Polynomial approximations built from derivatives at a point. 32 problems with worked solutions; in 32 of them every
equation is proved by a computer algebra system.
Try each one before opening the solution. Every line is labeled with what checked it. How we verify
Find the Maclaurin polynomial of degree 5 for \( \displaystyle f(x) = \sqrt{x + 1} \).
Show the first step
Show the whole solution
p_n(x) = Σ f⁽ᵏ⁾(0)/k! · xᵏ, for k = 0 to n.
Reviewed
\[ \left. \sqrt{x + 1} \right|_{\substack{ x=0 }} = 1 \]
f⁽0⁾(0).✓ Proved
\[ \left. \frac{d}{d x} \sqrt{x + 1} \right|_{\substack{ x=0 }} = \frac{1}{2} \]
f⁽1⁾(0).✓ Proved
\[ \left. \frac{d^{2}}{d x^{2}} \sqrt{x + 1} \right|_{\substack{ x=0 }} = - \frac{1}{4} \]
f⁽2⁾(0).✓ Proved
\[ \left. \frac{d^{3}}{d x^{3}} \sqrt{x + 1} \right|_{\substack{ x=0 }} = \frac{3}{8} \]
f⁽3⁾(0).✓ Proved
\[ \left. \frac{d^{4}}{d x^{4}} \sqrt{x + 1} \right|_{\substack{ x=0 }} = - \frac{15}{16} \]
f⁽4⁾(0).✓ Proved
\[ \left. \frac{d^{5}}{d x^{5}} \sqrt{x + 1} \right|_{\substack{ x=0 }} = \frac{105}{32} \]
f⁽5⁾(0).✓ Proved
\[ 105 \cdot \frac{1}{32} x^{5} \cdot 1 \cdot \frac{1}{120} - 15 \cdot \frac{1}{16} x^{4} \cdot 1 \cdot \frac{1}{24} + 3 \cdot \frac{1}{8} x^{3} \cdot 1 \cdot \frac{1}{6} - 1 \cdot \frac{1}{4} x^{2} \cdot 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} x 1 + 1 = \frac{7 x^{5}}{256} - \frac{5 x^{4}}{128} + \frac{x^{3}}{16} - \frac{x^{2}}{8} + \frac{x}{2} + 1 \]
Assemble the polynomial.✓ Proved
Answer \( p_{5}(x) = \frac{7 x^{5}}{256} - \frac{5 x^{4}}{128} + \frac{x^{3}}{16} - \frac{x^{2}}{8} + \frac{x}{2} + 1 \)
✓ Nihil obstat Lines: 7 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the Maclaurin polynomial of degree 5 for \( \displaystyle f(x) = \frac{1}{1 - x} \).
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p_n(x) = Σ f⁽ᵏ⁾(0)/k! · xᵏ, for k = 0 to n.
Reviewed
\[ \left. 1 \frac{1}{1 - x} \right|_{\substack{ x=0 }} = 1 \]
f⁽0⁾(0).✓ Proved
\[ \left. \frac{d}{d x} 1 \frac{1}{1 - x} \right|_{\substack{ x=0 }} = 1 \]
f⁽1⁾(0).✓ Proved
\[ \left. \frac{d^{2}}{d x^{2}} 1 \frac{1}{1 - x} \right|_{\substack{ x=0 }} = 2 \]
f⁽2⁾(0).✓ Proved
\[ \left. \frac{d^{3}}{d x^{3}} 1 \frac{1}{1 - x} \right|_{\substack{ x=0 }} = 6 \]
f⁽3⁾(0).✓ Proved
\[ \left. \frac{d^{4}}{d x^{4}} 1 \frac{1}{1 - x} \right|_{\substack{ x=0 }} = 24 \]
f⁽4⁾(0).✓ Proved
\[ \left. \frac{d^{5}}{d x^{5}} 1 \frac{1}{1 - x} \right|_{\substack{ x=0 }} = 120 \]
f⁽5⁾(0).✓ Proved
\[ 120 x^{5} \cdot 1 \cdot \frac{1}{120} + 24 x^{4} \cdot 1 \cdot \frac{1}{24} + 6 x^{3} \cdot 1 \cdot \frac{1}{6} + 2 x^{2} \cdot 1 \cdot \frac{1}{2} + x 1 + 1 = x^{5} + x^{4} + x^{3} + x^{2} + x + 1 \]
Assemble the polynomial.✓ Proved
Answer \( p_{5}(x) = x^{5} + x^{4} + x^{3} + x^{2} + x + 1 \)
✓ Nihil obstat Lines: 7 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the Maclaurin polynomial of degree 5 for \( \displaystyle f(x) = \operatorname{atan}{\left(x \right)} \).
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p_n(x) = Σ f⁽ᵏ⁾(0)/k! · xᵏ, for k = 0 to n.
Reviewed
\[ \left. \operatorname{atan}{\left(x \right)} \right|_{\substack{ x=0 }} = 0 \]
f⁽0⁾(0).✓ Proved
\[ \left. \frac{d}{d x} \operatorname{atan}{\left(x \right)} \right|_{\substack{ x=0 }} = 1 \]
f⁽1⁾(0).✓ Proved
\[ \left. \frac{d^{2}}{d x^{2}} \operatorname{atan}{\left(x \right)} \right|_{\substack{ x=0 }} = 0 \]
f⁽2⁾(0).✓ Proved
\[ \left. \frac{d^{3}}{d x^{3}} \operatorname{atan}{\left(x \right)} \right|_{\substack{ x=0 }} = -2 \]
f⁽3⁾(0).✓ Proved
\[ \left. \frac{d^{4}}{d x^{4}} \operatorname{atan}{\left(x \right)} \right|_{\substack{ x=0 }} = 0 \]
f⁽4⁾(0).✓ Proved
\[ \left. \frac{d^{5}}{d x^{5}} \operatorname{atan}{\left(x \right)} \right|_{\substack{ x=0 }} = 24 \]
f⁽5⁾(0).✓ Proved
\[ 24 x^{5} \cdot 1 \cdot \frac{1}{120} - 2 x^{3} \cdot 1 \cdot \frac{1}{6} + x 1 = \frac{x^{5}}{5} - \frac{x^{3}}{3} + x \]
Assemble the polynomial.✓ Proved
Answer \( p_{5}(x) = \frac{x^{5}}{5} - \frac{x^{3}}{3} + x \)
✓ Nihil obstat Lines: 7 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the Maclaurin polynomial of degree 5 for \( \displaystyle f(x) = \ln{\left(x + 1 \right)} \).
Show the first step
Show the whole solution
p_n(x) = Σ f⁽ᵏ⁾(0)/k! · xᵏ, for k = 0 to n.
Reviewed
\[ \left. \ln{\left(x + 1 \right)} \right|_{\substack{ x=0 }} = 0 \]
f⁽0⁾(0).✓ Proved
\[ \left. \frac{d}{d x} \ln{\left(x + 1 \right)} \right|_{\substack{ x=0 }} = 1 \]
f⁽1⁾(0).✓ Proved
\[ \left. \frac{d^{2}}{d x^{2}} \ln{\left(x + 1 \right)} \right|_{\substack{ x=0 }} = -1 \]
f⁽2⁾(0).✓ Proved
\[ \left. \frac{d^{3}}{d x^{3}} \ln{\left(x + 1 \right)} \right|_{\substack{ x=0 }} = 2 \]
f⁽3⁾(0).✓ Proved
\[ \left. \frac{d^{4}}{d x^{4}} \ln{\left(x + 1 \right)} \right|_{\substack{ x=0 }} = -6 \]
f⁽4⁾(0).✓ Proved
\[ \left. \frac{d^{5}}{d x^{5}} \ln{\left(x + 1 \right)} \right|_{\substack{ x=0 }} = 24 \]
f⁽5⁾(0).✓ Proved
\[ 24 x^{5} \cdot 1 \cdot \frac{1}{120} - 6 x^{4} \cdot 1 \cdot \frac{1}{24} + 2 x^{3} \cdot 1 \cdot \frac{1}{6} - 1 x^{2} \cdot 1 \cdot \frac{1}{2} + x 1 = \frac{x^{5}}{5} - \frac{x^{4}}{4} + \frac{x^{3}}{3} - \frac{x^{2}}{2} + x \]
Assemble the polynomial.✓ Proved
Answer \( p_{5}(x) = \frac{x^{5}}{5} - \frac{x^{4}}{4} + \frac{x^{3}}{3} - \frac{x^{2}}{2} + x \)
✓ Nihil obstat Lines: 7 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the Maclaurin polynomial of degree 5 for \( \displaystyle f(x) = \sin{\left(x \right)} \).
Show the first step
Show the whole solution
p_n(x) = Σ f⁽ᵏ⁾(0)/k! · xᵏ, for k = 0 to n.
Reviewed
\[ \left. \sin{\left(x \right)} \right|_{\substack{ x=0 }} = 0 \]
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
\[ \left. \frac{d^{4}}{d x^{4}} \sin{\left(x \right)} \right|_{\substack{ x=0 }} = 0 \]
f⁽4⁾(0).✓ Proved
\[ \left. \frac{d^{5}}{d x^{5}} \sin{\left(x \right)} \right|_{\substack{ x=0 }} = 1 \]
f⁽5⁾(0).✓ Proved
\[ x^{5} \cdot 1 \cdot \frac{1}{120} - 1 x^{3} \cdot 1 \cdot \frac{1}{6} + x 1 = \frac{x^{5}}{120} - \frac{x^{3}}{6} + x \]
Assemble the polynomial.✓ Proved
Answer \( p_{5}(x) = \frac{x^{5}}{120} - \frac{x^{3}}{6} + x \)
✓ Nihil obstat Lines: 7 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the Maclaurin polynomial of degree 5 for \( \displaystyle f(x) = \cos{\left(x \right)} \).
Show the first step
Show the whole solution
p_n(x) = Σ f⁽ᵏ⁾(0)/k! · xᵏ, for k = 0 to n.
Reviewed
\[ \left. \cos{\left(x \right)} \right|_{\substack{ x=0 }} = 1 \]
f⁽0⁾(0).✓ Proved
\[ \left. \frac{d}{d x} \cos{\left(x \right)} \right|_{\substack{ x=0 }} = 0 \]
f⁽1⁾(0).✓ Proved
\[ \left. \frac{d^{2}}{d x^{2}} \cos{\left(x \right)} \right|_{\substack{ x=0 }} = -1 \]
f⁽2⁾(0).✓ Proved
\[ \left. \frac{d^{3}}{d x^{3}} \cos{\left(x \right)} \right|_{\substack{ x=0 }} = 0 \]
f⁽3⁾(0).✓ Proved
\[ \left. \frac{d^{4}}{d x^{4}} \cos{\left(x \right)} \right|_{\substack{ x=0 }} = 1 \]
f⁽4⁾(0).✓ Proved
\[ \left. \frac{d^{5}}{d x^{5}} \cos{\left(x \right)} \right|_{\substack{ x=0 }} = 0 \]
f⁽5⁾(0).✓ Proved
\[ x^{4} \cdot 1 \cdot \frac{1}{24} - 1 x^{2} \cdot 1 \cdot \frac{1}{2} + 1 = \frac{x^{4}}{24} - \frac{x^{2}}{2} + 1 \]
Assemble the polynomial.✓ Proved
Answer \( p_{5}(x) = \frac{x^{4}}{24} - \frac{x^{2}}{2} + 1 \)
✓ Nihil obstat Lines: 7 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the Maclaurin polynomial of degree 5 for \( \displaystyle f(x) = e^{x} \).
Show the first step
Show the whole solution
p_n(x) = Σ f⁽ᵏ⁾(0)/k! · xᵏ, for k = 0 to n.
Reviewed
\[ \left. e^{x} \right|_{\substack{ x=0 }} = 1 \]
f⁽0⁾(0).✓ Proved
\[ \left. \frac{d}{d x} e^{x} \right|_{\substack{ x=0 }} = 1 \]
f⁽1⁾(0).✓ Proved
\[ \left. \frac{d^{2}}{d x^{2}} e^{x} \right|_{\substack{ x=0 }} = 1 \]
f⁽2⁾(0).✓ Proved
\[ \left. \frac{d^{3}}{d x^{3}} e^{x} \right|_{\substack{ x=0 }} = 1 \]
f⁽3⁾(0).✓ Proved
\[ \left. \frac{d^{4}}{d x^{4}} e^{x} \right|_{\substack{ x=0 }} = 1 \]
f⁽4⁾(0).✓ Proved
\[ \left. \frac{d^{5}}{d x^{5}} e^{x} \right|_{\substack{ x=0 }} = 1 \]
f⁽5⁾(0).✓ Proved
\[ x^{5} \cdot 1 \cdot \frac{1}{120} + x^{4} \cdot 1 \cdot \frac{1}{24} + x^{3} \cdot 1 \cdot \frac{1}{6} + x^{2} \cdot 1 \cdot \frac{1}{2} + x 1 + 1 = \frac{x^{5}}{120} + \frac{x^{4}}{24} + \frac{x^{3}}{6} + \frac{x^{2}}{2} + x + 1 \]
Assemble the polynomial.✓ Proved
Answer \( p_{5}(x) = \frac{x^{5}}{120} + \frac{x^{4}}{24} + \frac{x^{3}}{6} + \frac{x^{2}}{2} + x + 1 \)
✓ Nihil obstat Lines: 7 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the Maclaurin polynomial of degree 5 for \( \displaystyle f(x) = e^{2 x} \).
Show the first step
Show the whole solution
p_n(x) = Σ f⁽ᵏ⁾(0)/k! · xᵏ, for k = 0 to n.
Reviewed
\[ \left. e^{2 x} \right|_{\substack{ x=0 }} = 1 \]
f⁽0⁾(0).✓ Proved
\[ \left. \frac{d}{d x} e^{2 x} \right|_{\substack{ x=0 }} = 2 \]
f⁽1⁾(0).✓ Proved
\[ \left. \frac{d^{2}}{d x^{2}} e^{2 x} \right|_{\substack{ x=0 }} = 4 \]
f⁽2⁾(0).✓ Proved
\[ \left. \frac{d^{3}}{d x^{3}} e^{2 x} \right|_{\substack{ x=0 }} = 8 \]
f⁽3⁾(0).✓ Proved
\[ \left. \frac{d^{4}}{d x^{4}} e^{2 x} \right|_{\substack{ x=0 }} = 16 \]
f⁽4⁾(0).✓ Proved
\[ \left. \frac{d^{5}}{d x^{5}} e^{2 x} \right|_{\substack{ x=0 }} = 32 \]
f⁽5⁾(0).✓ Proved
\[ 32 x^{5} \cdot 1 \cdot \frac{1}{120} + 16 x^{4} \cdot 1 \cdot \frac{1}{24} + 8 x^{3} \cdot 1 \cdot \frac{1}{6} + 4 x^{2} \cdot 1 \cdot \frac{1}{2} + 2 x 1 + 1 = \frac{4 x^{5}}{15} + \frac{2 x^{4}}{3} + \frac{4 x^{3}}{3} + 2 x^{2} + 2 x + 1 \]
Assemble the polynomial.✓ Proved
Answer \( p_{5}(x) = \frac{4 x^{5}}{15} + \frac{2 x^{4}}{3} + \frac{4 x^{3}}{3} + 2 x^{2} + 2 x + 1 \)
✓ Nihil obstat Lines: 7 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify