Math 150B Exam 2 Spring 2018

Math 150 B Exam 2 Spring 2018name

Find the sum of the following series: a) ∑ from n=2 to ∞ of 1 / [n(n−1)] b) ∑ from n=1 to ∞ of (5n−1) / (6n). Determine whether each of the following infinite series converges or diverges, stating the test(s) used and showing work: a) ∑ from n=1 to ∞ of (3 + 8 sin n) / (n² + 1) b) ∑ from n=1 to ∞ of n⁴ / (n⁴ + 7) c) ∑ from n=1 to ∞ of (5 + 2n) / (1 + 2n² + n⁴). Determine if the following series converge absolutely, conditionally, or diverge: ∑ from n=1 to ∞ of (−1)ⁿ⁺¹ 3√5n. Find the interval and radius of convergence for the power series: ∑ from n=1 to ∞ of (−1)ⁿ⁻¹ n−1/2 (4x²)ⁿ, checking endpoints. Recall that 1 / (1−x) = ∑ from n=0 to ∞ of xⁿ for |x|

Paper For Above instruction

The given problems span fundamental concepts in calculus, including series summation, convergence tests, power series, and Taylor polynomials. Each problem demands a precise approach grounded in mathematical principles, aiming to deepen understanding of infinite series, convergence criteria, power series representations, and polynomial approximations.

Sum of Series

Firstly, consider the sum of the series ∑ₙ=2^∞ 1/[n(n−1)]. Recognizing the telescoping nature, rewrite the general term as 1/[n(n−1)] = 1/(n−1) - 1/n. The series telescopes as follows:

∑_{n=2}^∞ (1/(n−1) - 1/n) = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ...

Most terms cancel out, leaving the sum as the limit when the partial sum up to N approaches infinity:

S_N = 1 - 1/N, and as N approaches infinity, S_N approaches 1. Therefore, the infinite sum converges to 1.

Next, evaluate ∑_{n=1}^∞ (5n−1)/ (6n). Rewrite it as:

(5n−1)/ (6n) = (5/6) - (1/6n). The series becomes:

∑_{n=1}^∞ [(5/6) - (1/6n)] = ∑_{n=1}^∞ (5/6) - (1/6) ∑_{n=1}^∞ (1/n).

Since ∑_{n=1}^∞ (1/n) diverges (harmonic series), the entire series diverges.

Convergence Tests

For the series ∑_{n=1}^∞ (3 + 8 sin n)/(n² + 1): As n→∞, the dominant term's behavior is determined by the numerator and denominator. Since |sin n| ≤ 1, numerator oscillates between 3−8=−5 and 3+8=11, but bounded. The denominator grows without bound as n², so the series is comparable to a constant over n², which converges by the p-series test with p=2 > 1. The series converges by comparison test.

Next, for ∑_{n=1}^∞ n⁴ / (n⁴ + 7): for large n, the terms approximate 1, so the series diverges by divergence test as the limit of the general term approaches 1, not zero.

For ∑_{n=1}^∞ (5 + 2n) / (1 + 2n² + n⁴): for large n, numerator ~ 2n, denominator ~ n⁴, so the general term behaves like 2n / n⁴ = 2 / n³, which converges by p-series with p=3 > 1.

Series Test for Absolute/Conditional Convergence

The series ∑_{n=1}^∞ (−1)^n+1 3√5n alternates in sign; its absolute value is ∑_{n=1}^∞ 3√5n. Since 3√5n ∼ n^{1/2}, which diverges as n→∞, the original series converges conditionally by the Alternating Series Test, but not absolutely.

Power Series and Radius of Convergence

For the power series ∑_{n=1}^∞ (−1)^{n−1} n−1/2 (4x²)^n, consider the ratio test:

a_{n} = (−1)^{n−1} n−1/2 (4x²)^n.

Taking absolute value and ratio:

|a_{n+1}|/|a_{n}| = [(n+1) - 1/2] / (n - 1/2) |4x²| = [(n + 1/2)/(n - 1/2)] |4x²|.

As n→∞, (n + 1/2)/(n - 1/2) → 1. So, the series converges when |4x²|

Power Series for a Function

The function f(x) = x² / (1 + x²) can be written as:

f(x) = x² / (1 + x²) = x² * (1 / (1 + x²))

Note that 1 / (1 + x²) can be expressed as a power series:

1 / (1 + x²) = ∑_{n=0}^∞ (−1)^n x^{2n}, valid for |x|

Multiplying by x², the series representation becomes:

f(x) = x² * ∑_{n=0}^∞ (−1)^n x^{2n} = ∑_{n=0}^∞ (−1)^n x^{2n + 2}. The radius of convergence is 1.

Taylor Polynomial Approximation

The Taylor polynomial T₄(x) for (9 + x)^{1/2} at x=0:

T₄(x) ≈ 3 + (1/6) x − x² + x³ − x⁴.

To find the polynomial for (9 + x)^{3/2} at a=0, differentiate the function and build the Taylor series or use known relationships. Alternatively, recognize that (9 + x)^{3/2} = [(9 + x)^{1/2}]^3, and approximate accordingly. The first few derivatives at zero will guide expansion, but a shortcut involves using the existing polynomial or applying the binomial theorem for fractional powers.

Approximation of (9 + x)^{3/2}

Using the binomial expansion:

(9 + x)^{3/2} = 9^{3/2} (1 + x/9)^{3/2}

With 9^{3/2} = (√9)^3 = 3^3 = 27, and expanding (1 + h)^{3/2} via binomial theorem for small h:

(1 + h)^{3/2} ≈ 1 + (3/2) h + (−1/8) h² + ...

Substituting h=x/9:

(1 + x/9)^{3/2} ≈ 1 + (3/2)(x/9) - (1/8)(x/9)^2.

Thus, the approximation for (9 + x)^{3/2} is:

27 [1 + (3/2)(x/9) - (1/8)(x/9)^2] = 27 + (3/2)27 (x/9) − (1/8)27 * (x/9)^2.

Calculating numerators:

(3/2)*27 = 40.5, and 27/8 = 3.375.

Therefore, the polynomial is approximately:

27 + 40.5 (x/9) - 3.375 (x/9)^2.

Simplify:

27 + (40.5/9) x − (3.375/81) x² = 27 + 4.5 x - 0.0416667 x².

Approximating √10^{3/2}

Using the Taylor polynomial, substitute x=1 into the polynomial for (9 + x)^{3/2}, giving an approximation of √10^{3/2}. Alternatively, use the previously determined polynomial for approximation at x=1:

f(x) ≈ 27 + 4.5(1) - 0.0416667(1)^2 = 27 + 4.5 - 0.0416667 ≈ 31.4583.

This provides an approximation of the value of √10^{3/2} near x=1, corresponding to x=10−9=1.

Approximate f(x) = x^{2/3} at a = 8

Compute Taylor polynomial of degree 2 at a=8:

f(x) = x^{2/3}, so f'(x) = (2/3) x^{-1/3} and f''(x) = - (2/9) x^{-4/3}.

Evaluating at x=8:

f(8) = 8^{2/3} = (2^3)^{2/3} = 2^2=4.

f'(8) = (2/3) 8^{-1/3} = (2/3) (1/2) = 1/3.

f''(8) = - (2/9) 8^{-4/3} = - (2/9)(1/8)^{4/3} =

Since 8^{1/3} = 2, then 8^{4/3} = (8^{1/3})^{4} = 2^{4} = 16.

Therefore, f''(8) = - (2/9) (1/16) = - (2/9) (1/16) = - 1 / 72.

The degree 2 Taylor polynomial at a=8:

T_2(x) = f(8) + f'(8)(x−8) + (1/2)f''(8)(x−8)^2 = 4 + (1/3)(x−8) - (1/144)(x−8)^2.

Maximum Error Estimation

For x in [7,9], the remainder R_2(x) for degree 2 Taylor polynomial is bounded by the next derivative's maximum. Using Taylor’s theorem, the Lagrange form:

|R_2(x)| ≤ (1/3!) max_{ξ in [7,9]} |f'''(ξ)| |x−8|^3.

f'''(x) = derivative of f''(x):

f'''(x) = derivative of (-2/9) x^{-4/3} = (8/27) x^{-7/3}.

On [7,9], maximum of |f'''(ξ)| occurs at the minimum ξ=7:

|f'''(7)| = (8/27) * 7^{-7/3}.

Estimate:

|f'''(7)| ≈ (8/27) * 7^{-7/3}. Since 7^{1/3} ≈ 1.913, then 7^{7/3} ≈ (1.913)^7 ≈ 97.1, so:

|f'''(7)| ≈ (8/27) / 97.1 ≈ 0.00304.

Therefore, maximum error when x in [7,9]:

|R_2(x)| ≤ (1/6) 0.00304 (|x−8|)^3.

At maximum, |x−8| = 1, so:

Maximum error ≈ (1/6) 0.00304 1 ≈ 0.0005067.

This tiny error indicates the Taylor polynomial provides a good approximation in the interval.

Conclusion

The series and functions examined demonstrate key techniques in advanced calculus, from telescoping series to power series expansions, and Taylor polynomial approximations. Applying convergence tests ensures the correct classification of series, while series manipulations reveal insights about the functions involved. Power series and Taylor approximations serve as vital tools for approximating functions near specific points, aiding in computational and analytical tasks across mathematics and applied sciences.

References

  • Rudin, W. (1976). Principles of Mathematical Analysis. McGraw-Hill.
  • Apostol, T. M. (1967). Calculus, Volumes 1 & 2. John Wiley & Sons.
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  • Strang, G. (2016). Introduction to Calculus. Wellesley-Cambridge Press.
  • Lay, D. C. (2012). Linear Algebra and Its Applications. Addison-Wesley.
  • Anton, H., Bivens, I., & Davis, S. (2013). Calculus: Early Transcendentals. Wiley.
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