Taylor series

Definition

About a general point

Suppose $f$ is a function that is infinitely differentiable at a point $x_0$ in its domain. The Taylor series of $f$ about $x_0$ is the power series given as follows:

$\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k$

Here's a version with the first few terms written explicitly:

$f(x_0)+f^{\prime }(x_0)(x-x_0)+\frac{f^{″}(x_0)}{2}(x-x_0{)}^2+\frac{f^{‴}(x_0)}{6}(x-x_0{)}^3+\dots$

About the point 0

In the special case of the above definition where $x_0=0$ (and in particular $f$ is infinitely differentiable at 0), the Taylor series is as follows:

$\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(0)}{k!}{x}^k$

Here's a version with the first few terms written explicitly:

$f(0)+f^{\prime }(0)x+\frac{f^{″}(0)}{2}{x}^2+\frac{f^{‴}(0)}{6}{x}^3+\dots +$

Well defined on germs of a functions

The Taylor series operator about a point $x_0$ can be thought of as a mapping:

(Germs of $C^{\mathrm{\infty }}$-functions defined about $x_0$) $\to$ (Formal power series centered at $x_0$)

In fact, this mapping is a $\mathbb{R}$-algebra homomorphism that commutes with the differential structure.

Here, two functions $f$ and $g$ are said to have the same germ about a point $x_0$ if there is an open interval $U$ containing $x_0$ such that $f(x)=g(x)\mathrm{\forall }x\in U$.

Relation with Taylor polynomials

The Taylor series can be viewed as a limit of Taylor polynomials. The $n^{th}$ Taylor polynomial for a function $f$ at a point $x_0$ in the domain is the truncation of the Taylor series to powers up to the $n^{th}$ power. If we denote the polynomial by $P_n(f;x_0)$, it is given as:

$P_n(f;x_0)=x\mapsto \sum _{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k$

Note that this is a polynomial of degree at most $n$. The degree is exactly $n$ if and only if $f^{(n)}(x_0)\ne 0$.

Whether or not the Taylor series of a function converges to the function is determined by whether or not the sequence of Taylor polynomials of the function converges to the function.

Computation of Taylor series

To calculate the Taylor series of a function $f$ at a point $x_0$, we use the following procedure:

• Compute formal expressions for $f,f^{\prime },f^{″},\dots$, i.e., $f$ and all its derivatives, at a generic point.
• Evaluate all these at $x_0$.
• Plug the result into the Taylor series formula.

Example of exponential function

For further information, refer: Exponential function#Taylor series

Consider the exponential function:

$f=\exp$

i.e., $f(x)=e^x$.

We want to compute the Taylor series of this function at 0.

Applying the procedure above, we get:

• Formal expressions for $f,f^{\prime },f^{″},\dots$: These are $\exp ,\exp ,\exp ,\dots$. The sequence is a constant sequence of functions with all its members equal to $\exp$.
• Evaluate at 0: Since $\exp (0)=1$, we get $1,1,1,\dots$. The sequence is a constant sequence with value 1 in all places.
• Plug the result into the Taylor series formula: We get:

Taylor series for $\exp (x)$ is $\sum _{k=0}^{\mathrm{\infty }}\frac{x^k}{k!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots$

Note that it is also true that the Taylor series for the exponential function converges to the exponential function everywhere; this is because the function is globally analytic. However, this fact is not a priori obvious, and we are not asserting it as part of the computation of the Taylor series.

Example of cosine function

For further information, refer: Cosine function#Taylor series

Consider the cosine function:

$f=\cos$

We want to compute the Taylor series of this function at 0.

Applying the procedure above, we get:

• Formal expressions for $f,f^{\prime },f^{″},\dots$: These are $\cos ,-\sin ,-\cos ,\sin ,\cos ,-\sin ,-\cos ,\sin ,\dots$. The sequence is periodic with period 4.
• Evaluate at 0: Since $\cos 0=1,\sin 0=0$, the values are $1,0,-1,0,1,0,-1,0,\dots$. The sequence is periodic with period 4.
• Plug the result into the Taylor series formula: We get:

Taylor series for $\cos x$ is $1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots$ which can be rewritten compactly as $\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k{x}^{2k}}{(2k)!}$. Note that the $k$ here is half the exponent on $x$, so this is a little different from the usual way of writing Taylor series.

Note that it is also true that the Taylor series for the cosine function converges to the cosine function everywhere; this is because the function is globally analytic. However, this fact is not a priori obvious, and we are not asserting it as part of the computation of the Taylor series.

Example of sine function

For further information, refer: Sine function#Computation of Taylor series

Consider the sine function:

$f=\sin$

We want to compute the Taylor series of this function at 0.

Applying the procedure above, we get:

• Formal expressions for $f,f^{\prime },f^{″},\dots$: These are $\sin ,\cos ,-\sin ,-\cos ,\sin ,\cos ,-\sin ,-\cos ,\dots$. The sequence is periodic with period 4.
• Evaluate at 0: Since $\sin 0=0,\cos 0=1$, the values are $0,1,0,-1,0,1,0,-1,\dots$. The sequence is periodic with period 4.
• Plug the result into the Taylor series formula: We get:

Taylor series for $\sin x$ is $x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots$ which can be rewritten compactly as $\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k{x}^{2k+1}}{(2k+1)!}$. Note that the $k$ here is roughly half the exponent on $x$, so this is a little different from the usual way of writing Taylor series.

Note that it is also true that the Taylor series for the sine function converges to the sine function everywhere; this is because the function is globally analytic. However, this fact is not a priori obvious, and we are not asserting it as part of the computation of the Taylor series.

Facts

Preservation of structure

Together, the first three facts show that the Taylor series operator is a homomorphism of $\mathbb{R}$-algebras that commutes with the differential structure. The fourth fact show that it preserves an additional structure:

• Taylor series operator is linear
• Taylor series operator commutes with differentiation
• Taylor series operator is multiplicative
• Taylor series operator commutes with composition

Convergence to the original function

• Composite of Taylor series operator and power series summation operator is identity map: What this essentially says is that if a power series centered at $x_0$ converges to $f$ on an open interval centered at $x_0$, then the power series must equal the Taylor series of $f$.
• Composite of power series summation operator and Taylor series operator is not identity map: It is possible to have an everywhere infinitely differentiable function $f$ and a point $x_0$ in the domain such that the sum of the Taylor series of $f$ at $x_0$ is not equal to $f$ on any interval of positive radius centered at $x_0$. In fact, we can arrange our example so that the power series sum agrees with $f$ only at $x_0$. Note that if this happens, then there cannot be any other power series centered at $x_0$ that converges to $f$ on a positive radius of convergence.

A function whose Taylor series at a point converges to the function in an open interval centered at the point is termed a locally analytic function at the point. If the Taylor series converges to the function everywhere, the function is termed a globally analytic function. We have that locally analytic not implies globally analytic.