Question

Question 2 (Essay Worth 30 points)

(10.08 HC)

The function h is defined by the power series =h of x is equal to x cubed plus the quantity x to the fourth power over 3 end quantity plus the quantity x to the fifth power over 5 end quantity plus x to the sixth power over 7 end quantity plus the quantity x to the seventh power over 9 end quantity plus dot dot dot equals the summation from n equals 0 to infinity of the quantity x to the power of n plus 3 end quantity over the quantity 2 times n plus 1 end quantity period

Part A: Determine the interval of convergence of the power series for h. (10 points)

Part B: Find h ′(−1) and determine if the new series converges or diverges. Justify your response and discuss the relationship between the radius of convergence and interval of convergence of h(x) and h ′(x). (10 points)

Part C: Determine if the power series h(x2) has any points of inflection. Justify your answer. (10 points)

Answer 1

I'm not sure if this what your looking for as an answer, but I hope it helped!

The function h(x) is given as the power series:

=h(x) = x³ + (x⁴/3) + (x⁵/5) + (x⁶/7) + (x⁷/9) + ...

Part A: To determine the interval of convergence, we use the ratio test:

lim(n→∞) |(x^(n+4))/(2(n+2)+1) * (2n+1)/(x^n+3)| = lim(n→∞) |x|^2/2 = |x|^2/2

The ratio test tells us that the series converges if |x|²/2 < 1 and diverges if |x|²/2 > 1. Therefore, the interval of convergence is -√2 < x < √2.

Part B: To find h′(−1), we differentiate the power series term by term:

h′(x) = 3x² + 4x³/3 + 5x⁴/5 + 6x⁵/7 + 7x⁶/9 + ...

h′(−1) = 3(−1)² + 4(−1)³/3 + 5(−1)⁴/5 + 6(−1)⁵/7 + 7(−1)⁶/9 + ...

= 3 − 4/3 + 1/5 − 6/7 + 1/9 − ...

= ∑(n=0 to ∞) (−1)^n * (2n+3)/(2n+1)

We can use the alternating series test to show that the series converges. The terms decrease in absolute value, and the limit of the terms approaches zero. Therefore, h′(−1) converges.

The relationship between the radius of convergence and interval of convergence of h(x) and h′(x) is that the radius of convergence of h′(x) is the same as that of h(x), but the interval of convergence may be different due to the behavior at the endpoints.

Part C: To determine if h(x²) has any points of inflection, we differentiate twice:

h(x²) = (x²)³ + ((x²)⁴/3) + ((x²)⁵/5) + ((x²)⁶/7) + ((x²)⁷/9) + ...

h′(x) = 3x⁴ + 4x⁵/3 + 5x⁶/5 + 6x⁷/7 + 7x⁸/9 + ...

h″(x) = 12x³ + 20x⁴/3 + 30x⁵/5 + 42x⁶/7 + 56x⁷/9 + ...

We can see that h″(x) is always positive, so h(x²) does not have any points of inflection.