Maximum error. Calculus for Taylor polynomials?
The series extends as cos(x) = 1 - x^2 / 2 + x^4/24 - x^6/720 + ... Note that, because on the given interval we have |x|<1, the remainder after two terms will be less than x^4/24 . So the maximum error will be pi^4/6144 ≈ 0.0159
Error estimates of Maclaurin/Taylor series?
e^x = 1 + x + x²/2! + x³/3! + . . . But you are using: e^h = 1 + h + R_1(h), where R_1(h) = M(h)²/2!. Remember that M is the maximum of the second derivative of e^x on the interval [0, 0.01]. Since the second derivative of e^x is still e^x and e^x is always increasing, M will be e^0.01. Thus, your error will be e^(0.01)*(0.01)²/2 = 5.05 x 10^(-5) This divided by 0.01 gives a percent error of 0.00505 which is less that 0.006.
Estimate the error by computing an upper bound on the reminder?
The Taylor polynomial is correct. ----------------------------- Using the Taylor series remainder formula, then |error| <= | f ''''(c) x^4 / 4!| for some c in (0, 1). For the upper bound on the error, now we just maximize each term. In this case f ''''(c) = e^(-c), which has maximum value e^(-0) = 1 on [0, 1] Likewise, x^4 has maximum value 1^4 = 1 on [0, 1]. Thus, |error| < 1 * x^4/24 <= 1^4/24 = 1/24. ---------------------------- I hope this helps!
Series Question Help! :)?
1) Let f(x) = e^x. The error is given by f^(5)(c) (0.28 - 0)^5/5! for some c in (0, 0.28). Since f^(5)(c) = e^c < e^(0.28), since f is an increasing function, the upper bound of the error is e^(0.28) * (0.28)^5 / 5!. --------------- 2) Note that this is the second order Maclaurin polynomial for f. So, the error is given by f '''(c) (x - 0)^3/3! for some c between 0 and x. However, f '''(c) = 8 * 7 * 6(1 + c)^5 = 336(1 + c)^5. So, |error| = |336(1 + c)^5 * x^3/3!| < 336 * (1 + 0.01)^5 * (0.01)^3 / 3!, since |x| ≤ 0.01 = 336 * (1.01)^5 * (0.01)^3 / 3!. ----------- 3) Use the geometric series. 1/(8+x) = 1/[8(1 + x/8)] ...........= (1/8) * 1/(1 - (-x/8)) ...........= (1/8) * Σ(n = 0 to ∞) (-x/8)^n ...........= (1/8) * Σ(n = 0 to ∞) (-1)^n x^n / 8^(n+1). ----------- 4) Σ(n = 1 to ∞) 8/(n(n+3)) = lim(t→∞) Σ(n = 1 to t) (8/3) [1/n - 1/(n+3)], by partial fractions = lim(t→∞) (8/3) [(1 - 1/4) + (1/2 - 1/5) + (1/3 - 1/6) + (1/4 - 1/7) + ... + (1/t - 1/(t+3))] = lim(t→∞) (8/3) (1 + 1/2 + 1/3 - 1/(t+1)) - 1/(t+2) - 1/(t+3)], since all other terms cancel in pairs = (8/3) (1 + 1/2 + 1/3 - 0) = 44/9. I hope this helps!
What makes this Taylor Polynomial wrong?
sin(6/7) = 0.755975 n = 3 question: Use an appropriate Taylor polynomial about 0 and the Lagrange Remainder Formula to approximate sin(6/7) with an error less than 0.0001. sin(6/7) ≈ ? What is the smallest value of n for which the approximation above is guaranteed to have an error less than 0.0001? n = ? Image: http://imgur.com/a/sWlxq Thank you
Taylor's inequality?
What estimate does Taylor’s Inequality provide for the error R2(x) = √x − T2(x) in using the degree 2 Taylor polynomial T2(x) you derived in part (i) as an approximation to √x on the interval [4, 4.05]? NOTE: this is part two of one question. Part 1 was finding the 2 degree Taylor polynomial for f centered at x=4 for the function f(x) = √x The answer was T2(x) = 2 + 1/4(x − 4) −1/64(x − 4)^2