...

Example

As a simple application of the above, consider the plane, parametrised by polar coordinates (R,θ). We can go to a new coordinate system (cartesian coordinates) by defining functions x(R,θ) = Rcosθ and y(R,θ) = Rsinθ. This makes is possible given any point (R,θ) to find corresponding cartesian coordinates (x,y). When can we go back, and converse cartesian into polar coordinates? By the previous example, we need , with This article describes some of the common coordinate systems that appear in elementary mathematics. ... Cartesian means relating to the French mathematician and philosopher Descartes, who, among other things, worked to merge algebra and Euclidean geometry. ...

Since detJ = R, the conversion back to polar coordinates is only possible if . This is a consequence of the fact that at that point polar coordinates are not good: at the origin the value of θ is not well-defined.

Banach space version

Based on Grave's inverse function theorem it is possible to extend the implicit function theorem to Banach space valued mappings. In mathematics, the inverse function theorem gives sufficient conditions for a vector-valued function to be invertible on an open region containing a point in its domain. ...

Let X, Y, Z be Banach spaces. Let the mapping be Fréchet differentiable. If , f(x₀,y₀) = 0, and is a Banach space isomorphism from Y onto Z. Then there exist neighborhoods U of x₀ and V of y₀ and a Frechet differentiable function such that f(x,g(x)) = 0 and f(x,y) = 0 if and only if y = g(x), for all . In mathematics, Banach spaces (pronounced ), named after Stefan Banach who studied them, are one of the central objects of study in functional analysis. ... Not to be confused with Differentiation in Fréchet spaces. ...