Lagrange Error Bound - AP Calculus BC

  1. Evaluate R2(x)R_2(x) for f(x)=exf(x) = e^x at x=0.2x = 0.2, a=0a = 0.
    R2(0.2)=e0.2(0.2)33!R_2(0.2) = \frac{e^{0.2} (0.2)^3}{3!}. The third derivative of exe^x is exe^x, maximum at x=0.2x = 0.2.
  2. What is the error bound for f(x)=x4f(x) = x^4 at x=0.1x = 0.1, a=0a = 0, n=3n = 3?
    R3(0.1)=M(0.1)44!R_3(0.1) = \frac{M(0.1)^4}{4!}. The fourth derivative of x4x^4 is 24, which is constant.
  3. What is the purpose of Lagrange Error Bound?
    Estimates the error of a Taylor polynomial approximation. This provides an upper bound on the approximation error.
  4. What must be true about f(n+1)(x)f^{(n+1)}(x) for Lagrange Error Bound?
    It must be continuous on the interval. Continuity ensures the maximum value MM exists.
  5. Compute R1(x)R_1(x) for f(x)=x2f(x) = x^2 at x=0.1x = 0.1, a=0a = 0.
    R1(0.1)=M(0.1)22R_1(0.1) = \frac{M(0.1)^2}{2}. The second derivative of x2x^2 is constant 2, so M=2M = 2.
  6. Compute MM for f(x)=sin⁡(x)f(x) = \sin(x) over [0,π4][0, \frac{\pi}{4}].
    M=1M = 1. All derivatives of sin⁡(x)\sin(x) have absolute value at most 1.
  7. Specify the degree of polynomial for R4(x)R_4(x) in Lagrange Error Bound.
    Fourth-degree polynomial. This refers to the Taylor polynomial of degree 4.
  8. What is required for f(x)f(x) in the Lagrange Error Bound context?
    f(x)f(x) must be (n+1)(n+1) times differentiable. This ensures the (n+1)(n+1)-th derivative exists for the error bound.
  9. What is the role of cc in the Lagrange Error Bound?
    cc is some value in the interval between aa and xx. This is where the (n+1)(n+1)-th derivative is evaluated in the error formula.
  10. Identify the variable xx in the Lagrange Error Bound formula.
    The point at which we evaluate the error. This is where we want to estimate the Taylor polynomial error.
  11. What is the maximum error for f(x)=x3f(x) = x^3 at x=0.1x = 0.1, a=0a = 0, n=2n = 2?
    R2(0.1)=M(0.1)33!R_2(0.1) = \frac{M(0.1)^3}{3!}. The third derivative of x3x^3 is 6, which is constant.
  12. Calculate R1(x)R_1(x) for f(x)=xf(x) = \sqrt{x} at x=0.5x = 0.5, a=0.4a = 0.4.
    R1(0.5)=M(0.1)22R_1(0.5) = \frac{M(0.1)^2}{2}. Distance is ∣0.5−0.4∣=0.1|0.5 - 0.4| = 0.1 from center a=0.4a = 0.4.
  13. Identify the purpose of the Taylor polynomial in Lagrange Error Bound.
    Approximates the function near aa. Taylor polynomials give local approximations centered at point aa.
  14. Evaluate R3(x)R_3(x) for f(x)=ln⁡(x)f(x) = \ln(x) at x=1.5x = 1.5, a=1a = 1.
    R3(1.5)=M(0.5)44!R_3(1.5) = \frac{M(0.5)^4}{4!}. Distance is ∣1.5−1∣=0.5|1.5 - 1| = 0.5 from the center a=1a = 1.
  15. Determine the significance of ∣x−a∣|x-a| in Lagrange Error Bound.
    Distance from center affects error size. Larger distances from the center increase the error bound exponentially.
  16. Identify MM for f(x)=cos⁡(x)f(x) = \cos(x) over [0, \frac{{\pi}{3}].
    M=1M = 1. All derivatives of cos⁡(x)\cos(x) have absolute value at most 1.
  17. How does the degree nn affect the Lagrange Error Bound?
    Higher nn provides better approximation. More terms in the polynomial reduce the approximation error.
  18. Calculate R1(x)R_1(x) for f(x)=ln⁡(1+x)f(x) = \ln(1+x) at x=0.1x = 0.1, a=0a = 0.
    R1(0.1)=M(0.1)22R_1(0.1) = \frac{M(0.1)^2}{2}. The second derivative of ln⁡(1+x)\ln(1+x) gives the maximum MM.
  19. Describe the behavior of Rn(x)R_n(x) as nn increases.
    Approaches zero, improving approximation. Higher-degree polynomials provide increasingly accurate approximations.
  20. Find R2(x)R_2(x) for f(x)=exf(x) = e^x at x=0.5x = 0.5, a=0a = 0, n=2n = 2.
    R2(0.5)=e0.5(0.5)33!R_2(0.5) = \frac{e^{0.5} (0.5)^3}{3!}. Since f′′′(x)=exf'''(x) = e^x, the maximum on [0,0.5][0, 0.5] is e0.5e^{0.5}.
  21. What is the significance of (n+1)!(n+1)! in Lagrange Error Bound?
    Factorial of (n+1)(n+1) in the denominator of the error term. This factorial makes the error bound decrease rapidly as nn increases.
  22. Describe the condition for xx in the Lagrange Error Bound.
    xx must be close to aa for best approximation. Closer values to aa give smaller error bounds.
  23. What is the general form of Rn(x)R_n(x)?
    Error term in Taylor approximation. This measures how much the polynomial differs from the function.
  24. Find MM for f(x)=ln⁡(x)f(x) = \ln(x) over [1,2][1, 2].
    M=11M = \frac{1}{1}. For ln⁡(x)\ln(x), the second derivative −1x2-\frac{1}{x^2} has maximum at x=1x = 1.
  25. What is the maximum error for f(x)=sin⁡(x)f(x) = \sin(x) at x=0.1x = 0.1, a=0a = 0, n=3n = 3?
    R3(0.1)=M(0.1)44!R_3(0.1) = \frac{M(0.1)^4}{4!}. The fourth derivative of sin⁡(x)\sin(x) is sin⁡(x)\sin(x), with maximum 1.
  26. State the interval condition for Lagrange Error Bound.
    xx must be within the interval where MM is maximum. This ensures MM is well-defined on the interval containing xx and aa.
  27. Define the term MM in the Lagrange Error Bound formula.
    Maximum value of ∣f(n+1)(c)∣|f^{(n+1)}(c)| on the interval. This is the maximum absolute value of the (n+1)(n+1)-th derivative.
  28. What does nn represent in the Lagrange Error Bound formula?
    Degree of the Taylor polynomial. This determines how many terms are in the Taylor polynomial.
  29. Find MM for f(x)=exf(x) = e^x over [0,2][0, 2].
    M=e2M = e^2. The function exe^x is increasing, so maximum occurs at x=2x = 2.
  30. What is the error estimate for f(x)=cos⁡(x)f(x) = \cos(x) with T3(x)T_3(x) at x=0.2x = 0.2?
    R3(0.2)=M(0.2)44!R_3(0.2) = \frac{M(0.2)^4}{4!}. The fourth derivative of cos⁡(x)\cos(x) is cos⁡(x)\cos(x), with maximum MM.
  31. What type of function is f(x)f(x) in Lagrange Error Bound?
    Differentiable function. The function must have enough derivatives for the error bound.
  32. What is needed to calculate MM in Lagrange Error Bound?
    The derivative f(n+1)f^{(n+1)} over the interval. We need the maximum of ∣f(n+1)(x)∣|f^{(n+1)}(x)| over the interval.
  33. What is the error estimate for f(x)=cos⁡(x)f(x) = \cos(x) with T3(x)T_3(x) at x=0.2x = 0.2?
    R3(0.2)=M(0.2)44!R_3(0.2) = \frac{M(0.2)^4}{4!}. The fourth derivative of cos⁡(x)\cos(x) is cos⁡(x)\cos(x), with maximum MM.
  34. Determine R2(x)R_2(x) for f(x)=sin⁡(x)f(x) = \sin(x) at x=0.1x = 0.1, a=0a = 0.
    R2(0.1)=M(0.1)33!R_2(0.1) = \frac{M(0.1)^3}{3!}. The third derivative of sin⁡(x)\sin(x) is −cos⁡(x)-\cos(x), with maximum 1.
  35. Which derivative is used in the Lagrange Error Bound?
    The (n+1)(n+1)-th derivative of the function. This derivative determines the maximum value MM in the error bound.
  36. What does the Lagrange Error Bound quantify?
    The error of a Taylor polynomial approximation. This gives the maximum possible difference from the true function value.
  37. What does the error bound depend on in Lagrange Error Bound?
    The (n+1)(n+1)-th derivative and distance ∣x−a∣|x-a|. Higher derivatives and distance from center increase the error.
  38. What is Rn(x)R_n(x) in the context of Lagrange Error Bound?
    The remainder or error term of Taylor polynomial. This represents the difference between f(x)f(x) and its polynomial approximation.
  39. Identify the variable aa in the Lagrange Error Bound formula.
    Center of the Taylor polynomial. This is the point around which the Taylor series is expanded.
  40. What is the Lagrange Error Bound formula for Taylor polynomials?
    Rn(x)=M∣x−a∣n+1(n+1)!R_n(x) = \frac{M|x-a|^{n+1}}{(n+1)!}. This bounds the error between the function and its Taylor polynomial.
  41. Identify the role of factorial in Lagrange Error Bound.
    Denominator scaling factor in error term. The factorial grows rapidly, making higher-order errors much smaller.

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