# Mean value theorem

Mean value theorem For the theorem in harmonic function theory, see Harmonic function § The mean value property. Part of a series of articles about Calculus Fundamental theorem Leibniz integral rule Limits of functionsContinuity Mean value theoremRolle's theorem Differential Integral Series Vector Multivariable Advanced Specialized Miscellaneous vte For any function that is continuous on {displaystyle [a,b]} and differentiable on {displaystyle (a,b)} there exists some {displaystyle c} in the interval {displaystyle (a,b)} such that the secant joining the endpoints of the interval {displaystyle [a,b]} is parallel to the tangent at {displaystyle c} .

In mathematics, the mean value theorem states, roughly, that for a given planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints. It is one of the most important results in real analysis. This theorem is used to prove statements about a function on an interval starting from local hypotheses about derivatives at points of the interval.

More precisely, the theorem states that if {displaystyle f} is a continuous function on the closed interval {displaystyle [a,b]} and differentiable on the open interval {displaystyle (a,b)} , then there exists a point {displaystyle c} in {displaystyle (a,b)} such that the tangent at {displaystyle c} is parallel to the secant line through the endpoints {displaystyle {big (}a,f(a){big )}} and {displaystyle {big (}b,f(b){big )}} , that is, {displaystyle f'(c)={frac {f(b)-f(a)}{b-a}}.} Contents 1 History 2 Formal statement 3 Proof 4 Implications 5 Cauchy's mean value theorem 6 Generalization for determinants 7 Mean value theorem in several variables 8 Mean value theorem for vector-valued functions 9 Cases where theorem cannot be applied (Necessity of conditions) 10 Mean value theorems for definite integrals 11 A probabilistic analogue of the mean value theorem 12 Mean value theorem in complex variables 13 See also 14 Notes 15 References 16 External links History A special case of this theorem for inverse interpolation of the sine was first described by Parameshvara (1380–1460), from the Kerala School of Astronomy and Mathematics in India, in his commentaries on Govindasvāmi and Bhāskara II.[1] A restricted form of the theorem was proved by Michel Rolle in 1691; the result was what is now known as Rolle's theorem, and was proved only for polynomials, without the techniques of calculus. The mean value theorem in its modern form was stated and proved by Augustin Louis Cauchy in 1823.[2] Many variations of this theorem have been proved since then.[3][4] Formal statement The function {displaystyle f} attains the slope of the secant between {displaystyle a} and {displaystyle b} as the derivative at the point {displaystyle xi in (a,b)} . It is also possible that there are multiple tangents parallel to the secant.

Let {displaystyle f:[a,b]to mathbb {R} } be a continuous function on the closed interval {displaystyle [a,b]} , and differentiable on the open interval {displaystyle (a,b)} , where {displaystyle asup _{(a,b)}|f'|} be some real number. Let {displaystyle E={0leq tleq 1mid |f(a+t(b-a))-f(a)|leq Mt(b-a)}.} We want to show {displaystyle 1in E} . By continuity of {displaystyle f} , the set {displaystyle E} is closed. It is also nonempty as {displaystyle 0} is in it. Hence, the set {displaystyle E} has the largest element {displaystyle s} . If {displaystyle s=1} , then {displaystyle 1in E} and we are done. Thus suppose otherwise. For {displaystyle 1>t>s} , {displaystyle {begin{aligned}&|f(a+t(b-a))-f(a)|\&leq |f(a+t(b-a))-f(a+s(b-a))-f'(a+s(b-a))(t-s)(b-a)|+|f'(a+s(b-a))|(t-s)(b-a)\&+|f(a+s(b-a))-f(a)|.end{aligned}}} Let {displaystyle epsilon >0} be such that {displaystyle M-epsilon >sup _{(a,b)}|f'|} . By the differentiability of {displaystyle f} at {displaystyle a+s(b-a)} (note {displaystyle s} may be 0), if {displaystyle t} is sufficiently close to {displaystyle s} , the first term is {displaystyle leq epsilon (t-s)(b-a)} . The second term is {displaystyle leq (M-epsilon )(t-s)(b-a)} . The third term is {displaystyle leq Ms(b-a)} . Hence, summing the estimates up, we get: {displaystyle |f(a+t(b-a))-f(a)|leq tM|b-a|} , a contradiction to the maximality of {displaystyle s} . Hence, {displaystyle 1=sin M} and that means: {displaystyle |f(b)-f(a)|leq M(b-a).} Since {displaystyle M} is arbitrary, this then implies the assertion. Finally, if {displaystyle f} is not differentiable at {displaystyle a} , let {displaystyle a'in (a,b)} and apply the first case to {displaystyle f} restricted on {displaystyle [a',b]} , giving us: {displaystyle |f(b)-f(a')|leq (b-a')sup _{(a,b)}|f'|} since {displaystyle (a',b)subset (a,b)} . Letting {displaystyle a'to a} finishes the proof. {displaystyle square } For some applications of mean value inequality to establish basic results in calculus, see also Calculus on Euclidean space#Basic notions.

A certain type of generalization of the mean value theorem to vector-valued functions is obtained as follows: Let f be a continuously differentiable real-valued function defined on an open interval I, and let x as well as x + h be points of I. The mean value theorem in one variable tells us that there exists some t* between 0 and 1 such that {displaystyle f(x+h)-f(x)=f'(x+t^{*}h)cdot h.} On the other hand, we have, by the fundamental theorem of calculus followed by a change of variables, {displaystyle f(x+h)-f(x)=int _{x}^{x+h}f'(u),du=left(int _{0}^{1}f'(x+th),dtright)cdot h.} Thus, the value f′(x + t*h) at the particular point t* has been replaced by the mean value {displaystyle int _{0}^{1}f'(x+th),dt.} This last version can be generalized to vector valued functions: Proposition — Let U ⊂ Rn be open, f : U → Rm continuously differentiable, and x ∈ U, h ∈ Rn vectors such that the line segment x + th, 0 ≤ t ≤ 1 remains in U. Then we have: {displaystyle f(x+h)-f(x)=left(int _{0}^{1}Df(x+th),dtright)cdot h,} where Df denotes the Jacobian matrix of f and the integral of a matrix is to be understood componentwise.

Proof. Let f1, …, fm denote the components of f and define: {displaystyle {begin{cases}g_{i}:[0,1]to mathbb {R} \g_{i}(t)=f_{i}(x+th)end{cases}}} Then we have {displaystyle {begin{aligned}f_{i}(x+h)-f_{i}(x)&=g_{i}(1)-g_{i}(0)=int _{0}^{1}g_{i}'(t),dt\&=int _{0}^{1}left(sum _{j=1}^{n}{frac {partial f_{i}}{partial x_{j}}}(x+th)h_{j}right)dt=sum _{j=1}^{n}left(int _{0}^{1}{frac {partial f_{i}}{partial x_{j}}}(x+th),dtright)h_{j}.end{aligned}}} The claim follows since Df is the matrix consisting of the components {displaystyle {tfrac {partial f_{i}}{partial x_{j}}}} . {displaystyle square } The mean value inequality can then be obtained as a corollary of the above proposition (though under the assumption the derivatives are continuous).[10] Cases where theorem cannot be applied (Necessity of conditions) Both conditions for Mean Value Theorem are necessary: f(x) is differentiable on (a,b) f(x) is continuous on [a,b] Where one of the above conditions is not satisfied, Mean Value Theorem is not valid in general, and so it cannot be applied.

Function is differentiable on open interval a,b The necessity of the first condition can be seen by the counterexample where the function {displaystyle f(x)=|x|} on [-1,1] is not differentiable.

Function is continuous on closed interval a,b The necessity of the second condition can be seen by the counterexample where the function {displaystyle f(x)={begin{cases}1,&{text{at }}x=0\0,&{text{if }}xin (0,1]end{cases}}} {displaystyle f(x)} satisfies criteria 1 since {displaystyle f'(x)=0} on {displaystyle (0,1)} But not criteria 2 since {displaystyle {frac {f(1)-f(0)}{1-0}}=-1} and {displaystyle -1neq 0=f'(x)} for all {displaystyle xin (0,1)} so no such {displaystyle c} exists Mean value theorems for definite integrals First mean value theorem for definite integrals Geometrically: interpreting f(c) as the height of a rectangle and b–a as the width, this rectangle has the same area as the region below the curve from a to b[11] Let f : [a, b] → R be a continuous function. Then there exists c in (a, b) such that {displaystyle int _{a}^{b}f(x),dx=f(c)(b-a).} Since the mean value of f on [a, b] is defined as {displaystyle {frac {1}{b-a}}int _{a}^{b}f(x),dx,} we can interpret the conclusion as f achieves its mean value at some c in (a, b).[12] In general, if f : [a, b] → R is continuous and g is an integrable function that does not change sign on [a, b], then there exists c in (a, b) such that {displaystyle int _{a}^{b}f(x)g(x),dx=f(c)int _{a}^{b}g(x),dx.} Proof that there is some c in [a, b][13] Suppose f : [a, b] → R is continuous and g is a nonnegative integrable function on [a, b]. By the extreme value theorem, there exists m and M such that for each x in [a, b], {displaystyle mleq f(x)leq M} and {displaystyle f[a,b]=[m,M]} . Since g is nonnegative, {displaystyle mint _{a}^{b}g(x),dxleq int _{a}^{b}f(x)g(x),dxleq Mint _{a}^{b}g(x),dx.} Now let {displaystyle I=int _{a}^{b}g(x),dx.} If {displaystyle I=0} , we're done since {displaystyle 0leq int _{a}^{b}f(x)g(x),dxleq 0} means {displaystyle int _{a}^{b}f(x)g(x),dx=0,} so for any c in (a, b), {displaystyle int _{a}^{b}f(x)g(x),dx=f(c)I=0.} If I ≠ 0, then {displaystyle mleq {frac {1}{I}}int _{a}^{b}f(x)g(x),dxleq M.} By the intermediate value theorem, f attains every value of the interval [m, M], so for some c in [a, b] {displaystyle f(c)={frac {1}{I}}int _{a}^{b}f(x)g(x),dx,} that is, {displaystyle int _{a}^{b}f(x)g(x),dx=f(c)int _{a}^{b}g(x),dx.} Finally, if g is negative on [a, b], then {displaystyle Mint _{a}^{b}g(x),dxleq int _{a}^{b}f(x)g(x),dxleq mint _{a}^{b}g(x),dx,} and we still get the same result as above.

QED Second mean value theorem for definite integrals There are various slightly different theorems called the second mean value theorem for definite integrals. A commonly found version is as follows: If G : [a, b] → R is a positive monotonically decreasing function and φ : [a, b] → R is an integrable function, then there exists a number x in (a, b] such that {displaystyle int _{a}^{b}G(t)varphi (t),dt=G(a^{+})int _{a}^{x}varphi (t),dt.} Here {displaystyle G(a^{+})} stands for {textstyle {lim _{xto a^{+}}G(x)}} , the existence of which follows from the conditions. Note that it is essential that the interval (a, b] contains b. A variant not having this requirement is:[14] If G : [a, b] → R is a monotonic (not necessarily decreasing and positive) function and φ : [a, b] → R is an integrable function, then there exists a number x in (a, b) such that {displaystyle int _{a}^{b}G(t)varphi (t),dt=G(a^{+})int _{a}^{x}varphi (t),dt+G(b^{-})int _{x}^{b}varphi (t),dt.} Mean value theorem for integration fails for vector-valued functions If the function {displaystyle G} returns a multi-dimensional vector, then the MVT for integration is not true, even if the domain of {displaystyle G} is also multi-dimensional.

For example, consider the following 2-dimensional function defined on an {displaystyle n} -dimensional cube: {displaystyle {begin{cases}G:[0,2pi ]^{n}to mathbb {R} ^{2}\G(x_{1},dots ,x_{n})=left(sin(x_{1}+cdots +x_{n}),cos(x_{1}+cdots +x_{n})right)end{cases}}} Then, by symmetry it is easy to see that the mean value of {displaystyle G} over its domain is (0,0): {displaystyle int _{[0,2pi ]^{n}}G(x_{1},dots ,x_{n})dx_{1}cdots dx_{n}=(0,0)} However, there is no point in which {displaystyle G=(0,0)} , because {displaystyle |G|=1} everywhere.