Distinguish coefficients from derivatives

A polynomial and its derivatives

The polynomial \(P(x)=7+5(x-2)\) \(+3(x-2)^2\) \(+4(x-2)^3\) has the center 2. Its coefficients are the numbers in front of the powers: 7, 5, 3 and 4.

Press "Differentiate once". The constant 7 disappears, and each other power drops by one while its exponent moves to the front. For example, \(3(x-2)^2\) becomes \(6(x-2)\).

At \(x=2\), every term that still contains \(x-2\) is 0, so only the constant term of \(P'\) remains. This gives \(P'(2)=5\).

Differentiate again to find \(P''(2)=6\), which is 2 times the coefficient 3. A third time gives \(P'''(2)=24\), which is 6 times the coefficient 4.

Each derivative is its coefficient times the factorial in the middle row. The first two factorials are 1, so the two numbers differ only from the second derivative on.

Any coefficient, any order

The coefficients are now letters, and the polynomial can have any degree. Press a button to choose the order \(k\).

Differentiating \((x-a)^k\) a total of \(k\) times brings down the factors \(k,\ k-1,\ \ldots,\ 1\), and their product is \(k!\).

Terms with a lower power are constants before the \(k\)-th derivative, so they become 0. Terms with a higher power keep a factor \(x-a\), so they are 0 at the center. Only \(k!\,c_k\) remains.

This means the \(k\)-th derivative at the center is \(k!\) times the coefficient, for every polynomial and every center.

Notation

1. The coefficient \(c_k\) is the number in front of \((x-a)^k\). The derivative \(\cy{f^{(k)}(a)}\) is the value of the \(k\)-th derivative at the center \(a\).

2. The derivative equals the coefficient times \(k!\). The coefficient equals the derivative divided by \(k!\).

   \[\begin{aligned}\cy{f^{(k)}(a)}&=k!\,c_k\\c_k&=\dfrac{\cy{f^{(k)}(a)}}{k!}\end{aligned}\]

   The symbol \(k!\) means \(1\cdot2\cdots k\), and \(0!=1\).

3. Both numbers use the same center. A polynomial with the center 2 uses derivatives at 2, not at 0.

   The polynomial must be the Taylor polynomial of \(f\) at \(a\). Then it has the same derivatives as \(f\) at \(a\), up to its degree.

When it fails

Without the factorial, the coefficient 4 is read as the third derivative, in red because it is wrong. The true value is \(3!\cdot4=24\), a factor of 6 larger.

The error grows with the order, because the factorials grow: 2, 6, 24, 120.

Example 1: from derivatives to coefficients

A function \(g\) has \(g(1)=4\), \(g'(1)=-3\), \(g''(1)=10\) and \(g'''(1)=-12\). Write its Taylor polynomial of degree 3 at \(x=1\).

1. The center is 1, so the polynomial uses powers of \(x-1\). The middle row shows the factorials.

   \[\begin{aligned}0!&=1,\quad 1!=1\\2!&=2,\quad 3!=6\end{aligned}\]

2. Divide each derivative by its factorial to find the coefficients.

   \[\begin{aligned}c_0&=4,\quad c_1=-3\\c_2&=\dfrac{10}{2}=5,\quad c_3=\dfrac{-12}{6}=-2\end{aligned}\]

3. Write each coefficient in front of its power of \(x-1\).

   \[\begin{aligned}P_3(x)&=4-3(x-1)\\&\quad+5(x-1)^2-2(x-1)^3\end{aligned}\]

4. To test the answer, differentiate three times. The third derivative of \(-2(x-1)^3\) is \(6\cdot(-2)=-12\), which is \(g'''(1)\).

   \[P_3'''(x)=-12\]

Example 2: from coefficients to derivatives

A function \(h\) has the Taylor polynomial \(Q(x)=2-(x+1)\) \(+3(x+1)^2-\tfrac52(x+1)^4\) at \(x=-1\). Find \(h''(-1)\) and \(h^{(4)}(-1)\).

1. The center is \(-1\), so both derivatives are taken at \(-1\). The coefficient of \((x+1)^2\) is 3, and the coefficient of \((x+1)^4\) is \(-\tfrac52\). The power \((x+1)^3\) is missing, so its coefficient is 0.

2. A tempting wrong move: Read the derivatives from the coefficients.

   \[\begin{aligned}\cr{h''(-1)}&\cr{=3}\\\cr{h^{(4)}(-1)}&\cr{=-\tfrac52}\end{aligned}\]

   Differentiate \(Q\) twice: \(3(x+1)^2\) becomes 6 and every higher term keeps a factor \(x+1\), so \(Q''(-1)=6\), not 3. The Taylor polynomial has the derivatives of \(h\), so \(h''(-1)=6\).

3. Multiply each coefficient by its factorial.

   \[\begin{aligned}h''(-1)&=2!\cdot3=6\\h^{(4)}(-1)&=4!\cdot\left(-\tfrac52\right)=-60\end{aligned}\]

Practice

A function \(f\) satisfies \(f^{(n)}(0)=3^n\) for every \(n\ge0\). Find the coefficient of \(x^4\) in its Taylor polynomial at 0.

The center is 0, so the coefficient of \(x^4\) uses the fourth derivative at 0. Divide it by the matching factorial.

Questions

A polynomial \(S\) has the center 3. The table shows its coefficients, the factorials and its derivatives at 3, with a question mark where a number is hidden. Questions 1 and 2 use the table.

1. What is \(S'''(3)\)?

   • \(-6\)
   • \(-1\)
   • 6
   Answer

   \(-6\)

2. The table shows \(S''(3)=4\). What is the coefficient of \((x-3)^2\)?

   • 2
   • 4
   • 8
   Answer

   2

3. A function \(f\) has a Taylor polynomial at 0 whose coefficient of \(x^5\) is \(\tfrac{3}{10}\). What is \(f^{(5)}(0)\)?

   • 36
   • \(\tfrac{3}{10}\)
   • \(\tfrac1{400}\)
   Answer

   36

Summary

You multiply a coefficient by \(k!\) to find the \(k\)-th derivative at the center. You divide the derivative by \(k!\) to find the coefficient.

This works because differentiating \((x-a)^k\) a total of \(k\) times gives \(k!\), and every other term is 0 at the center.

You need this conversion when a problem gives the derivatives at the center and asks for the polynomial, or the reverse.

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