Proof of Taylor’s Theorem

Comments on notation: Suppose α = (α_{1}, α_{2}, …, α_{n}) is a multi-index. The length of α is |α| = α_{1} + ⋯ + α_{n}, and α! is defined to be α_{1}! ⋯ α_{n}!. For h ∈ ℝ^{n}, h^{α} is the monomial h_{1}^{α_{1}} ⋯ h_{n}^{α_{n}}. (By calling h^{α} a “monomial”, we mean in particular that α_{i} = 0 implies h_{i}^{α_{i}} = 1, even if h_{i} = 0.)

If f : U ⊆ ℝ^{n} → ℝ is a C^{k}-function and |α| ≤ k, then we use ∂^{α}f to denote the (mixed) partial derivative

\frac{∂^{α_{1}}}{∂x_{1}^{α_{1}}} \frac{∂^{α_{2}}}{∂x_{2}^{α_{2}}} … \frac{∂^{α_{n}}}{∂x_{n}^{α_{n}}} f.

We also define ∂^{(0,...,0)}f = f.

For notational convenience, we make use of the fact that the operators ∂/∂x_{i} can be used like monomials (as long as we put them to the left of the function they’re operating on); e.g.,

(\frac{∂}{∂x_{i}} + \frac{∂}{∂x_{j}} ) f = \frac{∂}{∂x_{i}} f + \frac{∂}{∂x_{j}} f.

Taylor’s theorem. Let U be an open subset of ℝ^{n} and let f ∈ C^{k} = C^{k}(U, ℝ). Let x ∈ U, and let h ∈ ℝ^{n} be any vector such that x + th ∈ U for all t ∈ [0, 1].

(i) f satisfies the Taylor formula with integral remainder term:

f(x + h) = ∑_{|α|<k} ∂^{α}f(x)\frac{h^{α}}{α!} + k ∑_{|α|=k} (∫_{0}^{1}(1 − t)^{k−1}∂^{α}f(x + th) dt) \frac{h^{α}}{α!}

Proof. Since C^{1} ⊃ C^{2} ⊃ C^{3} ⊃ ⋯, we use induction on k.

For f ∈ C^{1}, set g(t) = f(x + th). Then the fundamental theorem of calculus gives

g(1) = g(0) + ∫_{0}^{1} \frac{d}{dt} g(t) dt.

By the chain rule, \frac{d}{dt}g(t) = (h_{1} \frac{∂}{∂x_{1}} + ⋯ + h_{n} \frac{∂}{∂x_{n}} ) f(x + th). We conclude that the formula is true, i.e.,

f(x + h) = f(x) + ∑_{i=1}^{n} (∫_{0}^{1} \frac{∂}{∂x_{i}} f(x + th) dt) h_{i}.

Now assume f ∈ C^{k+1} and that the formula is true for f in C^{k}. Integration by parts gives

∫_{0}^{1}(1 − t)^{k−1}∂^{α}f(x + th) dt
= − \frac{(1 − t)^{k}}{k} ∂^{α}f(x + th)|_{t=0}^{1} + ∫_{0}^{1} \frac{(1 − t)^{k}}{k} \frac{d}{dt} ∂^{α}f(x + th) dt
= \frac{1}{k} ∂^{α}f(x) + \frac{1}{k} ∫_{0}^{1}(1 − t)^{k} (h_{1} \frac{∂}{∂x_{1}} + ⋯ + h_{n} \frac{∂}{∂x_{n}} ) ∂^{α}f(x + th) dt.

It is consistent with our previous notation to let

(h∂)^{α}f = (h_{1}^{α_{1}} \frac{∂^{α_{1}}}{∂x_{1}^{α_{1}}} h_{2}^{α_{2}} \frac{∂^{α_{2}}}{∂x_{2}^{α_{2}}} ⋯ h_{n}^{α_{n}} \frac{∂^{α_{n}}}{∂x_{n}^{α_{n}}} ) f = h^{α}∂^{α}f,

so we adopt this convention. We recall the multinomial formula

(a_{1} + a_{2} + ⋯ + a_{n})^{m} = ∑_{b_{1}+···+b_{n}=m} \binom{m}{b_{1}, …, b_{n}}a_{1}^{b_{1}} ⋯ a_{n}^{b_{n}}

which, if we let a = (a_{1}, …, a_{n}) and β = (b_{1}, …, b_{n}), we can write as

\frac{1}{m!}(a_{1} + ⋯ + a_{n})^{m} = ∑_{|β|=m} \frac{a^{β}}{β!} .

Using the multinomial formula and the fact that mixed partials of order ≤ k + 1 commute for functions in C^{k+1}, we have

(h_{1} \frac{∂}{∂x_{1}} + ⋯ + h_{n} \frac{∂}{∂x_{n}} ) ∑_{|α|=k} \frac{h^{α}}{α!} ∂^{α}f(x + th)

= (h_{1} \frac{∂}{∂x_{1}} + ⋯ + h_{n} \frac{∂}{∂x_{n}} ) ∑_{|α|=k} \frac{(h∂)^{α}}{α!} f(x + th)

= \frac{1}{k!} (h_{1} \frac{∂}{∂x_{1}} + ⋯ + h_{n} \frac{∂}{∂x_{n}} ) (h_{1} \frac{∂}{∂x_{1}} + ⋯ + h_{n} \frac{∂}{∂x_{n}} )^{k} f(x + th)

= \frac{k + 1}{(k + 1)!} (h_{1} \frac{∂}{∂x_{1}} + ⋯ + h_{n} \frac{∂}{∂x_{n}} )^{k+1} f(x + th)
= (k + 1) ∑_{|α|=k+1} \frac{(h∂)^{α}}{α!} f(x + th) = (k + 1) ∑_{|α|=k+1} \frac{h^{α}}{α!} ∂^{α}f(x + th).

Now we apply the gathered equations to the formula in the case f ∈ C^{k}, and get

f(x + h)
= ∑_{|α|<k} ∂^{α}f(x)\frac{h^{α}}{α!} + k ∑_{|α|=k} (∫_{0}^{1}(1 − t)^{k−1}∂^{α}f(x + th) dt) \frac{h^{α}}{α!}

= ∑_{|α|<k} ∂^{α}f(x)\frac{h^{α}}{α!} + k ∑_{|α|=k} (\frac{1}{k} ∂^{α}f(x) + \frac{1}{k} ∫_{0}^{1}(1 − t)^{k} \frac{d}{dt}∂^{α}f(x + th) dt) \frac{h^{α}}{α!}

= ∑_{|α|≤k} ∂^{α}f(x)\frac{h^{α}}{α!} + (k + 1) ∑_{|α|=k+1} (∫_{0}^{1}(1 − t)^{k}∂^{α}f(x + th) dt) \frac{h^{α}}{α!}

which is what we wanted. (Note that we used the fact that integrals commute with finite sums.) □

(ii) The Taylor formula in (i) implies that

f(x + h) = ∑_{|α|≤k} ∂^{α}f(x)\frac{h^{α}}{α!} + o(|h|^{k})

as h → 0.

Proof. The above equation means

f(x + h) − ∑_{|α|≤k} ∂^{α}f(x)\frac{h^{α}}{α!} = o(|h|^{k}) as h → 0,

i.e., that the ratio of the left-hand side to |h|^{k} goes to zero as h → 0. Using the Taylor formula from (i), this means we need to show that

lim_{h→0} ∑_{|α|=k} \frac{h^{α}}{|h|^{k}α!} [k (∫_{0}^{1}(1 − t)^{k−1}∂^{α}f(x + th) dt) − ∂^{α}f(x)] = 0.

It suffices to prove this limit for each term of the sum, because there are finitely many terms. Because α! is a constant, we can ignore it. Also, |h^{α}| = |h_{1}|^{α_{1}} ⋯ |h_{n}|^{α_{n}} ≤ h_{0}^{k} ≤ |h|^{k}, where h_{0} = max{|h_{i}| : 1 ≤ i ≤ n}, because α_{1} + ⋯ + α_{n} = k. Thus we only need to show that the limit of the expression in brackets is zero. Set F = ∂^{α}f. Then F is continuous on U, because f ∈ C^{k}. The expression in brackets becomes

k (∫_{0}^{1}(1 − t)^{k−1}F (x + th) dt) − F (x).

By the continuity of F, this expression is continuous in h (as long as h satisfies x + th ∈ U for all t ∈ [0, 1]). Thus to find the limit we can simply evaluate at h = 0:

k (∫_{0}^{1}(1 − t)^{k−1}F (x + t0) dt) − F (x)
= F (x) (∫_{0}^{1} k(1 − t)^{k−1} dt) − F (x)
= F (x) − F (x) = 0.

This proves the desired formula. □