Math · Calculus II · Worked example

Approximate e^0.2 with a Taylor polynomial

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

Find the coefficients

Every derivative of eˣ is eˣ, so each derivative at 0 is 1, and the kth coefficient is 1/k!.

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

Evaluate

Substitute 0.2.

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

Bound the error

The third derivative is eˣ, which on [0, 0.2] is at most e^0.2. Since e < 3, e^0.2 < 3^0.2 < 1.25 (as 1.25⁵ ≈ 3.05), so M = 1.25 works and the error is below 1.25(0.2)³/3! = 1/600 ≈ 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.

Result

e^0.2 ≈ 1.22, with error below 0.00167; the actual error is 0.0014.

Your turn

Use the degree-three Taylor polynomial of eˣ at 0 to approximate e^0.2.

Show the answer and explanation

1.221333…

P₃(0.2) = 1 + 0.2 + 0.02 + 0.008/6 = 1.221333…, within 0.00007 of e^0.2 = 1.221403.

1 + 0.2 + 0.02 + \frac{0.008}{6} = \frac{458}{375}

Keep exploring

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

Return to the concept →

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