Read the information a polynomial provides

A polynomial with a hidden function

The orange curve is the Taylor polynomial \(P(x)=5+3(x-2)\) \(-2(x-2)^2+\tfrac13(x-2)^3\) of a function \(f\), centered at 2. The faint curve is \(f\), which in a real problem you do not see.

At the center \(x=2\), the orange dot lies on both curves, and the two curves have the same slope and the same bend.

Drag the input away from 2. A red segment opens between the curves and grows with the distance from the center.

The polynomial gives exact facts about \(f\) at the center, but away from the center it gives only an approximation.

What a polynomial of degree n knows

The hidden function is \(f(x)=P(x)+k(x-a)^4\), and the slider changes \(k\). The faint curve moves, but the orange polynomial stays still, so many different functions have the same Taylor polynomial.

If the polynomial has degree \(n\), all of these functions have the same height and the same derivatives up to order \(n\) at the center.

They can differ in the higher derivatives and away from the center, and the polynomial cannot show these differences.

Notation

1. The constant term \(c_0\) is the height of \(f\) at the center, so the orange line is flat at that height.

   \[\cy{f(a)}=c_0\]

2. The coefficient \(c_1\) of \((x-a)\) is the slope at the center. If it is positive, \(f\) rises at \(a\). If it is negative, \(f\) falls at \(a\).

   \[\cy{f'(a)}=c_1\]

3. The coefficient \(c_2\) of \((x-a)^2\) is \(f''(a)/2!\), so it gives the bend. If \(f''(a)\) is positive, \(f\) is concave up at \(a\). If it is negative, \(f\) is concave down.

   \[\cy{f''(a)}=2!\,c_2\]

4. For every order \(k\) up to the degree \(n\), the derivative at the center is \(k!\) times the coefficient \(c_k\) of \((x-a)^k\). The polynomial gives no derivative of order above \(n\).

   \[\begin{aligned}f^{(k)}(a)&=k!\,c_k,\\k&=0,1,\ldots,n\end{aligned}\]

When it fails

The slider changes \(k\) in \(f(x)=P(x)+k(x-a)^4\). The polynomial \(P\) of degree 3 stays the same, but \(f^{(4)}(a)=24k\) changes, so \(P\) does not contain the fourth derivative.

The values of \(f\) away from \(a\) also change with \(k\), so there \(P(x)\) is only an approximation.

Example 1: reading height, slope and bend

A function \(f\) has the Taylor polynomial \(P_2(x)=7-3(x-1)+\tfrac52(x-1)^2\) at \(x=1\). Find \(f(1)\), \(f'(1)\) and \(f''(1)\), and describe the graph of \(f\) at \(x=1\).

1. The constant term is the height at the center.

   \[f(1)=7\]

2. The coefficient of \((x-1)\) is the slope. It is negative, so \(f\) falls at \(x=1\).

   \[f'(1)=-3\]

3. The coefficient of \((x-1)^2\) is \(f''(1)/2!\), so the second derivative is twice the coefficient. It is positive, so \(f\) is concave up at \(x=1\).

   \[f''(1)=2\cdot\tfrac52=5\]

Example 2: what the polynomial does not say

A function \(g\) has the Taylor polynomial \(Q(x)=6+2(x-3)\) \(+(x-3)^2-\tfrac12(x-3)^3\) at \(x=3\). Find \(g'''(3)\) and \(g^{(4)}(3)\), or say that \(Q\) does not give them.

1. The coefficient of \((x-3)^3\) is \(-\tfrac12\), so the third derivative is \(3!\) times it.

   \[g'''(3)=3!\cdot\left(-\tfrac12\right)=-3\]

2. A tempting wrong move: The polynomial has no term in \((x-3)^4\), so the fourth derivative is 0.

   \[\cr{g^{(4)}(3)=0}\]

   Each faint curve \(Q+k(x-3)^4\) has the Taylor polynomial \(Q\), but its fourth derivative at 3 is \(24k\). The red labels show three different values, so \(Q\) cannot give the fourth derivative.

3. The polynomial \(Q\) stops at degree 3, so it does not give \(g^{(4)}(3)\).

Practice

A function \(f\) has the Taylor polynomial \(P_2(x)=2-(x+3)^2\) at \(x=-3\). What does \(P_2\) say about a local maximum or minimum of \(f\) at \(x=-3\)?

Read the slope and the bend at the center. Is the tangent flat, and does the curve bend up or down?

Questions

The solid orange curve is the Taylor polynomial of degree 2 of a function \(f\) at \(x=1\). The faint curve is \(f\). Questions 1 and 2 use the figure.

1. What is \(f'(1)\)?

   • 1
   • 3
   • \(-\dfrac12\)
   Answer

   1

2. The orange curve is at 3 when \(x=3\). Which statement is justified?

   • The value \(f(3)\) is approximately 3. The polynomial cannot give it exactly.
   • The value \(f(3)\) is exactly 3.
   • The polynomial says nothing about \(f(3)\).
   Answer

   The value \(f(3)\) is approximately 3. The polynomial cannot give it exactly.

3. A function \(h\) has the Taylor polynomial \(2+5x-3x^2+2x^3+\tfrac12x^4\) at 0. Which statement does the polynomial justify?

   • \(h'''(0)=12\)
   • \(h'''(0)=2\)
   • \(h(1)=6.5\)
   Answer

   \(h'''(0)=12\)

Summary

You read the height, the slope and the bend of a function at the center from the first three coefficients of its Taylor polynomial.

The polynomial and the function have the same derivatives at the center, up to the degree, so these readings are exact.

You treat these readings as exact, and the values of the polynomial away from the center as approximations.

This is one of four skills in topic 10.11 of Which & Why: AP Calculus.