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Taylor polynomials

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52766712
For \(-1<x\leq1\), the natural logarithm has the series representation \(\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\). 1. Use the first four nonzero terms to approximate \(\ln(1.2)\). Round to four decimal places. 2. Explain why this series cannot be used by direct substitution to calculate \(\ln(4)\).

Hints

- Solve \(1+x=1.2\) before substituting. - Keep track of the alternating signs. - Compare the required value of \(x\) for \(\ln(4)\) with the given interval.

Solution

1. To approximate \(\ln(1.2)\), set \(1+x=1.2\), so \(x=0.2\). 2. The fourth-degree approximation is \(0.2-\frac{0.2^2}{2}+\frac{0.2^3}{3}-\frac{0.2^4}{4}=0.182266\ldots\). 3. Therefore, \(\ln(1.2)\approx0.1823\). 4. For \(\ln(4)\), direct substitution would require \(1+x=4\), or \(x=3\). Since \(3\) is outside the stated interval \((-1,1]\), the series representation does not apply there.

Answer

1. \(\ln(1.2)\approx0.1823\) 2. Direct substitution would require \(x=3\), which is outside \((-1,1]\).
52993712
Let \(f(x)=e^{-2x}\). Find the cubic polynomial \(p(x)=a_3x^3+a_2x^2+a_1x+a_0\) whose value and first three derivatives at \(x=0\) match those of \(f\).

Hints

- Compute the first three derivatives of the exponential function at \(0\). - Write the corresponding derivatives of a general cubic polynomial. - Evaluate the polynomial and its derivatives at \(0\). - Match corresponding values to determine the coefficients.

Solution

1. Compute the derivatives of \(f\): \(f'(x)=-2e^{-2x}\), \(f''(x)=4e^{-2x}\), and \(f'''(x)=-8e^{-2x}\). At \(x=0\), the values are \(f(0)=1\), \(f'(0)=-2\), \(f''(0)=4\), and \(f'''(0)=-8\). 2. For the cubic polynomial, \(p(0)=a_0\), \(p'(0)=a_1\), \(p''(0)=2a_2\), and \(p'''(0)=6a_3\). Matching the values gives \(a_0=1\), \(a_1=-2\), \(2a_2=4\), and \(6a_3=-8\). Therefore, \(a_2=2\) and \(a_3=-\frac{4}{3}\), so \(p(x)=-\frac{4}{3}x^3+2x^2-2x+1\).

Answer

\(p(x)=-\frac{4}{3}x^3+2x^2-2x+1\)

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