Taylor’s theorem

Example 1: $f(x)=e^x$

How to use

• Visualize Taylor’s formula for $f(x)=\sin(x)$ by comparing the function with its Taylor polynomial $T_n$ around a movable expansion point $x_0$.
• See the remainder term at a chosen point $x$: $R_n(x)=f(x)-T_n(x).$
• Drag the blue point $x_0$ on the curve to choose the expansion point.
• Move the slider $n$ to select the degree of the Taylor polynomial $T_n$.
• Drag the orange point $x$ on the curve to choose where you evaluate the approximation.
• The blue curve is $f(x)=\sin(x)$.
• The pink curve is the Taylor polynomial $T_n(x)$ centered at $x_0$.
• The green vertical dashed segment at the current $x$ is the remainder $R_n(x)$: its length is $|f(x)-T_n(x)|$.
• The endpoints of the green segment are $(x,f(x))$ and $(x,T_n(x))$.
• Fix $x_0$ and drag $x$ close to $x_0$: the green segment becomes small, illustrating that $T_n(x)$ approximates $f(x)$ well near the expansion point.
• Keep $x_0$ fixed and increase $n$: the pink curve typically matches the blue curve better near $x_0$, and the remainder at the chosen $x$ usually decreases.
• For a fixed degree $n$, move $x$ farther away from $x_0$: the remainder generally grows, showing that Taylor polynomials are local approximations.

Example 2: $g(x)=x^5|x|$

How to use

• See that Taylor’s theorem needs enough differentiability at the expansion point.
• For $g(x)=x^5|x|,$ the function is $5$-times differentiable at $x_0=0$, but not $6$-times differentiable at $x_0=0$. The applet shows what goes wrong when you try to use degree $n=6$ near $0$.
• Drag the blue point $x_0$ on the curve to choose the expansion point.
• Move the slider $n$ (allowed values $0,\dots,6$).
• Drag the orange point $x$ on the curve to choose where you evaluate the approximation.
• The blue curve is $g(x)$.
• The pink curve is the Taylor polynomial $T_n(x)$ around the chosen $x_0$ (only shown when the required derivatives exist).
• The green vertical dashed segment at the current $x$ is the remainder $R_n(x)=g(x)-T_n(x)$ (its length is $|g(x)-T_n(x)|$).
• If the Taylor polynomial is not defined (the “failure case”), the applet hides $T_n$ and $R_n$, and the graph of $g$ turns red as a warning.
• First set $n=5$ and move $x_0$ close to $0$: the Taylor polynomial is still defined, and near $x_0$ the remainder segment tends to get small.
• Now set $n=6$ and drag $x_0$ toward $0$:
  • as $x_0$ gets sufficiently close to $0$, the applet declares the Taylor polynomial undefined (pink curve and green remainder disappear, and $g$ turns red).
  • this corresponds to the fact that $g^{(6)}(0)$ does not exist, so the degree-6 Taylor polynomial at $x_0=0$ is not well-defined.
• Even before the polynomial disappears, you can observe instability for $n=6$ when $x_0$ is near $0$:
  • keep $n=6$,
  • choose $x_0$ close to $0$,
  • move the evaluation point $x$ a bit away from $x_0$,
  • notice that the remainder segment $R_6(x)$ can become very large.

Description

Reference: Lecture Notes Calculus 1 ( May22,2022 ) Definition6.6 and Theorem6.19 page 115

In both applets:

• Drag the blue point $x_0$ to choose the expansion point.
• Use the slider $n$ to choose the Taylor degree.
• Drag the orange point $x$ to evaluate the approximation at a point.
• The vertical segment labelled $R_n(x)$ visualizes the approximation error at the current $x$.

Example 1: $f(x)=e^x$

Link to code.

For the exponential function, all derivatives exist everywhere and satisfy

Hence the Taylor polynomial is well-defined for every degree $n$ and every expansion point $x_0$.

In the code, the Taylor polynomial is implemented as in the definition:

const T = (x: number) => {
  const x0 = getX0();
  const n = getN();
  let s = 0;
  const dx = x - x0;

  for (let k = 0; k <= n; k++) {
    s += (fDerivAt(k, x0) / factorial(k)) * Math.pow(dx, k);
  }
  return s;
};

The remainder is drawn as a vertical segment at the current x:

createSegment(board, [
  [() => getX(), () => f(getX())],
  [() => getX(), () => T(getX())],
], {
  name: "Rₙ(x)",
  dash: 2,
  strokeWidth: 4,
  withLabel: true,
}, COLORS.green);

Example 2: $g(x)=x^5|x|$

Link to code.

In this applet we use

Equivalently,

so it coincides with $x^6$ for $x>0$ and with $-x^6$ for $x<0$.

The $n$-th Taylor polynomial of a function $f$ at the expansion point $x_0$ is defined by

const T = (x: number) => {
  const x0 = getX0();
  const n = getN();
  const dx = x - x0;

  let s = 0;
  for (let k = 0; k <= n; k++) {
    const dk = gDerivAt(k, x0); // g^(k)(x0)
    s += (dk / factorial(k)) * Math.pow(dx, k);
  }
  return s;
};

For $x \neq 0$, since

the function is a polynomial on each side of $0$.

At $x = 0$ we have

but the sixth derivative does not exist, because the left and right limits differ:

Therefore, $g$ is five times differentiable at $0$, but not six times differentiable at $0$.

In the applet, this situation is handled by declaring the Taylor polynomial undefined when $x_0$ is sufficiently close to $0$ and $n \ge 6$:

const EPS = 0.05;
const taylorIsDefined = () =>
  !(Math.abs(getX0()) < EPS && getN() >= 6);

The Taylor graph and the remainder segment are hidden whenever this condition fails:

const updateVisibility = () => {
  const v = taylorIsDefined();
  tGraph.setAttribute({ visible: v });
  rSeg.setAttribute({ visible: v });
};

board.on("update", updateVisibility);
updateVisibility();

To observe the failure, drag the blue point to $x_0 \approx 0$ and set the slider to $n = 6$. The Taylor polynomial disappears because $g$ is not six times differentiable at $0$.