Taylor and Maclaurin polynomials
Problem 7.43 · hard
Find the Maclaurin polynomial of degree 3 for \( \displaystyle f(x) = \sin{\left(x \right)} \).
- p_n(x) = Σ f⁽ᵏ⁾(0)/k! · xᵏ, for k = 0 to n.
- \[ \left. \sin{\left(x \right)} \right|_{\substack{ x=0 }} = 0 \]f⁽0⁾(0).✓ Proved
- \[ \left. \frac{d}{d x} \sin{\left(x \right)} \right|_{\substack{ x=0 }} = 1 \]f⁽1⁾(0).✓ Proved
- \[ \left. \frac{d^{2}}{d x^{2}} \sin{\left(x \right)} \right|_{\substack{ x=0 }} = 0 \]f⁽2⁾(0).✓ Proved
- \[ \left. \frac{d^{3}}{d x^{3}} \sin{\left(x \right)} \right|_{\substack{ x=0 }} = -1 \]f⁽3⁾(0).✓ Proved
- \[ \left(-1\right) x^{3} \cdot 1 \cdot \frac{1}{6} + x 1 = - \frac{x^{3}}{6} + x \]Assemble the polynomial.✓ Proved
Answer \( p_{3}(x) = - \frac{x^{3}}{6} + x \)
Lines: 5 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | sympy.series expands f on its own and matches |
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/taylor_polynomial, checked 2026-09-26 with SymPy 1.14.0.