Math · Calculus II · Worked example

Approximate e to the power 0.2

Use the degree-two Taylor polynomial of $e^{x}$ at $0$ to approximate $e^{0.2}$ , and bound the error.

e^{0.2} \approx P_{2} (0.2)

Find the derivatives at $0$

Every derivative of $e^{x}$ is $e^{x}$ , and $e^{0} = 1$ , so the value and the first two derivatives at $0$ are all $1$ .

f (0) = f^{'} (0) = f^{′′} (0) = 1

Divide by the factorials

The coefficients are $1 / 0! = 1$ , $1 / 1! = 1$ and $1 / 2! = 1 / 2$ . Multiply each by the matching power of $x$ and add.

P_{2} (x) = 1 + x + \frac{x^{2}}{2}

Evaluate

Substitute $x = 0.2$ . The terms are $1$ , $0.2$ and $0.04 / 2 = 0.02$ .

P_{2} (0.2) = 1 + 0.2 + 0.02 = 1.22

Bound the error

The third derivative is $e^{x}$ , which on $[0, 0.2]$ is at most $e^{0.2}$ . Since $e < 3$ , $e^{0.2} < 3^{0.2} < 1.25$ , because ${1.25}^{5} \approx 3.05$ . With $M = 1.25$ the error is below $1.25 {(0.2)}^{3} / 3!$ $= 1 / 600$ $\approx 0.00167$ .

\frac{1.25 {(0.2)}^{3}}{6} = \frac{1}{600}

Compare with the true value

$e^{0.2} = 1.2214$ , so the actual error is $0.0014$ , inside the bound. The bound is a guarantee, and the actual error is smaller.

Result

$e^{0.2} \approx 1.22$ , with error below $0.00167$ . The actual error is $0.0014$ .

e^{0.2} \approx 1.22

Your turn

1. One like the example

Use the degree-three Taylor polynomial of $e^{x}$ at $0$ to approximate $e^{0.2}$ .

Show the answer and explanation

$1.221333 \dots$

The next coefficient is $1 / 3! = 1 / 6$ , so $P_{3} (x)$ $= 1 + x + x^{2} / 2 + x^{3} / 6$ . Then $P_{3} (0.2)$ $= 1 + 0.2 + 0.02 + 0.008 / 6$ $= 1.221333 \dots$ , within $0.00007$ of $e^{0.2} = 1.221403$ .

\begin{array}{l} P (x) = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} \\ P (0.2) = \frac{458}{375} \\ P (0.2) \approx 1.221333 \end{array}

2. A variation

Use the degree-two Taylor polynomial of $e^{x}$ at $0$ to approximate $e^{- 0.1}$ , and bound the error.

Show the answer and explanation

$e^{- 0.1} \approx 0.905$ , with error below $1 / 6000 \approx 0.000167$ .

$P_{2} (- 0.1)$ $= 1 - 0.1 + 0.01 / 2$ $= 0.905$ . Between $- 0.1$ and $0$ the third derivative $e^{x}$ is at most $e^{0} = 1$ , so $M = 1$ and the error is below ${(0.1)}^{3} / 3! = 1 / 6000$ . The true value is $0.904837$ , which is $0.000163$ away.

\begin{array}{l} P (x) = 1 + x + \frac{x^{2}}{2} \\ P (- 0.1) = 0.905 \\ \frac{{(0.1)}^{3}}{6} = \frac{1}{6000} \end{array}

3. Somewhere new

Use the degree-two Taylor polynomial of $\sqrt{x}$ centered at $4$ to approximate $\sqrt{4.1}$ .

Show the answer and explanation

$\sqrt{4.1} \approx 2.0248438$ , within $0.000002$ of $\sqrt{4.1} = 2.0248457$ .

$f (4) = 2$ . $f^{'} (x) = 1 / (2 \sqrt{x})$ gives $f^{'} (4) = 1 / 4$ , and $f^{''} (x) = - 1 / (4 x^{3 / 2})$ gives $f^{''} (4) = - 1 / 32$ , which divided by $2!$ is $- 1 / 64$ . $P_{2} (x)$ $= 2 + (x - 4) / 4 - {(x - 4)}^{2} / 64$ , and $P_{2} (4.1)$ $= 2 + 0.025 - 0.00015625$ $= 2.02484375$ .

\begin{array}{l} f (x) = \sqrt{x} \\ f^{'} (x) = \frac{1}{2 \sqrt{x}} \\ f^{′′} (x) = - \frac{1}{4 x^{3 / 2}} \\ f^{'} (4) = \frac{1}{4} \\ f^{′′} (4) = - \frac{1}{32} \\ P (x) = 2 + \frac{x - 4}{4} - \frac{{(x - 4)}^{2}}{64} \\ P (4.1) = 2.02484375 \end{array}

Keep exploring

In Taylor Approximation, raise the degree to $3$ . The sampled error on $[0, 0.2]$ drops from $0.0014$ to $0.00007$ .

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Find the derivatives at 00

Divide by the factorials

Evaluate

Bound the error

Compare with the true value

Result

Your turn

1. One like the example

1.221333⁢…1.221333…

2. A variation

e−0.1≈0.905e−0.1≈0.905, with error below 1⁢/6000≈0.0001671/6000≈0.000167.

3. Somewhere new

4.1≈2.0248438√4.1≈2.0248438, within 0.0000020.000002 of 4.1=2.0248457√4.1=2.0248457.

Keep exploring

Study it, then try it

Find the derivatives at $0$

$1.221333 \dots$

$e^{- 0.1} \approx 0.905$ , with error below $1 / 6000 \approx 0.000167$ .

$\sqrt{4.1} \approx 2.0248438$ , within $0.000002$ of $\sqrt{4.1} = 2.0248457$ .