Free math worksheets

- Use derivative order \(n+1=3\). - Substitute the distance \(0.2\) into \(\frac{M|x-a|^{n+1}}{(n+1)!}\).

1. Here, \(n=2\), \(M=6\), and \(|x-a|=0.2\). 2. The Lagrange error bound is \( |R_2(x)|\leq\frac{6(0.2)^3}{3!}=0.008\).

\( |R_2(x)|\leq0.008\).

- The remainder uses the fourth derivative because the polynomial has degree \(3\). - Keep the factorial \(4!\) in the denominator.

1. Here, \(n=3\), \(M=4\), and \(|x-1|=0.1\). 2. Therefore, \( |R_3(1.1)|\leq\frac{4(0.1)^4}{4!}=\frac{1}{60{,}000}\approx0.0000166667\).

\( |R_3(1.1)|\leq\frac{1}{60{,}000}\approx0.0000166667\).

- The first omitted derivative is the second derivative. - Square the distance from the center before dividing by \(2!\).

1. Here, \(n=1\), \(M=0.5\), and \(|x-2|=0.3\). 2. Thus, \( |R_1(2.3)|\leq\frac{0.5(0.3)^2}{2!}=0.0225\).

\( |R_1(2.3)|\leq0.0225\).

- Evaluate the polynomial before working with the remainder. - A degree-\(1\) approximation uses a bound on the second derivative. - Add and subtract a rounded-up error bound from the polynomial value.

1. Substituting \(x=0.1\) into the polynomial gives \(P_1(0.1)=1.1\). 2. For a degree-\(1\) Maclaurin polynomial, the remainder uses the second derivative. Since the second derivative of \(e^x\) is \(e^x\), the stated bound is \(M=e^{0.1}\) on \([0, 0.1]\). 3. The Lagrange bound is \( |R_1(0.1)|\leq\frac{e^{0.1}(0.1)^2}{2!}\approx0.00552585<0.005526\). Thus \(0.005526\) is a guaranteed decimal upper bound. 4. Therefore, \( 1.1-0.005526\leq e^{0.1}\leq1.1+0.005526\), so a guaranteed interval is \([1.094474, 1.105526]\).

a) \(1.1\). b) \( |R_1(0.1)|\leq\frac{e^{0.1}(0.1)^2}{2}<0.005526\). c) \(e^{0.1}\in[1.094474, 1.105526]\).

- Differentiate twice before looking for \(M\). - Determine whether \(\frac{1}{x^2}\) increases or decreases across the interval. - Use the full distance from \(2\) to \(1.5\).

1. Since \(f''(x)=-\frac{1}{x^2}\), \( |f''(x)|=\frac{1}{x^2}\). 2. This function decreases as \(x\) increases, so its maximum on \([1.5, 2]\) occurs at \(x=1.5\): \( M=\frac{1}{(1.5)^2}=\frac49\). 3. With \(|1.5-2|=0.5\), \( |R_1(1.5)|\leq\frac{\left(\frac49\right)(0.5)^2}{2!}=\frac{1}{18}\approx0.0555556\).

\( M=\frac49\) at \(x=1.5\), and \( |R_1(1.5)|\leq\frac{1}{18}\approx0.0555556\).

- Differentiate through order \(3\), not just through the polynomial degree. - The smallest positive \(x\) gives the largest value of \(\frac{6}{x^4}\). - Use \(\frac{(0.5)^3}{3!}\) in the remainder formula.

1. The third derivative is \( f'''(x)=-\frac{6}{x^4}\), so \(|f'''(x)|=\frac{6}{x^4}\). 2. This magnitude decreases as \(x\) increases. Therefore, the maximum occurs at \(x=1.5\): \( M=\frac{6}{(1.5)^4}=\frac{32}{27}\). 3. With distance \(0.5\), \( |R_2(1.5)|\leq\frac{\left(\frac{32}{27}\right)(0.5)^3}{3!}=\frac{2}{81}\approx0.0246914\).

\( M=\frac{32}{27}\) at \(x=1.5\), and \( |R_2(1.5)|\leq\frac{2}{81}\approx0.0246914\).

- Evaluate the cubic polynomial first. - Match the polynomial degree with the next derivative order in the remainder formula. - Use the supplied maximum together with the fourth power of the distance from the center.

1. Substitution gives \( P_3(0.2)=0.2-\frac{(0.2)^3}{6}\approx0.1986666667\). 2. For a degree-\(3\) polynomial, the Lagrange remainder uses the fourth derivative. 3. With \(M=0.2\), the error satisfies \( |R_3(0.2)|\leq\frac{0.2(0.2)^4}{4!}=\frac{1}{75000}\approx0.0000133333<0.000013334\).

a) \(\sin(0.2)\approx0.1986666667\). b) \( |R_3(0.2)|\leq\frac{1}{75000}<0.000013334\).

- The remainder derivative order is one greater than the polynomial degree. - The exponent and factorial in the remainder use that same next order. - The exponential function reaches its largest derivative value at the right endpoint.

1. The degree-\(2\) polynomial is \( P_2(x)=1+x+\frac{x^2}{2}\), so \(P_2(0.2)=1.22\). 2. A degree-\(2\) Taylor polynomial requires the third derivative, the third power of the distance, and the divisor \(3!\). The student used the degree-\(1\) remainder pattern. 3. Since every derivative of \(e^x\) is \(e^x\), a valid maximum on \([0, 0.2]\) is \(M=e^{0.2}\). 4. The correct bound is \( |R_2(0.2)|\leq\frac{e^{0.2}(0.2)^3}{3!}\approx0.00162854\).

a) \(1.22\). b) The remainder must use power \(3\) and divisor \(3!\), not power \(2\) and \(2!\). c) \( |R_2(0.2)|\leq\frac{e^{0.2}(0.2)^3}{3!}<0.00162854\).

- Differentiate \(x^{\frac12}\) three times. - Check which endpoint makes the denominator smallest. - The input is one unit from the center.

1. The third derivative is \( f'''(x)=\frac{3}{8x^{\frac52}}\). 2. This positive function decreases as \(x\) increases, so the maximum occurs at \(x=3\): \( M=\frac{3}{8\cdot3^{\frac52}}=\frac{1}{24\sqrt{3}}\). 3. The distance from the center is \(1\), so \( |R_2(3)|\leq\frac{M}{3!}=\frac{1}{144\sqrt{3}}\approx0.00400938\).

\( M=\frac{1}{24\sqrt{3}}\) at \(x=3\), and \( |R_2(3)|\leq\frac{1}{144\sqrt{3}}\approx0.00400938\).

- Identify the derivative order required by a quadratic approximation. - Use the interval itself to bound \(\sin x\). - Form the interval by subtracting and adding a rounded-up remainder bound.

1. The polynomial value is \( P_2(0.4)=1-\frac{(0.4)^2}{2}=0.92\). 2. For \(f(x)=\cos x\), \(f'''(x)=\sin x\). On \([0, 0.4]\), \(0\leq\sin x\leq x\leq0.4\), so \(M=0.4\) is valid. 3. The remainder satisfies \( |R_2(0.4)|\leq\frac{0.4(0.4)^3}{3!}=\frac{8}{1875}\approx0.0042666667\). Rounding this bound up to ten decimal places preserves the guarantee. 4. Therefore, \(\cos(0.4)\) lies in \([0.9157333333, 0.9242666667]\).

a) \(0.92\). b) \(|\sin x|\leq0.4\) on \([0, 0.4]\). c) \( |R_2(0.4)|\leq\frac{8}{1875}<0.0042666667\), so \(\cos(0.4)\in[0.9157333333, 0.9242666667]\).

- Use the signed displacement \(-0.2\) in the polynomial. - Determine how \(6/x^4\) changes as \(x\) moves across the interval. - A valid \(M\) must work at every point between the center and the input. - In part d), compare the center value with the larger derivative magnitude found elsewhere on the interval.

1. At \(x=0.8\), \(x-1=-0.2\). Therefore, \( P_2(0.8)=1+0.2+0.04=1.24\). 2. Since \(f'''(x)=-6/x^4\), its magnitude is \(6/x^4\). This decreases as \(x\) increases, so the maximum on \([0.8, 1]\) is \( M=\frac{6}{(0.8)^4}=14.6484375\). 3. The remainder satisfies \( |R_2(0.8)|\leq\frac{M(0.2)^3}{3!}=0.01953125\). 4. The value \(6\) is only the derivative magnitude at the center. It does not bound \(6/x^4\) at points closer to \(0.8\), where the derivative magnitude is larger.

a) \(1.24\). b) \(M=14.6484375\). c) \( |R_2(0.8)|\leq0.01953125\). d) \(6\) is not an upper bound on the entire interval \([0.8, 1]\).

- Use the fourth derivative because the polynomial has degree \(3\). - Solve the error inequality for the distance from the center. - Confirm that the resulting interval stays inside the interval where the derivative bound is known.

1. For a degree-\(3\) Taylor polynomial, the error bound is \( |R_3(x)|\leq\frac{12|x-2|^4}{4!}=\frac12|x-2|^4\). 2. Require \( \frac12|x-2|^4\leq0.0008\), so \(|x-2|^4\leq0.0016\). 3. Since \((0.2)^4=0.0016\), the condition is \(|x-2|\leq0.2\). 4. This interval lies inside the stated derivative-bound interval, so the largest guaranteed interval is \([1.8, 2.2]\).

\([1.8, 2.2]\).

- Differentiate \(1/(1-x)\) enough times to identify the derivative pattern. - Find where the denominator \((1-x)^{n+2}\) is smallest on the interval. - Check the two consecutive degrees around the first successful bound.

1. Since \( f^{(n+1)}(x)=\frac{(n+1)!}{(1-x)^{n+2}}\), its magnitude increases on \([0, 0.25]\). Therefore, the maximum occurs at \(x=0.25\): \( M=\frac{(n+1)!}{(0.75)^{n+2}}\). 2. The distance from the center is \(0.25\), so \( |R_n(0.25)|\leq\frac{M(0.25)^{n+1}}{(n+1)!}=\frac{(0.25)^{n+1}}{(0.75)^{n+2}}=\frac{4}{3^{n+2}}\). 3. For \(n=9\), \( \frac{4}{3^{11}}\approx0.0000225803>10^{-5}\). For \(n=10\), \( \frac{4}{3^{12}}\approx0.00000752671<10^{-5}\). Thus the smallest degree is \(10\). 4. The approximation is \( P_{10}(0.25)=\sum_{k=0}^{10}(0.25)^k=\frac{1398101}{1048576}\approx1.333333015\).

a) \( M=\frac{(n+1)!}{(0.75)^{n+2}}\), attained at \(x=0.25\). b) \( |f(0.25)-P_n(0.25)|\leq\frac{4}{3^{n+2}}\). c) The least degree is \(n=10\), and \( P_{10}(0.25)=\frac{1398101}{1048576}\approx1.333333015\).

- The input to the Maclaurin polynomial is \(0.1\), not \(1.1\). - Determine where \(\frac{2}{(1+x)^3}\) is largest on the interval. - Add and subtract the exact remainder bound from the approximation.

1. The approximation is \( P_2(0.1)=0.1-\frac{(0.1)^2}{2}=0.095\). 2. Since \( f'''(x)=\frac{2}{(1+x)^3}\), its maximum on \([0, 0.1]\) occurs at \(x=0\). Thus \(M=2\). 3. The remainder satisfies \( |R_2(0.1)|\leq\frac{2(0.1)^3}{3!}=\frac{1}{3000}\approx0.0003333333\). 4. Therefore, \( \ln(1.1)\in\left[\frac{71}{750}, \frac{143}{1500}\right]\).

a) \(0.095\). b) \(M=2\). c) Error at most \( \frac{1}{3000}\); interval \( \left[\frac{71}{750}, \frac{143}{1500}\right]\).

- Each Taylor polynomial must be written in powers of its own displacement from the center. - The same derivative bound leaves the distance term as the decisive difference. - Compare \((0.9)^3\) with \((0.1)^3\).

1. Centered at \(0\), \( P_{2,0}(0.9)=1+0.9+\frac{0.9^2}{2}=2.305\). 2. Centered at \(1\), the polynomial is \( P_{2,1}(x)=e\left[1+(x-1)+\frac{(x-1)^2}{2}\right]\). Thus \( P_{2,1}(0.9)=0.905e\approx2.460045\). 3. The Maclaurin error bound is \( \frac{e(0.9)^3}{3!}\approx0.330271\). 4. The error bound from center \(1\) is \( \frac{e(0.1)^3}{3!}\approx0.000453\). 5. The center \(1\) is much better because \(0.9\) is only \(0.1\) unit from that center.

a) \(2.305\). b) \(0.905e\approx2.460045\). c) Bounds: approximately \(0.330271\) from center \(0\) and \(0.000453\) from center \(1\); center \(1\) is better.

- Build the tangent-line approximation from the value and slope at \(4\). - Read whether the displayed derivative magnitude rises or falls across the interval. - The distance from \(3.5\) to the center is \(0.5\).

1. Since \(f(4)=2\) and \(f'(4)=1/4\), \( P_1(x)=2+\frac14(x-4)\). Thus \( P_1(3.5)=1.875\). 2. The graph shows that \(m(x)\) decreases across the interval, so its maximum occurs at \(x=3.5\). 3. Therefore, \( M=\frac{1}{4(3.5)^{3/2}}\approx0.038180\). 4. The Lagrange bound is \( |R_1(3.5)|\leq\frac{M(0.5)^2}{2!}\approx0.004773\).

a) \(\sqrt{3.5}\approx1.875\). b) The maximum occurs at \(x=3.5\), with \(M\approx0.038180\). c) Error at most \(0.004773\).

- Build each tangent-line polynomial using the value and derivative at its own center. - The second derivative magnitude is largest at the smaller positive input in each interval. - Compare the final numerical bounds, not only the distances from the centers.

1. Centered at \(1\), the linear polynomial is \(P_{1,1}(x)=x-1\), so \(P_{1,1}(1.5)=0.5\). 2. Since \(|f''(x)|=\frac{1}{x^2}\), the maximum on \([1, 1.5]\) is \(M=1\). The bound is \( |R_{1,1}|\leq\frac{1(0.5)^2}{2}=0.125\). 3. Centered at \(2\), the polynomial is \( P_{1,2}(x)=\ln 2+\frac12(x-2)\), so \(P_{1,2}(1.5)=\ln 2-0.25\). 4. On \([1.5, 2]\), the maximum of \(\frac{1}{x^2}\) is \(\frac{4}{9}\). Thus \( |R_{1,2}|\leq\frac{(4/9)(0.5)^2}{2}=\frac{1}{18}\approx0.055556\). 5. Center \(2\) gives the smaller guaranteed error.

a) Approximation \(0.5\), error at most \(0.125\). b) Approximation \(\ln 2-0.25\), error at most \( \frac{1}{18}\approx0.055556\). c) Center \(2\).

- Evaluate the cubic polynomial before applying the remainder formula. - Form the full guaranteed interval by adding and subtracting the error bound. - Compare that interval with the rounding cutoffs, not only with the central approximation.

1. The polynomial value is \( P_3(0.3)=0.3-\frac{0.3^3}{6}=0.2955\). 2. The Lagrange error bound is \( |R_3(0.3)|\leq\frac{0.3(0.3)^4}{4!}=0.00010125\). Therefore, \( 0.29539875\leq\sin(0.3)\leq0.29560125\). 3. Every number in this interval rounds to \(0.30\) to the nearest hundredth, so that rounding is guaranteed. 4. The interval crosses the midpoint \(0.2955\) between \(0.295\) and \(0.296\), so rounding to the nearest thousandth is not guaranteed.

a) \( P_3(0.3)=0.2955\), with error at most \(0.00010125\). b) Yes. The guaranteed hundredth is \(0.30\). c) No. The guaranteed interval crosses the thousandth-rounding boundary \(0.2955\).

- Write the absolute value of the next derivative and find its maximum on the entire interval. - The distance from the center is largest at the right endpoint. - Check the first degree that succeeds and the degree immediately below it.

1. The next derivative is \( f^{(n+1)}(x)=(-1)^n\frac{n!}{(1+x)^{n+1}}\). 2. On \([0,0.25]\), its absolute value is at most \(n!\), so the Lagrange bound gives \( |R_n(x)|\leq\frac{n!|x|^{n+1}}{(n+1)!}=\frac{|x|^{n+1}}{n+1}\). 3. The largest bound on the interval occurs at \(x=0.25\). 4. For \(n=2\), \( \frac{0.25^3}{3}\approx0.00520833\), which is too large. 5. For \(n=3\), \( \frac{0.25^4}{4}=0.0009765625\), which is below \(0.001\). 6. Therefore, the least degree is \(3\).

\(n=3\).

- The order of the remainder depends on the Taylor degree, including a degree with coefficient \(0\). - Use \(4!\) for the degree-\(3\) bound and \(5!\) for the degree-\(4\) bound. - Compare both the power of \(0.3\) and the factorial in the two expressions.

1. As a degree-\(3\) Taylor polynomial, the remainder uses the fourth derivative. The given bound gives \( |R_3(0.3)|\leq\frac{0.3(0.3)^4}{4!}=0.00010125\). 2. As a degree-\(4\) Taylor polynomial, the remainder uses the fifth derivative. The given bound gives \( |R_4(0.3)|\leq\frac{(0.3)^5}{5!}=0.00002025\). 3. The coefficient of \(x^4\) in the sine series is \(0\), so the degree-\(4\) polynomial has the same displayed terms as the degree-\(3\) polynomial. 4. Declaring the correct higher degree moves the remainder to the fifth derivative and produces the tighter valid bound.

a) \( |R_3(0.3)|\leq0.00010125\). b) \( |R_4(0.3)|\leq0.00002025\). c) The fourth-degree coefficient is zero, but the degree-\(4\) remainder begins with the fifth derivative.

- Evaluate the same linear Taylor model at both endpoints. - Add and subtract the full remainder bound to determine the sign at each endpoint. - Once opposite signs are guaranteed, apply the Intermediate Value Theorem.

1. The degree-\(1\) Taylor polynomial centered at \(a\) is \( P_1(a+h)=0.01-0.2h\). 2. At the left endpoint, \(h=-0.1\), so \( P_1(a-0.1)=0.03\). 3. At the right endpoint, \(h=0.1\), so \( P_1(a+0.1)=-0.01\). 4. At either endpoint, the Lagrange error is at most \( \frac{1(0.1)^2}{2!}=0.005\). 5. Therefore, \( f(a-0.1)\in[0.025, 0.035]\), so it is positive, while \( f(a+0.1)\in[-0.015, -0.005]\), so it is negative. 6. Since \(f\) is continuous and changes sign, the Intermediate Value Theorem guarantees at least one zero in \((a-0.1, a+0.1)\).

The Lagrange bounds guarantee \(f(a-0.1)>0\) and \(f(a+0.1)<0\). Therefore, continuity guarantees at least one zero in \((a-0.1, a+0.1)\).

- The sign of the next derivative determines whether the Taylor polynomial is above or below the function. - Build a one-sided interval around each polynomial value rather than a symmetric interval. - Compare the widths of the two valid intervals.

1. For \(f(x)=\ln(1+x)\), the degree-\(1\) polynomial is \(P_1(x)=x\), so \(P_1(0.2)=0.2\). 2. Because \(f''(x)<0\), the remainder after \(P_1\) is negative. Its magnitude is at most \( \frac{1(0.2)^2}{2!}=0.02\). Therefore, \( 0.18\leq\ln(1.2)\leq0.20\). 3. The degree-\(2\) polynomial is \( P_2(x)=x-\frac{x^2}{2}\), so \(P_2(0.2)=0.18\). 4. Because \(f'''(x)>0\), the remainder after \(P_2\) is positive. Its magnitude is at most \( \frac{2(0.2)^3}{3!}=0.0026666667\). 5. Therefore, \( 0.18\leq\ln(1.2)\leq0.1826666667\), which is the tighter interval.

a) \( 0.18\leq\ln(1.2)\leq0.20\). b) \( 0.18\leq\ln(1.2)\leq0.1826666667\). c) Tighter interval: \([0.18, 0.1826666667]\).

- Use the given derivative values to build the polynomial before comparing it with the actual value. - In this problem, the remainder is known rather than merely bounded. - Rearrange the Lagrange remainder equation to solve for the derivative value.

1. The degree-\(2\) Maclaurin polynomial is \( P_2(x)=1-2x+\frac32x^2\). 2. At \(x=0.2\), \( P_2(0.2)=1-0.4+\frac32(0.04)=0.66\). The actual remainder is \( R_2(0.2)=0.6624-0.66=0.0024\). 3. Taylor's theorem says that, for some \(\xi\in(0, 0.2)\), \( R_2(0.2)=\frac{f'''(\xi)}{3!}(0.2)^3\). Therefore, \( f'''(\xi)=\frac{6(0.0024)}{(0.2)^3}=1.8\). 4. Since the third derivative equals \(1.8\) somewhere in the interval, any valid uniform bound for \(|f'''|\) must be at least \(1.8\). Thus \(1.5\) cannot be valid.

a) \(P_2(0.2)=0.66\); actual error \(0.0024\). b) There is a \(\xi\in(0, 0.2)\) such that \(f'''(\xi)=1.8\). c) Any valid bound must satisfy \(M\geq1.8\), so \(M=1.5\) is impossible.

- Isolate the exponential containing the unknown point before taking a logarithm. - Check that each resulting value lies between \(0\) and \(1\). - Distinguish the existence statement in Taylor's theorem from the uniform maximum used to create a bound.

1. From the first remainder equation, \( e^{\xi_1}=2(e-2)\), so \( \xi_1=\ln[2(e-2)]\approx0.3623\). 2. From the second equation, \( e^{\xi_2}=6\left(e-\frac52\right)\), so \( \xi_2=\ln\left[6\left(e-\frac52\right)\right]\approx0.2698\). 3. The exact Lagrange remainder uses the derivative at some intermediate point determined by the actual error. A standard bound replaces that unknown value with a maximum valid on the whole interval. On \([0,1]\), that maximum is \(e\), attained at the endpoint rather than at either intermediate point.

a) \( \xi_1=\ln[2(e-2)]\approx0.3623\). b) \( \xi_2=\ln\left[6\left(e-\frac52\right)\right]\approx0.2698\). c) The exact remainder uses an interior derivative value; the error bound uses a worst-case maximum over the interval.

- Find the farthest point from each center on its assigned interval. - Keep the derivative bound the same in both comparisons. - Use the exponent on \(|x-c|\) to compare the two guarantees without decimal approximations.

1. For a cubic Taylor polynomial centered at \(c\), \( |R_3(x)|\leq\frac{e|x-c|^4}{4!}\). 2. For Strategy A, the greatest distance from \(0.5\) is \(0.5\). Thus \( |R_3(x)|\leq\frac{e(0.5)^4}{24}=\frac{e}{384}\). 3. For each half of Strategy B, the greatest distance from its center is \(0.25\). Thus \( |R_3(x)|\leq\frac{e(0.25)^4}{24}=\frac{e}{6144}\). 4. The improvement factor is \( \frac{e/384}{e/6144}=16\). 5. Strategy B halves the maximum distance to a center. Since the remainder contains the fourth power of that distance, the guarantee is multiplied by \((1/2)^4=1/16\).

a) Strategy A: \( \frac{e}{384}\). Strategy B: \( \frac{e}{6144}\). b) Strategy B improves the guarantee by a factor of \(16\). c) The maximum distance is halved, and a cubic Taylor remainder depends on the fourth power of the distance.

- Use \(M=1\) for the third derivative on the whole interval. - The worst-case distance from the center occurs at an endpoint. - Balance the two endpoint distances before constructing the polynomial.

1. The third derivative of \(\sin x\) is \(-\cos x\), and \(|-\cos x|\leq1\) on the interval. The Lagrange bound is therefore \( |R_2(x)|\leq\frac{|x-c|^3}{3!}\). 2. The farthest point in \(\left[0, \frac{\pi}{2}\right]\) from \(c\) is one of the endpoints. Thus \( |x-c|\leq\max\left(c,\frac{\pi}{2}-c\right)\), which proves the stated uniform bound. 3. The larger endpoint distance is smallest when the two endpoint distances are equal: \( c=\frac{\pi}{2}-c\). Hence the best center is \(c=\frac{\pi}{4}\). 4. The minimum guaranteed error is \( \frac{\left(\frac{\pi}{4}\right)^3}{6}=\frac{\pi^3}{384}\approx0.08075\). 5. Since \(\sin\left(\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}\), \( P_{2,\frac{\pi}{4}}(x)=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\left(x-\frac{\pi}{4}\right)-\frac{\sqrt2}{4}\left(x-\frac{\pi}{4}\right)^2\). 6. Because \(0.08075<0.081\), this bound guarantees error below \(0.081\) throughout the interval.

a) \( |\sin x-P_{2,c}(x)|\leq\frac{\left[\max\left(c,\frac{\pi}{2}-c\right)\right]^3}{6}\). b) \( c=\frac{\pi}{4}\); minimum bound \( \frac{\pi^3}{384}\approx0.08075\). c) \( P_{2,\frac{\pi}{4}}(x)=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}\left(x-\frac{\pi}{4}\right)-\frac{\sqrt2}{4}\left(x-\frac{\pi}{4}\right)^2\). Yes, the error is guaranteed to be below \(0.081\).

- Apply Taylor's theorem at the starting value but evaluate it at the unknown zero. - Substitute the Newton formula before estimating the remainder term. - Use the given bracket to control both the distance and the derivative ratio.

1. Since \(f'(x)=e^x\), \( x_1=0.70-\frac{e^{0.70}-2}{e^{0.70}}=-0.30+2e^{-0.70}\approx0.6931706\). 2. Taylor's theorem gives, for some \(\xi\) between \(\alpha\) and \(x_0\), \( 0=f(\alpha)=f(x_0)+f'(x_0)(\alpha-x_0)+\frac{f''(\xi)}{2}(\alpha-x_0)^2\). 3. By the definition of \(x_1\), \( x_1-\alpha=\frac{f''(\xi)}{2f'(x_0)}(\alpha-x_0)^2\). 4. Here \(f''(\xi)=e^\xi>0\), so \(x_1\geq\alpha\). Also, \(\xi\leq0.70\), so \( \frac{f''(\xi)}{f'(x_0)}=\frac{e^\xi}{e^{0.70}}\leq1\). Because \(|\alpha-x_0|<0.01\), \( 0\leq x_1-\alpha\leq\frac12(0.01)^2=5\times10^{-5}\). 5. Therefore, \( \alpha\in[x_1-0.00005, x_1]\subset[0.6931206, 0.6931707]\).

a) \( x_1=-0.30+2e^{-0.70}\approx0.6931706\). b) \( 0\leq x_1-\alpha\leq5\times10^{-5}\). c) \( \alpha\in[x_1-0.00005, x_1]\subset[0.6931206, 0.6931707]\); \(x_1\) is an overestimate.

- Turn the absolute derivative bound into the one-sided remainder estimate needed for a lower bound. - Minimize the polynomial lower bound rather than the unknown function. - A positive lower bound on the entire interval rules out every zero.

1. The degree-\(2\) Maclaurin polynomial is \( P_2(x)=1-2x+2x^2\). 2. Taylor's theorem gives \( f(x)=P_2(x)+\frac{f'''(\xi)}{3!}x^3\) for some \(\xi\in(0,x)\). Since \(f'''(\xi)\geq-3\), \( f(x)\geq1-2x+2x^2-\frac{x^3}{2}\). 3. Let \( q(x)=1-2x+2x^2-\frac{x^3}{2}\). Then \( q'(x)=-2+4x-\frac32x^2\). The zeros of \(q'\) are \(2/3\) and \(2\), both outside \([0, 0.5]\). Since \(q'(0)<0\), the polynomial decreases on the whole interval. 4. Its minimum is therefore \( q(0.5)=\frac{7}{16}\). Hence \( f(x)\geq\frac{7}{16}>0\) on \([0, 0.5]\), so \(f\) has no zero there.

a) \( P_2(x)=1-2x+2x^2\). b) \( f(x)\geq1-2x+2x^2-\frac{x^3}{2}\). c) The lower bound is at least \(\frac{7}{16}>0\), so \(f\) has no zero on \([0, 0.5]\).

- Locate the maximum derivative magnitude on the entire interval from the center to the evaluation point. - For the direct tail, replace every omitted denominator by the smallest one in the tail. - A bound that fails to approach \(0\) is inconclusive; it is not evidence that the actual error fails to approach \(0\).

1. For \(f(x)=\ln(1+x)\), \( |f^{(N+1)}(t)|=\frac{N!}{(1+t)^{N+1}}\). On \([-0.8, 0]\), this is largest at \(t=-0.8\), so \( M=N!5^{N+1}\). 2. The Lagrange bound is \( |R_N(-0.8)|\leq\frac{M(0.8)^{N+1}}{(N+1)!}=\frac{4^{N+1}}{N+1}\). 3. At \(x=-0.8\), all omitted terms have the same sign, so \( |R_N(-0.8)|=\sum_{n=N+1}^{\infty}\frac{0.8^n}{n}\). Since \(\frac1n\leq\frac{1}{N+1}\) for every omitted term, \( |R_N(-0.8)|\leq\frac{1}{N+1}\sum_{n=N+1}^{\infty}0.8^n=\frac{5(0.8)^{N+1}}{N+1}\). 4. The Lagrange bound grows and therefore is too weak to establish convergence. A valid upper bound need not be sharp. The direct tail bound approaches \(0\), so it proves that the Taylor partial sums converge to \(\ln(0.2)\).

a) \( |R_N(-0.8)|\leq\frac{4^{N+1}}{N+1}\). b) \( |R_N(-0.8)|\leq\frac{5(0.8)^{N+1}}{N+1}\to0\). c) The first bound is valid but too large; the second tends to \(0\) and therefore establishes convergence.

- Add a cubic term so the value and first two derivatives at the center remain unchanged. - Choose the cubic coefficient to make the third derivative equal to the allowed maximum magnitude. - A universal bound is sharp when admissible examples attain both ends of its interval.

1. The Lagrange bound is \( |f(0.2)-P_2(0.2)|\leq\frac{6(0.2)^3}{3!}=0.008\). 2. For \(f_\pm(x)=P_2(x)\pm ax^3\), the third derivative of the added term is \(\pm6a\). To attain the allowed magnitude \(6\), choose \(a=1\). 3. Thus, \( f_+(x)=1-x+x^2+x^3\) and \( f_-(x)=1-x+x^2-x^3\). At \(x=0.2\), their errors are \(0.008\) and \(-0.008\), respectively. 4. Since admissible functions attain both endpoints of the error interval, any smaller universal bound would fail for at least one of them.

a) \( |f(0.2)-P_2(0.2)|\leq0.008\). b) \(a=1\), so \( f_\pm(x)=1-x+x^2\pm x^3\); their exact errors are \(\pm0.008\). c) These functions attain equality, so the bound cannot be reduced without additional information.

- Differentiate the given local polynomial before applying an error bound. - Find the worst-case lower estimate at the largest distance from the center. - Use the mean value theorem to turn a derivative lower bound into a function-difference bound.

1. Differentiating the given Taylor polynomial gives \( Q_2(x)=1-\frac{x^2}{2}\) as the degree-\(2\) Maclaurin polynomial for \(f'\). 2. The next derivative for this Taylor approximation is \(f^{(4)}\), so \( |f'(x)-Q_2(x)|\leq\frac{|x|^3}{6}\). Therefore, \( f'(x)\geq1-\frac{x^2}{2}-\frac{|x|^3}{6}\). 3. On \(|x|\leq\frac12\), the right side is at least \( 1-\frac18-\frac1{48}=\frac{41}{48}>0\). Thus \(f\) is strictly increasing on the interval. 4. By the mean value theorem, for some \(c\in\left(-\frac12, \frac12\right)\), \( f\left(\frac12\right)-f\left(-\frac12\right)=f'(c)\). Hence \( f\left(\frac12\right)-f\left(-\frac12\right)\geq\frac{41}{48}\).

a) \( Q_2(x)=1-\frac{x^2}{2}\). b) \( f'(x)\geq1-\frac{x^2}{2}-\frac{|x|^3}{6}\geq\frac{41}{48}>0\). c) \(f\) is strictly increasing, and \( f\left(\frac12\right)-f\left(-\frac12\right)\geq\frac{41}{48}\).

- Find the sign and largest possible value of the third derivative on the interval from \(0\) to \(x\). - Apply the logarithm bound to \(x=\frac1n\) before using the exponential function. - For the limit, compare the two exponents rather than expanding the original power directly. - For part d), show that both bounding exponents approach \(1\) before exponentiating.

1. For \(f(x)=\ln(1+x)\), the degree-\(2\) Maclaurin polynomial is \(P_2(x)=x-\frac{x^2}{2}\). The first omitted derivative is \(f'''(t)=\frac{2}{(1+t)^3}\). 2. When \(x\geq0\) and \(0\leq t\leq x\), \(0<f'''(t)\leq2\). Therefore, the Lagrange remainder satisfies \(0\leq R_2(x)\leq\frac{2x^3}{3!}=\frac{x^3}{3}\). Adding this remainder to \(P_2(x)\) proves \(x-\frac{x^2}{2}\leq\ln(1+x)\leq x-\frac{x^2}{2}+\frac{x^3}{3}\). 3. Substitute \(x=\frac1n\) and multiply every part by \(n\): \(1-\frac{1}{2n}\leq n\ln\left(1+\frac1n\right)\leq1-\frac{1}{2n}+\frac{1}{3n^2}\). Since the exponential function is increasing, exponentiating gives \(e^{1-\frac{1}{2n}}\leq\left(1+\frac1n\right)^n\leq e^{1-\frac{1}{2n}+\frac{1}{3n^2}}\). 4. For \(n=20\), the exact interval is \(\left[e^{\frac{39}{40}}, e^{\frac{1171}{1200}}\right]\). Numerically, this is approximately \([2.651167, 2.653378]\), using outward rounding. 5. As \(n\to\infty\), both bounding exponents approach \(1\). Therefore, both bounding expressions approach \(e\). The squeeze theorem gives \(\left(1+\frac1n\right)^n\to e\).

a) \(x-\frac{x^2}{2}\leq\ln(1+x)\leq x-\frac{x^2}{2}+\frac{x^3}{3}\) for \(x\geq0\). b) \(e^{1-\frac{1}{2n}}\leq\left(1+\frac1n\right)^n\leq e^{1-\frac{1}{2n}+\frac{1}{3n^2}}\). c) \(\left(1+\frac1{20}\right)^{20}\in\left[e^{\frac{39}{40}}, e^{\frac{1171}{1200}}\right]\approx[2.651167, 2.653378]\). d) Both bounds approach \(e\), so the sequence between them also approaches \(e\).