Find the Maclaurin series for the given function and the interval of convergence

Question

Find the Maclaurin series for the given function and the interval of convergence. Find a power series solution of the form to the differential equation yn = y with initial condition Find the volume of the solid obtained by rotating the region enclosed by y = x2 - 1 and y = 2x - 1 about the axis y = 4 Sketch the region bounded by y = 4 / x + 1 and y = 1 for x [0, 4]find its centroid. Calculate the surface area of the solid obtained by rotating y = 2 / 3 x3 / 4 about the x-axis in the interval [0, 1] use Simpson's Rule to approximate sin(x2)dx with n = 8 partitions (i.e find S8).

Explanation / Answer

5 The n'th derivative of e^x is e^x, so the Maclaurin series is the sum for n=0 to infinity of x^n/n!

Thus, the Maclaurin series for 2e^2x is, multiplying by 2 and substituting 2x for x,

the sum for n=0 to infinity 2*(2x)^n/n! or 2^(n+1)x^n/n!

As the ratio of successive terms is 2x/n+1, the limit is 0 for all x, so the interval of convergence is (-inf, inf)

For ln(1-x^2), the Maclaurin series for ln(1 - x) = - the sum for n=0 to inf(x^n/n), the Maclaurin series for

ln(1-x^2) = the sum for n= 0 to inf (x^2n/n)

The ratio is x^2 *n/n+1, which has limit x^2

Thus, this converges when x^2 < 1, or -1 < x < 1

For x = +- 1, the sum of -(1/n) is a p-series with p = 1, so the series does not converge.

Convergence for x in (-1, 1)

6. f(x) = the sum of cn x^n

y(0) = 1, so c0 = 1

y'(0) = -1, so c1 = -1

y'' = y

f''(x) = the sum of cn n(n-1)x^n-2

Then, y'' = y, so the sum of cn n(n-1)x^n-2 = the sum of cn x^n

Then, cn+2 (n+2)(n+1)x^n = the sum of cnx^n

Then, cn+2 = cn(n+2)(n+1)

Thus, by induction, we see that, for n even, cn = c0/n! = 1/n!

For cn odd, cn = c1/n!

cn even = 1/n!

cn odd = -1/n!

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