IMPLICIT FUNCTION

(Redirected from Implicit function theorem)

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. To give a function ''f'' 'explicitly' is to provide a prescription for calculating the output value of the function ''y'' in terms of the input value ''x'' of the function
:: ''y'' = ''f''(''x'')
By contrast, the function is 'implicit' if the value of ''y'' is obtained from ''x'' by solving the equation
::''R''(''x'',''y'') = 0
Implicit functions can often be useful in situations where it is inconvenient to solve explicitly an equation of the form ''R''(''x'',''y'') = 0 for ''y'' in terms of ''x''. Even if it is possible to rearrange this equation to obtain ''y'' as an explicit function ''f''(''x''), it may not be desirable to do so since the expression of ''f'' may be much more complicated than the expression of ''R''. In other situations, the equation ''R''(''x'',''y'') = 0 may fail to define a function at all, and rather defines a kind of multiple-valued function. Nevertheless, in many situations, it is still possible to work with functions given implicitly. Some techniques from calculus, such as differentiation, can be performed with relative ease using ''implicit differentiation''.
The ''implicit function theorem'' provides a link between implicit and explicit functions. It states that if the equation ''R''(''x'', ''y'') = 0 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'').

Contents

Examples

Inverse functions

Algebraic functions

Caveats

Implicit differentiation

Examples

Formula for two variables

Implicit function theorem

Example

Statement of the theorem

Example

Example

Example

References

Examples

Inverse functions

Implicit functions commonly arise as one way of describing the notion of an inverse function. If ''f'' is a function, then the inverse function of ''f'' is a solution of the equation
:$x=f(y)\; implies\; y\; =\; f^\{-1\}(x)$
for ''y'' in terms of ''x''. Intuitively, an inverse function is obtained from ''f'' by interchanging the roles of the dependent and independent variables. Stated another way, the inverse function is the solution ''y'' of the equation
:$R(x,y)\; =\; x-f(y)\; =\; 0.$
'Examples.'
# The natural logarithm ''y'' = ln(''x'') is the solution of the equation ''x'' - ''e''^''y'' = 0.
# The product log is an implicit function given by ''x'' - ''y'' ''e''^''y'' = 0.

Algebraic functions

Main articles: Algebraic function

An 'algebraic function' is a solution ''y'' for an equation ''R''(''x'',''y'') = 0 where ''R'' is a polynomial of two variables. Algebraic functions play an important role in mathematical analysis and algebraic geometry. A simple example of an algebraic function is given by the unit circle:
:$x^2+y^2-1=0.$
Solving for ''y'' gives
:$y=pmsqrt\{1-x^2\}.$
Note that there are two "branches" to the implicit function: one where the sign is positive and the other where it is negative. Both branches are thought of belonging to the implicit function. In this way, implicit functions can be ''multiple-valued.''

Caveats

Not every equation $R(x,y)\; =\; 0$ has a graph that is the graph of a function, the circle equation being one prominent example. Another example is an implicit function given by ''x'' - ''C''(''y'') = 0 where ''C'' is a cubic polynomial having a "hump" in its graph. Thus, for an implicit function to be a true ''function'' it might be necessary to use just part of the graph. An implicit function can sometimes be successfully defined as a true function only after "zooming in" on some part of the ''x''-axis and "cutting away" some unwanted function branches. A resulting formula may only then qualify as a legitimate explicit function.
The defining equation ''R'' = 0 can also have other pathologies. For example, the implicit equation ''x'' = 0 does not define a function at all; it is a vertical line. In order to avoid a problem like this, various constraints are frequently imposed on the allowable sorts of equations or on the domain. The implicit function theorem provides a uniform way of handling these sorts of pathologies.

Implicit differentiation

In calculus, a method called 'implicit differentiation' can be applied to implicitly defined functions. This method is an application of the chain rule allowing one to calculate the derivative of a function given implicitly.
As explained in the introduction, $y$ can be given as a function of $x$ implicitly rather than
explicitly. When we have an equation $R(x,y)=0$, we may be able to solve it for $y$ and then
differentiate. However, sometimes it is simpler to differentiate $R(x,y)$ with respect to $x$ and
then solve for $dy/dx$.

Examples

'1.' Consider for example
:$y\; +\; x\; =\; -4\; ,$
This function normally can be manipulated by using algebra to change this equation to an explicit function:
:$f(x)\; =\; y\; =\; -x\; -\; 4\; ,$
Differentiation then gives . Alternatively, one can differentiate the equation:
:
:
Solving for :
:
'2.' An example of an implicit function, for which implicit differentiation might be easier than attempting to use explicit differentiation, is
:$x^4\; +\; 2y^2\; =\; 8\; ,$
In order to differentiate this explicitly, one would have to obtain (via algebra)
:,
and then differentiate this function. This creates two derivatives: one for $y\; >\; 0$ and another for .
One might find it substantially easier to implicitly differentiate the implicit function;
:
thus,
:
'3.' Sometimes standard explicit differentiation cannot be used. And, in order to obtain the derivative, another method such as implicit differentiation must be employed. An example of such a case is the implicit function $y^3\; -\; y\; =\; x$. It is impossible to express $y$ explicitly as a function of $x$ (at least using elementary means, although the cubic formula will suffice for restricted values of $x$ and $y$). Meaning, cannot be found by explicit differentiation. Using the implicit method, can be expressed:
:
factoring out shows that
: which yields the final answer
:

Formula for two variables

Suppose that $y$ is bound to $x$ by the equation $F(x,\; y)\; =\; 0$ and that y is a differentiable function of x. If F is differentiable, using the generalized chain rule on it yields
:.
Thus,
:.
"The Implicit Function Theorem states that if $F$ is defined on an open disk containing $(a,b)$, where $F(a,b)=0$, $F\_y\; (a,b)$
ot = 0, and $F\_x$ and $F\_y$ are continuous on the disk, then the equation $F(x,y)\; =\; 0$ defines $y$ as a function of $x$ near the point $(a,b)$ and the derivative of this function is given by..."

Implicit function theorem

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.

Example

Consider the unit circle. If we define the function $f$ as $f(x,y)\; =\; x^2\; +\; y^2\; -\; 1$, then the relation $f(x,y)\; =\; 0$ cuts out the unit circle. Explicitly, the unit circle is the set $\{\; (x,y)\; |\; f(x,y)\; =\; 0\; \}$. There is no way to represent the unit circle as the graph of a function $y\; =\; g(x)$ because for each choice of $x\; in\; (-1,1)$, there are two choices of $y$. Namely, $sqrt\{1-x^2\}$ and $-sqrt\{1-x^2\}$.
However, it is possible to represent part of the circle as a function. If we let $g\_1(x)\; =\; sqrt\{1-x^2\}$ for , then the graph of $y\; =\; g\_1(x)$ provides the upper half of the circle. Similarly, if $g\_2(x)\; =\; -sqrt\{1-x^2\}$, then the graph of $y\; =\; g\_2(x)$ 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 $g\_1(x)$ and $g\_2(x)$ in situations where we cannot write down explicit formulas. It guarantees that $g\_1(x)$ and $g\_2(x)$ 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 'R'^''n'' × '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'' : 'R'^''n'' → '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.
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 'R'^''n'', 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,
:$\{\; (x\_1,\; ldots,\; x\_n,\; g(x\_1,\; ldots,\; x\_n))\; \}\; =\; \{\; (x\_1,\; ldots,\; x\_n,\; y\_1,\; ldots,\; y\_m)\; |\; f(x\_1,\; ldots,\; x\_n,\; y\_1,\; ldots,\; y\_m)\; =\; 0\; \}\; cap\; (U\; imes\; V)$
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 $(a\_1,\; ldots,\; a\_n,\; b\_1,\; ldots,\; b\_m)$ to $(a,\; b)$, the Jacobian matrix is
:
(Df)(a,b) & = & egin{bmatrix}
rac{partial f_1}{partial x_1}(a,b) & cdots & rac{partial f_1}{partial x_n}(a,b) & rac{partial f_1}{partial y_1}(a,b) & cdots & rac{partial f_1}{partial y_m}(a,b)\
dots & ddots & dots & dots & ddots & dots\
rac{partial f_m}{partial x_1}(a,b) & cdots & rac{partial f_m}{partial x_n}(a,b) & rac{partial f_m}{partial y_1}(a,b) & cdots & rac{partial f_m}{partial y_m}(a,b)\
end{bmatrix}\
& = & egin{bmatrix} X & | & Y end{bmatrix}\
end{matrix}
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 (''x''₁,...,''x''_n, ''y''₁, ..., ''y''_m). Fix a point (''a''₁,...,''a''_n,''b''₁,...,''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 (''a''₁,..., ''a''_n), an open set ''V'' containing (''b''₁,...,''b''_m), and a differentiable function ''g'':''U'' → ''V'' such that
::$\{\; (x\_1,\; ldots,\; x\_n,\; g(x\_1,\; ldots,\; x\_n))\; \}\; =\; \{\; (x\_1,\; ldots,\; x\_n,\; y\_1,\; ldots,\; y\_m)\; |\; f(x\_1,\; ldots,\; x\_n,\; y\_1,\; ldots,\; y\_m)\; =\; c\; \}\; cap\; (U\; imes\; V).$

Example

Lets go back to the example of the unit circle. In this case $n=m=1$ and $f(x,y)\; =\; x^2\; +\; y^2\; -\; 1$. The matrix of partial derivatives is just a $1\; imes\; 2$-matrix, given by
:
(Df)(a,b) & = & egin{bmatrix}
rac{partial f}{partial x}(a,b) & rac{partial f}{partial y}(a,b)\
end{bmatrix}\
& = & egin{bmatrix} 2a & 2b end{bmatrix}\
end{matrix}
Thus, here, Y is just a number; the linear map defined by it is invertible iff $b$
eq 0 . By the implicit function theorem we see that we can write the circle in the form $y=g(x)$ for all points where $y$
eq 0 . 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 $(x\_1,ldots,x\_m)$. We can introduce a new coordinate system by giving m functions $x\text{'}\_1(x\_1,ldots,x\_m),\; ldots,\; x\text{'}\_m(x\_1,ldots,x\_m)$. These functions allow to calculate the new coordinates $(x\text{'}\_1,ldots,x\text{'}\_m)$ of a point, given the old coordinates $(x\_1,ldots,x\_m)$. One might want to verify if the opposite is possible: given coordinates $(x\text{'}\_1,ldots,x\text{'}\_m)$, can we 'go back' and calculate $(x\_1,ldots,x\_m)$? The implicit function theorem will provide an answer to this question. The (new and old) coordinates $(x\text{'}\_1,ldots,x\text{'}\_m,\; x\_1,ldots,x\_m)$ are related by $f=0$, with
:$$
f(x'_1,ldots,x'_m,x_1,ldots x_m)=(x'_1(x_1,ldots x_m)-x'_1,ldots , x'_m(x_1,ldots, x_m)-x'_m).

Now the Jacobian matrix of ''f'' at a certain point $(a,b)$ is given by
:
(Df)(a,b) & = & egin{bmatrix}
1 & cdots & 0 & rac{partial x'_1}{partial x_1}(a,b) & cdots & rac{partial x'_1}{partial x_m}(a,b)\
dots & ddots & dots & dots & ddots & dots\
0 & cdots & 1 & rac{partial x'_m}{partial x_1}(a,b) & cdots & rac{partial x'_m}{partial x_m}(a,b)\
end{bmatrix}\
& = & egin{bmatrix} 1_m & | & J end{bmatrix}.\
end{matrix}
Where $1\_m$ denotes the $m\; imes\; m$ identity matrix, and J is the $m\; imes\; m$ 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 $(x\_1,ldots,x\_m)$ as a function of $(x\text{'}\_1,ldots,x\text{'}\_m)$ if J is invertible. Demanding J is invertible is equivalent to $det\; J$
eq 0 , 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.

Example

As a simple application of the above, consider the plane, parametrised by polar coordinates $(R,\; heta)$. We can go to a new coordinate system (cartesian coordinates) by defining functions $x(R,\; heta)=R\; cos\; heta$ and
$y(R,\; heta)=R\; sin\; heta$. This makes is possible given any point $(R,\; heta)$ to find corresponding cartesian coordinates $(x,y)$. When can we go back, and converse cartesian into polar coordinates? By the previous example, we need $det\; J$
eq 0 , with
:$$
J =egin{bmatrix}
rac{partial x(R, heta)}{partial R} & rac{partial x(R, heta)}{partial heta} \
rac{partial y(R, heta)}{partial R} & rac{partial y(R, heta)}{partial heta} \
end{bmatrix}=
egin{bmatrix}
cos heta & -R sin heta \
sin heta & R cos heta
end{bmatrix}.

Since $det\; J\; =\; R$, the conversion back to polar coordinates is only possible if $R$
eq 0 . This is a consequence of the fact that at that point polar coordinates are not good: at the origin the value of $heta$
is not well-defined.

References

★ Principles of Mathematical Analysis, , Walter, Rudin, McGraw-Hill, , ISBN 0-07-054235-X

★ Calculus on Manifolds, , Michael, Spivak, HarperCollins, , ISBN 0-8053-9021-9

★ Foundations of Differentiable Manifolds and Lie Groups, , Frank, Warner, Springer, , ISBN 0-387-90894-3

★ Calculus Concepts And Contexts, , James, Stewart, Brooks/Cole Publishing Company, 1998,