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Amy Ferguson Moncure

Calculus BC ยท Unit 10

Taylor Polynomials

Match the derivatives at the center, and divide the kth derivative by k!. Maclaurin polynomials are centered at 0.

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Key Ideas Easy Medium Hard

\(P_n(x)=\sum_{k=0}^{n}\dfrac{f^{(k)}(a)}{k!}(x-a)^k\)

What to Remember

The polynomial

\(P_n(x)=\sum_{k=0}^{n}\dfrac{f^{(k)}(a)}{k!}(x-a)^k\)

The first two terms

The constant term is \(f(a)\). The linear term is \(f'(a)(x-a)\).

Maclaurin means center 0

\(e^x\), \(\sin x\), and \(\cos x\) have familiar polynomials at 0. Divide each derivative by the factorial.

A table is enough

If the derivatives at \(a\) are listed, you do not need a formula for \(f\).

Level 1

Easy

Memorized Maclaurin polynomials, through a small degree.

Example
Write the degree-3 Maclaurin polynomial for \(e^x\).

\(e^x\) and every derivative equal 1 at \(x=0\).

\(P_3(x)=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}\)

\(P_3(x)=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}\)

Answer: \(1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}\)

1
Write the degree-3 Maclaurin polynomial for \(\sin x\).
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The even derivatives at 0 are 0.

\(\sin 0=0\), \(\cos 0=1\), and the third derivative is \(-\cos x\), which is \(-1\) at 0.

\(P_3(x)=x-\dfrac{x^3}{6}\)

Answer: \(x-\dfrac{x^3}{6}\)

2
Write the degree-4 Maclaurin polynomial for \(\cos x\).
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\(\cos 0=1\), the first derivative is 0 at 0, and \(f''(0)=-1\).

The degree-4 derivative at 0 is 1.

\(P_4(x)=1-\dfrac{x^2}{2}+\dfrac{x^4}{24}\)

Answer: \(1-\dfrac{x^2}{2}+\dfrac{x^4}{24}\)

3
Write the degree-1 Maclaurin polynomial for \(e^x\).
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\(P_1(x)=f(0)+f'(0)x\)

\(P_1(x)=1+x\)

Answer: \(1+x\)

4
What is the coefficient of \(x^2\) in the Maclaurin polynomial for \(e^x\)?
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\(f''(0)=1\)

Divide by \(2!\).

The coefficient is \(\dfrac{1}{2}\).

Answer: \(\dfrac{1}{2}\)

5
Which polynomial is the degree-2 Maclaurin polynomial for \(e^x\)?
A. \(1+x+x^2\)
B. \(1+x+\dfrac{x^2}{2}\)
C. \(1+\dfrac{x^2}{2}\)
D. \(x+\dfrac{x^2}{2}\)
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The constant is 1 and the linear coefficient is 1.

The quadratic coefficient is \(1/2!\), which is \(1/2\).

B is the polynomial.

Answer: B. \(1+x+\dfrac{x^2}{2}\)

Level 2

Medium

A center that is not zero, or a short table of derivatives.

Example
Write the degree-3 Taylor polynomial for \(\ln x\) centered at \(1\).

\(f(1)=0\), \(f'(x)=1/x\), \(f''(x)=-1/x^2\), \(f'''(x)=2/x^3\)

\(f'(1)=1\), \(f''(1)=-1\), \(f'''(1)=2\)

\(P_3(x)=(x-1)+\dfrac{-1}{2}(x-1)^2+\dfrac{2}{6}(x-1)^3\)

\(P_3(x)=(x-1)-\dfrac{(x-1)^2}{2}+\dfrac{(x-1)^3}{3}\)

Answer: \((x-1)-\dfrac{(x-1)^2}{2}+\dfrac{(x-1)^3}{3}\)

1
A function satisfies \(f(2)=3\), \(f'(2)=-1\), \(f''(2)=4\), and \(f'''(2)=6\). Write \(P_2(x)\) centered at 2, and find \(P_2(3)\).
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The coefficient of \((x-2)^2\) is \(4/2!=2\).

\(P_2(x)=3-(x-2)+2(x-2)^2\)

At \(x=3\), \(x-2=1\).

\(P_2(3)=3-1+2\cdot 1\)

\(P_2(3)=4\)

Answer: \(P_2(x)=3-(x-2)+2(x-2)^2\), and \(P_2(3)=4\)

2
Write the degree-2 Taylor polynomial for \(\sqrt{x}\) centered at 4.
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\(f(4)=2\)

\(f'(x)=\dfrac{1}{2}x^{-1/2}\), so \(f'(4)=\dfrac{1}{4}\)

\(f''(x)=-\dfrac{1}{4}x^{-3/2}\), so \(f''(4)=-\dfrac{1}{32}\)

Divide \(f''(4)\) by \(2\).

\(P_2(x)=2+\dfrac{1}{4}(x-4)-\dfrac{1}{64}(x-4)^2\)

Answer: \(2+\dfrac{1}{4}(x-4)-\dfrac{1}{64}(x-4)^2\)

3
Write the degree-2 Maclaurin polynomial for \(\ln(1+x)\).
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At 0, \(\ln 1=0\), the first derivative \(1/(1+x)\) equals 1, and the second derivative \(-1/(1+x)^2\) equals \(-1\).

\(P_2(x)=x-\dfrac{x^2}{2}\)

Answer: \(x-\dfrac{x^2}{2}\)

4
Write the degree-2 Maclaurin polynomial for \(e^{2x}\).
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Substitute \(u=2x\) into \(1+u+u^2/2\).

\(1+2x+\dfrac{(2x)^2}{2}\)

\(1+2x+2x^2\)

Answer: \(1+2x+2x^2\)

5
Use the degree-1 Maclaurin polynomial for \(\sin x\) to approximate \(\sin 0.1\).
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\(P_1(x)=x\)

\(P_1(0.1)=0.1\)

Answer: \(0.1\)

Level 3

Hard

Build a polynomial from a longer list, or evaluate the square-root polynomial.

Example
Derivatives at 0 satisfy \(f(0)=1\), \(f'(0)=0\), \(f''(0)=-2\), and \(f'''(0)=6\). Write \(P_3(x)\).

The linear term is 0.

The coefficient of \(x^2\) is \(-2/2!=-1\).

The coefficient of \(x^3\) is \(6/6=1\).

\(P_3(x)=1-x^2+x^3\)

Answer: \(1-x^2+x^3\)

1
Write the degree-3 Maclaurin polynomial for \(\dfrac{1}{1-x}\).
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The geometric series begins \(1+x+x^2+x^3+\cdots\) for \(|x|<1\).

Stop after the cubic term.

\(P_3(x)=1+x+x^2+x^3\)

Answer: \(1+x+x^2+x^3\)

2
Write the degree-4 Maclaurin polynomial for \(\cos(2x)\).
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Substitute \(u=2x\) into \(1-u^2/2+u^4/24\).

\(1-\dfrac{(2x)^2}{2}+\dfrac{(2x)^4}{24}\)

\(1-\dfrac{4x^2}{2}+\dfrac{16x^4}{24}\)

\(1-2x^2+\dfrac{2}{3}x^4\)

Answer: \(1-2x^2+\dfrac{2}{3}x^4\)

3
Use the degree-2 polynomial for \(\sqrt{x}\) centered at 4 to approximate \(\sqrt{4.4}\).
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\(x-4=\dfrac{2}{5}\)

\(P_2(4.4)=2+\dfrac{1}{4}\cdot\dfrac{2}{5}-\dfrac{1}{64}\cdot\left(\dfrac{2}{5}\right)^2\)

\(=2+\dfrac{1}{10}-\dfrac{1}{64}\cdot\dfrac{4}{25}\)

\(=2+\dfrac{1}{10}-\dfrac{1}{400}\)

\(=\dfrac{800}{400}+\dfrac{40}{400}-\dfrac{1}{400}\)

\(=\dfrac{839}{400}\)

Answer: \(\dfrac{839}{400}\)

4
In the table \(f''(2)=4\), why is the coefficient of \((x-2)^2\) equal to 2 rather than 4?
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The Taylor formula divides the \(k\)th derivative by \(k!\).

Here \(k=2\) and \(2!=2\).

\(4/2=2\)

Answer: divide by \(2!\), so the coefficient is 2

5
Write \(P_2(x)\) for a function with \(f(0)=5\), \(f'(0)=-3\), and \(f''(0)=8\).
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The quadratic coefficient is \(8/2=4\).

\(P_2(x)=5-3x+4x^2\)

Answer: \(5-3x+4x^2\)

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