Question

Question 3 (Essay Worth 40 points)

(10.01, 10.09 HC)

The power series for f of x is equal to 1 over the quantity 1 minus x end quantity is defined as 1 plus x plus x squared plus x cubed plus dot dot dot equals the summation from n equals 0 to infinity of x to the nth power comma and the power series for −sinx is defined as negative x plus the quantity x cubed over 3 factorial end quantity minus the quantity x to the fifth power over 5 factorial end quantity plus x to the seventh power over 7 factorial plus dot dot dot equals the summation from n equals 0 to infinity of negative 1 to the nth power time the quantity negative x to the 2 times n minus 1 power end quantity over the quantity 2 times n minus 1 end quantity factorial period

Part A: Find the general term of the power series for g of x is equal to 4 over the quantity x squared minus 4 end quantity and evaluate the infinite sum when x = 1. Justify your solution. (15 points)

Part B: Find an upper bound for the error of the approximation sin of zero point 3 is approximately zero point 3 minus the quantity zero point 3 to the third power over 3 factorial end quantity period Round your final answer to five decimal places. (15 points)

Part C: Find a power series for h(x) = ln(1 + x) centered at x = 0 and show the work that leads to your conclusion. (10 points)

Answer 1

centered at x = 0.

Explanation:

Part A: The power series for g(x) can be obtained by using the formula for a geometric series with a first term of 1 and a common ratio of (x/2). Then we have:

g(x) = 4/((x+2)(x-2)) = 4/(4*(1 + x/2)*(1 - x/2))

= 1/(1 - x/2) - 1/(1 + x/2)

We can then use the power series for 1/(1-x) to find the power series for g(x):

g(x) = 1/(1 - x/2) - 1/(1 + x/2)

= (1/2) * (1 + x/2 + (x/2)^2 + (x/2)^3 + ...) - (1/2) * (1 - x/2 + (x/2)^2 - (x/2)^3 + ...)

= x + (3/4)*x^2 + (5/8)*x^3 + (35/64)*x^4 + ...

To evaluate the infinite sum when x = 1, we can substitute x = 1 into the power series and use the formula for an infinite geometric series:

g(1) = 1 + (3/4) + (5/8) + (35/64) + ...

= 1/(1 - 1/2) - 1/(1 + 1/2)

= 2 - 2/3

= 4/3

Therefore, the infinite sum when x = 1 is 4/3.

Part B: To find an upper bound for the error of the approximation sin(0.3) ≈ 0.3 - 0.3^3/3!, we can use the formula for the remainder term in a Taylor series:

Rn(x) = f^(n+1)(c) * (x-a)^(n+1) / (n+1)!

where f(x) = sin(x), a = 0.3, n = 3, and c is some number between a and x. We want to find an upper bound for |R3(0.3)|.

Taking the fourth derivative of f(x) = sin(x), we get:

f^(4)(x) = -sin(x)

Since |sin(c)| ≤ 1 for any c, we have:

|R3(0.3)| ≤ |f^(4)(c)| * (0.3-0)^4 / 4!

≤ 1 * 0.3^4 / 24

≤ 0.000625

Therefore, an upper bound for the error is 0.000625, rounded to five decimal places.

Part C: We can find the power series for h(x) by differentiating the power series for ln(1+x) term by term. The power series for ln(1+x) is:

ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ...

Taking the derivative, we get:

h(x) = ln(1+x)'

= 1 - x + x^2 - x^3 + ...

which is the power series for (-1)^n x^n. Therefore, the power series for h(x) is:

h(x) = ∑(-1)^n x^n

centered at x = 0.