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Taylor-series representation by vanishing remainders

Statement

Let I be an open interval, let fC(I), and let aI. For each xI, f(x) equals the sum of the Taylor series of f at a if and only if

Rn,af(x)0.

Moreover, let r>0 and suppose that [ar,a+r]I. For every n0, set

Mn+1:=maxtarf(n+1)(t).

If

Mn+1rn+1(n+1)!0,

then the Taylor polynomials Tn,af converge uniformly to f on the compact interval [ar,a+r].

Facts & Assumptions

Given: An open interval I, a function fC(I), a point aI, and the Taylor polynomials and remainders of f at a.

[A1]

For the uniform assertion, r>0, [ar,a+r]I, and Mn+1rn+1/(n+1)!0, where Mn+1=maxtarf(n+1)(t).

[L1]

The nth partial sum of the Taylor series at a is Tn,af, and Rn,af(x)=f(x)Tn,af(x) (Taylor and Maclaurin series, Taylor polynomials and their remainders).

[L2]

A continuous real-valued function on a nonempty compact set attains its maximum and minimum (Extreme value theorem: a continuous real function on a nonempty compact subset of R attains a greatest and a least value).

[L3]

If f(n+1)(t)M throughout the closed interval between a and x, then Rn,af(x)Mxan+1(n+1)! (A uniform derivative bound gives a uniform Taylor remainder bound).

[L4]

A sequence (gn) converges uniformly to g on a set E exactly when, for every ε>0, there is N such that gn(x)g(x)<ε for every xE and every nN (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

[L5]

A function is of class Ck on an interval when f(j) exists there for every jk and each such f(j) is continuous there, and it is smooth, or C, when it is Ck for every kN (Higher derivatives and the classes Ck and C).

[L7]

For reals uv, the closed bounded interval [u,v] is compact (Heine-Borel by bisection: every closed bounded interval [a,b] is compact).

Proof

technique · direct
1.1

Fix xI. If Tn,af(x)f(x), then Rn,af(x)=f(x)Tn,af(x)0.

L1assume-hypalgebra
1.2

Conversely, if Rn,af(x)0, then Tn,af(x)=f(x)Rn,af(x)f(x).

L1assume-hypalgebra
1.3

Let n0. Since fC(I), the derivative f(n+1) exists and is continuous on I, so f(n+1) is continuous on I; and [ar,a+r] is a closed bounded interval, hence compact, and it is a nonempty subset of I because r>0 and a belongs to it. Applying the extreme value theorem to f(n+1) on [ar,a+r] therefore produces the displayed maximum Mn+1.

A1L2L5L6L7
1.4

For every y[ar,a+r] and every n0, f(y)Tn,af(y)=Rn,af(y)Mn+1yan+1/(n+1)!Mn+1rn+1/(n+1)!.

A1L1L3
1.5

Given ε>0, choose N such that Mn+1rn+1/(n+1)!<ε whenever nN.

A1choose
2.1

Thus, for this arbitrary x, the Taylor series sums to f(x) if and only if Rn,af(x)0.

step 1.1step 1.2
2.2

For every nN and every y[ar,a+r], f(y)Tn,af(y)<ε.

step 1.4step 1.5
3.1

Hence Tn,aff uniformly on [ar,a+r].

L4step 2.2

Depends on

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